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Numerical Study of Laser Interaction with Solid Materials. University of Sulaimani, College of Science, Physics Department, 2011
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Kurdistan Iraqi Region Ministry of Higher Education University of Sulaimani College of Science Physics Department
Numerical Study of Laser Interaction with
Solid Materials
Prepared by
Chia HQadr Shiraz Q Ghafur Hewar AAbdul
Supervised by
Dr Omed Ghareb Abdullah
2010 - 2011
ii
Acknowledgments
First of all we would like to say Alhamdulillah for giving us the
strength and health to do this project work until it done and not forgotten
to our family for providing with everything that are related to this project
work and their advice They also supported us and encouraged us to
complete this task so that we will not procrastinate in doing it
Then we would like to thank our Supervisor Dr Omed Ghareb
Abdullah for guiding us throughout this project We had some difficulties
in doing this task but he taught us patiently until we knew what to do He
tried and tried to teach us until we understand what we supposed to do with
the project work
Last but not least those who were helping us in doing this project by
sharing ideas They were helpful that when we combined and discussed
together we had this task done
Chia Shiraz amp Hewar
iii
Contents
Chapter One Basic Concepts
11 Introduction
12 Definition of the Laser
13 Active laser medium or gain medium
14 A Survey of Laser Types
141 Gas Lasers
142 Solid Lasers
143 Molecular Lasers
144 Free-Electron Lasers
15 Pulsed operation
16 Heat and heat capacity
17 Thermal conductivity
18 Derivation in one dimension
19 Aim of present work
Chapter Two Theoretical Aspects
21 Introduction
22 One dimension laser heating equation
23 Numerical solution of Initial value problems
24 Finite Difference Method
241 First Order Forward Difference
iv
242 First Order Backward Difference
242 First Order Central Difference
25 Procedures
Chapter three Results and Discussion
31 Introduction
32 Numerical solution with constant laser power density and
constant thermal properties
33 Evaluation of function 119920119920(119957119957) of laser flux density
34 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and constant thermal properties
35 Evaluation the Thermal Conductivity as functions of
temperature
36 Evaluation the Specific heat as functions of temperature
37 Evaluation the Density as functions of temperature
38 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =
119914119914(119931119931)120646120646 = 120646120646(119931119931)
39 Laser interaction with copper material
310 Conclusions
References
v
Abstract
In recent years much effort has gone into the understanding of the
interaction of short laser pulses with matter The present works have
typically involved studying the interaction of high intensity laser pulses with
high-density solid target In this study the NDYAG pulsed laser with
maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for
laser energy with time as well as a function of laser intensity with time are
presented in this study
The finite difference method was used to calculate the temperature
distribution as a function of laser depth penetration in lead and copper
materials
The best polynomial fits for thermal conductivity specific heat capacity
and density of metals as a function of temperature was obtained using
Matlab software At the first all these properties were assumed to be
constants and then the influence of varying these properties with
temperature was tacked in to account The temperature gradient of lead
shows to be greater than that of copper this may be due to the high thermal
conductivity and high specific heat capacity of copper with that of lead
1
Chapter One
Basic concepts
11 Introduction
Laser is a mechanism for emitting light with in electromagnetic radiation
region of the spectrum with different output intensity Max Plank published
work in 1900 that provided the understanding that light is a form of
electromagnetic radiation without this understanding the laser would have
been invented The principle of the laser was first known in 1917 when Albert
Einstein describe the theory of stimulated emission and Theodor Maiman in
1960 invent the first laser using a lasing medium of ruby that was stimulated
by using high energy flash of intense light
We have four types of laser according to their gain medium which are
(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)
So laser is provided a controlled source of atomic and electronic excitations
involving non equilibrium phenomena that lend themselves to processing of
novel material and structure because laser used in wide range application in
our life such as welding cutting drilling industrial and medical field Maiman
and other developer of laser weapons sighting system and powerful laser for
use in surgery and other areas where moderated powerful pinpoint source of
heat was needed And today laser are used in corrective eye surgery and
providing apprecise source of heat for cutting and cauterizing tissue
12 Definition of the Laser
The word laser is an acronym for Light Amplification by Stimulated Emission
of Radiation The laser makes use of processes that increase or amplify light
signals after those signals have been generated by other means These
processes include (1) stimulated emission a natural effect that was deduced
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
ii
Acknowledgments
First of all we would like to say Alhamdulillah for giving us the
strength and health to do this project work until it done and not forgotten
to our family for providing with everything that are related to this project
work and their advice They also supported us and encouraged us to
complete this task so that we will not procrastinate in doing it
Then we would like to thank our Supervisor Dr Omed Ghareb
Abdullah for guiding us throughout this project We had some difficulties
in doing this task but he taught us patiently until we knew what to do He
tried and tried to teach us until we understand what we supposed to do with
the project work
Last but not least those who were helping us in doing this project by
sharing ideas They were helpful that when we combined and discussed
together we had this task done
Chia Shiraz amp Hewar
iii
Contents
Chapter One Basic Concepts
11 Introduction
12 Definition of the Laser
13 Active laser medium or gain medium
14 A Survey of Laser Types
141 Gas Lasers
142 Solid Lasers
143 Molecular Lasers
144 Free-Electron Lasers
15 Pulsed operation
16 Heat and heat capacity
17 Thermal conductivity
18 Derivation in one dimension
19 Aim of present work
Chapter Two Theoretical Aspects
21 Introduction
22 One dimension laser heating equation
23 Numerical solution of Initial value problems
24 Finite Difference Method
241 First Order Forward Difference
iv
242 First Order Backward Difference
242 First Order Central Difference
25 Procedures
Chapter three Results and Discussion
31 Introduction
32 Numerical solution with constant laser power density and
constant thermal properties
33 Evaluation of function 119920119920(119957119957) of laser flux density
34 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and constant thermal properties
35 Evaluation the Thermal Conductivity as functions of
temperature
36 Evaluation the Specific heat as functions of temperature
37 Evaluation the Density as functions of temperature
38 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =
119914119914(119931119931)120646120646 = 120646120646(119931119931)
39 Laser interaction with copper material
310 Conclusions
References
v
Abstract
In recent years much effort has gone into the understanding of the
interaction of short laser pulses with matter The present works have
typically involved studying the interaction of high intensity laser pulses with
high-density solid target In this study the NDYAG pulsed laser with
maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for
laser energy with time as well as a function of laser intensity with time are
presented in this study
The finite difference method was used to calculate the temperature
distribution as a function of laser depth penetration in lead and copper
materials
The best polynomial fits for thermal conductivity specific heat capacity
and density of metals as a function of temperature was obtained using
Matlab software At the first all these properties were assumed to be
constants and then the influence of varying these properties with
temperature was tacked in to account The temperature gradient of lead
shows to be greater than that of copper this may be due to the high thermal
conductivity and high specific heat capacity of copper with that of lead
1
Chapter One
Basic concepts
11 Introduction
Laser is a mechanism for emitting light with in electromagnetic radiation
region of the spectrum with different output intensity Max Plank published
work in 1900 that provided the understanding that light is a form of
electromagnetic radiation without this understanding the laser would have
been invented The principle of the laser was first known in 1917 when Albert
Einstein describe the theory of stimulated emission and Theodor Maiman in
1960 invent the first laser using a lasing medium of ruby that was stimulated
by using high energy flash of intense light
We have four types of laser according to their gain medium which are
(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)
So laser is provided a controlled source of atomic and electronic excitations
involving non equilibrium phenomena that lend themselves to processing of
novel material and structure because laser used in wide range application in
our life such as welding cutting drilling industrial and medical field Maiman
and other developer of laser weapons sighting system and powerful laser for
use in surgery and other areas where moderated powerful pinpoint source of
heat was needed And today laser are used in corrective eye surgery and
providing apprecise source of heat for cutting and cauterizing tissue
12 Definition of the Laser
The word laser is an acronym for Light Amplification by Stimulated Emission
of Radiation The laser makes use of processes that increase or amplify light
signals after those signals have been generated by other means These
processes include (1) stimulated emission a natural effect that was deduced
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
iii
Contents
Chapter One Basic Concepts
11 Introduction
12 Definition of the Laser
13 Active laser medium or gain medium
14 A Survey of Laser Types
141 Gas Lasers
142 Solid Lasers
143 Molecular Lasers
144 Free-Electron Lasers
15 Pulsed operation
16 Heat and heat capacity
17 Thermal conductivity
18 Derivation in one dimension
19 Aim of present work
Chapter Two Theoretical Aspects
21 Introduction
22 One dimension laser heating equation
