One-dimensional conduction-with_no_heat_generation

Lectures on Heat Transfer --One-Dimensional, Steady-State Heat Conduction without Heat Generation

Dr. M. Thirumaleshwarformerly:

Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,

Mangalore

Preface

• This file contains slides on One-dimensional, steady state heat conduction without heat generation.

• The slides were prepared while teaching • The slides were prepared while teaching Heat Transfer course to the M.Tech. students in Mechanical Engineering Dept. of St. Joseph Engineering College, Vamanjoor, Mangalore, India, during Sept. – Dec. 2010.

Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful to teachers, students, researchers and professionals working in this field.

• For students, it should be particularly useful to study, quickly review the subject, useful to study, quickly review the subject, and to prepare for the examinations.

• ��

Aug. 2016 3MT/SJEC/M.Tech.

References• 1. M. Thirumaleshwar: Fundamentals of Heat &

Mass Transfer, Pearson Edu., 2006• 2. Cengel Y. A. Heat Transfer: A Practical

Approach, 2nd Ed. McGraw Hill Co., 2003• 3. Cengel, Y. A. and Ghajar, A. J., Heat and

Aug. 2016 MT/SJEC/M.Tech. 4

• 3. Cengel, Y. A. and Ghajar, A. J., Heat and Mass Transfer - Fundamentals and Applications, 5th Ed., McGraw-Hill, New York, NY, 2014.

• 4. Incropera , Dewitt, Bergman, Lavine: Fundamentals of Heat and Mass Transfer, 6th

Ed., Wiley Intl.

References… contd.

• 5. M. Thirumaleshwar: Software Solutions to Problems on Heat Transfer – CONDUCTION-Part-I, Bookboon, 2013

• http://bookboon.com/en/software-solutions-to-problems-on-•heat-transfer-ebook

One-Dimensional, Steady-State Heat Conduction without Heat

Generation

• Plane slab - composite slabs – contact resistance – cylindrical Systems – composite cylinders - spherical systems – composite

cylinders - spherical systems – composite spheres - critical thickness of insulation –optimum thickness – systems with variable thermal conductivity

1. Plane Slab• Assumptions:• One dimensional conduction i.e.

thickness L is small compared to the dimensions in the y and z directions

• Steady state conduction i.e. temperature at any point within the

temperature at any point within the slab does not change with time; of course, temperatures at different points within the slab will be different.

• No internal heat generation• Material of the slab is

homogeneous (i.e. constant density) and isotropic (i.e. value of k is same in all directions).

Rslab = L/(kA)

Fig. 4.1 Plane slab and Thermal circuit

Plane Slab (contd.)

• We start with the general differential equation in Cartesian coordinates:

With these assumptions:

• So, the governing equation for the plane slab with the above-mentioned assumptions becomes:

Temperature field is obtained by solving equation (4.2):

C1 and C2 are obtained from Boundary conditions:

• B.C.(i): T = T1 at x = 0• B.C.(ii): T = T2 at x = L• Solving:

Equation (4.4) can be written in non-dimensional form as follows:

Next, to find the heat flux: Apply Fourier’s Law:

Also, i.e.

Heat transfer through composite slabs:

• Assumptions:• Steady state, one

dimensional heat conduction

• No internal heat generation

• Constant thermal

Tb, hb

Fluid flow

Ta, ha

k1 k2 k3

T4T1 T2 T3

Temp. Profile

• Constant thermal conductivities k1, k2 and k3

• There is ‘perfect thermal contact’ between layers i.e. there is no temperature drop at the interface and the temperature profile is continuous

L1 L2 L3

Fig. 4.2 Composite slab with three layers and the Thermal resistance network

TbT1 T2 T3 T4

Ra R1 R2 R3 Rb

• Observe that heat flows from the fluid at temperature Ta to the left surface of slab 1 by convection, then by conduction through slabs 1, 2 and 3, and then, by convection from the right surface of slab 3 to the fluid at temperature Tb.

• Now, considering each case by turn:• Convection at the left surface of slab 1:

• Convection at the left surface of slab 1: • Q = ha A (Ta – T1), from Newton’s Law of

cooling; i.e.

