Lecture Notes_Heat and Mass Transfer

8/20/2019 Lecture Notes_Heat and Mass Transfer

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LECTURE NOTES

ON

HEAT & MASS TRANSFER

BY

DR. T.R.SEETHARAM(PESIT, BANGALORE)

1


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CHAPTER 1

INTRODUCTORY CONCEPTS AND BASIC LAWSOF HEAT TRANSFER

1.1. Intro!"t#on$% We recall from our knowledge of thermodynamics that heat is a form of energy transfer that takes place from a region of higher temperature to a region oflower temperature solely due to the temperature difference between the two regions. Withthe knowledge of thermodynamics we can determine the amount of heat transfer for anysystem undergoing any process from one equilibrium state to another. Thus thethermodynamics knowledge will tell us only how much heat must be transferred to

achieve a specified change of state of the system. But in practice we are more interestedin knowing the rate of heat transfer (i.e. heat transfer per unit time) rather than theamount. This knowledge of rate of heat transfer is necessary for a design engineer todesign all types of heat transfer equipments like boilers condensers furnaces coolingtowers dryers etc.The sub!ect of heat transfer deals with the determination of the rate ofheat transfer to or from a heat e"change equipment and also the temperature at anylocation in the device at any instant of time. The basic requirement for heat transfer is the presence of a#temperature difference$. The temperature difference is the driving force for heat transfer !ust as the voltage difference for electric current flow and pressure difference for fluidflow. %ne of the parameters on which the rate of heat transfer in a certain direction

depends is the magnitude of the temperature gradient in that direction. The larger thegradient higher will be the rate of heat transfer.

1.. H't Trn*'r M'"+n#$% There are three mechanisms by which heat transfercan take place. &ll the three modes require the e"istence of temperature difference. Thethree mechanisms are' (i) conduction (ii) convection and (iii) radiation

1..1Con!"t#on$% t is the energy transfer that takes place at molecular levels.onduction is the transfer of energy from the more energetic molecules of a substance tothe ad!acent less energetic molecules as a result of interaction between the molecules. nthe case of liquids and gases conduction is due to collisions and diffusion of the

molecules during their random motion. n solids it is due to the vibrations of themolecules in a lattice and motion of free electrons.Fourier’s Law of Heat Conduction:- The empirical law of conduction based one"perimental results is named after the *rench +hysicist ,oseph *ourier. The law statesthat the rate of heat flow by conduction in any medium in any direction is proportional tothe area normal to the direction of heat flow and also proportional to the temperaturegradient in that direction. *or e"ample the rate of heat transfer in "-direction can bewritten according to *ouriers law as

/


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1.

0" 2 & (dT 3 d") 44444444.(1.1)

%r 0" 5 2 k & (dT 3 d") W4444444.. ..(1./)n equation (1./) 0" is the rate of heat transfer in positive "-direction through area & ofthe medium normal to "-direction (dT3d") is the temperature gradient and k is theconstant of proportionality and is a material property called “thermal conductivity”.6ince heat transfer has to take place in the direction of decreasing temperature (dT3d")has to be negative in the direction of heat transfer. Therefore negative sign has to beintroduced in equation (1./) to make 0" positive in the direction of decreasingtemperature thereby satisfying the second law of thermodynamics. f equation (1./) isdivided throughout by & we have

q" 5 (0" 3 &) 5 2 k (dT 3 d") W3m

/

444..(1.7)q" is called the heat flux.

Thermal Conductivity:- The constant of proportionality in the equation of *ouriers lawof conduction is a material property called the thermal conductivity.The units of thermalconductivity can be obtained from equation (1./) as follows'

6olving for k from 8q. (1./) we have k 5 2 q" 3 (dT3d")

Therefore units of k 5 (W3m/ ) (m3 9) 5 W 3 (m : 9) or W 3 (m : ; ). Thermalconductivity is a measure of a materials ability to conduct heat. The thermalconductivities of materials vary over a wide range as shown in *ig. 1.1. t can be seen from this figure that the thermal conductivities of gases such asair vary by a factor of 1;


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1.-

F#. 1.1$ T/0#" rn' o* t+'r "on!"t#2#t#' o* 2r#o! t'r#

n the case of solids heat conduction is due to two effects' the vibration of latticeinduced by the vibration of molecules positioned at relatively fi"ed positions andenergy transported due to the motion of free electrons. The relatively high thermalconductivities of pure metals are primarily due to the electronic component. The latticecomponent of thermal conductivity strongly depends on the way the molecules arearranged. *or e"ample diamond which is highly ordered crystalline solid has thehighest thermal conductivity at room temperature.

@nlike metals which are good electrical and heat conductors crystalline solidssuch as diamond and semiconductors such as silicon are good heat conductors but poorelectrical conductors. =ence such materials find widespread use in electronic industry.espite their high price diamond heat sinks are used in the cooling of sensitive electroniccomponents because of their e"cellent thermal conductivity. 6ilicon oils and gaskets arecommonly used in the packaging of electronic components because they provide bothgood thermal contact and good electrical insulation.

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133

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1.3

3.1

3.31

4 (W5%6)

So#'t

L#7!#'t

Non%M't#"o#

Non%M't#"#7!#

In!t#nMt'r#

Non%M't#"'

E2"!t'In!t#nt'r#

S#2'r

Co00'r

So#!

St''

M'r"!r/

O8#'

Pt#"

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Wt'r

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F#9r'

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1.: %ne would e"pect that metal alloys will have high thermalconductivities because pure metals have high thermal conductivities. *or e"ample onewould e"pect that the value of the thermal conductivity k of a metal alloy made of two

metals with thermal conductivities k 1 and k / would lie between k 1 and k /.But this is notthe case. n fact k of a metal alloy will be less than that of either metal.

The thermal conductivities of materials vary with temperature. Butfor some materials the variation is insignificant even for wide temperature range.&ttemperatures near absolute Cero the thermal conductivities of certain solids are e"tremelylarge. *or e"ample copper at /; 9 will have a thermal conductivity of /;;;; W 3 (m-9)which is about D; times the conductivity at room temperature. The temperaturedependence of thermal conductivity makes the conduction heat transfer analysis morecomple" and involved. &s a first appro"imation analysis for solids with variableconductivity is carried out assuming constant thermal conductivity which is an average

value of the conductivity for the temperature range of interest.Thermal Diffusivity:- This is a property which is very helpful in analyCing transient heatconduction problem and is normally denoted by the symbol . t is defined as follows. =eat conducted k 5 -------------------------------------- 5 -------- (m/3s) 44(1.


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1.;

7 < + =T >>>>>>>>>>>>>>.. (1.;)

where q is the heat flu" GT is the temperature difference between the bulk fluid and thesurface which is in contact with the fluid and Hh$ is called the “convective heat transfer coefficient” or “surface film coefficient ” . 8q.(1.D) is generally referred to as the Iewtons law of cooling.f Ts is the surface temperature Tf is the temperature of the bulk fluid and if Ts J Tf then 8q. (1.D) in the direction of heat transfer can be written as

7 5 + ?T @ T* >>>>>>>>>>>>...(1.)

and if Ts K Tf the equation reduces to

7 5 + ?T* @ T >>>>>>>>>>>>...(1.9)

The heat transfer coefficient h depends on (i) the type of flow (i.e. whether theflow is laminar or turbulent) (ii) the geometry of the body and flow passage area (iii)the thermo-physical properties of the fluid namely the density E viscosity L specific heatat constant pressure p and the thermal conductivity of the fluid k and (iv) whether themechanism of convection is forced convection or free convection. The heat transfercoefficient for free convection will be generally lower than that for forced convection asthe fluid velocities in free convection are much lower than those in forced convection.The heat transfer coefficients for some typical applications are given in table 1./.

T9' 1.$ T/0#" 2!' o* t+' "on2'"t#2' +'t trn*'r "o'**#"#'nt +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%Type of flow h W 3 (m/ : 9)Free convection

Mases / : /D Aiquids D; : 1;;;Forced Convection

Mases /D : /D; Aiquids D; : /;;;;Convection with change of phase

Boiling or condensation /D;; : 1;;;;;

1..-. T+'r R#t#on$% Thermal radiation is the energy emitted by matter (solidliquid or gas) by virtue of its temperature. This energy is transported by electromagneticwaves (or alternatively photons).While the transfer of energy by conduction andconvection requires the presence of a material medium radiation does not require.nfactradiation transfer occurs most effectively in vacuum.

onsider radiation transfer process for the surface shown in *ig.1./a.Fadiation that

N


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1.

is emitted by the surface originates from the thermal energy of matter bounded by thesurface and the rate at which this energy is released per unit area is called as the surfaceemissive power E.&n ideal surface is one which emits ma"imum emissive power and iscalled an ideal radiator or a black body.6tefan-BoltCmans law of radiation states that theemissive power of a black body is proportional to the fourth power of the absolutetemperature of the body. Therefore if 8 b is the emissive power of a black body attemperature T ;9 then

8 b T <

%r E9 < T : >>>>>>>>>>>>>.(1.)

O is the Stefan!olt"man constant (O 5 D.NP " 1; 2 Q W 3 (m/ : 9 >>>>>>>>>>>>(1.)

where R is called the emissivity of the surface (; S R S 1).The emissivity provides ameasure of how efficiently a surface emits radiation relative to a black body. Theemissivity strongly depends on the surface material and finish. Fadiation may also incident on a surface from its surroundings. The rate at whichthe radiation is incident on a surface per unit area of the surface is calle the “irradiation”of the surface and is denoted by M. The fraction of this energy absorbed by the surface iscalled “absorptivity” of the surface and is denoted by the symbol . The fraction of the

G E

6urface of emissivity R absorptivity and

temperature T

s

6urface of emissivity R area &

and temperature T

s

6urroundings (black) at T

surr () (9)

F#.1.$ R#t#on '8"+n'$ () t !r*"' n (9) 9't''n !r*"'n r' !rro!n#n

q

surr

q

sG

P


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1.

incident energy is reflected and is called the “reflectivity” of the surface denoted by E andthe remaining fraction of the incident energy is transmitted through the surface andis called the “transmissivity” of the surface denoted by . t follows from the definitions

of E and that J J K < 1 >>>>>>>>>>>>>>.(1.) Therefore the energy absorbed by a surface due to any radiation falling on it is given by

G9 < G >>>>>>>>>>>>>(1.13)

The absorptivity of a body is generally different from its emissivity. =owever inmany practical applications to simplify the analysis is assumed to be equal to itsemissivity R.

adiation !"change:- When two bodies at different temperatures #see$ each other heatis e"changed between them by radiation. f the intervening medium is filled with asubstance like air which is transparent to radiation the radiation emitted from one bodytravels through the intervening medium without any attenuation and reaches the other body and vice versa. Then the hot body e"periences a net heat loss and the cold body anet heat gain due to radiation heat e"change between the two. The analysis of radiationheat e"change among surfaces is quite comple" which will be discussed in chapter 1;.=ere we shall consider two simple e"amples to illustrate the method of calculating theradiation heat e"change between surfaces. &s the first e"ample let us consider a small opaque plate (for an opaquesurface 5 ;) of area & emissivity R and maintained at a uniform temperature Ts. Aetthis plate is e"posed to a large surroundings of area &su (&su JJ &) whish is at a uniformtemperature Tsur as shown in *ig. 1./b.The space between them contains air which istransparent to thermal radiation.