23 Numerical solution of Initial value problems
24 Finite Difference Method
241 First Order Forward Difference
iv
242 First Order Backward Difference
242 First Order Central Difference
25 Procedures
Chapter three Results and Discussion
31 Introduction
32 Numerical solution with constant laser power density and
constant thermal properties
33 Evaluation of function 119920119920(119957119957) of laser flux density
34 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and constant thermal properties
35 Evaluation the Thermal Conductivity as functions of
temperature
36 Evaluation the Specific heat as functions of temperature
37 Evaluation the Density as functions of temperature
38 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =
119914119914(119931119931)120646120646 = 120646120646(119931119931)
39 Laser interaction with copper material
310 Conclusions
References
v
Abstract
In recent years much effort has gone into the understanding of the
interaction of short laser pulses with matter The present works have
typically involved studying the interaction of high intensity laser pulses with
high-density solid target In this study the NDYAG pulsed laser with
maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for
laser energy with time as well as a function of laser intensity with time are
presented in this study
The finite difference method was used to calculate the temperature
distribution as a function of laser depth penetration in lead and copper
materials
The best polynomial fits for thermal conductivity specific heat capacity
and density of metals as a function of temperature was obtained using
Matlab software At the first all these properties were assumed to be
constants and then the influence of varying these properties with
temperature was tacked in to account The temperature gradient of lead
shows to be greater than that of copper this may be due to the high thermal
conductivity and high specific heat capacity of copper with that of lead
1
Chapter One
Basic concepts
11 Introduction
Laser is a mechanism for emitting light with in electromagnetic radiation
region of the spectrum with different output intensity Max Plank published
work in 1900 that provided the understanding that light is a form of
electromagnetic radiation without this understanding the laser would have
been invented The principle of the laser was first known in 1917 when Albert
Einstein describe the theory of stimulated emission and Theodor Maiman in
1960 invent the first laser using a lasing medium of ruby that was stimulated
by using high energy flash of intense light
We have four types of laser according to their gain medium which are
(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)
So laser is provided a controlled source of atomic and electronic excitations
involving non equilibrium phenomena that lend themselves to processing of
novel material and structure because laser used in wide range application in
our life such as welding cutting drilling industrial and medical field Maiman
and other developer of laser weapons sighting system and powerful laser for
use in surgery and other areas where moderated powerful pinpoint source of
heat was needed And today laser are used in corrective eye surgery and
providing apprecise source of heat for cutting and cauterizing tissue
12 Definition of the Laser
The word laser is an acronym for Light Amplification by Stimulated Emission
of Radiation The laser makes use of processes that increase or amplify light
signals after those signals have been generated by other means These
processes include (1) stimulated emission a natural effect that was deduced
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
iv
242 First Order Backward Difference
242 First Order Central Difference
25 Procedures
Chapter three Results and Discussion
31 Introduction
32 Numerical solution with constant laser power density and
constant thermal properties
33 Evaluation of function 119920119920(119957119957) of laser flux density
34 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and constant thermal properties
35 Evaluation the Thermal Conductivity as functions of
temperature
36 Evaluation the Specific heat as functions of temperature
37 Evaluation the Density as functions of temperature
38 Numerical solution with variable laser power density (
119920119920 = 119920119920 (119957119957) ) and variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 =
119914119914(119931119931)120646120646 = 120646120646(119931119931)
39 Laser interaction with copper material
310 Conclusions
References
v
Abstract
In recent years much effort has gone into the understanding of the
interaction of short laser pulses with matter The present works have
typically involved studying the interaction of high intensity laser pulses with
high-density solid target In this study the NDYAG pulsed laser with
maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for
laser energy with time as well as a function of laser intensity with time are
presented in this study
The finite difference method was used to calculate the temperature
distribution as a function of laser depth penetration in lead and copper
materials
The best polynomial fits for thermal conductivity specific heat capacity
and density of metals as a function of temperature was obtained using
Matlab software At the first all these properties were assumed to be
constants and then the influence of varying these properties with
temperature was tacked in to account The temperature gradient of lead
shows to be greater than that of copper this may be due to the high thermal
conductivity and high specific heat capacity of copper with that of lead
1
Chapter One
Basic concepts
11 Introduction
Laser is a mechanism for emitting light with in electromagnetic radiation
region of the spectrum with different output intensity Max Plank published
work in 1900 that provided the understanding that light is a form of
electromagnetic radiation without this understanding the laser would have
been invented The principle of the laser was first known in 1917 when Albert
Einstein describe the theory of stimulated emission and Theodor Maiman in
1960 invent the first laser using a lasing medium of ruby that was stimulated
by using high energy flash of intense light
We have four types of laser according to their gain medium which are
(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)
So laser is provided a controlled source of atomic and electronic excitations
involving non equilibrium phenomena that lend themselves to processing of
novel material and structure because laser used in wide range application in
our life such as welding cutting drilling industrial and medical field Maiman
and other developer of laser weapons sighting system and powerful laser for
use in surgery and other areas where moderated powerful pinpoint source of
heat was needed And today laser are used in corrective eye surgery and
providing apprecise source of heat for cutting and cauterizing tissue
12 Definition of the Laser
The word laser is an acronym for Light Amplification by Stimulated Emission
of Radiation The laser makes use of processes that increase or amplify light
signals after those signals have been generated by other means These
processes include (1) stimulated emission a natural effect that was deduced
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
v
Abstract
In recent years much effort has gone into the understanding of the
interaction of short laser pulses with matter The present works have
typically involved studying the interaction of high intensity laser pulses with
high-density solid target In this study the NDYAG pulsed laser with
maximum energy 119864119864119898119898119898119898119898119898 = 02403 119869119869 was used The mathematical function for
laser energy with time as well as a function of laser intensity with time are
presented in this study
The finite difference method was used to calculate the temperature
distribution as a function of laser depth penetration in lead and copper
materials
The best polynomial fits for thermal conductivity specific heat capacity
and density of metals as a function of temperature was obtained using
Matlab software At the first all these properties were assumed to be
constants and then the influence of varying these properties with
temperature was tacked in to account The temperature gradient of lead
shows to be greater than that of copper this may be due to the high thermal
conductivity and high specific heat capacity of copper with that of lead
1
Chapter One
Basic concepts
11 Introduction
Laser is a mechanism for emitting light with in electromagnetic radiation
region of the spectrum with different output intensity Max Plank published
work in 1900 that provided the understanding that light is a form of
electromagnetic radiation without this understanding the laser would have
been invented The principle of the laser was first known in 1917 when Albert
Einstein describe the theory of stimulated emission and Theodor Maiman in
1960 invent the first laser using a lasing medium of ruby that was stimulated
by using high energy flash of intense light
We have four types of laser according to their gain medium which are
(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)
So laser is provided a controlled source of atomic and electronic excitations
involving non equilibrium phenomena that lend themselves to processing of
novel material and structure because laser used in wide range application in
our life such as welding cutting drilling industrial and medical field Maiman
and other developer of laser weapons sighting system and powerful laser for
use in surgery and other areas where moderated powerful pinpoint source of
heat was needed And today laser are used in corrective eye surgery and
providing apprecise source of heat for cutting and cauterizing tissue
12 Definition of the Laser
The word laser is an acronym for Light Amplification by Stimulated Emission
of Radiation The laser makes use of processes that increase or amplify light
signals after those signals have been generated by other means These
processes include (1) stimulated emission a natural effect that was deduced
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
1
Chapter One
Basic concepts
11 Introduction
Laser is a mechanism for emitting light with in electromagnetic radiation
region of the spectrum with different output intensity Max Plank published
work in 1900 that provided the understanding that light is a form of
electromagnetic radiation without this understanding the laser would have
been invented The principle of the laser was first known in 1917 when Albert
Einstein describe the theory of stimulated emission and Theodor Maiman in
1960 invent the first laser using a lasing medium of ruby that was stimulated
by using high energy flash of intense light
We have four types of laser according to their gain medium which are
