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PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
Solutions Manual
for
Heat and Mass Transfer: Fundamentals & Applications
Fourth Edition
Yunus A. Cengel & Afshin J. Ghajar
McGraw-Hill, 2011
Chapter 2
HEAT CONDUCTION EQUATION
PROPRIETARY AND CONFIDENTIAL
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Introduction
2-1C Heat transfer is a vector quantity since it has direction as well as magnitude. Therefore, we must specify both direction
and magnitude in order to describe heat transfer completely at a point. Temperature, on the other hand, is a scalar quantity.
2-2C The heat transfer process from the kitchen air to the refrigerated space is
transient in nature since the thermal conditions in the kitchen and the
refrigerator, in general, change with time. However, we would analyze this
problem as a steady heat transfer problem under the worst anticipated conditions
such as the lowest thermostat setting for the refrigerated space, and the
anticipated highest temperature in the kitchen (the so-called design conditions).
If the compressor is large enough to keep the refrigerated space at the desired
temperature setting under the presumed worst conditions, then it is large enough
to do so under all conditions by cycling on and off. Heat transfer into the
refrigerated space is three-dimensional in nature since heat will be entering
through all six sides of the refrigerator. However, heat transfer through any wall
or floor takes place in the direction normal to the surface, and thus it can be
analyzed as being one-dimensional. Therefore, this problem can be simplified
greatly by considering the heat transfer to be onedimensional at each of the four
sides as well as the top and bottom sections, and then by adding the calculated
values of heat transfer at each surface.
2-3C The term steady implies no change with time at any point within the medium while transient implies variation with
time or time dependence. Therefore, the temperature or heat flux remains unchanged with time during steady heat transfer
through a medium at any location although both quantities may vary from one location to another. During transient heat
transfer, the temperature and heat flux may vary with time as well as location. Heat transfer is one-dimensional if it occurs
primarily in one direction. It is two-dimensional if heat tranfer in the third dimension is negligible.
2-4C Heat transfer through the walls, door, and the top and bottom sections of an oven is transient in nature since the thermal
conditions in the kitchen and the oven, in general, change with time. However, we would analyze this problem as a steady
heat transfer problem under the worst anticipated conditions such as the highest temperature setting for the oven, and the
anticipated lowest temperature in the kitchen (the so called “design” conditions). If the heating element of the oven is large
enough to keep the oven at the desired temperature setting under the presumed worst conditions, then it is large enough to do
so under all conditions by cycling on and off.
Heat transfer from the oven is three-dimensional in nature since heat will be entering through all six sides of the
oven. However, heat transfer through any wall or floor takes place in the direction normal to the surface, and thus it can be
analyzed as being one-dimensional. Therefore, this problem can be simplified greatly by considering the heat transfer as
being one- dimensional at each of the four sides as well as the top and bottom sections, and then by adding the calculated
values of heat transfers at each surface.
2-5C Heat transfer to a potato in an oven can be modeled as one-dimensional since temperature differences (and thus heat
transfer) will exist in the radial direction only because of symmetry about the center point. This would be a transient heat
transfer process since the temperature at any point within the potato will change with time during cooking. Also, we would
use the spherical coordinate system to solve this problem since the entire outer surface of a spherical body can be described
by a constant value of the radius in spherical coordinates. We would place the origin at the center of the potato.
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2-14C Heat transfer to a canned drink can be modeled as two-dimensional since temperature differences (and thus heat
transfer) will exist in the radial and axial directions (but there will be symmetry about the center line and no heat transfer in
the azimuthal direction. This would be a transient heat transfer process since the temperature at any point within the drink
will change with time during heating. Also, we would use the cylindrical coordinate system to solve this problem since a
cylinder is best described in cylindrical coordinates. Also, we would place the origin somewhere on the center line, possibly
at the center of the bottom surface.
2-15 A certain thermopile used for heat flux meters is considered. The minimum heat flux this meter can detect is to be
determined.
Assumptions 1 Steady operating conditions exist.
Properties The thermal conductivity of kapton is given to be 0.345 W/m⋅K.