The radiation energy emitted by the plate is given by

0em 5 & R O Ts<

The large surroundings can be appro"imated as a black body in relation to the small plate.Then the radiation flu" emitted by the surroundings is O Tsur


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1.&ssuming 5 R for the plate the above e"pression for 0net reduces to

r < A ?T: @ T!r: >>>>>>.(1.11)

The above e"pression can be used to calculate the net radiation heat e"change between asmall area and a large surroundings. &s the second e"ample consider two finite surfaces &1 and &/ as shown in *ig. 1.7.

The surfaces are maintained at absolute temperatures T1 and T/ respectively and haveemissivities R1 and R/. +art of the radiation leaving &1 reaches &/ while the remainingenergy is lost to the surroundings. 6imilar considerations apply for the radiation leaving&/.f it is assumed that the radiation from the surroundings is negligible when comparedto the radiation from the surfaces &1 and &/ then we can write the e"pression for theradiation emitted by &1 and reaching &/ as

01U/ 5 *12 / &1R1O T1


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1.

5 V*12 / &1R1O T1>>>>>.(1.1-)

adiation Heat Transfer Coefficient:- @nder certain restrictive conditions it is possibleto simplify the radiation heat transfer calculations by defining a radiation heat transfercoefficient hr analogous to convective heat transfer coefficient as

0r 5 hr & XT

*or the e"ample of radiation e"change between a surface and the surroundingsV8q. (1. 11) using the concept of radiation heat transfer coefficient we can write

0r 5 hr &VTs : Tsur 5 & R O VTs>>>>>(1.1:)

1.-.F#rt L o* T+'ro/n#" (L o* "on'r2t#on o* 'n'r/) 00#' toH't Trn*'r Pro9' $% The first law of thermodynamics is an essential tool for

solving many heat transfer problems. =ence it is necessary to know the generalformulation of the first law of thermodynamics.First law e#uation for a control volume:- & control volume is a region in space bounded by a control surface through which energy and matter may pass.There are two options offormulating the first law for a control volume. %ne option is formulating the law on arate basis. That is at any instant there must be a balance between all ener#y rates. &lternatively the first law must also be satisfied over any time interval Xt. *or such aninterval there must be a balance between the amounts of all energy changes.

First Law on rate basis :- $he rate at which thermal and mechanical ener#y enters acontrol volume, plus the rate at which thermal ener#y is #enerated within the control

volume, minus the rate at which thermal and mechanical ener#y leaves the control

volume must be equal to the rate of increase of stored ener#y within the control volume. onsider a control volume shown in *ig. 1.< which shows that thermal and

1;


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1.13 .

mechanical energy are entering the control volume at a rate denoted by E#n thermal and

.

mechanical energy are leaving the control volume at a rate denoted by Eo!t. The rate at.

which energy is generated within the control volume is denoted by E and the rate at.

which energy is stored within the control volume is denoted by Et. The general form ofthe energy balance equation for the control volume can be written as follows' . . . . E#n J E Eo!t < Et >>>>>>>>>>>(1.1;) .Et is nothing but the rate of increase of energy within the control volume and hence can be written as equal to Et 5 t.

First Law over a Time Interval Δt :- %ver a time interval &t, the amount of thermal

and mechanical ener#y that enters a control volume, plus the amount of thermal ener#y

#enerated within the control volume minus the amount of thermal ener#y that leaves the

control volume is equal to the increase in the amount of ener#y stored within the controlvolume.The above statement can be written symbolically as

E#n J E @ Eo!t < Et >>>>>>>>>>..(1.1)

.

E#n

.

Eo!t

.

E

.

Et

F#. 1.:$ Con'r2t#on o* 'n'r/ *or "ontro 2o!' on rt' 9#

11


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1.11The inflow and outflow energy terms are surface phenomena. That is they are associatede"clusively with the processes occurring at the boundary surface and are proportional tothe surface area. The energy generation term is associated with conversion from some other

form (chemical electrical electromagnetic or nuclear) to thermal energy. t is avolumetric phenomenon.That is it occurs within the control volume and is proportional tothe magnitude of this volume. *or e"ample e"othermic chemical reaction may be taking place within the control volume. This reaction converts chemical energy to thermalenergy and we say that energy is generated within the control volume. onversion ofelectrical energy to thermal energy due to resistance heating when electric current is passed through an electrical conductor is another e"ample of thermal energy generation 8nergy storage is also a volumetric phenomenon and energy change withinthe control volume is due to the changes in kinetic potential and internal energy of matter within the control volume.

1.:. I!trt#2' E80'$ A. Conduction

!"ample $%$:- 'eat flux throu#h a wood slab () mm thick, whose inner and outer surface temperatures are *) ) + and ) ) + respectively, has been determined to

be *) -m. -hat is the thermal conductivity of the wood slab/

Solution:

&ssuming steady state conduction across the thickness of the slab and noting that the slabis not generating any thermal energy the first law equation for the slab can be written as

Fate at which thermal energy (conduction) is entering the slab at the surface " 5 ;

L

8

T1

T

1/

&iven:- T1 5


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1.1

is equal to the rate at which thermal energy is leaving the slab at the surface " 5 AThat is

0"[" 5 ; 5 0"[" 5 A 5 0" 5 constantBy *ouriers law we have 0" 5 2 k& (dT 3 d").

6eparating the variables and integrating both sides w.r.t. H" we have A T/ 0" \d" 5 2 k& \dT . %r 0" 5 k& (T1 : T/) 3 A

; T1

=eat flu" 5 q 5 0" 3 & 5 k(T1 : T/) 3 A

=ence k 5 q A 3 (T1 : T/) 5


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1.1-

!"ample $%(:--hat is the thickness required of a masonry wall havin# a thermal conductivity of ).:( -4m56, if the heat transfer rate is to be 9) ; of the rate throu#h

another wall havin# thermal conductivity of ).( -4m56 and a thickness of 3)) mm/

!oth walls are sub


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1.1:

per unit len#th, > required to maintain the surface temperature of the cylinder at 2)) ) +

for different stream velocities ? of the air. $he results are as follows0

1ir velocity, ? 4ms6 0 3 * 9 3

>ower, > 4-m6 0 *() @(9 A92 3(): 3A@2

4a6 7etermine the convective heat transfer coefficient for each velocity and display

your results #raphically. 4h B > ).*26

4b61ssumin# the dependence of the heat transfer coefficient on velocity to be of the form h B +? n , determine the parameters + and n from the results of part 4a6.

Solution:-

f h is the surface heat transfer coefficient then the power dissipated by the cylinder byconvection is given by

+ 5 h&s (Ts - T])

Where &s is the area of contact between the fluid and the surface of the cylinder.Therefore + 5 h ^A (Ts - T])

%r h 5 + 3 V^A(Ts - T]) 5 + 3 V^ " ;.;/D " 1 "(7;; :


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1.1;

&ir _elocity _ (m3s) ' 1 / < Q 1/

+ower+ (W3m) '


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1.1

ln 5 7.1 or 5 //

(ln h : ln ) (


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1.1

!"ample $%+:- 1 ( cm diameter sphere at 3) ) + is suspended in air at ) ) +. =f the

convective heat transfer coefficient between the surface and air is 3( -4m56,

determine the heat loss from the sphere.

Solution:-

T


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1.1

C. Radiation:

!"ample $%,:- 1 sphere 3) cm in diameter is suspended inside a lar#e evacuated

chamber whose walls are kept at 2)) 5. =f the surface of the sphere is black and

maintained at ()) 5 what would be the radiation heat loss from the sphere to the wallsof the chamber/. -hat would be the heat loss if the surface of the sphere has an

emissivity of ).9/

Solution:

!"ample $%:- 1 vacuum system as used in sputterin# conductin# thin films on micro

circuits, consists of a base plate maintained at a temperature of 2)) 5 by an

electric heater and a shroud within the enclosure maintained at :: 5 by circulatin#liquid nitro#en. $he base plate insulated on the lower side is ).2 m in diameter and has

an emissivity of ).(.

4a6 'ow much electrical power must be provided to the base plate heater/

4b6 1t what rate must liquid nitro#en be supplied to the shroud if its latent heat of

vapori"ation is 3( kCk#/

Solution:- T1 5 7;; 9 Z T/ 5 PP 9 Z d 5 ;.7 m Z R1 5 ;./D

6urface area of the top surface of the base plate 5 &s 5 (^ 3


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1.1

5 ;.;P;P m/

(a) 0r 5 R1O &s (T1


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1.3

Trial $:- &ssume Ts 5 7D; 9. Then A=6 of 8q. (1) 5 ?P/.N which is more than F=6 of

8q.(1). =ence Ts K 7D; 9.Trial ' :- &ssume Ts 5 7


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1.1

8q.(1) has to be solved by trial and error.

Trail $:- &ssume Ts 5 7;; 9. Then A=6 5 1D?Q1 which is J F=6.

Trail ' :- &ssume Ts 5 /?D 9. Then A=6 5 1DP1;.P7 which is K F=6. =ence Ts shouldlie between 7;;9 and /?D 9.

Trial ( :- &ssume Ts 5 /?P 9 . Then A=6 5 1DQ1? which is almost equal to F=6 (Within;.7< )Therefore Ts 5 /?P 9..