(solid liquid gas and plasma) such as (ruby dye He-Ne and X-ray lasers)
So laser is provided a controlled source of atomic and electronic excitations
involving non equilibrium phenomena that lend themselves to processing of
novel material and structure because laser used in wide range application in
our life such as welding cutting drilling industrial and medical field Maiman
and other developer of laser weapons sighting system and powerful laser for
use in surgery and other areas where moderated powerful pinpoint source of
heat was needed And today laser are used in corrective eye surgery and
providing apprecise source of heat for cutting and cauterizing tissue
12 Definition of the Laser
The word laser is an acronym for Light Amplification by Stimulated Emission
of Radiation The laser makes use of processes that increase or amplify light
signals after those signals have been generated by other means These
processes include (1) stimulated emission a natural effect that was deduced
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
2
by considerations relating to thermodynamic equilibrium and (2) optical
feedback (present in most lasers) that is usually provided by mirrors
Thus in its simplest form a laser consists of a gain or amplifying medium
(where stimulated emission occurs) and a set of mirrors to feed the light back
into the amplifier for continued growth of the developing beam as seen in
Fig(11) Laser light differs from ordinary light in four ways Briefly it is much
more intense directional monochromatic and coherent Most lasers consist
of a column of active material with a partly reflecting mirror at one end and a
fully reflecting mirror at the other The active material can be solid (ruby
crystal) liquid or gas (HeNe COR2R etc)
Fig(11) Simplified schematic of typical laser
13 Active laser medium or gain medium
Laser medium is the heart of the laser system and is responsible for
producing gain and subsequent generation of laser It can be a crystal solid
liquid semiconductor or gas medium and can be pumped to a higher energy
state The material should be of controlled purity size and shape and should
have the suitable energy levels to support population inversion In other
words it must have a metastable state to support stimulated emission Most
lasers are based on 3 or 4 level energy level systems which depends on the
lasing medium These systems are shown in Figs (12) and (13)
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
3
In case of a three-level laser the material is pumped from level 1 to level 3
which decays rapidly to level 2 through spontaneous emission Level 2 is a
metastable level and promotes stimulated emission from level 2 to level 1
Fig(12) Energy states of Three-level active medium
On the other hand in a four-level laser the material is pumped to level 4
which is a fast decaying level and the atoms decay rapidly to level 3 which is
a metastable level The stimulated emission takes place from level 3 to level 2
from where the atoms decay back to level 1 Four level lasers is an
improvement on a system based on three level systems In this case the laser
transition takes place between the third and second excited states Since
lower laser level 2 is a fast decaying level which ensures that it rapidly gets
empty and as such always supports the population inversion condition
Fig(13) Energy states of Four-level active medium
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
4
14 A Survey of Laser Types
Laser technology is available to us since 1960rsquos and since then has been
quite well developed Currently there is a great variety of lasers of different
output power operating voltages sizes etc The major classes of lasers
currently used are Gas Solid Molecular and Free Electron lasers Below we
will cover some most popular representative types of lasers of each class and
describe specific principles of operation construction and main highlights
141 Gas Lasers
1 Helium-Neon Laser
The most common and inexpensive gas laser the helium-neon laser is
usually constructed to operate in the red at 6328 nm It can also be
constructed to produce laser action in the green at 5435 nm and in the
infrared at 1523 nm
One of the excited levels of helium at 2061 eV is very close to a level in
neon at 2066 eV so close in fact that upon collision of a helium and a neon
atom the energy can be transferred from the helium to the neon atom
Fig (14) The components of a Hilium-Neon Laser
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
5
Fig(15) The lasing action of He-Ne laser
Helium-Neon lasers are common in the introductory physics laboratories
but they can still be quite dangerous An unfocused 1-mW HeNe laser has a
brightness equal to sunshine on a clear day (01 wattcmP
2P) and is just as
dangerous to stare at directly
2- Carbon Dioxide Laser
The carbon dioxide gas laser is capable of continuous output powers above
10 kilowatts It is also capable of extremely high power pulse operation It
exhibits laser action at several infrared frequencies but none in the visible
spectrum Operating in a manner similar to the helium-neon laser it employs
an electric discharge for pumping using a percentage of nitrogen gas as a
pumping gas The COR2R laser is the most efficient laser capable of operating at
more than 30 efficiency
The carbon dioxide laser finds many applications in industry particularly for
welding and Cutting
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
6
3- Argon Laser
The argon ion laser can be operated as a continuous gas laser at about 25
different wavelengths in the visible between (4089 - 6861) nm but is best
known for its most efficient transitions in the green at 488 nm and 5145 nm
Operating at much higher powers than the Helium-Neon gas laser it is not
uncommon to achieve (30 ndash 100) watts of continuous power using several
transitions This output is produced in hot plasma and takes extremely high
power typically (9 ndash 12) kW so these are large and expensive devices
142 Solid Lasers
1 Ruby Laser
The ruby laser is the first type of laser actually constructed first
demonstrated in 1960 by T H Maiman The ruby mineral (corundum) is
aluminum oxide with a small amount (about 005) of Chromium which gives
it its characteristic pink or red color by absorbing green and blue light
The ruby laser is used as a pulsed laser producing red light at 6943 nm
After receiving a pumping flash from the flash tube the laser light emerges for
as long as the excited atoms persist in the ruby rod which is typically about a
millisecond
A pulsed ruby laser was used for the famous laser ranging experiment which
was conducted with a corner reflector placed on the Moon by the Apollo
astronauts This determined the distance to the Moon with an accuracy of
about 15 cm
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
7
Fig (16) Principle of operation of a Ruby laser
2- Neodymium-YAG Laser
An example of a solid-state laser the neodymium-YAG uses the NdP
3+P ion to
dope the yttrium-aluminum-garnet (YAG) host crystal to produce the triplet
geometry which makes population inversion possible Neodymium-YAG lasers
have become very important because they can be used to produce high
powers Such lasers have been constructed to produce over a kilowatt of
continuous laser power at 1065 nm and can achieve extremely high powers in
a pulsed mode
Neodymium-YAG lasers are used in pulse mode in laser oscillators for the
production of a series of very short pulses for research with femtosecond time
resolution
Fig(17) Construction of a Neodymium-YAG laser
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
8
3- Neodymium-Glass Lasers
Neodymium glass lasers have emerged as the design choice for research in
laser-initiated thermonuclear fusion These pulsed lasers generate pulses as
short as 10-12 seconds with peak powers of 109 kilowatts
143 Molecular Lasers
Eximer Lasers
Eximer is a shortened form of excited dimer denoting the fact that the
lasing medium in this type of laser is an excited diatomic molecule These
lasers typically produce ultraviolet pulses They are under investigation for use
in communicating with submarines by conversion to blue-green light and
pulsing from overhead satellites through sea water to submarines below
The eximers used are typically those formed by rare gases and halogens in
electron excited Gas discharges Molecules like XeF are stable only in their
excited states and quickly dissociate when they make the transition to their
ground state This makes possible large population inversions because the
ground state is depleted by this dissociation However the excited states are
very short-lived compared to other laser metastable states and lasers like the
XeF eximer laser require high pumping rates
Eximer lasers typically produce high power pulse outputs in the blue or
ultraviolet after excitation by fast electron-beam discharges
The rare-gas xenon and the highly active fluorine seem unlikely to form a
molecule but they do in the hot plasma environment of an electron-beam
initiated gas discharge They are only stable in their excited states if stable
can be used for molecules which undergo radioactive decay in 1 to 10
nanoseconds This is long enough to achieve pulsed laser action in the blue-
green over a band from 450 to 510 nm peaking at 486 nm Very high power
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
9
pulses can be achieved because the stimulated emission cross-sections of the
laser transitions are relatively low allowing a large population inversion to
build up The power is also enhanced by the fact that the ground state of XeF
quickly dissociates so that there is little absorption to quench the laser pulse
action
144 Free-Electron Lasers
The radiation from a free-electron laser is produced from free electrons
which are forced to oscillate in a regular fashion by an applied field They are
therefore more like synchrotron light sources or microwave tubes than like
other lasers They are able to produce highly coherent collimated radiation
over a wide range of frequencies The magnetic field arrangement which
produces the alternating field is commonly called a wiggler magnet
Fig(18) Principle of operation of Free-Electron laser
The free-electron laser is a highly tunable device which has been used to
generate coherent radiation from 10-5 to 1 cm in wavelength In some parts of
this range they are the highest power source Applications of free-electron
lasers are envisioned in isotope separation plasma heating for nuclear fusion
long-range high resolution radar and particle acceleration in accelerators
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
10
15 Pulsed operation
Pulsed operation of lasers refers to any laser not classified as continuous
wave so that the optical power appears in pulses of some duration at some
repetition rate This encompasses a wide range of technologies addressing a