Analysis The minimum heat flux can be determined from
2 = 17.3W/m
- 002 m
0. 1 C
( 0. 345 W/m C) L
t q & k
2-16 The rate of heat generation per unit volume in the uranium rods is given. The total rate of heat generation in each rod is
to be determined.
g = 2× 10
8 W/m
3 Assumptions Heat is generated uniformly in the uranium rods.
Analysis The total rate of heat generation in the rod is determined D = 5 cm
by multiplying the rate of heat generation per unit volume by the
volume of the rod
L = 1 m
( / 4 ) ( 2 10 W/m)[ ( 0. 05 m) / 4 ]( 1 m) 3 .93 10 W= 393 kW
2 8 3 2 5 E (^) gen = e &gen rod= e &genπ D L = × π = ×
& V
2-17 The variation of the absorption of solar energy in a solar pond with depth is given. A relation for the total rate of heat
generation in a water layer at the top of the pond is to be determined.
Assumptions Absorption of solar radiation by water is modeled as heat generation.
Analysis The total rate of heat generation in a water layer of surface area A and thickness L at the top of the pond is
determined by integration to be
b
Ae (1 e )
bL 0
− −
=
∫ ∫
L L bx
x
bx
b
e E e d ee Adx Ae
0
0 0
gen gen 0 ( ) V
V
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2-18 The rate of heat generation per unit volume in a stainless steel plate is given. The heat flux on the surface of the plate is
to be determined.
Assumptions Heat is generated uniformly in steel plate.
e
L
Analysis We consider a unit surface area of 1 m
2
. The total rate of
heat generation in this section of the plate is
( ) ( 5 10 W/m )( 1 m )(0.03m) 1.5 10 W
6 3 2 5 E (^) gen = e gen plate= e gen A × L = × = ×
& & V &
Noting that this heat will be dissipated from both sides of the plate, the heat flux on
either surface of the plate becomes
2 = = 75 kW/m ×
×
2 2
5
plate
gen 75 , 000 W/m 2 1 m
1. 5 10 W
A
E
q
2-19E The power consumed by the resistance wire of an iron is given. The heat generation and the heat flux are to be
determined.
Assumptions Heat is generated uniformly in the resistance wire.
Analysis An 800 W iron will convert electrical energy into^ q^ = 800 W
heat in the wire at a rate of 800 W. Therefore, the rate of heat
generation in a resistance wire is simply equal to the power
rating of a resistance heater. Then the rate of heat generation in
the wire per unit volume is determined by dividing the total
rate of heat generation by the volume of the wire to be
D = 0.08 in
L = 15 in
7 3 ⎟= 6.256^ ×^10 Btu/h ⋅ ft ⎠
1 W
- 412 Btu/h
[ ( 0. 08 / 12 ft) / 4 ]( 15 / 12 ft)
800 W
2 2
gen
wire
gen gen π D L π
E E
e
V
Similarly, heat flux on the outer surface of the wire as a result of this heat generation is determined by dividing the total rate
of heat generation by the surface area of the wire to be
5 2 ⎟= 1.043 × 10 Btu/h ⋅ ft ⎠
1 W
- 412 Btu/h
( 0. 08 / 12 ft)( 15 / 12 ft)
gen^800 W
wire
gen
π DL π
E
A
E
q
Discussion Note that heat generation is expressed per unit volume in Btu/h⋅ft
3 whereas heat flux is expressed per unit surface
area in Btu/h⋅ft
2 .
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Heat Conduction Equation
2-21C The one-dimensional transient heat conduction equation for a plane wall with constant thermal conductivity and heat
generation is t
T
k α
e
x
T
∂ (^) gen 1
2
. Here T is the temperature, x is the space variable, is the heat generation per unit
volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the time.
e & gen
2-22C The one-dimensional transient heat conduction equation for a long cylinder with constant thermal conductivity and
heat generation is t
T
k
e
r
T
r r r ∂
α
1 ∂ &gen^1
. Here T is the temperature, r is the space variable, g is the heat generation per
unit volume, k is the thermal conductivity, α is the thermal diffusivity, and t is the time.