7 or

7r

7"1

7"

Therefore qa 5 R O T

s


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CHAPTER

GOERNING EUATIONS OF CONDUCTION

.1.Intro!"t#on$ n this chapter the governing basic equations for conduction in

artesian coordinate system is derived. The corresponding equations in cylindrical andspherical coordinate systems are also mentioned. >athematical representations ofdifferent types of boundary conditions and the initial condition required to solveconduction problems are also discussed. &fter studying this chapter the student will beable to write down the governing equation and the required boundary conditions andinitial condition if required for any conduction problem.

.. On' @ D#'n#on Con!"t#on E7!t#on $ n order to derive the one-dimensionalconduction equation let us consider a volume element of the solid of thickness X" along" : direction at a distance H" from the origin as shown in *ig. /.1.0" represents the rate

/7


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of heat transfer in " : direction entering into the volume element at " &(") area of heatflow at the section " q is the thermal energy generation within the element per unitvolume and 0"YX" is the rate of conduction out of the element at the section " Y X". Theenergy balance equation per unit time for the element can be written as follows'

.

V Fate of heat conduction into the element at " Y Fate of thermal energy generationwithin the element 2 Fate of heat conduction out of the element at " Y X"

5 Fate of increase of internal energy of the element.

i.e. 0" Y 0g : 0"YX" 5 8 3 t

or 0" Y q &(") X" : 0" Y (0" 3 ")X" Y (/0" 3 "/)(X")/ 3 / Y 44.

5 3 t (E&(")X" pT)

Ieglecting higher order terms and noting that E and p are constants the above equationsimplifies to

0" Y q &(") X" : 0" Y (0" 3 ")X" 5 E&(")X" p (T3 t)

%r 2 (0" 3 ") Y q &(") 5 E&(") p (T3 t)

8

8

8 J 8

7

F#. .1$ No'n"t!r' *or on' #'n#on "on!"t#on '7!t#on

O

A(8)

/


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@sing *ouriers law of conduction 0" 5 2 k &(") (T 3 ") the above equationsimplifies to 2 3 " 2 k &(") (T 3 ") Y q &(") 5 E&(") p (T3 t)

%r 15A(8) V5 V8 4 A(8) (VT 5 V8) J 7 < C0 (VT5 Vt) >>>>>(.1)

8q. (/.1) is the most general form of conduction equation for one-dimensional unsteadystate conduction.

'%'%$%!#uation for one-dimensional conduction in plane walls :- *or plane walls thearea of heat flow &(") is a constant. =ence 8q. (/.1) reduces to the form

V5 V8 4 (VT 5 V8) J 7 < C0 (VT5 Vt) >>>>>>>(.)

(i) f the thermal conductivity of the solid is constant then the above equation reduces to

(V

T 5 V8

) J (7 5 4) < (15 )(VT5 Vt) >>>>>>>>>(.-)(ii) *or steady state conduction problems in solids of constant thermal conductivitytemperature within the solid will be independent of time (i.e.(T3 t) 5 ;)and hence 8q. (/.7) reduces to

(T 5 8 )J (7 5 4) < 3>>>>>>>>>>>>>.(.:)

.-

(iii) *or a solid of constant thermal conductivity for which there is no thermal energygeneration within the solid q 5 ; and the governing for steady state conduction isobtained by putting q 5 ; in 8q. (/.>>>>>>>>>>>(.:)

'%'%'%!#uation for one-dimensional radial conduction in cylinders:-

/D


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.*or radial conduction in cylinders by convention the radial coordinate is denoted by Hrinstead of H" and the area of heat flow through the cylinder of length Aat any radius r isgiven by &(") 5 &(r) 5 /^rA. =ence substituting this e"pression for &(") and replacing " by r in 8q. (/.1) we have

13(/^rA)3 r k /^rA (T 3 r) Y q 5 E p (T3t)

%r (15r) V5 Vr 4 r (VT 5 Vr) J 7 < C0 (VT5 Vt)>>>>>.(.;)

(i) *or cylinders of constant thermal conductivity the above equation reduces to

(15r) V5 Vr r (VT 5 Vr) J 7 5 4 < (1 5 ) (VT5 Vt)>>>>>.(.)

.:

(ii) *or steady state radial conduction (i.e. (T3 t) 5 ; ) in cylinders of constant k theabove equation

reduces to (15r) 5 r r (T 5 Vr) J 7 5 4 < 3 >>>>>>>>>>.(.)

(iii) *or steady state radial conduction in cylinders of constant k and having no thermalenergy generation (i.e. q 5 ;) the above equation reduces to

R

r

L

r

r

/N


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5 r r (T 5 Vr) < 3 >>>>>>>>>>>>(.)

'%'%(%!#uation for one-dimensional radial conduction in spheres:- *or one-dimensionalradial conduction in spheres the area of heat flow at any radius r is given by &(r) 5 >>>>(.)

(i) *or spheres of constant thermal conductivity the above equation reduce to

15r V5 Vr r (VT 5 Vr) J 7 5 4 < (1 5 ) (VT5 Vt) >>>>>..(.13)

(ii) *or steady state conduction in spheres of constant k the above equation further reduce

to 15r V5 Vr r (VT 5 Vr) J 7 5 4 < 3 >>>>>>>>>>>(.11)

(iii) *or steady state conduction in spheres of constant k and without any thermal energygeneration the above equation further reduces to

15r 5 r r (T 5 r) < 3 >>>>>>>>>>>>>>(.1)

!#uation in compact form:- The general form of one : dimensional conductionequations for plane walls cylinders and spheres equations (/../) (/.D) and (/.?) can bewritten in a compact form as follows'

15rn V5 Vr 4 rn (VT 5 Vr) J 7 < C0 (VT5 Vt) >>>>.(.1-)

Where n 5 ; for plane walls n 5 1 for radial conduction in cylinders n 5 / for radial conduction in spheresand for plane walls it is customary to replace the Hr variable by H" variable.

.;

.-.T+r'' #'n#on "on!"t#on '7!t#on$ While deriving the one : dimensionalconduction equation we assumed that conduction heat transfer is taking place only alongone direction. By allowing conduction along the remaining two directions and followingthe same procedure we obtain the governing equation for conduction in three dimenions.

'%(%$% Three dimensional conduction e#uation in Cartesian coordinate system: Aet usconsider a volume element of dimensions X" Xy and XC in " y and C directions

/P


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respectively. The conduction heat transfer across the si" surfaces of the element is shownin *ig. /.7.

Iet Fate of conduction into the element in "-direction 5 0" : 0" Y X"

5 0" : V0" Y (0"3") X" Y (/0"3"/)(X")/ 3 / Y 4.

5 2 (0"3") X" by neglecting higher order terms.

5 2 3 " V2 k " Xy XC(T 3 ") X"

5 3 "Vk " (T 3 ") X" Xy XC

6imilarly the net rate of conduction into the elementin y : direction 5 3 yVk y (T 3 y) X" Xy XC

and in C : direction 5 3 CVk C (T 3 C) X" Xy XC.

.

=ence the net rate of conduction into the element from all the three directions

0in 5 3 "Vk " (T 3 ") Y 3 yVk y (T 3 y) Y 3 CVk C (T 3 C) X" Xy XC

Fate of heat thermal energy generation in the element 5 0g 5 q X" Xy XC

Fate of increase of internal energy within the element 5 8 3 t 5 E X" Xy XC p (T 3 t)

X"

Xy

XC

8

8 J 8

/

8

/

F#. .-$ Con!"t#on +'t trn*'r "ro t+' #8 *"' o* 2o!' '''nt

/Q

J / J /


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&pplying law of thermodynamics for the volume element we have

0in Y 0g 5 8 3 t

6ubstituting the e"pressions for 0in 0g and 8 3 t and simplifying we getV 5 V8?4 8 (VT 5 V8) J V 5 V/?4 / (VT 5 V/) J V 5 V?4 (VT 5 V) J 7 < C0 (VT 5 Vt)

>>>>>>>>(.1:)

8quation (/.1


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.-.. T+r'' #'n#on "on!"t#on '7!t#on #n "/#nr#" "oor#nt' /t'$

t is convenient to e"press the governing conduction equation in cylindrical coordinatesystem when we want to analyse conduction in cylinders. &ny point + in space can be

located by using the cylindrical coordinate system r and C and its relation to theartesian coordinate system (6ee *ig. /.


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@sing the relation between " y C and r and the conduction equation (/.1D) can betransformed into the equation in terms of r and as follows.

1 V VT 1 V V T 1 V V T%%% %%%% ? 4r %%%% J %%%%%%%%%% %%%%%?4 %%%%%%% J %%%%%%%%%%% %%% ?4 #n %%%%% r Vr Vr r #n V X V X r #n V V

V T

J 7 < C0 %%%%% >>>>>.(.1). V t

.:.Bo!nr/ n In#t# Con#t#on$

The temperature distribution within any solid is obtained by integrating the aboveconduction equation with respect to the space variable and with respect to time.Thesolution thus obtained is called the “#eneral solution” involving arbitrary constants ofintegration. The solution to a particular conduction problem is arrived by obtaining theseconstants which depends on the conditions at the bounding surfaces of the solid as well as

.the initial condition. The thermal conditions at the boundary surfaces are called the“boundary conditions” . Boundary conditions normally encountered in practice are'4i6 Specified temperature 4 also called as boundary condition of the first kind6,4ii6 Specified heat flux (also known as boundary condition of the second kind6,

P(8,/,)

r

8

/

X

F#$ .;$ S0+'r#" "oor#nt' /t'

P

O

%+ 5 r sin .=ence

" 5 r sin cos Z

y 5 r sin sin Z

C 5 r cos

71


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4iii6 +onvective boundary condition (also known as boundary condition of the third kind6and 4iv6 radiation boundary condition. The mathematical representations of these boundary conditions are illustrated by means of a few e"amples below.

.:.1. S0'"#*#' T'0'rt!r' t t+' Bo!nr/$% onsider a plane wall of thickness A

whose outer surfaces are maintained at temperatures T; and TA as shown in *ig./.N. *orone-dimensional unsteady state conduction the boundary conditions can be written as

(i) at " 5 ; T(;t) 5 T; Z (ii) at " 5 A T(At) 5 TA.

onsider another e"ample of a rectangular plate as shown in *ig. /.P. The boundaryconditions for the four surfaces to determine two-dimensional steady state temperaturedistribution T("y) can be written as follows.