number of different motivations Some lasers are pulsed simply because they
cannot be run in continuous mode
In other cases the application requires the production of pulses having as
large an energy as possible Since the pulse energy is equal to the average
power divided by the repitition rate this goal can sometimes be satisfied by
lowering the rate of pulses so that more energy can be built up in between
pulses In laser ablation for example a small volume of material at the surface
of a work piece can be evaporated if it is heated in a very short time whereas
supplying the energy gradually would allow for the heat to be absorbed into
the bulk of the piece never attaining a sufficiently high temperature at a
particular point
Other applications rely on the peak pulse power (rather than the energy in
the pulse) especially in order to obtain nonlinear optical effects For a given
pulse energy this requires creating pulses of the shortest possible duration
utilizing techniques such as Q-switching
16 Heat and heat capacity
When a sample is heated meaning it receives thermal energy from an
external source some of the introduced heat is converted into kinetic energy
the rest to other forms of internal energy specific to the material The amount
converted into kinetic energy causes the temperature of the material to rise
The amount of the temperature increase depends on how much heat was
added the size of the sample the original temperature of the sample and on
how the heat was added The two obvious choices on how to add the heat are
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
11
to add it holding volume constant or to add it holding pressure constant
(There may be other choices but they will not concern us)
Lets assume for the moment that we are going to add heat to our sample
holding volume constant that is 119889119889119889119889 = 0 Let 119876119876119889119889 be the heat added (the
subscript 119889119889 indicates that the heat is being added at constant 119889119889) Also let 120549120549120549120549
be the temperature change The ratio 119876119876119889119889∆120549120549 depends on the material the
amount of material and the temperature In the limit where 119876119876119889119889 goes to zero
(so that 120549120549120549120549 also goes to zero) this ratio becomes a derivative
lim119876119876119889119889rarr0
119876119876119889119889∆120549120549119889119889
= 120597120597119876119876120597120597120549120549119889119889
= 119862119862119889119889 (11)
We have given this derivative the symbol 119862119862119889119889 and we call it the heat
capacity at constant volume Usually one quotes the molar heat capacity
119862119862119889 equiv 119862119862119889119889119881119881 =119862119862119889119889119899119899
(12)
We can rearrange Equation (11) as follows
119889119889119876119876119889119889 = 119862119862119889119889 119889119889120549120549 (13)
Then we can integrate this equation to find the heat involved in a finite
change at constant volume
119876119876119889119889 = 119862119862119889119889
1205491205492
1205491205491
119889119889120549120549 (14)
If 119862119862119889119889 R Ris approximately constant over the temperature range then 119862119862119889119889 comes
out of the integral and the heat at constant volume becomes
119876119876119889119889 = 119862119862119889119889(1205491205492 minus 1205491205491) (15)
Let us now go through the same sequence of steps except holding pressure
constant instead of volume Our initial definition of the heat capacity at
constant pressure 119862119862119875119875 R Rbecomes
lim119876119876119875119875rarr0
119876119876119875119875∆120549120549119875119875
= 120597120597119876119876120597120597120549120549119875119875
= 119862119862119875119875 (16)
The analogous molar heat capacity is
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
12
119862119862119875 equiv 119862119862119875119875119881119881 =119862119862119875119875119899119899
(17)
Equation (16) rearranges to
119889119889119876119876119875119875 = 119862119862119875119875 119889119889120549120549 (18)
which integrates to give
119876119876119875119875 = 119862119862119875119875
1205491205492
1205491205491
119889119889120549120549 (19)
When 119862119862119875119875 is approximately constant the integral in Equation (19) becomes
119876119876119875119875 = 119862119862119875119875(1205491205492 minus 1205491205491) (110)
Very frequently the temperature range is large enough that 119862119862119875119875 cannot be
regarded as constant In these cases the heat capacity is fit to a polynomial (or
similar function) in 120549120549 For example some tables give the heat capacity as
119862119862119901 = 120572120572 + 120573120573120549120549 + 1205741205741205491205492 (111)
where 120572120572 120573120573 and 120574120574 are constants given in the table With this temperature-
dependent heat capacity the heat at constant pressure would integrate as
follows
119876119876119901119901 = 119899119899 (120572120572 + 120573120573120549120549 + 1205741205741205491205492)
1205491205492
1205491205491
119889119889120549120549 (112)
119876119876119901119901 = 119899119899 120572120572(1205491205492 minus 1205491205491) + 119899119899 12057312057321205491205492
2 minus 12054912054912 + 119899119899
1205741205743
12054912054923 minus 1205491205491
3 (113)
Occasionally one finds a different form for the temperature dependent heat
capacity in the literature
119862119862119901 = 119886119886 + 119887119887120549120549 + 119888119888120549120549minus2 (114)
When you do calculations with temperature dependent heat capacities you
must check to see which form is being used for 119862119862119875119875 We are using the
convention that 119876119876 will always designate heat absorbed by the system 119876119876 can
be positive or negative and the sign indicates which way heat is flowing If 119876119876 is
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
13
positive then heat was indeed absorbed by the system On the other hand if
119876119876 is negative it means that the system gave up heat to the surroundings
17 Thermal conductivity
In physics thermal conductivity 119896119896 is the property of a material that indicates
its ability to conduct heat It appears primarily in Fouriers Law for heat conduction
Thermal conductivity is measured in watts per Kelvin per meter (119882119882 middot 119870119870minus1 middot 119881119881minus1)
The thermal conductivity predicts the rate of energy loss (in watts 119882119882) through
a piece of material The reciprocal of thermal conductivity is thermal
resistivity
18 Derivation in one dimension
The heat equation is derived from Fouriers law and conservation of energy
(Cannon 1984) By Fouriers law the flow rate of heat energy through a
surface is proportional to the negative temperature gradient across the
surface
119902119902 = minus119896119896 120571120571120549120549 (115)
where 119896119896 is the thermal conductivity and 120549120549 is the temperature In one
dimension the gradient is an ordinary spatial derivative and so Fouriers law is
119902119902 = minus119896119896 120549120549119909119909 (116)
where 120549120549119909119909 is 119889119889120549120549119889119889119909119909 In the absence of work done a change in internal
energy per unit volume in the material 120549120549119876119876 is proportional to the change in
temperature 120549120549120549120549 That is
∆119876119876 = 119888119888119901119901 120588120588 ∆120549120549 (117)
where 119888119888119901119901 is the specific heat capacity and 120588120588 is the mass density of the
material Choosing zero energy at absolute zero temperature this can be
rewritten as
∆119876119876 = 119888119888119901119901 120588120588 120549120549 (118)
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
14
The increase in internal energy in a small spatial region of the material
(119909119909 minus ∆119909119909) le 120577120577 le (119909119909 + 120549120549119909119909) over the time period (119905119905 minus ∆119905119905) le 120591120591 le (119905119905 + 120549120549119905119905) is
given by
119888119888119875119875 120588120588 [120549120549(120577120577 119905119905 + 120549120549119905119905) minus 120549120549(120577120577 119905119905 minus 120549120549119905119905)] 119889119889120577120577119909119909+∆119909119909
119909119909minus∆119909119909
= 119888119888119875119875 120588120588 120597120597120549120549120597120597120591120591
119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(119)
Where the fundamental theorem of calculus was used Additionally with no
work done and absent any heat sources or sinks the change in internal energy
in the interval [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909] is accounted for entirely by the flux of heat
across the boundaries By Fouriers law this is
119896119896 120597120597120549120549120597120597119909119909
(119909119909 + 120549120549119909119909 120591120591) minus120597120597120549120549120597120597119909119909
(119909119909 minus 120549120549119909119909 120591120591) 119889119889120591120591119905119905+∆119905119905
119905119905minus∆119905119905
= 119896119896 12059712059721205491205491205971205971205771205772 119889119889120577120577 119889119889120591120591
119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
(120)
again by the fundamental theorem of calculus By conservation of energy
119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 119889119889120577120577 119889119889120591120591119909119909+∆119909119909
119909119909minus∆119909119909
119905119905+∆119905119905
119905119905minus∆119905119905
= 0 (121)
This is true for any rectangle [119905119905 minus 120549120549119905119905 119905119905 + 120549120549119905119905] times [119909119909 minus 120549120549119909119909 119909119909 + 120549120549119909119909]
Consequently the integrand must vanish identically 119888119888119875119875 120588120588 120549120549120591120591 minus 119896119896 120549120549120577120577120577120577 = 0
Which can be rewritten as
120549120549119905119905 =119896119896119888119888119875119875 120588120588
120549120549119909119909119909119909 (122)
or
120597120597120549120549120597120597119905119905
=119896119896119888119888119875119875 120588120588
12059712059721205491205491205971205971199091199092 (123)
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
15
which is the heat equation The coefficient 119896119896(119888119888119875119875 120588120588) is called thermal
diffusivity and is often denoted 120572120572
19 Aim of present work
The goal of this study is to estimate the solution of partial differential
equation that governs the laser-solid interaction using numerical methods
The solution will been restricted into one dimensional situation in which we
assume that both the laser power density and thermal properties are
functions of time and temperature respectively In this project we attempt to
