Chapter 2 Heat Conduction Equation 4 Notes | EduRev

: Chapter 2 Heat Conduction Equation 4 Notes | EduRev

 Page 1


Chapter 2 Heat Conduction Equation 
 2-58 
 
Variable Thermal Conductivity 
 
2-94C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with 
constant thermal conductivity and no heat generation, the temperature in only the plane wall will vary 
linearly. 
 
2-95C  The thermal conductivity of a medium, in general, varies with temperature.  
 
2-96C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity 
varies linearly, the error involved in heat transfer calculation by assuming constant thermal conductivity at 
the average temperature is (a) none.  
 
2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies 
with temperature. 
 
2-98C  Yes, when the thermal conductivity of a medium varies linearly with temperature, the average 
thermal conductivity is always equivalent to the conductivity value at the average temperature.  
 
2-99 A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate is to 
be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 
Thermal conductivity varies quadratically.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
2
1 ? . 
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between T T
1 2
  and   can be determined from 
? ? ? ?
? ?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1 2 1
2
2 0
1 2
3
1
3
2 1 2 0
1 2
3
0
1 2
2
0
1 2
ave
3
1
3
3
) 1 ( ) (
2
1
2
1
2
1
T T T T k
T T
T T T T k
T T
T T k
T T
dT T k
T T
dT T k
k
T
T
T
T
T
T
?
?
?
?
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant 
average thermal conductivity k
ave
  equals the rate of heat transfer through the same medium with variable 
conductivity k(T). Then the rate of heat conduction through the plate can be determined  to be 
 ? ?
L
T T
A T T T T k
L
T T
A k Q
2 1 2
1 2 1
2
2 0
2 1
ave
3
1
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of 
Eq. 2-76, and performed the indicated integration. 
T 2 
x 
k(T) 
 L 
T 1 
Page 2


Chapter 2 Heat Conduction Equation 
 2-58 
 
Variable Thermal Conductivity 
 
2-94C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with 
constant thermal conductivity and no heat generation, the temperature in only the plane wall will vary 
linearly. 
 
2-95C  The thermal conductivity of a medium, in general, varies with temperature.  
 
2-96C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity 
varies linearly, the error involved in heat transfer calculation by assuming constant thermal conductivity at 
the average temperature is (a) none.  
 
2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies 
with temperature. 
 
2-98C  Yes, when the thermal conductivity of a medium varies linearly with temperature, the average 
thermal conductivity is always equivalent to the conductivity value at the average temperature.  
 
2-99 A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate is to 
be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 
Thermal conductivity varies quadratically.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
2
1 ? . 
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between T T
1 2
  and   can be determined from 
? ? ? ?
? ?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1 2 1
2
2 0
1 2
3
1
3
2 1 2 0
1 2
3
0
1 2
2
0
1 2
ave
3
1
3
3
) 1 ( ) (
2
1
2
1
2
1
T T T T k
T T
T T T T k
T T
T T k
T T
dT T k
T T
dT T k
k
T
T
T
T
T
T
?
?
?
?
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant 
average thermal conductivity k
ave
  equals the rate of heat transfer through the same medium with variable 
conductivity k(T). Then the rate of heat conduction through the plate can be determined  to be 
 ? ?
L
T T
A T T T T k
L
T T
A k Q
2 1 2
1 2 1
2
2 0
2 1
ave
3
1
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of 
Eq. 2-76, and performed the indicated integration. 
T 2 
x 
k(T) 
 L 
T 1 
Chapter 2 Heat Conduction Equation 
 2-59 
2-100  A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. 
The variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is 
expressed as 
             
?
ln( / )
Q k L
T T
r r
cylinder ave
?
?
2
1 2
2 1
?    
  
where L is the length of the cylinder, r 1  is the inner radius, and 
r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 2 ?rL is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k L
r
dr
Q
1 1
) ( 2 ?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 
?
ln [( ) ( ) / ] Q
r
r
Lk T T T T
1
0 1
2
1
2
2 2 ? ? ? ? ? ? ? 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r
r r
T T T T
2
0
1
2 1
1 2 1
2
1
2 2 2
0 ? ? ? ? ? ?
? ? ?
ave
ln( / )
ln( / )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r
r r
T T T T ( )
ln( / )
ln( / )
( ) ? ? ? ? ? ? ?
1 1 2 2
2
0
1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 2 
T 2 
r 
r 1 
T 1 
k(T) 
Page 3


Chapter 2 Heat Conduction Equation 
 2-58 
 
Variable Thermal Conductivity 
 
2-94C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with 
constant thermal conductivity and no heat generation, the temperature in only the plane wall will vary 
linearly. 
 
2-95C  The thermal conductivity of a medium, in general, varies with temperature.  
 
2-96C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity 
varies linearly, the error involved in heat transfer calculation by assuming constant thermal conductivity at 
the average temperature is (a) none.  
 
2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies 
with temperature. 
 
2-98C  Yes, when the thermal conductivity of a medium varies linearly with temperature, the average 
thermal conductivity is always equivalent to the conductivity value at the average temperature.  
 