(i) at " 5 ; T(;y) 5 j(y) Z (ii) at y 5 ; T(";) 5 T1 for all values of y

(iii) at " 5 a T(ay) 5 T/ for all values of yZ (iv) at y 5 b T("b) 5 (")

.:.. S0'"#*#' +'t *!8 t t+' 9o!nr/$% onsider a rectangular plate as shown in*ig. /.Q and whose boundaries are sub!ected to the prescribed heat flu" conditions asshown in the figure. Then the boundary conditions can be mathematically e"pressed as

follows..13

8

L

TL

T3

T(8,t)

F#. .$ Bo!nr/ "on#t#on F#..$ Bo!nr/ "on#t#on o*o* *#rt 4#n *or 0n' *#rt 4#n *or r'"tn!r 0t'

8

/

9

T < (8)

T1

Z(/)

T

T(8,/)

7/


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(i) at " 5 ; 2 k (T 3 ")[" 5 ; 5 q ; for ; S y S b Z

(ii) at y 5 ; (T 3 y)[y 5 ; 5 ; for ; S " S a Z

(iii) at " 5 a k (T 3 ")[" 5 a 5 q a for ; S y S b Z

(iv) at y 5 b 2 k (T 3 y)[y 5 b 5 ; for ; S " S a Z

.:.-. Bo!nr/ !r*"' !9['"t' to "on2'"t#2' +'t trn*'r$- *ig. /.? shows a planewall whose outer surfaces are sub!ected to convective boundary conditions.The surface at" 5 ; is in contact with a fluid which is at a uniform temperature Ti and the surface heattransfer coefficient is hi. 6imilarly the other surface at " 5 A is in contact with anotherfluid at a uniform temperature T; with a surface heat transfer coefficient h;. This type of boundary condition is encountered in heat e"changer wherein heat is transferred from hotfluid to the cold fluid with a metallic wall separating the two fluids. This type of boundary condition is normally referred to as the boundary condition of third kind. The

mathematical representation of the boundary conditions for the two surfaces of the planewall can be written as follows.

(i) at " 5 ; qconvection 5 q conductionZ i.e. hiVTi 2 T[" 5 ; 5 2 k(dT 3 d")[" 5 ;

.11

(ii) at " 5 A 2 k(dT 3 d")[" 5 A 5 h; VT[" 5 A 2 T;

9

8

/

73

7

79

#n!t'

T(8,/)

77

F#..$ Pr'"r#9' +'t *!8 9o!nr/ "on#t#on


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.:.:.R#t#on Bo!nr/ Con#t#on$*ig. /.1; shows a plane wall whose surface at "5A is having an emissivity HR and is radiating heat to the surroundings at a uniformtemperature Ts. The mathematical e"pression for the boundary condition at " 5 A can bewritten as follows'

(i) at " 5 A qconduction 5 qradiation Z i.e. 2 k (dT 3 d")[ " 5 A 5 O R V( T[ " 5 A)


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.:.;. G'n'r *or o* 9o!nr/ "on#t#on ("o9#n' "on!"t#on, "on2'"t#on nr#t#on 9o!nr/ "on#t#on)$ There are situations where the boundary surface issub!ected to combined conduction convection and radiation conditions as illustrated in*ig. /.11.t is a south wall of a house and the outer surface of the wall is e"posed to solar

radiation. The interior of the room is at a uniform temperature Ti. The outer air is atuniform temperature T; . The sky the ground and the surfaces of the surroundingstructures at this location is modeled as a surface at an effective temperature of Tsky.

8nergy balance for the outer surface is given by the equation

qconduction Y qsolar 5 qradiation Y qconvection

2 k (dT 3 d")[" 5 A Y qsolar 5 R O V(T[" 5 A)


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A. Derivation o conduction !"uations:

2.1. By writing an energy balance for a differential cylindrical volume element inthe ‘r’ variable (r is any radius), derive the one-dimensional time deendentheat conduction e!uation with internal heat generation and variable thermal

conductivity in the cylindrical coordinate system.

2.2. By writing an energy balance for a differential sherical volume element in the ‘r’ variable (r is any radius), derive the one-dimensional time deendent heatconduction e!uation with internal heat generation and variable thermalconductivity in the sherical coordinate system.

2.". By simlifying the three-dimensional heat conduction e!uation, obtain one-dimensional steady-state conduction e!uation with heat generation andconstant thermal conductivity for the following coordinate systems#

(a) $ectangular coordinate in the ‘%’ variable.(b) &ylindrical coordinate in the r variable.(c) 'herical coordinates in the ‘r’ variable

B. #at$ematical Formulation o Boundar% conditions:

2.. lane wall of thic*ness + is subected to a heat suly at a rate of ! /m2

at one boundary surface and dissiates heat from the surface by convectionto the ambient which is at a uniform temerature of 0 with a surface heattransfer coefficient of h.rite the mathematical formulation of the boundaryconditions for the lane wall.

2.. &onsider a solid cylinder of radius $ and height 3. 0he outer curved surface of the cylinder is subected to a uniform heating electrically at a rate of ! /m2.Both the circular surfaces of the cylinder are e%osed to an environment at

a uniform temeerature 0 with a surface heat transfer coefficient h.rite themathematical formulation of the boundary conditions for the solid cylinder.

2.4. hollow cylinder of inner radius ri, outer radius r and height 5 is subected tothe following boundary conditions.

(a) 0he inner curved surface is heated uniformly with an electric heater at aconstant rate of ! /m2,

(b) the outer curved surface dissiates heat by convection into an ambient at auniform temerature, 0 with a convective heat transfer coefficient,h

(c) the lower flat surface of the cylinder is insulated, and(d) the uer flat surface of the cylinder dissiates heat by convection into the

ambient at 0 with surface heat transfer coefficient h. rite the mathematicalformulation of the boundary conditions for the hollow cylinder.

.1:

C. Formulation o &eat Conduction 'roblems:

7N


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2.6. lane wall of thic*ness + and with constant thermal roerties is initially ata uniform temerature 0i. 'uddenly one of the surfaces of the wall issubected to heating by the flow of a hot gas at temerature 0 and the othersurface is *et insulated. 0he heat transfer coefficient between the hot gasand the surface e%osed to it is h. 0here is no heat generation in the wall.rite the mathematical formulation of the roblem to determine the one-

dimensional unsteady state temerature within the wall.

2.7. coer bar of radius $ is initially at a uniform temerature 0 i. 'uddenly theheating of the rod begins at time t8 by the assage of electric current,which generates heat at a uniform rate of !’’’ /m2. 0he outer surface of thedissiates heat into an ambient at a uniform temerature 0 with a convectiveheat transfer coefficient h. ssuming that thermal conductivity of the bar tobe constant, write the mathematical formulation of the heat conductionroblem to determine the one-dimensional radial unsteady state temeraturedistribution in the rod.

2.9. &onsider a solid cylinder of radius $ and height 5. 5eat is generated in thesolid at a uniform rate of !’’’ /m". :ne of the circular faces of the cylinder is

insulated and the other circular face dissiates heat by convection into amedium at a uniform temerature of 0 with a surface heat transfercoefficient of h. 0he outer curved surface of the cylinder is maintained at auniform temerature of 0. rite the mathematical formulation to determinethe two-dimensional steady state temerature distribution 0(r,;) in thecylinder.

2.1. &onsider a rectangular late as shown in


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2.12. &onsider a medium in which heat the heat conduction e!uation in its simlestform is given as (1/r2) >/>r (r2 >0 />r) 8 (1/?) (>0/>t)

(a) @s heat transfer steady state or unsteady stateA(b) @s heat transfer one-, two- or three-dimensionA(c) @s there heat generation in the mediumA(d) @s the thermal conductivity constant or variableA

2.1". &onsider a medium in which the heat conduction e!uation is given in itssimlest form as (1/r) >/>r (* r >0 />r) >/>; (* >0 />;) !’’’ 8

(a) @s heat transfer steady state or unsteady stateA(b) @s heat transfer one-, two- or three-dimensionA(c)@s there heat generation in the mediumA(d)@s the thermal conductivity constant or variableA

2.1. &onsider a medium in which heat the heat conduction e!uation in its simlestform is given as 1 >20 (1/r2) >/>r (r2 >0 />r) ---------- C -------D 8 (1/?) (>0/>t) r2 sin 2 E >F2

(a)@s heat transfer steady state or unsteady stateA(b)@s heat transfer one-, two- or three-dimensionA(c)@s there heat generation in the mediumA(d)@s the thermal conductivity constant or variableA

2.1.&onsider the north wall of a house of thic*ness +. 0he outer surface of the walle%changes heat by both convection and radiation.0he interior of the house ismaintained at a uniform temerature of 0i, while the e%terior of the house is at a

.1

"

T;

nsulated

hT]

a

b b

a

*ig. + /.1; ' 6chematic for problem /.1;

7Q

y

q; W3m/


39/370

uniform temerature 0. 0he s*y, the ground, and the surfaces of the surroundingstructures at this location can be modeled as a surface at an effective temerature of 0s*y for radiation heat e%change on the outer surface.0he radiation heat e%changebetween the inner surface of the wall and the surfaces of the other walls, floor andceiling are negligible.0he convective heat transfer coefficient for the inner and outersurfaces of the wall under consideration are h i and h resectively.0he thermal

conductivity of the wall material is G and the emissivity of the outer surface of thewall is ‘H’. ssuming the heat transfer through the wall is steady and onedimensional, e%ress the mathematical formulation (differential e!uation andboundary conditions) of the heat conduction roblem

So!t#on to T!tor# Pro9'

A. D'r#2t#on o* Con!"t#on E7!t#on$

.1. So!t#on$%

& cylindrical element of thickness dr in the radial direction at a radius r is shown in thefigure above. *or unsteady state one dimensional radial conduction with heat generationis given by 0r Y 0g : 0rYdr 5 (8 3 t)

%r 0r Y 0g : V0r Y (0r r)dr 5 (8 3 t)

%r (0r r)dr Y 0g 5 (8 3 t) 44444444444..(1)

where 0r is the rate of conduction into the element at radius r 5 k /^rA (T 3r)

.1

r

r

rJr

r

rJr

r

L

7?