2-23 We consider a thin element of thickness ∆ x in a large plane wall (see Fig. 2-12 in the text). The density of the wall is ρ,
the specific heat is c, and the area of the wall normal to the direction of heat transfer is A. In the absence of any heat
generation, an energy balance on this thin element of thickness ∆ x during a small time interval ∆ t can be expressed as
t
E
Q (^) x Qx x ∆
& & element
where
∆ E element = Et +∆ t − Et = mc ( Tt +∆ t − Tt )= ρ cA ∆ x ( Tt +∆ t − Tt )
Substituting,
t
T T
Q Q cA x
t t t x x x ∆
+∆ +∆ ρ
Dividing by A ∆ x gives
t
T T
c x
Q Q
A
x x x t t t
∆
+∆ +∆
Taking the limit as ∆ x → 0 and ∆ t → 0 yields
t
T
ρc x
T
kA A x ∂
since from the definition of the derivative and Fourier’s law of heat conduction,
∆ → x
T
kA x x
Q
x
Q (^) x x Qx
x
0
lim
Noting that the area A of a plane wall is constant, the one-dimensional transient heat conduction equation in a plane wall with
constant thermal conductivity k becomes
t
T
x^ α
T
2
2
where the property α = k / ρ c is the thermal diffusivity of the material.
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2-24 We consider a thin cylindrical shell element of thickness ∆ r in a long cylinder (see Fig. 2-14 in the text). The density of
the cylinder is ρ, the specific heat is c, and the length is L. The area of the cylinder normal to the direction of heat transfer at
any location is A =^2 π rL where^ r^ is the value of the radius at that location.^ Note that the heat transfer area^ A^ depends on^ r^ in
this case , and thus it varies with location. An energy balance on this thin cylindrical shell element of thickness ∆ r during a
small time interval ∆ t can be expressed as
t
E
Q (^) r Qr r E ∆
element element
where
∆ E element = Et +∆ t − Et = mc ( Tt +∆ t − Tt )= ρ cA ∆ r ( Tt +∆ t − Tt )
E & (^) element = e &gen Velement= e &gen A ∆ r
Substituting,
t
T T
Q Q e A r cA r
t t t r r r ∆
+∆ +∆ gen ρ
where A = 2 π rL. Dividing the equation above by A ∆ r gives
t
T T
e c r
Q Q
A
r r r t t t
∆
+∆ +∆
gen ρ
Taking the limit as ∆ r → 0 and ∆ t → 0 yields
t
T
e c r
T
kA A r ∂
⎟+ =ρ ⎠
gen
since, from the definition of the derivative and Fourier’s law of heat conduction,
∆ → r
T
kA r r
Q
r
Q (^) r r Qr
r
0
lim
Noting that the heat transfer area in this case is A = 2 π rL and the thermal conductivity is constant, the one-dimensional
transient heat conduction equation in a cylinder becomes
t
T
e r
T
r r r ∂
α
gen
where α = k / ρ c is the thermal diffusivity of the material.
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2-27 For a medium in which the heat conduction equation is given by t
T
y
T
x
T
α
2
2
2
2
:
( a ) Heat transfer is transient, ( b ) it is two-dimensional, ( c ) there is no heat generation, and ( d ) the thermal conductivity is
constant.
2-28 For a medium in which the heat conduction equation is given by 0
⎟+ gen = ⎠
e z
T
k r z
T
kr r r
( a ) Heat transfer is steady, ( b ) it is two-dimensional, ( c ) there is heat generation, and ( d ) the thermal conductivity is variable.
2-29 For a medium in which the heat conduction equation is given in its simplest by 0
⎟+ gen = ⎠
e dr
dT rk dr
d
r
( a ) Heat transfer is steady, ( b ) it is one-dimensional, ( c ) there is heat generation, and ( d ) the thermal conductivity is variable.
2-30 For a medium in which the heat conduction equation is given by t
T
r α
T
r r r ∂
2
( a ) Heat transfer is transient, ( b ) it is one-dimensional, ( c ) there is no heat generation, and ( d ) the thermal conductivity is
constant.
2-31 For a medium in which the heat conduction equation is given in its simplest by 0 2
2
dT
dr
dT r :
( a ) Heat transfer is steady, ( b ) it is one-dimensional, ( c ) there is no heat generation, and ( d ) the thermal conductivity is
constant.