Conduction through slab 1 :

Conduction through slab 2 :

• Conduction through slab 3 :

Convection at the right surface of slab 3:

• Adding (a), (b) , (c) , (d) and (e):

where Ra = convective resistance at left surface of slab 1,R1 = conductive resistance of slab 1,R2 = conductive resistance of slab 2,R3 = conductive resistance of slab 3, andRb = convective resistance at right surface of slab 3

• So, we write equation (g) as:

Now, observe the analogy with Ohm’s Law.

For thermal resistances in series, we have:

• When the resistances are in parallel, the total resistance is given by:

Insulated

T1 T2R1= L/(k1A)

R2= L/(k2A)

Fig. 4.3 Composite slab with parallel resistances

For thermal resistances in series and parallel:

• Applying the rules of electrical circuit for series and parallel resistances, we have:

Insulation

T4T1T2 T3

Insulation

Fig. 4.4 Composite slab with series-parallel resistancesand the equivalent Thermal circuit

Note: Conditions to be satisfied to apply this concept are: (i) steady state heat transfer and (ii) no internal heat generation.

Overall heat transfer coefficient, U (W/(m2.C)):

• We would like to have the heat transfer given by a simple relation of the form:

• Q = U A (Ta – Tb) = U A ∆T……(4.21)• Now, we have from eqn. (4.16):

• Now, we have from eqn. (4.16):

• Comparing eqn (4.16) and eqn. (4.21), we can write:

Remember the expression for U as given by eqn. (4.23); it is easierand is applicable when we deal with other geometries too.

Thermal contact resistance:• Physical reasoning for contact

resistance is explained with reference to Fig. 4.5:

• Physical contact between A and B occurs only at a few points i.e. at the ‘peaks’ as

Interface

Air gap

points i.e. at the ‘peaks’ as shown.

• Therefore, heat transfer occurs by conduction through this solid contact area and also by ‘gas conduction’ through the gas filling the interfacial voids.

TaTc1 T2

RcontactR1 R2

Fig. 4.5 Thermal contact resistance

• In effect, resistance to heat transfer is by two mechanisms:

• by solid conduction at the peaks, and • by gas conduction through the interfacial gas in the

voids. • Of these two, solid conduction is usually negligible.• Note that that there is a temperature drop at the

interface, (Tc1 – Tc2) and the temperature profile is

interface, (Tc1 – Tc2) and the temperature profile is not continuous.

• Thermal contact resistance is defined as the temperature drop at the interface divided by the heat transfer rate per unit area:

• Interface ‘thermal contact conductance’ is defined as the inverse of the contact resistance, and is given by:

Thermal contact resistance depends on:

•surface roughness — smoother the surface, lesser the resistance .interface temperature — higher the temperature, lesser the

resistance•interface pressure — higher the pressure, lesser the resistance

•type of material – softer the material, lesser the resistance

• Thermal contact resistance may be reduced by:

• making the mating surfaces very smooth• inserting a layer of conducting grease at the interface• inserting a ‘shim’ (thin foil) made of a soft material

such as indium, lead, tin or silver between the surfaces

• filling the interstitial voids with a gas of higher thermal conductivity than that of air (ex: helium)

conductivity than that of air (ex: helium)• increasing the interface pressure• in case of permanently bonded joints, contact

resistance can be reduced by using an epoxy or soft solder rich in lead, or a hard solder of gold/tin alloy

• Values of Rc are given in Heat Transfer Handbooks.

2. Cylindrical systems

Cylindrical systems (contd.)

• Assumptions:• Steady state conduction• One dimensional conduction, in the r-

direction only

direction only• Homogeneous, isotropic material with

constant k• No internal heat generation

• In cylindrical coordinates, we have the general diff. eqn. for conduction:

In this case:

• Therefore, the controlling differential equation for the cylindrical system becomes:

Integrating above eqn. twice:

• C1 and C2 are found out by applying the two B.C’s:at r = ri, T = Ti

at r = ro, T = To

Solving, we get:

Eqn. (4.34) is written in non-dimensional form as follows:

Note: For thin cylinders, ri ≈ ro , and then the temperature distribution within the shell is almost linear.

• Next, to find the heat transfer rate, Q:• We apply the Fourier’s Law. • Considering the inner radius ri,

Equation (4.36) gives the desired expression for rate of heat transfer through the cylindrical system.