investigate the laser interaction with both lead and copper materials by
predicting the temperature gradient with the depth of the metals
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
16
Chapter Two
Theoretical Aspects
21 Introduction
When a laser interacts with a solid surface a variety of processes can
occur We are mainly interested in the interaction of pulsed lasers with a
solid surface in first instance a metal When such a laser interacts with a
copper surface the laser energy will be transformed into heat The
temperature of the solid material will increase leading to melting and
evaporation of the solid material
The evaporated material (vapour atoms) will expand Depending on the
applications this can happen in vacuum (or very low pressure) or in a
background gas (helium argon air)
22 One dimension laser heating equation
In general the one dimension laser heating processes of opaque solid slab is
represented as
120588120588 119862119862 1198791198791 = 120597120597120597120597120597120597
( 119870119870 119879119879120597120597 ) (21)
With boundary conditions and initial condition which represent the pre-
vaporization stage
minus 119870119870 119879119879120597120597 = 0 119891119891119891119891119891119891 120597120597 = 119897119897 0 le 119905119905 le 119905119905 119907119907
minus 119870119870 119879119879120597120597 = 120572120572 119868119868 ( 119905119905 ) 119891119891119891119891119891119891 120597120597 = 00 le 119905119905 le 119905119905 119907119907 (22)
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
17
119879119879 ( 120597120597 0 ) = 119879119879infin 119891119891119891119891119891119891 119905119905 = 0 0 le 120597120597 le 119897119897
where
119870119870 represents the thermal conductivity
120588120588 represents the density
119862119862 represents the specific heat
119879119879 represents the temperature
119879119879infin represents the ambient temperature
119879119879119907119907 represents the front surface vaporization
120572120572119868119868(119905119905) represents the surface heat flux density absorbed by the slab
Now if we assume that 120588120588119862119862 119870119870 are constant the equation (21) becomes
119879119879119905119905 = 119889119889119889119889 119879119879120597120597120597120597 (23)
With the same boundary conditions as in equation (22)
where 119889119889119889119889 = 119870119870120588120588119862119862
which represents the thermal diffusion
But in general 119870119870 = 119870119870(119879119879) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879) there fore the derivation
equation (21) with this assuming implies
119879119879119905119905 = 1
120588120588(119879119879) 119862119862(119879119879) [119870119870119879119879 119879119879120597120597120597120597 + 119870119870 1198791198791205971205972] (24)
With the same boundary and initial conditions in equation (22) Where 119870119870
represents the derivative of K with respect the temperature
23 Numerical solution of Initial value problems
An immense number of analytical solutions for conduction heat-transfer
problems have been accumulated in literature over the past 100 years Even so
in many practical situations the geometry or boundary conditions are such that an
analytical solution has not been obtained at all or if the solution has been
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
18
developed it involves such a complex series solution that numerical evaluation
becomes exceedingly difficult For such situation the most fruitful approach to
the problem is numerical techniques the basic principles of which we shall
outline in this section
One way to guarantee accuracy in the solution of an initial values problems
(IVP) is to solve the problem twice using step sizes h and h2 and compare
answers at the mesh points corresponding to the larger step size But this requires
a significant amount of computation for the smaller step size and must be
repeated if it is determined that the agreement is not good enough
24 Finite Difference Method
The finite difference method is one of several techniques for obtaining
numerical solutions to differential equations In all numerical solutions the
continuous partial differential equation (PDE) is replaced with a discrete
approximation In this context the word discrete means that the numerical
solution is known only at a finite number of points in the physical domain The
number of those points can be selected by the user of the numerical method In
general increasing the number of points not only increases the resolution but
also the accuracy of the numerical solution
The discrete approximation results in a set of algebraic equations that are
evaluated for the values of the discrete unknowns
The mesh is the set of locations where the discrete solution is computed
These points are called nodes and if one were to draw lines between adjacent
nodes in the domain the resulting image would resemble a net or mesh Two key
parameters of the mesh are ∆120597120597 the local distance between adjacent points in
space and ∆119905119905 the local distance between adjacent time steps For the simple
examples considered in this article ∆120597120597 and ∆119905119905 are uniform throughout the mesh
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
19
The core idea of the finite-difference method is to replace continuous
derivatives with so-called difference formulas that involve only the discrete
values associated with positions on the mesh
Applying the finite-difference method to a differential equation involves
replacing all derivatives with difference formulas In the heat equation there are
derivatives with respect to time and derivatives with respect to space Using
different combinations of mesh points in the difference formulas results in
different schemes In the limit as the mesh spacing (∆120597120597 and ∆119905119905) go to zero the
numerical solution obtained with any useful scheme will approach the true
solution to the original differential equation However the rate at which the
numerical solution approaches the true solution varies with the scheme
241 First Order Forward Difference
Consider a Taylor series expansion empty(120597120597) about the point 120597120597119894119894
empty(120597120597119894119894 + 120575120575120597120597) = empty(120597120597119894119894) + 120575120575120597120597 120597120597empty1205971205971205971205971205971205971
+1205751205751205971205972
2 1205971205972empty1205971205971205971205972
1205971205971
+1205751205751205971205973
3 1205971205973empty1205971205971205971205973
1205971205971
+ ⋯ (25)
where 120575120575120597120597 is a change in 120597120597 relative to 120597120597119894119894 Let 120575120575120597120597 = ∆120597120597 in last equation ie
consider the value of empty at the location of the 120597120597119894119894+1 mesh line
empty(120597120597119894119894 + ∆120597120597) = empty(120597120597119894119894) + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (26)
Solve for (120597120597empty120597120597120597120597)120597120597119894119894
120597120597empty120597120597120597120597120597120597119894119894
=empty(120597120597119894119894 + ∆120597120597) minus empty(120597120597119894119894)
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (27)
Notice that the powers of ∆120597120597 multiplying the partial derivatives on the right
hand side have been reduced by one
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
20
Substitute the approximate solution for the exact solution ie use empty119894119894 asymp empty(120597120597119894119894)
and empty119894119894+1 asymp empty(120597120597119894119894 + ∆120597120597)
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (28)
The mean value theorem can be used to replace the higher order derivatives
∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ =∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (29)
where 120597120597119894119894 le 120585120585 le 120597120597119894119894+1 Thus
120597120597empty120597120597120597120597120597120597119894119894asympempty119894119894+1 minus empty119894119894
∆120597120597+∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (210)
120597120597empty120597120597120597120597120597120597119894119894minusempty119894119894+1 minus empty119894119894
∆120597120597asymp∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585 (211)
The term on the right hand side of previous equation is called the truncation
error of the finite difference approximation
In general 120585120585 is not known Furthermore since the function empty(120597120597 119905119905) is also
unknown 1205971205972empty1205971205971205971205972 cannot be computed Although the exact magnitude of the
truncation error cannot be known (unless the true solution empty(120597120597 119905119905) is available in
analytical form) the big 119978119978 notation can be used to express the dependence of
the truncation error on the mesh spacing Note that the right hand side of last
equation contains the mesh parameter ∆120597120597 which is chosen by the person using
the finite difference simulation Since this is the only parameter under the users
control that determines the error the truncation error is simply written
∆1205971205972
2 1205971205972empty1205971205971205971205972
120585120585= 119978119978(∆1205971205972) (212)
The equals sign in this expression is true in the order of magnitude sense In
other words the 119978119978(∆1205971205972) on the right hand side of the expression is not a strict
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
21
equality Rather the expression means that the left hand side is a product of an
unknown constant and ∆1205971205972 Although the expression does not give us the exact
magnitude of (∆1205971205972)2(1205971205972empty1205971205971205971205972)120597120597119894119894120585120585 it tells us how quickly that term
approaches zero as ∆120597120597 is reduced
Using big 119978119978 notation Equation (28) can be written
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597+ 119978119978(∆120597120597) (213)
This equation is called the forward difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 because
it involves nodes 120597120597119894119894 and 120597120597119894119894+1 The forward difference approximation has a
truncation error that is 119978119978(∆120597120597) The size of the truncation error is (mostly) under
our control because we can choose the mesh size ∆120597120597 The part of the truncation
error that is not under our control is |120597120597empty120597120597120597120597|120585120585
242 First Order Backward Difference
An alternative first order finite difference formula is obtained if the Taylor series
like that in Equation (4) is written with 120575120575120597120597 = minus∆120597120597 Using the discrete mesh
variables in place of all the unknowns one obtains
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (214)