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2-32 We consider a small rectangular element of length ∆ x , width ∆ y , and height ∆ z = 1 (similar to the one in Fig. 2-20). The
density of the body is ρ and the specific heat is c. Noting that heat conduction is two-dimensional and assuming no heat
generation, an energy balance on this element during a small time interval ∆ t can be expressed as
ofthe element
theenergycontent
Rateofchangeof
at thesurfacesat
Rateofheatconduction
surfacesat and
conductionat the
Rateofheat
x y x x y y
or t
E
Q (^) x Qy Qx x Qy y ∆
& & & & element
Noting that the volume of the element is V element = ∆ x ∆ y ∆ z =∆ x ∆ y × 1 , the change in the energy content of the element can
be expressed as
∆ E element = Et +∆ t − Et = mc ( Tt +∆ t − Tt )= ρ c ∆ x ∆ y ( Tt +∆ t − Tt )
Substituting, t
T T
Q Q Q Q c x y
t t t x y x x y y ∆
+∆ +∆ +∆ ρ
Dividing by ∆ x ∆ y gives
t
T T
c y
Q Q
x x
Q Q
y
x x x y y y t t t
∆
+∆ +∆ +∆
& & &^ &
Taking the thermal conductivity k to be constant and noting that the heat transfer surface areas of the element for heat
conduction in the x and y directions are A (^) x = ∆ y × 1 and Ay =∆ x × 1 ,respectively, and taking the limit as ∆ x , ∆ y ,and∆ t → 0
yields
t
T
y^ α
T
x
T
2
2
2
2
since, from the definition of the derivative and Fourier’s law of heat conduction,
2
2
0
lim x
T
k x
T
k x x
T
k y z x y z x
Q
x y z
Q Q
y z
x x x x
x (^) ∂
+∆
∆→
2
2
0
lim y
T
k y
T
k y y
T
k x z y x z y
Q
y x z
Q Q
x z
y y y y
y (^) ∂
+∆
∆→
Here the property α = k / ρ c is the thermal diffusivity of the material.
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2-34 Consider a thin disk element of thickness ∆ z and diameter D in a long cylinder (Fig. P2-34). The density of the cylinder
is ρ, the specific heat is c, and the area of the cylinder normal to the direction of heat transfer is , which is
constant. An energy balance on this thin element of thickness ∆ z during a small time interval ∆ t can be expressed as
2
A = π D
ofthe element
theenergycontent
Rateofchangeof
theelement
generationinside
Rateofheat
surfaceat +
conductionat the
Rateofheat
thesurfaceat
conductionat
Rateofheat
z z z
or,
t
E
Q (^) z Qz z E ∆
element element
But the change in the energy content of the element and the rate of heat generation within the element can be expressed as
∆ E element = Et +∆ t − Et = mc ( Tt +∆ t − Tt )= ρ cA ∆ z ( Tt +∆ t − Tt )
and
E & (^) element = e &gen Velement= e &gen A ∆ z
Substituting,
t
T T
Q Q e A z cA z
t t t z z z ∆
+∆ +∆ &gen ρ
Dividing by A ∆ z gives
t
T T
e c z
Q Q
A
z z z t t t
∆
+∆ +∆
gen ρ
Taking the limit as ∆ z → 0 and ∆ t → 0 yields
t
T
e c z
T
kA A z ∂
⎟+ =ρ ⎠
gen
since, from the definition of the derivative and Fourier’s law of heat conduction,
∆ → z
T
kA z z
Q
z
Q (^) z z Qz
z
0
lim
Noting that the area A and the thermal conductivity k are constant, the one-dimensional transient heat conduction equation in
the axial direction in a long cylinder becomes
t
T
k
e
z
T
α
∂ (^) gen 1
2
where the property α = k / ρ c is the thermal diffusivity of the material.
2-35 For a medium in which the heat conduction equation is given by t
T T
r r
T
r r r ∂
α
∂φ
θ
sin
2
2
2 2
2 2
( a ) Heat transfer is transient, ( b ) it is two-dimensional, ( c ) there is no heat generation, and ( d ) the thermal conductivity is
constant.