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0g is the rate of heat generation within the element 5 /^ rA dr q

(8 3 t) is the rate of increase of the energy of the element. 5 /^ rA dr E p (T 3 t)where d_5/^rAdr-------- volume

6ubstituting these e"pressions in 8q.(1) we get

V 3r ( /^ rAk (T 3r) )dr Y /^ rA dr q 5 /^ rA dr E p (T 3 t)6implifying we get

(1 3 r) 3r Vkr(T3r) Y q 5 E p (T 3 t)

.. So!t#on$

onsider a spherical element of thickness dr at any radius r as shown in the figure above.The energy balance equation for one : dimensional radial unsteady state conduction withheat generation is given by

0r Y 0g : 0rYdr 5 (8 3 t)

%r 0r Y 0g : V0r Y (0r 3 r) dr 5 (8 3 t)%r - (0r 3 r) dr Y 0g 5 (8 3 t) 44444444(1)

Where 0r 5 rate of heat conducted in to the element at radius r 5 - k


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(8 3 t) 5 rate of increase of energy of the element 5 E (


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*or steady state conduction ( T 3 t) 5 ; Z *or one-dimensional radial conduction wehave T 3 5 ; and T 3 C 5 ;. Therefore T 3 r 5 dT 3 dr. With these simplificationsthe general form of conduction equation reduces to

(1 3 r) d 3 dr (kr dT3dr) Y q 5 ;

*or constant thermal conductivity the above equation reduces to

(1 3 r) d 3 dr (r dT3dr) Y q3 k 5 ;.

The general form of conduction equation in spherical coordinate system is given by

(13r /) 3 r(kr / T 3 r) Y 13(r / sin / ) 3 (k T3)

Y 13(r

/

sin ) 3 (k sin T3) Y q

5 E p (T3 t)444..(1)*or steady state conduction (T t) 5 ; Z *or one dimensional radial conduction we have

T3 5 ; and T3 5 ;. Therefore T 3 r 5 dT 3 dr. 6ubstituting these conditions in

8q. (1) we have (13r /) d 3 dr (kr / dT 3 dr) Y q 5 ;.

*or constant thermal conductivity the above equation reduces to

(13r /) d 3 dr (r / dT 3 dr) Y q 3 k 5 ;.

.3


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B. Mt+'t#" For!t#on o* t+' Bo!nr/ Con#t#on$

.:. So!t#on$%

.1

+,TQ

73

L

Boundary conditions are '(i)at " 5 ;Z k (dT 3 d")

" 5 ; 5 q

;

(ii) at " 5 AZ k(dT 3 d")" 5 A 5 h(T["5A - T])

8


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.;. So!t#on$%

.. So!t#on$%

.

r

\

R

73

h T]

73

Boundary conditions are'(i) at r 5 ;Z (T3r) 5 ; (a"is of symmetry)

(ii) at r 5 FZ k(T3r) 5 q;

(iii) at C 5 ;Z hVT[C5;

- T] Y k(T3C)

C5;5;

(iv) at C 5 Z k(T3r) 5 h V T[C 5 C T]

r1

H7

3

In!t'

+,T

Q

+r,T

Q Boundary conditions are'(i) at r 5 r

1 k(T3r) 5 q

; for all CZ

(ii) at r 5 r / k(T3r) 5 h

r VT[

r5r/ - T

]

for all C

(iii) at C 5 ; (T3C) 5 ; for all r.

(iv) at C 5 =

k(T3C)C 5=

5 hCVT[

C5= - T

]

for all r

from the problem' hC5h

r 5h

r

73


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C. Mt+'t#" For!t#on o* Con!"t#on Pro9'$

.. So!t#on$%

.. So!t#on$%

.-

L

In!t' +Q,T

QT <T(8,t)

T < T# t t < 3

Moverning differential equation todetermine T("t) is given by

( /T 3 "/) 5 (1 3 ) (T 3 t)where is the thermal diffusivity of thewall. nitial condition isat time t 5 ; T 5 T

i for all ".

The boundary conditions are ' (i) at " 5 ; (T 3 ")"5; 5 ;. (nsulated) for all t J;

(ii) at " 5 A k ( T 3 ")"5A

5 h]

VT["5A

T] for all tJ;

F

hT]

T 5 Ti at t S ;q for t S ;


46/370

.. So!t#on$%

.13. So!t#on$%

The governing differential equation to determine T("y) is given by

/T 3 "/ Y /T 3 y/ Y q3k 5 ;

R

H

In!t'

+, TQ

T3

The governing differential equationto determine T(rC) is given by

(13r) 3r(rT3r) Y /

T3C/

Y q3k 5 ;Boundary conditions are'(i) at r 5 ; T3r[

r5; 5 ; for all C

(a"is of symmetry).(ii) at r 5 F T 5 T

; for all C.

(iii) at C 5 ; T3C[C5;

5 ; for all r.

(iv) at C 5 =

k (T3C)C5=

5 h (T [C5=

T])

for all r.

r

8

/

9

In!t'

+, TQ

T3

73

7

for all ".


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.:

Boundary conditions are'

(i) at "5; k(T 3 ")["5; 5 q; for all y Z (ii) at " 5 a T 5 T; for all y

(iii) at y 5 ; T 3 y 5 ; for all " Z (iv) at y 5 b k(T 3 y)[y5b 5 hVT [y5b T].

.11. So!t#on$ The given differential equation is

>20 / >%2 8 (1/?) (>0 / >t)

t can be seen from this equation that T depends on one space variable " and the timevariable t. =ence the problem is one dimensional transient conduction problem. Io heatgeneration term appears in the equation indicating that the medium is not generating anyheat.The thermal conductivity of the medium does not appear within the differentialsymbol indicating that the conductivity of the medium is constant.

.1. So!t#on$ The given differential equation is

(1/r) d / dr(r * d0/dr) !’’’ 8 .

t can be seen from this equation that the temperature T depends only on one spacevariable Hr and it does not depend on time t. &lso the heat generation term q appears inthe differential equation.=ence the problem is a one-dimensional steady state conduction problem with heat generation. 6ince the thermal conductivity appears within thedifferential symbol it follows that the thermal conductivity of the medium is not aconstant but varies with temperature.

.1-. So!t#on$ The given differential equation is

(1/r) >/>r (* r >0 />r) >/>; (* >0 />;) !’’’ 8

t can be seen from the above equation that the temperature T depends on two spacevariables r and C and does not depend on time. There is the heat generation termappearing in the equation and the thermal conductivity k appears within the differentialsymbol >/>r and >/>;. =ence the problem is two-dimensional steady state conductionwith heat generation in a medium of variable thermal conductivity.

.1:. So!t#on$ The given differential equation is 1 >20 (1/r2) >/>r (r2 >0 />r) ---------- C -------D 8 (1/?) (>0/>t) r2 sin 2 E >F2

t can be seen from the given equation that the temperature T depends two space variablesr and and it also depends on the time variable t. There is no heat generation termappearing in the given equation . &lso the thermal conductivity k do not appear


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.;

within the differential symbol. =ence the given equation represents two-dimensionalsteady state conduction in a medium of constant thermal conductivity and the medium isnot generating any heat.

.1;. So!t#on$

The problem is one-dimensional steady state conduction without any heat generation andthe wall is of constant thermal conductivity. =ence the governing differential equation is d/T 3 d"/ 5 ;.

The boundary conditions are'

(i) at " 5 ; hi VTi : T [" 5 ; 5 2 k (dT3d")[" 5; Z

(ii) at " 5 A qconduction 5 qconvection Y qradiation

%r 2 k (dT3d")[" 5A 5 h;VT[" 5 A 2 T; Y R; O VT[" 5 A


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CHAPTER -

ONE DIMENSIONAL STEADY STATECONDUCTION

-.1. Intro!"t#on$% n this chapter the problems of one-dimensional steady stateconduction without and with thermal energy generation in slabs cylinders and spheresand sub!ected to different types of boundary conditions are analyCed to determine thetemperature distribution and rate of heat flow. The concept of thermal resistance isintroduced and the use of this concept for solving conduction in composite layers isillustrated. The problem of critical thickness of insulation for cylinder and sphere are also

analyCed. The effects of variable thermal conductivity on temperature distribution andrate of heat transfer are also studied. *inally the problems of one dimensional heatconduction in e"tended surfaces (fins) sub!ected to different types of boundary conditionsare e"amined.-.. Con!"t#on W#t+o!t H't G'n'rt#on

(%'%$% The 0lane 1all 2The 3la45:- The statement of the problem is to determine thetemperature distribution and rate of heat transfer for one dimensional steady stateconduction in a plane wall without heat generation sub!ected to specified boundaryconditions.

The governing equation for one 2 dimensional steady state conduction without heatgeneration is given by

8

L

T < T(8)

F#. -.1$ On' #'n#on t'/ tt' "on!"t#on #n 9

8 R < L 5(A4)

T

T1


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-. d/T ----- 5 ; 44444444444444(7.1) d"/

ntegrating 8q.(7.1) twice with respect to " we get T 5 1" Y / 444444444444(7./)

where 1 and / are constants which can be evaluated by knowing the boundaryconditions.

0lane wall with specified 4oundary surface temperatures:- f the surface at " 5 ; ismaintained at a uniform temperature T1 and the surface at " 5 A is maintained at anotheruniform temperature T/ then the boundary conditions can be written as follows'(i) at " 5 ; T(") 5 T1 Z (ii) at " 5 A T(") 5 T/.

ondition (i) in 8q.(7./) gives T1 5 /.

ondition (ii) in 8q. (7./) gives T/ 5 1A Y T1

T/ : T1%r 1 5 -------------. A6ubstituting for 1 and / in 8q. (7./) we get the temperature distribution in the planewall as " T(") 5 (T/ : T1) --- -- Y T1 A

%r T(8) @ T1 8 %%%%%%%%%%%% < %%%%%%%% >>>>>>>>>>>..(-.-) (T @ T1) L

!"pression for ate of Heat Transfer:

The rate of heat transfer at any section " is given by *ouriers law as

0" 5 2 k &(") (dT 3 d")

*or a plane wall &(") 5 constant 5 &. *rom 8q. (7.7) dT3d" 5 (T/ : T1) 3 A.

=ence 0" 5 2 k & (T/ : T1) 3 A.

D;


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-.-

4A(T1 @ T)Or 8 < %%%%%%%%%%%%%%%% >>>>>>>>>>>>..(-.:) L

Concept of thermal resistance for heat flow:

t can be seen from the above equation that 0" is independent of " and is a constant.8q.(7.