Notice the alternating signs of terms on the right hand side Solve for (120597120597empty120597120597120597120597)120597120597119894119894
to get
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894
∆120597120597minus∆1205971205972
1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (215)
Or using big 119978119978 notation
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894 minus empty119894119894minus1
∆120597120597+ 119978119978(∆120597120597) (216)
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
22
This is called the backward difference formula because it involves the values of
empty at 120597120597119894119894 and 120597120597119894119894minus1
The order of magnitude of the truncation error for the backward difference
approximation is the same as that of the forward difference approximation Can
we obtain a first order difference formula for (120597120597empty120597120597120597120597)120597120597119894119894 with a smaller
truncation error The answer is yes
242 First Order Central Difference
Write the Taylor series expansions for empty119894119894+1 and empty119894119894minus1
empty119894119894+1 = empty119894119894 + ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
+∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (217)
empty119894119894minus1 = empty119894119894 minus ∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+∆1205971205972
2 1205971205972empty1205971205971205971205972
120597120597119894119894
minus∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (218)
Subtracting Equation (10) from Equation (9) yields
empty119894119894+1 minus empty119894119894minus1 = 2∆120597120597 120597120597empty120597120597120597120597120597120597119894119894
+ 2∆1205971205973
3 1205971205973empty1205971205971205971205973
120597120597119894119894
+ ⋯ (219)
Solving for (120597120597empty120597120597120597120597)120597120597119894119894 gives
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597minus∆1205971205972
3 1205971205973empty1205971205971205971205973
120597120597119894119894
minus ⋯ (220)
or
120597120597empty120597120597120597120597120597120597119894119894
=empty119894119894+1 minus empty119894119894minus1
2∆120597120597+ 119978119978(∆1205971205972) (221)
This is the central difference approximation to (120597120597empty120597120597120597120597)120597120597119894119894 To get good
approximations to the continuous problem small ∆120597120597 is chosen When ∆120597120597 ≪ 1
the truncation error for the central difference approximation goes to zero much
faster than the truncation error in forward and backward equations
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
23
25 Procedures
The simple case in this investigation was assuming the constant thermal
properties of the material First we assumed all the thermal properties of the
materials thermal conductivity 119870119870 heat capacity 119862119862 melting point 119879119879119898119898 and vapor
point 119879119879119907119907 are independent of temperature About laser energy 119864119864 at the first we
assume the constant energy after that the pulse of special shapes was selected
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) was investigated using Matlab program as shown in Appendix
The equation of thermal conductivity and specific heat capacity of metal as a
function of temperature was obtained by best fitting of polynomials using
tabulated data in references
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
24
Chapter Three
Results and Discursion
31 Introduction
The development of laser has been an exciting chapter in the history of
science and engineering It has produced a new type of advice with potential for
application in an extremely wide variety of fields Mach basic development in
lasers were occurred during last 35 years The lasers interaction with metal and
vaporize of metals due to itrsquos ability for welding cutting and drilling applicable
The status of laser development and application were still rather rudimentary
The light emitted by laser is electro magnetic radiation this radiation has a wave
nature the waves consists of vibrating electric and magnetic fields many studies
have tried to find and solve models of laser interactions Some researchers
proposed the mathematical model related to the laser - plasma interaction and
the others have developed an analytical model to study the temperature
distribution in Infrared optical materials heated by laser pulses Also an attempt
have made to study the interaction of nanosecond pulsed lasers with material
from point of view using experimental technique and theoretical approach of
dimensional analysis
In this study we have evaluate the solution of partial difference equation
(PDE) that represent the laser interaction with solid situation in one dimension
assuming that the power density of laser and thermal properties are functions
with time and temperature respectively
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
25
32 Numerical solution with constant laser power density and constant
thermal properties
First we have taken the lead metal (Pb) with thermal properties
119870119870 = 22506 times 10minus5 119869119869119898119898119898119898119898119898119898119898 119898119898119898119898119870119870
119862119862 = 014016119869119869119892119892119870119870
120588120588 = 10751 1198921198921198981198981198981198983
119879119879119898119898 (119898119898119898119898119898119898119898119898119898119898119898119898119892119892 119901119901119901119901119898119898119898119898119898119898) = 600 119870119870
119879119879119907119907 (119907119907119907119907119901119901119901119901119907119907 119901119901119901119901119898119898119898119898119898119898) = 1200 119870119870
and we have taken laser energy 119864119864 = 3 119869119869 119860119860 = 134 times 10minus3 1198981198981198981198982 where 119860119860
represent the area under laser influence
The numerical solution of equation (23) with boundary and initial conditions
in equation (22) assuming (119868119868 = 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) with the thermal properties
of lead metal by explicit method using Matlab program give us the results as
shown in Fig (31)
Fig(31) Depth dependence of the temperature with the laser power density
1198681198680 = 76 times 106 119882119882119898119898119898119898 2
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
26
33 Evaluation of function 119920119920(119957119957) of laser flux density
From following data that represent the energy (119869119869) with time (millie second)
Time 0 001 01 02 03 04 05 06 07 08
Energy 0 002 017 022 024 02 012 007 002 0
By using Matlab program the best polynomial with deduced from above data
was
119864119864(119898119898) = 37110 times 10minus4 + 21582 119898119898 minus 57582 1198981198982 + 36746 1198981198983 + 099414 1198981198984
minus 10069 1198981198985 (31)
As shown in Fig (32)
Fig(32) Laser energy as a function of time
Figure shows the maximum value of energy 119864119864119898119898119907119907119898119898 = 02403 119869119869 The
normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) is deduced by dividing 119864119864(119898119898) by the
maximum value (119864119864119898119898119907119907119898119898 )
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 =119864119864(119898119898)
(119864119864119898119898119907119907119898119898 ) (32)
The normalized function ( 119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899 ) was shown in Fig (33)
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
27
Fig(33) Normalized laser energy as a function of time
The integral of 119864119864(119898119898) normalized over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 = 08 (119898119898119898119898119898119898119898119898) must
equal to 3 (total laser energy) ie
119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (33)
Therefore there exist a real number 119875119875 such that
119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (34)
that implies 119875119875 = 68241 and
119864119864(119898119898) = 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
08
00
119899119899119898119898 = 3 (35)
The integral of laser flux density 119868119868 = 119868119868(119898119898) over 119898119898 from 119898119898 = 00 119898119898119901119901 119898119898 =
08 ( 119898119898119898119898119898119898119898119898 ) must equal to ( 1198681198680 = 76 times 106 119882119882119898119898119898119898 2) there fore
119868119868 (119898119898)119899119899119898119898 = 1198681198680 11986311986311989811989808
00
(36)
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
28
Where 119863119863119898119898 put to balance the units of equation (36)
But integral
119868119868 = 119864119864119860119860
(37)
and from equations (35) (36) and (37) we have
119868119868 (119898119898)11989911989911989811989808
00
= 119899119899 int 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
0800 119899119899119898119898
119860119860 119863119863119898119898 (38)
Where 119899119899 = 395 and its put to balance the magnitude of two sides of equation
(38)
There fore
119868119868(119898119898) = 119899119899 119875119875119864119864119898119898119901119901119907119907119898119898119907119907119898119898119898119898119899119899119898119898119899119899
119860119860 119863119863119898119898 (39)
As shown in Fig(34) Matlab program was used to obtain the best polynomial
that agrees with result data
119868119868(119898119898) = 26712 times 101 + 15535 times 105 119898119898 minus 41448 times 105 1198981198982 + 26450 times 105 1198981198983
+ 71559 times 104 1198981198984 minus 72476 times 104 1198981198985 (310)
Fig(34) Time dependence of laser intensity
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
29
34 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
constant thermal properties
With all constant thermal properties of lead metal as in article (23) and
119868119868 = 119868119868(119898119898) we have deduced the numerical solution of heat transfer equation as in
equation (23) with boundary and initial condition as in equation (22) and the
depth penetration is shown in Fig(35)
Fig(35) Depth dependence of the temperature when laser intensity function
of time and constant thermal properties of Lead
35 Evaluation the Thermal Conductivity as functions of temperature
The best polynomial equation of thermal conductivity 119870119870(119879119879) as a function of
temperature for Lead material was obtained by Matlab program using the
experimental data tabulated in researches
119870119870(119879119879) = minus17033 times 10minus3 + 16895 times 10minus5 119879119879 minus 50096 times 10minus8 1198791198792 + 66920
times 10minus11 1198791198793 minus 41866 times 10minus14 1198791198794 + 10003 times 10minus17 1198791198795 (311)
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
30
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