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Boundary and Initial Conditions; Formulation of Heat Conduction Problems
2-36C The mathematical expressions of the thermal conditions at the boundaries are called the boundary conditions. To
describe a heat transfer problem completely, two boundary conditions must be given for each direction of the coordinate
system along which heat transfer is significant. Therefore, we need to specify four boundary conditions for two-dimensional
problems.
2-37C The mathematical expression for the temperature distribution of the medium initially is called the initial condition.
We need only one initial condition for a heat conduction problem regardless of the dimension since the conduction equation
is first order in time (it involves the first derivative of temperature with respect to time). Therefore, we need only 1 initial
condition for a two-dimensional problem.
2-38C A heat transfer problem that is symmetric about a plane, line, or point is said to have thermal symmetry about that
plane, line, or point. The thermal symmetry boundary condition is a mathematical expression of this thermal symmetry. It is
equivalent to insulation or zero heat flux boundary condition, and is expressed at a point x 0 as ∂ T ( x 0 , t )/∂ x = 0.
2-39C The boundary condition at a perfectly insulated surface (at x = 0, for example) can be expressed as
0 or
x
T t
x
T t k which indicates zero heat flux.
2-40C Yes, the temperature profile in a medium must be perpendicular to an insulated surface since the slope ∂ T / ∂ x = 0 at
that surface.
2-41C We try to avoid the radiation boundary condition in heat transfer analysis because it is a non-linear expression that
causes mathematical difficulties while solving the problem; often making it impossible to obtain analytical solutions.
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2-45 A long pipe of inner radius r 1 , outer radius r 2 , and thermal conductivity
k is considered. The outer surface of the pipe is subjected to convection to a
medium at with a heat transfer coefficient of h. Assuming steady one-
dimensional conduction in the radial direction, the convection boundary
condition on the outer surface of the pipe can be expressed as
T ∞
r 2
r 1
h , T ∞
[ ( ) ]
2
2 − = hTr − T ∞ dr
dT r k
2-46 A spherical shell of inner radius r 1 , outer radius r 2 , and thermal
conductivity k is considered. The outer surface of the shell is subjected to
radiation to surrounding surfaces at. Assuming no convection and
steady one-dimensional conduction in the radial direction, the radiation
boundary condition on the outer surface of the shell can be expressed as
T surr
k
T surr r (^1) r 2
[ ]
4 surr
4 2
2 ( )
Tr T dr
dT r − k =εσ −
2-47 A spherical container consists of two spherical layers A and B that are at
perfect contact. The radius of the interface is ro. Assuming transient one-
dimensional conduction in the radial direction, the boundary conditions at the
interface can be expressed as
ro
T (^) A ( ro , t )= TB ( ro , t )
and r
T r t k r
T r t k
B o B
A o A ∂
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2-48 Heat conduction through the bottom section of a steel pan that is used to boil water on top of an electric range is
considered. Assuming constant thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the
differential equation and the boundary conditions) of this heat conduction problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3
There is no heat generation in the medium. 4 The top surface at x = L is subjected to convection and the bottom surface at x =
0 is subjected to uniform heat flux.
Analysis The heat flux at the bottom of the pan is
2 2 2
gen 33 , 820 W/m ( 0. 20 m) / 4
0. 85 ( 1250 W)
×
π D π
E
A
Q
q s
s s
&^ &
Then the differential equation and the boundary conditions for this heat conduction problem can be expressed as
2
2
= dx
dT
[ ( ) ]
33 , 280 W/m
hTL T dx
dTL k
q dx
dT k & s
2-49E A 2-kW resistance heater wire is used for space heating. Assuming constant thermal conductivity and one-
dimensional heat transfer, the mathematical formulation (the differential equation and the boundary conditions) of this heat
conduction problem is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3
Heat is generated uniformly in the wire.