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-.:

The e"pression for rate of heat transfer 0" can be written as follows'

0" 5 hi & VTi : T1

(Ti : T1) (Ti : T1)or 0" 5 --------------- 5 ---------------- 444444444(7.Na) 1 3 (hi &) F ci

F ci 5 1 3 (hi&) is called thermal resistance for convection at the surface at " 5 ;

(T1 : T/)6imilarly 0" 5 --------------- 4444444444444444(7.Nb) F

where F 5 A 3(&k) is the thermal resistance offered by the wall for conduction and(T/ : To) 0" 5 --------------- 444444444444444..(7.Nc) F co

Where F co 5 1 3 (ho&) is the thermal resistance offered by the fluid at the surface at " 5 Afor convection. t follows from 8quations (7.Na) (7.Nb) and (7.Nc) that

(Ti : T1) (T1 : T/) (T/ : T;)0" 5 --------------- 5 ------------------ 5 --------------

F ci F F co

(T# @ To)%r 8 < %%%%%%%%%%%%%%%%%%% >>>>>>>>>>>>>>(-.) ?R "# J R J R "o

-... R# Con!"t#on #n Hoo C/#n'r$

The governing differential equation for one-dimensional steady state radial conduction ina hollow cylinder of constant thermal conductivity and without thermal energy generationis given by 8q.(/.1;b) with n 5 1' i.e. d --- Vr (dT 3 dr) 5 ; 4444444444.(7.Q) dr

ntegrating the above equation once with respect to Hr we get

r (dT 3 dr) 5 1

or (dT 3 dr) 5 13 r

D/


53/370

-.;ntegrating once again with respect to Hr we get

T(r) 5 1 ln r Y / 444444444..(7.?)

where 1 and / are constants of integration which can be determined by knowing the boundary conditions of the problem.

Hollow cylinder with prescri4ed surface temperatures: Aet the inner surface at r 5 r 1 bemaintained at a uniform temperature T1 and the outer surface at r 5 r / be maintained atanother uniform temperature T/ as shown in *ig. 7.7.

6ubstituting the condition at r 1 in 8q.(7.?) we get

T1 5 1 ln r 1 Y / 4444444444.(7.1;a)

and the condition at r / in 8q. (7.?) we get T/ 5 1 ln r / Y / 4444444444.(7.1;b)

6olving for 1 and / from the above two equations we get

(T1 : T/) (T1 : T/) 1 5 ---------------- 5 ------------------- Vln r 1 : ln r / ln (r 1 3 r /)

(T1 : T/)and / 5 T1 2 ------------------ ln r 1 ln (r 1 3 r /)

6ubstituting these e"pressions for 1 and / in 8q. (7.?) we have

(T1 : T/) (T1 : T/) T(r) 5 -------------- ln r Y T1 2 ---------------- ln r 1 ln (r 1 3 r /) ln (r 1 3 r /)

or ?T(r) @ T1 n (r 5 r1) %%%%%%%%%%%%%%% < %%%%%%%%%%%%%%%%%%% >>>>>>>>>>>>>>>>(-.11) ? T @ T1 n (r 5 r1)

D7


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-.

8q. (7.11) gives the temperature distribution with respect to the radial direction in ahollow cylinder. The plot of 8q. (7.11) is shown in *ig. 7.


55/370

-.

ln (r / 3 r 1) 1where F 5 ----------------- 5 --------- 44444444444..(7.1


56/370

where F ci 5 1 3 (/^ r 1Ahi)444444444444444444..(7.1Db)

(T1 : T/)&lso 0r 5 -------------- 4444444444444444444..(7.1Dc)

F -.

where F 5 ln (r / 3 r 1) 3 (/^Ak)44444444444.(7.1Dd)

(T/ : To)&nd 0r 5 --------------- 4444444444444444444.(7.1De) F co 1Where F co 5 --------------4444444444444..(7.1Df) (/^r /Aho)

*rom 8qs.(7.1Da) (7.1Dc) and (7.1De) we have

(Ti : T1) (T1 : T/) (T/ : To)0r 5 ------------- 5 -------------- 5 ---------------- F ci F F co

(T# @ To)%r r < %%%%%%%%%%%%%%%%%%%%%% >>>>>>>>>.(-.1) R "# J R J R "o

6urface incontact withfluid at T

o and

heat transfercoefficient ho

6urface in contactWith fluid at T

i

and

6urface heat transfer oefficient h

i

r1

r

F co

F F ci

0r

DN


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where F ci F and F co are given by 8qs.(7.1Db) (7.1Dd) and (7.1Df) respectively.

-..-. R# Con!"t#on #n Hoo S0+'r'$

The governing differential equation for one-dimensional steady state radial conduction in

a hollow sphere without thermal energy generation is given by 8q.(/.1;b) with n 5 1' i.e.-.

d --- Vr / (dT 3 dr) 5 ; 4444444444.(7.1P) dr

ntegrating the above equation once with respect to Hr we get

r / (dT 3 dr) 5 1

or (dT 3 dr) 5 13 r /

ntegrating once again with respect to Hr we get

T(r) 5 2 1 3 r Y / 444444444..(7.1Q)

where 1 and / are constants of integration which can be determined by knowing the boundary conditions of the problem.

Hollow sphere with prescri4ed surface temperatures: 4i6 Expression for temperature distribution0Aet the inner surface at r 5 r 1 be maintainedat a uniform temperature T1 and the outer surface at r 5 r / be maintained at anotheruniform temperature T/ as shown in *ig. 7.N.

The boundary conditions for this problem can be written as follows'

(i) at r 5 r 1 T(r) 5 T1 and (ii) at r 5 r / T(r) 5 T/.

ondition (i) in 8q. (7.1Q) gives T1 5 2 1 3 r 1 Y / 4444444444.(7.1?a)

ondition (ii) in 8q. (7.1Q) gives T/ 5 2 1 3 r / Y / 4444444444.(7.1?b)

6olving for 1 and / from 8qs. (7.1?a) and (7.1?b) we have

(T1 : T/) (T1 : T/)1 5 ------------------- and / 5 T1 Y --------------------------

V1 3 r / : 1 3 r 1 r 1V1 3 r / : 1 3 r 1

6ubstituting these e"pressions for 1 and / in 8q. (7.1Q) we get

DP


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(T1 : T/) 3 r (T1 : T/) 3 r 1T(r) 5 2 ----------------------- Y T1 Y ---------------------- V1 3 r / : 1 3 r 1 V1 3 r / : 1 3 r 1

-.13

%r T(r) @ T1 ?1 5 r @ 1 5 r%%%%%%%%%%%%%%%%% < %%%%%%%%%%%%%%%%%%%%%% >>>>>>>>>>>(-.3)

?T1 @ T ?1 5 r @ 1 5 r1

4ii6 Expression for Gate of 'eat $ransfer0 The rate of heat transfer for the hollow sphereis given by

0r 5 2k &(r)(d T 3 dr) 4444444444444444..(7./1)

Iow at any radius for a sphere &(r) 5


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8q.(7.//) can be written as r < ?T1 @ T 5 R >>>>>>>>>>>..(-.-)

Where F is the thermal resistance for the hollow sphere and is given by

R < (r @ r1) 5 : ] 4 r1 r >>>>>>>>>>>>>>.(-.-9)-.11

Hollow sphere with convective conditions at the surfaces:- *ig. 7.P shows a hollowsphere whose boundary surfaces at radii r 1 and r / are in contact with fluids attemperatures Ti and T; with surface heat transfer coefficients hi and h; respectively.

The thermal resistance network for the above problem is shown in *ig.7.Q

r

r1

S!r*"' #n "ont"t #t+

*!# t T3 n !r*"'+'t trn*'r "o'**#"#'nt +

3

S!r*"' #n "ont"t #t+*!# t T

# n !r*"' +'t

trn*'r "o'**#"#'nt +#

F#. -.$ R# "on!"t#on #n +oo 0+'r' #t+ "on2'"t#2'"on#t#on t t+' to 9o!nr/ !r*"'

D?


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.F#. -.$ T+'r "#r"!#t *or +oo 0+'r' #t+ "on2'"t#2' 9o!nr/ "on#t#on-.1

When T1 5 the inside surface temperature of the sphere and

F ci 5 1 3 (hi&i) 5 the thermal resistance for convection for the inside surface

%r F ci 5 1 3 (< ^ r 1/ hi) 444444444444444444444.(7./Db)

0r 5 Fate of heat transfer by conduction through the hollow sphere

5 VT1 : T/ 3 F with F 5 (r / : r 1) 3 < ^ k r 1 r /

&nd 0co 5 Fate of heat transfer by convection from the outer surface of the sphere to theouter fluid and is given by VT/ : T;

0co 5 ho &o VT/ : To 5 --------------- 44444(7./Na) F co

Where T/ 5 outside surface temperature of the sphere and

&o 5 outside surface area of the sphere 5 < ^ r // so that

F co 5 1 3 < ^ r // ho44444444444.(7./Nb)

Iow 8q.(7./


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VTi : T1 VT1 : T/ VT/ : T10r 5 hi &i VTi : T1 5 -------------- 5 ---------------- 5 ----------------

F ci F F co

?T# @ To r < %%%%%%%%%%%%%%%%%%%%%% >>>>>>>>>>>>>>>>(-.) ?R "# J R J R "o

-..:. St'/ Stt' "on!"t#on #n "o0o#t' '#!$

There are many engineering applications in which heat transfer takes place through amedium composed of several different layers each having different thermal conductivity.These layers may be arranged in series or in parallel or they may be arranged withcombined series-parallel arrangements. 6uch problems can be conveniently solved using

electrical analogy as illustrated in the following sections.Composite 0lane wall:- (#) L/'r #n 'r#'$ onsider a plane wall consisting of threelayers in series with perfect thermal contact as shown in *ig. 7.1;.The equivalent thermal

-.1-

resistance network is also shown. f 0 is the rate of heat transfer through an area & of thecomposite wall then we can write the e"pression for 0 as follows'

F#. -.13$ A "o0o#t' 0n' #t+ t+r'' /'r #n 'r#' n t+' '7!#2'ntt+'r r'#tn"' n'tor4

S!r*"' #n"ont"t#t+ *!#t T

# n

!r*"'+'ttrn*'r"o'**#"#'nt+

#

S!r*"' #n "ont"t #t+ *!#t T

3 n !r*"' +'t trn*'r

"o'**#"#'nt +o

L

L1

L-

4 1 4

4

-

T1 T

T

- T

:

R "# R 1 R R - R "o

T# T

1 T

T

- T

: T

o

N1


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(T/ : T7) (T1 : T/) (T1 : T/) (T/ : T7) (T7 : Tco)0 5 -------------- 5 --------------- 5 ------------- 5 ------------ 5 ---------------- F co F 1 F / F 7 F co