300 353 400 332 500 315 600 190 673 1575 773 152 873 150 973 150 1073 1475 1173 1467 1200 1455
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (36)
Fig(36) The best fitting of thermal conductivity of Lead as a function of
temperature
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
31
36 Evaluation the Specific heat as functions of temperature
The equation of specific heat 119862119862(119879119879) as a function of temperature for Lead
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
300 01287 400 0132 500 0136 600 01421 700 01465 800 01449 900 01433 1000 01404 1100 01390 1200 01345
The best polynomial fitted for these data was
119862119862(119879119879) = minus46853 times 10minus2 + 19426 times 10minus3 119879119879 minus 86471 times 10minus6 1198791198792
+ 19546 times 10minus8 1198791198793 minus 23176 times 10minus11 1198791198794 + 13730
times 10minus14 1198791198795 minus 32083 times 10minus18 1198791198796 (312)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (37)
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
32
Fig(37) The best fitting of specific heat capacity of Lead as a function of
temperature
37 Evaluation the Density as functions of temperature
The density of Lead 120588120588(119879119879) as a function of temperature tacked from literature
was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
300 11330 400 11230 500 11130 600 11010 800 10430
1000 10190 1200 9940
The best polynomial of this data was
120588120588(119879119879) = 10047 + 92126 times 10minus3 119879119879 minus 21284 times 10minus3 1198791198792 + 167 times 10minus8 1198791198793
minus 45158 times 10minus12 1198791198794 (313)
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
33
The density of Lead as a function of temperature and the best polynomial fitting
are shown in Fig (38)
Fig(38) The best fitting of density of Lead as a function of temperature
38 Numerical solution with variable laser power density ( 119920119920 = 119920119920 (119957119957) ) and
variable thermal properties 119922119922 = 119922119922(119931119931)119914119914 = 119914119914(119931119931)120646120646 = 120646120646(119931119931)
We have deduced the solution of equation (24) with initial and boundary
condition as in equation (22) using the function of 119868119868(119898119898) as in equation (310)
and the function of 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as in equation (311 312 and 313)
respectively then by using Matlab program the depth penetration is shown in
Fig (39)
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
34
Fig(39) Depth dependence of the temperature for pulse laser on Lead
material
39 Laser interaction with copper material
The same time dependence of laser intensity as shown in Fig(34) with
thermal properties of copper was used to calculate the temperature distribution as
a function of depth penetration
The equation of thermal conductivity 119870119870(119879119879) as a function of temperature for
copper material was obtained from the experimental data tabulated in literary
The Matlab program used to obtain the best polynomial equation that agrees
with the above data
119870119870(119879119879) = 602178 times 10minus3 minus 166291 times 10minus5 119879119879 + 506015 times 10minus8 1198791198792
minus 735852 times 10minus11 1198791198793 + 497708 times 10minus14 1198791198794 minus 126844
times 10minus17 1198791198795 (314)
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
35
119879119879 ( 119870119870) 119870119870 119869119869
119898119898119898119898119898119898119898119898 1198981198981198981198981198701198700 times 10minus5
100 482 200 413 273 403 298 401 400 393 600 379 800 366 1000 352 1100 346 1200 339 1300 332
The previous thermal conductivity data and the best fitting of the data are
shown in Fig (310)
Fig(310) The best fitting of thermal conductivity of Copper as a function of
temperature
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
36
The equation of specific heat 119862119862(119879119879) as a function of temperature for Copper
material was obtained from the following experimental data tacked from
literatures
119879119879 (119870119870) 119862119862 (119869119869119892119892119870119870)
100 0254
200 0357
273 0384
298 0387
400 0397
600 0416
800 0435
1000 0454
1100 0464
1200 0474
1300 0483
The best polynomial fitted for these data was
119862119862(119879119879) = 61206 times 10minus4 + 36943 times 10minus3 119879119879 minus 14043 times 10minus5 1198791198792
+ 27381 times 10minus8 1198791198793 minus 28352 times 10minus11 1198791198794 + 14895
times 10minus14 1198791198795 minus 31225 times 10minus18 1198791198796 (315)
The specific heat capacity data and the best polynomial fitting of the data are
shown in Fig (311)
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
37
Fig(311) The best fitting of specific heat capacity of Copper as a function of
temperature
The density of copper 120588120588(119879119879) as a function of temperature tacked from
literature was used to find the best polynomial fitting
119879119879 (119870119870) 120588120588 (1198921198921198981198981198981198983)
100 9009 200 8973 273 8942 298 8931 400 8884 600 8788 800 8686
1000 8576 1100 8519 1200 8458 1300 8396
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
38
The best polynomial of this data was
120588120588(119879119879) = 90422 minus 29641 times 10minus4 119879119879 minus 31976 times 10minus7 1198791198792 + 22681 times 10minus10 1198791198793
minus 76765 times 10minus14 1198791198794 (316)
The density of copper as a function of temperature and the best polynomial
fitting are shown in Fig (312)
Fig(312) The best fitting of density of copper as a function of temperature
The depth penetration of laser energy for copper metal was calculated using
the polynomial equations of thermal conductivity specific heat capacity and
density of copper material 119870119870(119879119879)119862119862(119879119879) and 120588120588(119879119879) as a function of temperature
(equations (314 315 and 316) respectively) with laser intensity 119868119868(119898119898) as a
function of time the result was shown in Fig (313)
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
39
The temperature gradient in the thickness of 0018 119898119898119898119898 was found to be 900119901119901119862119862
for lead metal whereas it was found to be nearly 80119901119901119862119862 for same thickness of
copper metal so the depth penetration of laser energy of lead metal was smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
Fig(313) Depth dependence of the temperature for pulse laser on Copper
material
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
40
310 Conclusions
The Depth dependence of temperature for lead metal was investigated in two
case in the first case the laser intensity assume to be constant 119868119868 = 1198681198680 and the
thermal properties (thermal conductivity specific heat) and density of metal are
also constants 119870119870 = 1198701198700 120588120588 = 1205881205880119862119862 = 1198621198620 in the second case the laser intensity
vary with time 119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity
specific heat) and density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 =
120588120588(119879119879) 119862119862 = 119862119862(119879119879) Comparison the results of this two cases shows that the
penetration depth in the first case is smaller than that of the second case about
(190) times
The temperature distribution as a function of depth dependence for copper
metal was also investigated in the case when the laser intensity vary with time
119868119868 = 119868119868(119898119898) and the thermal properties (thermal conductivity specific heat) and
density of metal are function of temperature 119870119870 = 119896119896(119898119898) 120588120588 = 120588120588(119879119879) 119862119862 = 119862119862(119879119879)
The depth penetration of laser energy of lead metal was found to be smaller
than that of copper metal this may be due to the high thermal conductivity and
high specific heat capacity of copper with that of lead metal
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
41
References [1] Adrian Bejan and Allan D Kraus Heat Transfer Handbook John Wiley amp
Sons Inc Hoboken New Jersey Canada (2003)
[2] S R K Iyengar and R K Jain Numerical Methods New Age International (P) Ltd Publishers (2009)
[3] Hameed H Hameed and Hayder M Abaas Numerical Treatment of Laser Interaction with Solid in One Dimension A Special Issue for the 2nd
[9]
Conference of Pure amp Applied Sciences (11-12) March p47 (2009)
[4] Remi Sentis Mathematical models for laser-plasma interaction ESAM Mathematical Modelling and Numerical Analysis Vol32 No2 PP 275-318 (2005)
[5] John Emsley ldquoThe Elementsrdquo third edition Oxford University Press Inc New York (1998)
[6] O Mihi and D Apostol Mathematical modeling of two-photo thermal fields in laser-solid interaction J of optic and laser technology Vol36 No3 PP 219-222 (2004)
[7] J Martan J Kunes and N Semmar Experimental mathematical model of nanosecond laser interaction with material Applied Surface Science Vol 253 Issue 7 PP 3525-3532 (2007)
[8] William M Rohsenow Hand Book of Heat Transfer Fundamentals McGraw-Hill New York (1985)
httpwwwchemarizonaedu~salzmanr480a480antsheatheathtml
[10] httpwwwworldoflaserscomlaserprincipleshtm
[11] httpenwikipediaorgwikiLaserPulsed_operation
[12] httpenwikipediaorgwikiThermal_conductivity
[13] httpenwikipediaorgwikiHeat_equationDerivation_in_one_dimension
[14] httpwebh01uaacbeplasmapageslaser-ablationhtml
[15] httpwebcecspdxedu~gerryclassME448codesFDheatpdf
42
Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
43
E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
44
title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
45
elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
46
r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
47
for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
48
for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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Appendix This program calculate the laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) plot(tEr) grid on hold on i=000208 E1=polyval(ui) plot(iE1-b) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the normalized laser energy as a function of time clear all clc t=[00112345678] E=[00217222421207020] u=polyfit(tE5) i=000108 E1=polyval(ui) m=max(E1) unormal=um E2=polyval(unormali) plot(iE2-b) grid on hold on Enorm=Em plot(tEnormr) xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the laser intensity as a function of time clear all clc Io=76e3 laser intensity Jmille sec cm^2 p=68241 this number comes from pintegration of E-normal=3 ==gt p=3integration of E-normal z=395 this number comes from the fact zInt(I(t))=Io in magnitude A=134e-3 this number represents the area through heat flow Dt=1 this number comes from integral of I(t)=IoDt its come from equality of units t=[00112345678] E=[00217222421207020] Energy=polyfit(tE5) this step calculate the energy as a function of time i=000208 loop of time E1=polyval(Energyi) m=max(E1) Enormal=Energym this step to find normal energy I=z((pEnormal)(ADt))