Thermal resistance of cyl. system

• Now, writing eqn. (4.36) in a form analogous to Ohm’s Law:

Therefore,

Composite cylinders:

k1k2 Q

Ta TbQQ

Ra R1 R2 Rb

(a) Composite cylinders and equiv. thermal ciruit

Temp. Profile

(b) Composite cylinders and temp. profile

TaT1T2T3Tb

Composite cylinders (contd.):

• Assumptions:• Steady state heat flow• One dimensional conduction in the r direction only• No internal heat generation• Perfect thermal contact between layers

• Perfect thermal contact between layers• Under these stipulations, we note that heat flow rate is

constant and the same through each layer.• Let us write separately the heat flow equations for

each layer:

• 1. Convection from the hot fluid to inner wall at T1:

2. Conduction through first cylindrical layer:

• 3. Conduction through second cylindrical layer:

4. Convection from the outer wall at T3 to cold fluid at Tb:

• Adding equations (a), (b), (c) and (d):(Ta – Tb) = Q (Ra + R1 + R2 + Rb)

If there are N concentric cylinders, we can write :

Overall heat transfer coefficient for the cylindrical system:

• We would like to write the heat transfer rate in terms of the known overall temperature difference, as follows:

Q = U A ∆Toverall = U A (Ta – Tb)

where U is an overall heat transfer coefficient and A is the area normal to the direction of heat flow.

the area normal to the direction of heat flow. • In the case of a cylindrical system, area normal to the

direction of heat flow is 2πrL, and clearly, this varies with r. Therefore, while dealing with cylindrical systems, we have to specify as to which area U is based on i.e. whether it is based on inside area or outside area.

• We write:

Comparing eqn. (4.44) with eqn. (4.41), we get:

Therefore,

• We can also write:

• i.e.

To calculate Ui or Uo while solving numerical problems, justremember equation (4.46), viz.:

• Prob. 1(c) - VTU - M.Tech.-Jan. 2010: An insulated steam pipe having OD of 30 mm is to be covered with 2 layers of insulation, each having a thickness of 20 mm. The thermal conductivity of one material is 5 times that of the other. Assuming that the inner and outer surface temperatures of

inner and outer surface temperatures of composite insulation are fixed, how much heat transfer will be increased when the better insulation material is next to the pipe than it is at the outer layer?

Data: d1 30 mm

d2 70 mm

d3 110 mm

Case 1: Insulation with smaller k applied on the surface of the pipe:

2 pi. k. L.

2 pi. 5. k. L.

Q1∆T

• Case 2: Insulation with higher k applied on the surface of the pipe: Case 2: Insulation with higher k applied on the surface of the pipe:

2 pi. 5. k. L.

2 pi. k. L.

Q2∆T

Therefore, Q2

Now, define: Q2byQ1

Q2byQ1 1.509=

i.e. heat transfer increases by 50.9% when insulation with higher k is applied immediately next to the pipe surface.

• To see variation of Q with the thermal cond. multiplier N:

Q2byQ1 N( )

N 1 1.5, 10..

3. Spherical Systems:

Spherical Systems (contd.):

• Assumptions:• Steady state conduction• One dimensional conduction, in the r direction only• Homogeneous, isotropic material with constant k• No internal heat generation

• No internal heat generation

For one dimensional conduction in r - direction only, we have:

• For the above mentioned assumptions, this reduces to:

Integrating:

• Integrating again,

where C1 and C2 are constants of integration, found out by applying the two B.C’s, viz.:at r = ri, T = Tiat r = ro, T = To

Getting and substituting C1 and C2 in eqn. (4.51), we have:

• Eqn. (4.52) is written in non-dimensional form as follows:

Next, to find the heat transfer rate, Q:We apply the Fourier’s Law. Since it is steady state conduction, with no heat generation, Q is the same through each layer.

no heat generation, Q is the same through each layer.Considering the outer surface, i.e. at r = ro:

Thermal resistance for conduction for a spherical shell:

• Now, writing eqn. (4.54) in a form analogous to Ohm’s Law:

Then, thermal resistance for conduction for a spherical shell is given by:

Composite Spheres:

k1k2 Q

Ta TbQQ

Ra R1 R2 Rb

Fig. 4.12(a) Composite spheres and equiv. thermal ciruit

r3 Temp. Profile

Fig. 4.12(b) Composite spheres and temp. profile

T2T3Tb

Composite Spheres:

• Assumptions:• Steady state heat flow• One dimensional conduction in the r direction only• No internal heat generation• Perfect thermal contact between layers

• Perfect thermal contact between layers• Under the given stipulations, it is clear that heat flow

rate, Q through each layer is the same. • Let us write separately the heat flow equations for

each layer:

• 1. Convection from the hot fluid to inner wall at T1:

2. Conduction through first spherical layer:

• 3. Conduction through second spherical layer:

• Adding equations (a), (b), (c) and (d):(Ta – Tb) = Q (Ra + R1 + R2 + Rb)

If there are N concentric spheres, we can write :

Overall heat transfer coefficient for the spherical system:

• We write:

Overall heat transfer coefficient for the spherical system (contd.):

• We can also write:

Overall heat transfer coefficient for the spherical system (contd.):

• Note: Equations (4.62 a) and (4.62 b) give Ui and Uo in terms of the inside and outside radii.

• You need not memorize them. • To calculate Ui or Uo while solving numerical problems,

just remember equation (4.46), viz.:

4. Critical thickness of insulation for cylindrical system:

• Generally, addition of insulation does reduce the heat loss;

• However, there are some interesting cases where the addition of insulation, in fact,

addition of insulation, in fact, increases the heat loss!

• Consider a cylinder with insulation where r1 is the inner radius of insulation layer (or, outer radius of pipe), r2 is the outer radius of insulation layer.

Critical thickness of insulation(contd.):

• Resistance to heat transfer is made up of two components viz. conductive resistance through the cylindrical insulation layer [ = ln(r2/r1)/ (2 π k L) ]and convective resistance between the wall surface and the surroundings [ = 1/(h. Ao )].

surface and the surroundings [ = 1/(h. Ao )].• As the insulation thickness is increased i.e. as

insulation radius r2 is increased, conductive resistance of insulation increases; however, convective resistance, given by [1/(h. Ao)] goes on decreasing since Ao, the outside surface area goes on increasing with increasing radius.

Critical thickness of insulation (contd.):

• Therefore, the total resistance may increase or decrease, depending on the relative rates of change of these two resistances.

• For the above case, the equivalent thermal resistance circuit is shown below:

T1 TaQQ

Rins Rconv

Fig. 4.13(a) Equiv. Thermal curcuit for a cylinder with insulation

• Consider any radius r of the insulation.

• We have:

Variation of Rins and Rconv and Rtot with r are shown in Fig.

Fig. 4.13(b) Variation of resistances with insulation radius for a cylinder

Note that Rtot passes through a minimum.

• The insulation radius at which the resistance to heat flow is minimum is called ‘critical radius’, rc; i.e. the heat flow is a maximum at the critical radius.

• Variation of heat flow per unit length, (Q/L), with r is shown in Fig. 4.13 (c):

Q/LQmax

Fig. 4.13(c) Heat transfer per unit length vs. insulation radius for a cylinder

r1 rc r

• Mathematically, to find out at what insulation radius r the Rtot becomes a minimum for the cylindrical system, differentiate the expression for Rtot and equate to zero.

• We have:

Equation (4.63) gives the expression for critical radius, rc for thecylindrical system.

• To confirm that at r = rc , Rtot indeed is minimum, find out the value of (d2 Rtot /dr2 ) at r = rc, and it will be found to be positive; i.e at r = rc , the Rtot is a minimum.

• There are two cases of practical interest, as shown in Fig. 4.14:

• (a) insulating current carrying wires, and

• (b) insulation of steam pipes and refrigerant lines.Q

Fig. 4.14(a) Heat transfer per unit length vs. insulation radius for a cylinder when r1 < r c

r1 rc r

Q/LQmax

r1rc r

Q/L Qmax

Fig. 4.14(b) Heat transfer per unit length vs. insulation radius for a cylinder when r1 > r c

Critical thickness of insulation for a sphere:

• We have, for the spherical system:

rInsulation

Sphere

Differentiating Rtot w.r.t. r and equating to zero, we get:

Fig. 4.15 Critical radius for a Sphere

equating to zero, we get:

Equation (4.66) gives the expression for critical radius, rc for thespherical system.