(Ti : T;) Ti : T;)%r 0 5 --------------------------------- 5 ------------44444444444.(7./Q) F ci Y F 1 Y F / Y F 7 Y F co F total

7verall heat transfer coefficient for a composite wall:- t is sometimes convenient toe"press the rate of heat transfer through a medium in a manner which is analogous to the Iewtons law of cooling as follows'

f @ is the overall heat transfer coefficient for the composite wall shown in *ig. (7.1;)then

0 5 @ & (Ti : To) 444444444444...(7./?)

omparing 8q. (7./Q) with 8q. (7./?) we have the e"pression for @ as 1 @ 5 --------------- 44444444444444..(7.7;) & F total

-.1: 1 1%r @ 5 ------------------------------------5 ----------------------------------------------------- & V F ci Y F 1 Y F / Y F 7 &V13(hi&) Y A13(&k 1) Y A/3(&k /) Y A73(&k 7)

1%r U < -------------------------------------------->>>>>>>>>>>>>>(-.-1) ? 15+# J L1 5 4 1 J L 5 4 J L- 5 4 -

N/


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(##) L/'r #n Pr'$% *ig.7.11 shows a composite plane wall in which three layers are

FF#. -.11$ S"+'t#" n '7!#2'nt t+'r "#r"!#t *or "o0o#t' #t+

/'r #n 0r'

arranged in parallel. Aet Hb be the dimension of these layers measured normal to the plane of the paper. Aet one surface of the composite wall be in contact with a fluid attemperature Ti and surface heat transfer coefficient hi and the other surface of the wall bein contact with another fluid at temperature To with surface heat transfer coefficient ho.The equivalent thermal circuit for the composite wall is also shown in *ig. 7.11. The rateof heat transfer through the composite wall is given by

0 5 01 Y 0/ Y 07 4444444444.(7.7/)-.1;

where 01 5 Fate of heat transfer through layer 1

0/ 5 Fate of heat transfer through layer / and

07 5 Fate of heat transfer through layer 7.

L

9

H1

H

H-

4 1

4

4 -

S!r*"' #n

"ont"t#t+ *!#t T

# #t+

+'ttrn*'r"o'**#"#'nt+

#

T#

R "#

R 1

R

R -

R "o

To

1

-

T1

T

N7

S!*"' #n "ont"t#t+ *!# t To n!r*"' +'t trn*'r"o'**#"#'nt +o


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F (T1 : T/) Iow 01 5 -------------- 44444444444444444444...(7.77a) F 1

Where F 1 5 A 3 (=1 bk 1)F (T1 : T/)6imilarly 0/ 5 -------------- 4444444444444444444(7.77b) F /

Where F / 5 A 3 (=/ bk /)

and(T1 : T/)

and 07 5 -------------- 5 444444444.. 4444444444.(7.77c)

F 7 Where F 7 5 A 3 (=7 bk 7)

6ubstituting these e"pressions in 8q. (7.7/) and simplifying we get

(T1 : T/) (T1 : T/) (T1 : T/) (T1 : T/)0 5 ------------- Y ---------------- Y ----------------- 5 -------------------- 444.(7.7


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transfer through the composite layer is given by

(Ti : T1) (T1 : T/) (T/ : T7) (T7 : To) (Ti : T;) Iow 0 5 ------------- 5 ------------ 5 ----------- 5 ------------- 5 ----------------------------- F ci F 1 F / F co VF ci Y F 1 Y F / Y F co

44444..(7.7N)

1 1Where F ci 5 1 3 Vhi&i 5 -------------- Z F 1 5 ---------- ln (r / 3 r 1) / ^ r 1A hi / ^ A k 1

1 1 F co 5 1 3 Vho&o 5 -------------- Z F / 5 ---------- ln (r 7 3 r /) / ^ r 7A ho / ^ A k /

The above e"pression for 0 can be e"tended to any number of layers.

-.1

r1

r r-

6urface in contact withfluid at T

o and surface heat

transfer coefficient h;

6urface in contact withfluid at T

i and surface

heat transfer coefficienth

i

4 1

4

R "#

R 1

R

R "o

T#

T1

T

T-

To

F#. -.1$ S"+'t#" n t+'r "#r"!#t #r *or "o0o#t'"/#n'r

ND


66/370

7verall Heat Transfer Coefficient for a Composite Cylinder:- *or a cylinder the area of heat flow in radial direction depends on the radius r we can define the overall heattransfer coefficient either based on inside surface area or based on outside surface area ofthe composite cylinder. Thus if @i is the overall heat transfer coefficient based on inside

surface area &i and @o is the overall heat transfer coefficient based on outside surface area&o then

0 5 @i&i (Ti : To) 4444444444444444444444444.(7.7P)

*rom equations (7.7N) and (7.7P) we have

(Ti : T;) Iow @i&i (Ti : To) 5 ----------------------------- VF ci Y F 1 Y F / Y F co

6ubstituting the e"pressions for &i F ciF 1F / and F co in the above equation we have

1/ ^ r 1A @i 5 -------------------------------------------------------------------------------------------- V1 3(/^r 1Ahi) Y 13(/^Ak 1)ln (r / 3 r 1) Y 13(/^Ak /)ln (r 7 3 r /) Y 13(/^r 7Aho)

1Or U# < %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >>..(-.-) ? 15+# J (r1 5 4 1) n (r5r1) J (r154 ) n (r-5r) J (r15r-) (15+o)

6imilarly it can be shown that

1 Uo < %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% >..(-.-) ?(r-5 r) (15+# ) J (r- 5 4 1) n (r5r1) J (r-54 ) n (r-5r) J (15+o)

Composite Concentric 3pheres:- *ig.7.17 shows a composite sphere having two layerswith the inner surface of the composite sphere in contact with fluid at a uniformtemperature Ti and surface heat transfer coefficient hi and the outer surface in contactwith another fluid at a uniform temperature To and surface heat transfer coefficient ho.The corresponding thermal circuit diagram is also shown in the figure.

-.1

NN


67/370

8q. (7.7N) is also applicable for the composite sphere of *ig. 7.17 e"cept that thee"pression for individual resistance will be different. Thus

(Ti : To)0 5 --------------------------- 44444444444444444.(7.


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!(am)le *.+:- &onsider a lane wall 1 mm thic* and of thermal conductivity 1/(m-G). 'teady state conditions are *nown to e%ist with 01 8 G and 028 4 G. Ietermine the heat flu% (magnitude and direction) and thetemerature gradient d0/d% for the coordinate system shown in


69/370

(9)

-.1

T(8), 4


70/370

(")

!(am)le *.,:-


71/370

ntegrating the above equation we have T " \dT 5 (


72/370

(T/ : T1) (13" : 13`1) T(") 5 T1 Y --------------------------------- 444444..(7) (13`/ : 13`1)

6ubstituting the given numerical values for `1 `/ T1 and T/ in 8q.(7) we get the

temperature distribution as follows' (N;; :


73/370

;.;DF 1 5 A1 3(k 1&1) 5 --------------- 5 ;.;;/D m/ : 9 3 W.

/; " 1

;.1;F / 5 A/ 3 (k /&/) 5 ---------------- 5 ;.;;/ m/ : 9 3 W. D; " 1

;.1D

F 7 5 A73 (k 7&7) 5 ------------------ 5 ;.;;1D m/

: 9 3 W. 1;; " 1

1F co 5 1 3 (h;&


74/370

0 1 1 @ 5 ---------------- 5 ------------- 5 -------------- (Ti : T;) F ;.117

5 Q.QD W 3 (m/

: 9).!(am)le *.:- comosite wall consisting of four different materials is shown in


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;.;DF


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So!t#on$

1 1 1F i 5 ------------- ln (r 13r i) Z F ci 5 ------------------ 5 --------------------- (/^ Ak &) hi &i (/^ r i A hi)

1 1 1F o 5 ------------- ln (r o3r 1) Z F co 5 ------------------ 5 ---------------------

(/^ Ak B) ho &o (/^ r o A ho) 0i 5 (Th : Ti) 3 VF i Y F ci Z 0o 5 (Th : To) 3 VF o Y F co Z

0total 5 0i Y 0o 5 (Th : Ti) 3 (F i Y F ci) Y (Th : To)3 (F o Y F co)

(Th : Ti) (Th : To)/^ r 1A q h 5 -------------------------- Y ----------------------- ln (r 13r i) 1 ln (r o3r 1) 1

---------- Y ------------- ---------- Y -------------/^ A k & /^ A r i hi /^ A k B /^ A r o ho

-.

r# r

1

ro

+'t'r

#

3

+#,T

#

+o,T

o

4 A

4 B

#

o

tot R # R "#

R o R

"o

PN


77/370

(Th : Ti) (Th : To)Therefore q h 5 ------------------------------- Y -------------------------------- V (r 13k &) ln (r 13r i) Y r 13(r ihi) V(r 13k B) ln (r o3r 1) Y r 13(r oho)

The temperature Th of the heater can be obtained from the above equation. 0o (Th : To) 3 (F o Y F co) (Th : To) " (F i Y F ci) Iow ------- 5 --------------------------- 5 -------------------------- 0i (Th : Ti) 3 (F i Y F ci) (Th : Ti) " (F o Y F co)

(Th : To) V 1 3 (r ihi) Y (1 3 k &) ln (r 13r i) 5 ----------- " ------------------------------------- (Th : Ti) V 1 3 (r oho) Y (1 3 k B) ln (r o3r 1)

!(am)le *./:- hollow aluminum shere with an electrical heater in the centre isused to determine the thermal conductivity of insulating materials. 0he inner andouter radii of the shere are 1 cm and 17 cm resectively and testing is done understeady state conditions with the inner surface of the aluminum maintained at 2 &.@n a articular test, a sherical shell of insulation is cast on the outer surface of thealuminum shere to a thi*ness of 12 cm. 0he system is in a room where the airtemerature is 2 & and the convection coefficient is " /(m2 L G). @f 7 aredissiated by the heater under steady state conditions, what is the thermalconductivity of the insulating materialA

(r / : r 1) (;.1Q : ;.1D)F 1 5 ---------------- 5 ------------------------------ 5


78/370

(r 7 : r /) (;.7; : ;.1Q)F / 5 ---------------- 5 ------------------------------ 5 ;.1PP 3 k / ; 3 W.