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E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
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title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
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elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
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r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
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for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
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6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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E2=polyval(Ii) plot(iE2-b) E2=polyval(It) hold on plot(tE2r) grid on xlabel(Time (m sec)) ylabel(Energy (J)) title(The time dependence of the energy) hold off
This program calculate the thermal conductivity as a function temperature clear all clc T=[300400500600673773873973107311731200] K=(1e-5)[353332531519015751525150150147514671455] KT=polyfit(TK5) i=300101200 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off
This program calculate the specific heat as a function temperature clear all clc T=[300400500600700800900100011001200] C=[12871321361421146514491433140413901345] u=polyfit(TC6) i=300101200 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off
This program calculate the dencity as a function temperature clear all clc T=[30040050060080010001200] P=[113311231113110110431019994] u=polyfit(TP4) i=300101200 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3))
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title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
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elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
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r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
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for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
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6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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title(Dencity as a function of temperature) hold off
This program calculate the vaporization time and depth peneteration when laser intensity vares as a function of time and thermal properties are constant ie I(t)=Io K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=1e-4 h=5e-6 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 t=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 Io=76e3 C=014016 rho=10751 du=K(rhoC) r1=(2alphaIoh)K for i=1N T1(i)=300 end for t=1100 t1=t1+1 dt=dt+2e-10 x=x+h x1(t1)=x r=(dtdu)h^2 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N
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elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
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r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
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for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
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6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=08 break end end if abs(Tv-T2(i))lt=08 break end dt=dt+2e-10 x=x+h t1=t1+1 x1(t1)=x r=(dtdu)h^2 This loop to calculate the next temperature depending on previous temperature for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=08 break end end if abs(Tv-T1(i))lt=08 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are constant ie I(t)=I(t) K(T)=Ko rho(T)=roho C(T)=Co a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec)
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r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
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for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
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6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=00175 h=00005 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 K=22506e-5 thermal conductivity of lead (JmSeccmK) alpha=1 C=014016 rho=10751 du=K(rhoC) for i=1N T1(i)=300 end for t=11200 t1=t1+1 dt=dt+5e-8 x=x+h x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(T1(i+1)-T1(i)+(r12)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+5e-8 x=x+h t1=t1+1 x1(t1)=x I=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 r=(dtdu)h^2 r1=(2alphaIh)K This loop to calculate the next temperature depending on previous temperature
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for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
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6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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for i=1N if i==1 T1(i)=T2(i)+2r(T2(i+1)-T2(i)+(r12)) elseif i==N T1(i)=T2(i)+2r(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1200 the vaporization temperature of lead dt=0 x=-h t1=0 alpha=1 for i=1N T1(i)=300 end
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
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6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
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for t=111000 t1=t1+1 dt=dt+9e-10 x=x+h x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+ 71510dt^4-72426dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+ 6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+9e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=26694+155258e5dt-41419e5dt^2+26432e5dt^3+71510dt^4-72426dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=-17033e-3+16895e-5T1(i)-50096e-8T1(i)^2+
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
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u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
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alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
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715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
49
6692e-11T1(i)^3-41866e-14T1(i)^4+10003e-17T1(i)^5 Kdash(i)=16895e-5-250096e-8T1(i)+ 36692e-11T1(i)^2-441866e-14T1(i)^3+510003e-17T1(i)^4 C(i)=-46853e-2+19426e-3T1(i)-86471e-6T1(i)^2+19546e-8T1(i)^3- 23176e-11T1(i)^4+1373e-14T1(i)^5-32089e-18T1(i)^6 rho(i)=10047+92126e-3T1(i)-21284e-5T1(i)^2+167e-8T1(i)^3- 45158e-12T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=iN T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
This program calculate the thermal conductivity as a function temperature clear all clc T=[1002002732984006008001000110012001300] K=(1e-5)[482413403401393379366352346339332] KT=polyfit(TK5) i=100101300 loop of temperature KT1=polyval(KTi) plot(TKr) hold on plot(iKT1-b) grid on xlabel(Temperature (K)) ylabel(Thermal conductivity (J m secK)) title(Thermal conductivity as a function of temperature) hold off This program calculate the specific heat as a function temperature clear all clc T=[1002002732984006008001000110012001300] C=[254357384387397416435454464474483]
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
50
u=polyfit(TC6) i=100101300 loop of temperature C1=polyval(ui) plot(TCr) hold on plot(iC1-b) grid on xlabel(Temperature (K)) ylabel(Specific heat capacity (JgmK)) title(Specific heat capacity as a function of temperature) hold off This program calculate the dencity as a function temperature clear all clc T=[1002002732984006008001000110012001300] P=[90098973894289318884878886868576851984588396] u=polyfit(TP4) i=100101300 loop of temperature P1=polyval(ui) plot(TPr) hold on plot(iP1-b) grid on xlabel(Temperature (K)) ylabel(Dencity (gcm^3)) title(Dencity as a function of temperature) hold off This program calculate the vaporization time and depth peneteration when laser intensity and thermal properties are variable ie I=I(t) K=K(T) rho=roh(T) C=C(T) a reprecent the first edge of metal plate b reprecent the second edge of metal plate h increment step N number of points Tv the vaporization temperature of metal dt iteration for time x iteration for depth K Thermal Conductivity alpha Absorption coefficient since the surface of metal is opaque =1 Io laser intencity with unit (JmSeccm^2) I laser intencity as a function of time with unit (JmSeccm^2) C Spescific heat with unit (JgK) rho Dencity with unit (gcm^3) du Termal diffusion with unit (cm^2mSec) r1 this element comes from finite difference method at boundary conditions T1 the initial value of temperature with unit (Kelvin) clear all clc a=0 b=0018 h=00009 N=round((b-a)h) Tv=1400 the vaporization temperature of copper dt=0 x=-h t1=0
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
51
alpha=1 for i=1N T1(i)=300 end for t=111000 t1=t1+1 dt=dt+1e-10 x=x+h x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+ 715599e4dt^4-724766e4dt^5 This loop calculate the temperature at second point of penetration depending on initial temperature T=300 for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) This equation to calculate the temperature at x=a if i==1 T2(i)=T1(i)+2r(i)(T1(i+1)-T1(i)+(r1(i)2)) This equation to calculate the temperature at x=N elseif i==N T2(i)=T1(i)+2r(i)(T1(i-1)-T1(i)) This equation to calculate the temperature at all other points else T2(i)=T1(i)+r(i)(T1(i+1)-2T1(i)+T1(i-1)) end To compare the calculated temperature with vaporization temperature if abs(Tv-T2(i))lt=04 break end end if abs(Tv-T2(i))lt=04 break end dt=dt+1e-10 x=x+h t1=t1+1 x1(t1)=x I(t1)=267126e1+155354e5dt-414184e5dt^2+264503e5dt^3+
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))
52
715599e4dt^4-724766e4dt^5 This loop to calculate the next temperature depending on previous temperature for i=1N The following equation represent the thermal conductivity specific heat and density as a function of temperature K(i)=602178e-3-16629e-5T1(i)+50601e-8T1(i)^2- 73582e-11T1(i)^3+4977e-14T1(i)^4-12684e-17T1(i)^5 Kdash(i)=-16629e-5+250601e-8T1(i)- 373582e-11T1(i)^2+44977e-14T1(i)^3-512684e-17T1(i)^4 C(i)=61206e-4+36943e-3T1(i)-14043e-5T1(i)^2+27381e-8T1(i)^3- 28352e-11T1(i)^4+14895e-14T1(i)^5-31225e-18T1(i)^6 rho(i)=90422-29641e-4T1(i)-31976e-7T1(i)^2+22681e-10T1(i)^3- 76765e-14T1(i)^4 du(i)=K(i)(rho(i)C(i)) r(i)=dt((h^2)rho(i)C(i)) r1(i)=((2alphaI(t1)h)K(i)) if i==1 T1(i)=T2(i)+2r(i)(T2(i+1)-T2(i)+(r1(i)2)) elseif i==N T1(i)=T2(i)+2r(i)(T2(i-1)-T2(i)) else T1(i)=T2(i)+r(i)(T2(i+1)-2T2(i)+T2(i-1)) end if abs(Tv-T1(i))lt=04 break end end if abs(Tv-T1(i))lt=04 break end end This loop to make the peneteration and temperature as matrices to polt them for i=1N T(i)=T1(i) x2(i)=x1(i) end plot(x2Trx2Tb-) grid on xlabel(Depth (cm)) ylabel(Temperature (K))