5. Optimum (or Economic) thickness of insulation :

• ‘Optimum’ or ‘Economic’ thickness of insulation is that thickness for which the combined cost of the amount of energy lost

amount of energy lost through the insulation and the total (material + labour) cost of insulation is a minimum.

Optimum (or Economic) thickness of insulation (contd.) :

• To compare three or four insulations for the same job, we can draw similar ‘Total cost curves’ for those insulations and the thickness of the insulation

thickness of the insulation having the lowest total cost is the optimum thickness.

• In Fig. 4.16(b), insulation D has optimum thickness.

6. Effect of variable thermal conductivity:

• When the k of a material varies rapidly with temperature or when the temperature range of operation is large, it becomes necessary to take into account the variation of k with temperature.

• Generally, k varies with temperature linearly as

• Generally, k varies with temperature linearly as follows:

Plane slab with variable thermal conductivity:

• We write from Fourier’s Law:

Substituting for k(T), separating the variables and

k = k(T)

Substituting for k(T), separating the variables and integrating from x = 0 to x = L (with T = T1 to T = T2):

Rslab = L/(kmA)

Fig. 4.17 Plane slab with k = k(T) and the thermal circuit

Plane slab with variable thermal conductivity:

Finally, we get:

Equation (c) gives the heat transfer rate for the plane slab, with variable thermal conductivity, k varying linearly with temperature.

Plane slab with variable thermal conductivity :

• Thermal resistance:

Temp. distribution:Eqn.(4.69) gives the temperature distribution within the slab, with the thermal conductivity varying linearly with temperature:

Plane slab with variable thermal conductivity :

• Temp. distribution:

Hollow cylinder with variable thermal conductivity:

• Let k vary with temperature linearly as follows:

• We write from Fourier’s Law:

• Substituting for k(T), separating the variables and integrating from r = ri to r = ro (with T = Ti to T = To), we get:

Finally:where

• To get temperature distribution:• Eqn.(4.72) below gives the temperature distribution

within the cylindrical shell, with the thermal conductivity varying linearly with temperature.

Hollow sphere with variable thermal conductivity:

• Let k of the material vary with linearly with temperature as given by eqn. (4.67): i.e. k(T) = ko (1 + βT).

• From Fourier’s Law:

• Substituting for k(T), separating the variables and integrating from r = ri to r = ro (with T = Ti to T = To), we get:

• Finally,

• Temperature distribution within the spherical shell:• Eqn.(4.75) gives the temperature distribution within

the spherical shell, with the thermal conductivity varying linearly with temperature.

Compare this with eqn.(4.69) for a slab, and (4.72) for a cylinder, with the k varying linearly with temperature.

Summary of Basic conduction relations:Table 4.1 Relations for steady state, one dimensional conduction with no

internal heat generation, and constant k

Summary of Basic conduction relations:Table 4.2(a)

Relations for steady state, one dimensional conduction with no internal heat generation and k varying linearly with temperature

as: k(T) = ko (1 + ββββ T) {km = ko (1 + ββββ Tm); Tm = (T1 + T2)/2}

Summary of Basic conduction relations:Table 4.2(b)

Summary of Basic conduction relations:Table 4.2(c)

Problem.. from Ref:2

Problem.. from Ref: 5

• "Prob. 1A.9: Consider an insulated spherical liquid oxygen (LOX) tank whose outside diameter is 0.5 m and the outer surface temperature (T_surface) is -10 deg.C. The tank is kept in surroundings at 25 deg.C. Latent heat is kept in surroundings at 25 deg.C. Latent heat of evaporation of LOX is 214kJ/kg. (i) If the emissivity of outer surface is 0.2 and convection heat transfer coeff (h)is 10 W/m^2.K, find out the boil off rate of LOX (ii) Plot the boil off rate as a function of emissivity, with the values of T_surface and h remaining the same as earlier."