79/370

The boundary conditions are ' (i) at r 5 F 1 k (dT3dr)[r5F1 5 q;

(ii) at r 5 F / T(r) 5 ;.

ntegrating 8q. (1) w.r.t. r once we get r / (dT3dr) 5 1

or dT 3 dr 5 1 3 r / 4444444444(/)

ntegrating once again w.r.t. r we get

T(r) 5 1 3 r Y / 444444.. (7)

*rom (/) (dT3dr)r 5 F1 5 1 3 F 1/

=ence condition (i) gives

k1 3 F 1/ 5 q;

%r 1 5 q; F 1/ 3 k

ondition (ii) in 8q.(/) gives ; 5 1 3 F / Y /

%r / 5 1 3 F / 5 (q;F 1/) 3 (kF /)

6ubstituting the e"pressions for 1 and / in 8q. (/) we have

q; F 1/ q; F 1/

T(r) 5 -------------- ------------------- k r k F /

6ubstituting the numerical values for q; k F 1 and F / we have

1.N " 1;D " ;.1/ 1.N " 1;D " ;.1/

T(r) 5 -------------------- 3 r --------------------


80/370

Therefore T(r) [r 5 F1 5 (


81/370

So!t#on$

F / 5 AB 3 (k B&B) 5 ;.;< 3 (;.;D " 1) 5 ;.Q m/ : 9 3 W.

F co 5 1 3 (ho&B) 5 1 3 ( /D " 1) 5 ;.;< m/

: 9 3 W.

F 5 F ci Y F 1 Y F cont Y F / Y F co 5 ;.;NP Y ;./ Y ;.77 Y ;.Q Y ;.;< 5 1.


82/370

!(am)le *.0:- very thin electric heater is wraed around the outer surface of along cylindrical tube whose inner surface is maintained at &. 0he tube wallhas inner and outer radii of 2 mm and 6 mm resectively and a thermalconductivity of 1 /(m-G). 0he thermal contact resistance between theheater and the outer surface of the tube er unit length is .1 (m-G) / .

0he outer surface of the heater is e%osed to a fluid with a temerature of L1 & and a convection coefficient of 1 /(m2-G). Ietermine the heaterower re!uired er length of the tube to maintain a heater temerature of 2&.

So!t#on$ ata' r 1 5 ;.;/D m Z r / 5 ;.;PD m Z k 5 1; W 3(m-9) Z T] 5 1; ; Z h 5 1;;

W3(m/ : 9)Z T1 5 D ; Z F cont 5 ;.;1 m : 9 3 W Z T/ 5 /D ; .

F#. P-.$ S"+'t#" n T+'r "#r"!#t *or E80' -.

1F co 5 1 3 (h &o) 5 ----------------------- 5 ;.;/1/ (m- 9) 3 W. 1;; " /^ " ;.;PD

1 1

F cond 5 ------------- ln (r / 3 r 1) 5 ------------------ ln (;.;PD 3 ;.;/D) 5 ;.;1PD (m-9) 3 W. /^ A k (/" ^ " 1 " 1;)

-.-:

r r

1

T1

H't'r

+,TQ

9

t!9'

tot

9

t!9'

R "o

R "ont

R "on

T

Q/


83/370

V/D : ( 1;)0amb 5 (T/ - T]) 3 F co 5 ------------------- 5 1ND1 W 3 m. ;.;/1/ (T/ : T1) V/D : D

0tube 5 ------------------ 5 --------------------- 5 P/P.7 W 3 m F cont Y F cond (;.;1 Y ;.;1PD)

+ower required 5 0total 5 0amb Y 0tube 5 1ND; Y P/P.7 5 /7PQ.7 W3m

-.-. On' D#'n#on St'/ Stt' Con!"t#on W#t+ H't G'n'rt#on$ Thegoverning equation for one : dimensional steady state conduction in solids which aregenerating is given as follows.

(i) +lane wall ' (d/T 3 d"/ ) Y q 3 k 5 ; 4444444444(7.


84/370

The boundary conditions are Z (i) at " 5 ; (dT3d") 5 ; (nsulated)

(ii) at " 5 A T 5 T1

ntegrating 8q.(1) once w.r.t. " we get

dT3d" 5 -(^3/A) (q;3k) sin (^"3/A) Y 1 44444(/)

ntegrating once again w.r.t. " we get

T(") 5 -(^3/A)/ (q;3k) cos (^"3/A) Y 1" Y / 44444.(7)

ondition (i) in 8q. (/) gives ; 5 ; Y 1 or 1 5 ;.

ondition (ii) in 8q. (7) gives T1 5 ; Y ; Y / or / 5 T1.

6ubstituting the values of 1 and / in 8q. (7) we get the temperature distribution as

T(") 5 -(^3/A)/ (q;3k) cos (^"3/A) Y T1 4444444..(


85/370

surface by ambient air at & with a heat transfer coefficient of 4 /(m 2-G).ssuming one-dimensional radial conduction determine the temerature atthe centre of the rod as well as at the outer surface of the rod.(16 &)

8q. (1) can be written asd 3 dr (r dT3dr) Y q r 3 k 5 ;.

ntegrating once w.r.t. r we get r dT3dr Y (q r /) 3 /k 5 1

or dT3dr Y (q r) 3 /k 5 1 3 r 4444444 (/)

ntegrating once again w.r.t. r we have q r /

T(r) 5 -------------- Y 1 ln r Y / 4444(7) < k ondition (i) in 8q. (/) gives ; Y ; 5 1 3 ; or 1 5 ;.

*rom 8q. (7) we have q F /

T(r)[r5F 5 -------------- Y / 4444444..(7)

< k

and from 8q. (/) we have (dT3dr) [r5F 5 (q F) 3 /k

Therefore condition (ii) gives

-.-

R

7 T < T(r)

The governing differential equation to determine one-dimensional steady stateradial conduction with heat generation is given by

(13r) d 3 dr (r dT3dr) Y q 3 k 5 ;444444444..(1)

The boundary conditions are ' (i) at r 5 ; dT3dr 5 ; (a"is of symmetry)

(ii; at r5 F k (dT3dr)[r5F

5 h VT [r5

T]

+, TQ

QD


86/370

k V (q3F) 3 /k 5 h V (q F / 3


87/370

q; r V 1 : (r3F)/

d3dr(r dT3dr) Y ---------------------- 5 ; 44444444..(1) k

Boundary conditions are' (i) at r 5 ; dT 3 dr 5 ; (a"is of symmetry)

(ii) at r 5 F T 5 T;

ntegrating 8q. (1) w.r.t. r once we get

q; r / r <

(r dT3dr) Y ---- V----- ----------- 5 1 9 /


88/370

ondition(i) in 8q.(/) gives

; Y ; 5 1 3 ; or 1 5 ;.

ondition (ii) in 8q.(7) gives

q; F / F <

T; 5 ------ V -------- ---------- Y / k < 1N F /

7 q; F /

%r / 5 T; Y ----------------1N k

6ubstituting the e"pressions for 1 and / in 8q. (7) we have

q; r / r < 7 q; F /

T(r) 5 ---------- V---------- --------- Y T; Y --------------- k < 1NF / 1N k

%r q; r / r < 7 q; F /

T(r) T; 5 ---------- V---------- --------- Y --------------- k < 1NF / 1N k

7 q; F / 7 " 1.1N " 1;D " (;.;/)/

Therefore T(r) [r5; T; 5 ------------------ 5 ------------------------------- 5 ;.QP; . 1N k 1N " 1;

!(am)le*.+3:-Ievelo an e%ression for one-dimensional radial steady statetemerature distribution in a solid shere of radius $ in which heat isgenerated at a rate given by

!’’’ 8 !o C 1 L (r/$)D /m".

ssume that the outer surface is maintained at a uniform temerature 0o.C0(r) 8 0 (! $2 / 12*) K 1 L 2 (r/$)2 (r/$)"MD

QQ


89/370

-.:3 So!t#on$%

q; r 7 r <

(r / dT3dr) Y ----- V ------ ---------- 5 1 k 7


90/370

q; r / r 7 q; F /

T(r) 5 ---- V ------ ---------- Y T; Y ----------- k N 1/F 1/ k

q; F /

or T(r) 5 T; Y ---------- V 1 / (r3F)/ Y (r3F)7 1/k

!(am)le *.++:- lane wall of thic*ness 2+ is generating heat according to the law

!’’’ 8 ! C1 L Q(0 L 0w)D

where !o, Q, and 0w are constants and 0 is the temerature at any section % from themid-lane of the wall. 0he two outer surfaces of the wall are maintained at a uniform

temerature 0w. Ietermine the one-dimensional steady state temeraturedistribution, 0(%) for the wall.

So!t#on$

Moverning differential equation for one-dimensional steady state conduction in a planewall which generating heat is given by

d/T 3 d"/ Y q 3 k 5 ;.

6ubstituting for q we have

-.:

L

8

T

T

"444 5 "3 6+ 7 89T 7 T

w;

?;


91/370

d/T 3 d"/ Y qo V 1 : (T : Tw) 3k 5 ;

efining a new variable 5 T : Tw the above equation can be written as

d/

3 d"/

Y qo V 1 : 3k 5 ;or d/ 3 d"/ 2 qo 3k 5 qo 3 k

or d/ 3 d"/ 2 m / 5 qo 3 k 44444444444..(1a)

where m/ 5 q; 3k 44444444444...(1b)

8q.(1a) is a second order linear non-homogeneous differential equation whose solution isgiven by

(") 5 h(") Y p(") 4444444444444.(/)where h(") satisfies the differential equation

d/h 3 d"/ 2 m /h 5 ; 444444444444.(7)

6olution to 8q.(7) is given by

h(") 5 &1 e m" Y &/ e 2 m"44444444444.(


92/370

-.:-

The constants &1 and &/ in 8q.(P) can be determined by using the two boundaryconditions which are'

(i)at " 5 ; dT 3 d" 5 ; (a"is of symmetry) i.e. d 3 d" 5 ;(ii) at " 5 A T 5 Tw Z i.e. 5 ;

*rom 8q.(P) d 3 d" 5 mV&1e m" : &/ e 2 m"

6ubstituting condition (i) we get mV&1 : &/ 5 ;

%r &1 5 &/.

6ubstituting condition (ii) in 8q.(P) we get &1Ve mA Y e 2 mA 5 1 3

(1 3 )%r &1 5 ---------------------------- Ve mA Y e 2 mA

6ubstituting the e"pressions for &1 and &/ in 8q. (P) we get the temperature distributionin the plane wall as (1 3 )

(") 5 T(") : Tw

Documents

Lecture Notes_Heat and Mass Transfer