2 kW Analysis The heat flux at the surface of the wire is
D = 0.12 in gen 2
s
353 .7W/in 2 ( 0. 06 in)(15in)
2000 W
π rL π
E
A
Q
q
o
s s
&^ &
L = 15 in
Noting that there is thermal symmetry about the center line and there is uniform heat flux at the outer surface, the differential
equation and the boundary conditions for this heat conduction problem can be expressed as
(^1) gen ⎟+ = ⎠
k
e
dr
dT r dr
d
r
2 353 .7W/in
s
o q dr
dTr k
dr
dT
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2-52 A spherical metal ball that is heated in an oven to a temperature of Ti throughout is allowed to cool in ambient air at T ∞
by convection and radiation. Assuming constant thermal conductivity and transient one-dimensional heat transfer, the
mathematical formulation (the differential equation and the boundary and initial conditions) of this heat conduction problem
is to be obtained.
Assumptions 1 Heat transfer is given to be transient and one-dimensional. 2 Thermal conductivity is given to be variable. 3
There is no heat generation in the medium. 4 The outer surface at r = ro is subjected to convection and radiation.
Analysis Noting that there is thermal symmetry about the midpoint and convection and radiation at the outer surface and
expressing all temperatures in Rankine, the differential equation and the boundary conditions for this heat conduction
problem can be expressed as
k r 2
T ∞
h
T surr
T i
t
T
c r
T
kr r r ∂
⎟=ρ ⎠
2
i
o o
o
Tr T
hTr T Tr T r
Tr t k
r
T t
= − +εσ − ∂
∞
[ ( ) ] [ ( ) ]
4 surr
4
2-53 Water flows through a pipe whose outer surface is wrapped with a thin electric heater that consumes 400 W per m
length of the pipe. The exposed surface of the heater is heavily insulated so that the entire heat generated in the heater is
transferred to the pipe. Heat is transferred from the inner surface of the pipe to the water by convection. Assuming constant
thermal conductivity and one-dimensional heat transfer, the mathematical formulation (the differential equation and the
boundary conditions) of the heat conduction in the pipe is to be obtained for steady operation.
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity is given to be constant. 3
There is no heat generation in the medium. 4 The outer surface at r = r 2 is subjected to uniform heat flux and the inner
surface at r = r 1 is subjected to convection.
Analysis The heat flux at the outer surface of the pipe is
r 1 r 2
Q = 400 W
h
T ∞
2
s 2
- 4 W/m 2 ( 0. 065 cm)(1m)
400 W
π rL π
Q
A
Q
q
s s s
Noting that there is thermal symmetry about the center line and there is
uniform heat flux at the outer surface, the differential equation and the
boundary conditions for this heat conduction problem can be expressed as
dr
dT r dr
d
2 2
1
- 6 W/m
[ ( ) ] 85 [ ( ) 90 ]
s
i i
q dr
dTr k
hTr T Tr dr
dTr k
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Solution of Steady One-Dimensional Heat Conduction Problems
2-54C Yes, this claim is reasonable since no heat is entering the cylinder and thus there can be no heat transfer from the
cylinder in steady operation. This condition will be satisfied only when there are no temperature differences within the
cylinder and the outer surface temperature of the cylinder is the equal to the temperature of the surrounding medium.
2-55C Yes, the temperature in a plane wall with constant thermal conductivity and no heat generation will vary linearly
during steady one-dimensional heat conduction even when the wall loses heat by radiation from its surfaces. This is because
the steady heat conduction equation in a plane wall is = 0 whose solution is regardless of the
boundary conditions. The solution function represents a straight line whose slope is C
2 2 d T / dx T ( x )= C 1 x + C 2
2-56C Yes, in the case of constant thermal conductivity and no heat generation, the temperature in a solid cylindrical rod
whose ends are maintained at constant but different temperatures while the side surface is perfectly insulated will vary
linearly during steady one-dimensional heat conduction. This is because the steady heat conduction equation in this case is
= 0 whose solution is
2 2 d T / dx T ( x )= C 1 x + C 2 which represents a straight line whose slope is C 1.
2-57C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transfer through a plain wall
in steady operation must be constant. But the value of this constant must be zero since one side of the wall is perfectly
insulated. Therefore, there can be no temperature difference between different parts of the wall; that is, the temperature in a
plane wall must be uniform in steady operation.