EES Solution:

"Data:"

h_fg = 214e03 [J/kg]D_tank = 0.5 [m]T_surface = -10 [C]T_amb = 25 [C]T_amb = 25 [C]epsilon = 0.2h = 10 [W/m^2-K]sigma = 5.67e-08 [W/m^2-K^4] “Stefan-Boltzmann constant”

"Calculations:"

A = pi * D_tank^2 "[m^2] ... surface area of the tank"Q_conv = h * A * (T_amb - T_surface) "[W] ... heat transfer to tank by

convection"Q_rad = epsilon * sigma * A * ((T_amb + 273)^4 - (T_surface + 273)^4)

"[W].... heat transfer to tank by radiation"

"Now, (Q_conv + Q_rad) reaches the LOX and results in boil off:"

m_boiloff = (Q_conv + Q_rad)/h_fg "[kg/s] .... boil off rate of LOX"

Results:

“To plot m_boiloff vs. emissivity with T_surface = -10 C and h = 10 W/m^2.K:”

First construct a parametric Table as shown below:

Then, create the plot:

Prob. 1C.8.[Ref: 5] A steel pipe of 220 mm OD is carrying steam at 280 C. It is insulated with a material of k = 0.06 . [ 1 + 0.0018.T ] where k is in W/m.C. Thickness of insulation is 50 mm and the outer surface temp is 50 C. Determine the heat flow per metre length of the pipe and the temp at the mid-thickness of pipe.[VTU-6th Sem-B.E.-June/July 2009]"

• "Data:"

• r_1 = 0.11 [m]• T_1 = 280 [C]• T_1 = 280 [C]• r_2 = 0.16 [m]• "Th. conductivity k is linearly varying: of the form k = k_0 ( 1 + beta.T),

where:"• k_0 =0.06• beta = 0.0018• T_2 = 50 [C]• L = 1 [m] "...length of pipe ... assumed"

• "Calculations:"

• T_m = (T_1+T_2)/2 "[C] ... mean temp of insulation"• k_m = k_0*(1+beta*T_m) "[W/m.C]] ... mean th. conductivity of

insulation"• R_ins = ln(r_2/r_1)/(2 * pi * k_m * L) "[C/W] ... thermal resist of cyl.

layer of insulation"• Q = (T_1-T_2)/R_ins "..finds Q (W), the heat transfer rate"

• "To find temp T(r) at any radius r :"• {

• r = 0.135 [m] "...starting with a value of r, this is the mid-thickness of insulation, but this is a variable to draw graph:"

• "Let temp be Tr at radius r:"

• T_m_r = (T_1 + Tr)/2• k_m_r = k_0*(1+beta*T_m_r)

• "Remember that in steady state, Q ( already calculated = 300.1 W) is the same through each layer:"

• Q = (2*pi*k_m_r*L)*(T_1-Tr)/ln(r/r_1) "..finds Tr (C) at r"• Q = (2*pi*k_m_r*L)*(T_1-Tr)/ln(r/r_1) "..finds Tr (C) at r"

"To draw the plot of variation of temp in the insulation with the radius:First, construct a parametric table, and then draw the graph:"

Prob. 1C.9.[Ref; 5] A long, hollow cylinder is constructed from a material whose k varies with temp as: k = 0.01 + 0.001.T, where k is in W/m.K and T is in deg.C. The inner and outer radii of the cylinder are 125 mm and 250 mm respectively. Under steady state conditions, the inner and outer surface temperatures are 698 K and 363 K respectively. Determine: (i) the rate of heat flow per metre length, (ii) temp of air on the outside of cylinder, if the surface heat transfer coeff on the exterior surface is 14.5 W/m^2.K [VTU - VI Sem. B.E. - Feb. 2002]

Now, consider the following extension of this problem:

Inside surface temp is fixed at 425 deg.C. Ambient temp T_amb is 35 deg. C. Calculate the outside

temp T_amb is 35 deg. C. Calculate the outside surface temp T2 as the heat transfer coeff. h varies from 4 W/m^2.C to 200 W/m^2.C. Other data remain the same.:

To get T2, apply the heat balance on the outer surface; i.e. heat reaching the surface from inside by conduction = heat lost to ambient by convection.

Use the Solve Block of Mathcad. Start with a trial value for T2:

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