2-99 A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate is to 
be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 
Thermal conductivity varies quadratically.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
2
1 ? . 
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between T T
1 2
  and   can be determined from 
? ? ? ?
? ?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1 2 1
2
2 0
1 2
3
1
3
2 1 2 0
1 2
3
0
1 2
2
0
1 2
ave
3
1
3
3
) 1 ( ) (
2
1
2
1
2
1
T T T T k
T T
T T T T k
T T
T T k
T T
dT T k
T T
dT T k
k
T
T
T
T
T
T
?
?
?
?
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant 
average thermal conductivity k
ave
  equals the rate of heat transfer through the same medium with variable 
conductivity k(T). Then the rate of heat conduction through the plate can be determined  to be 
 ? ?
L
T T
A T T T T k
L
T T
A k Q
2 1 2
1 2 1
2
2 0
2 1
ave
3
1
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of 
Eq. 2-76, and performed the indicated integration. 
T 2 
x 
k(T) 
 L 
T 1 
Chapter 2 Heat Conduction Equation 
 2-59 
2-100  A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. 
The variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is 
expressed as 
             
?
ln( / )
Q k L
T T
r r
cylinder ave
?
?
2
1 2
2 1
?    
  
where L is the length of the cylinder, r 1  is the inner radius, and 
r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 2 ?rL is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k L
r
dr
Q
1 1
) ( 2 ?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 
?
ln [( ) ( ) / ] Q
r
r
Lk T T T T
1
0 1
2
1
2
2 2 ? ? ? ? ? ? ? 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r
r r
T T T T
2
0
1
2 1
1 2 1
2
1
2 2 2
0 ? ? ? ? ? ?
? ? ?
ave
ln( / )
ln( / )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r
r r
T T T T ( )
ln( / )
ln( / )
( ) ? ? ? ? ? ? ?
1 1 2 2
2
0
1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 2 
T 2 
r 
r 1 
T 1 
k(T) 
Chapter 2 Heat Conduction Equation 
 2-60 
2-101  A spherical shell with variable conductivity is subjected to specified temperatures on both sides. The 
variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is expressed as 
             
?
Q k r r
T T
r r
sphere ave
?
?
?
4
1 2
1 2
2 1
?    
  
where  r 1 is the inner radius, r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 4 ?r
2
 is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k
r
dr
Q
1 1
) ( 4
2
?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 ] 2 / ) ( ) [( 4
1 1
2
1
2
1 0
1
T T T T k
r r
Q ? ? ? ? ?
?
?
?
?
?
?
?
?
? ? ?
?
 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r r
r r r
T T T T
2
0
2 1
2 1
1 2 1
2
1
2 2 2
0 ? ?
?
?
? ? ? ?
? ? ?
ave
( )
( )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r r
r r r
T T T T ( )
( )
( )
( ) ? ? ? ?
?
?
? ? ?
1 1 2 2
2
0
2 1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 1 r 2 
T 1 k(T) 
r 
T 2 
Page 4


Chapter 2 Heat Conduction Equation 
 2-58 
 
Variable Thermal Conductivity 
 
2-94C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with 
constant thermal conductivity and no heat generation, the temperature in only the plane wall will vary 
linearly. 
 
2-95C  The thermal conductivity of a medium, in general, varies with temperature.  
 
2-96C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity 
varies linearly, the error involved in heat transfer calculation by assuming constant thermal conductivity at 
the average temperature is (a) none.  
 
2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies 
with temperature. 
 
2-98C  Yes, when the thermal conductivity of a medium varies linearly with temperature, the average 
thermal conductivity is always equivalent to the conductivity value at the average temperature.  
 
2-99 A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate is to 
be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 
Thermal conductivity varies quadratically.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
2
1 ? . 
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between T T
1 2
  and   can be determined from 
? ? ? ?
? ?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1 2 1
2
2 0
1 2
3
1
3
2 1 2 0
1 2
3
0
1 2
2
0
1 2
ave
3
1
3
3
) 1 ( ) (
2
1
2
1
2
1
T T T T k
T T
T T T T k
T T
T T k
T T
dT T k
T T
dT T k
k
T
T
T
T
T
T
?
?
?
?
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant 
average thermal conductivity k
ave
  equals the rate of heat transfer through the same medium with variable 
conductivity k(T). Then the rate of heat conduction through the plate can be determined  to be 
 ? ?
L
T T
A T T T T k
L
T T
A k Q
2 1 2
1 2 1
2
2 0
2 1
ave
3
1
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of 
Eq. 2-76, and performed the indicated integration. 
T 2 
x 
k(T) 
 L 
T 1 
Chapter 2 Heat Conduction Equation 
 2-59 
2-100  A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. 
The variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is 
expressed as 
             
?
ln( / )
Q k L
T T
r r
cylinder ave
?
?
2
1 2
2 1
?    
  
where L is the length of the cylinder, r 1  is the inner radius, and 
r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 2 ?rL is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k L
r
dr
Q
1 1
) ( 2 ?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 
?
ln [( ) ( ) / ] Q
r
r
Lk T T T T
1
0 1
2
1
2
2 2 ? ? ? ? ? ? ? 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r
r r
T T T T
2
0
1
2 1
1 2 1
2
1
2 2 2
0 ? ? ? ? ? ?
? ? ?
ave
ln( / )
ln( / )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r
r r
T T T T ( )
ln( / )
ln( / )
( ) ? ? ? ? ? ? ?
1 1 2 2
2
0
1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 2 
T 2 
r 
r 1 
T 1 
k(T) 
Chapter 2 Heat Conduction Equation 
 2-60 
2-101  A spherical shell with variable conductivity is subjected to specified temperatures on both sides. The 
variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is expressed as 
             
?
Q k r r
T T
r r
sphere ave
?
?
?
4
1 2
1 2
2 1
?    
  
where  r 1 is the inner radius, r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 4 ?r
2
 is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k
r
dr
Q
1 1
) ( 4
2
?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 ] 2 / ) ( ) [( 4
1 1
2
1
2
1 0
1
T T T T k
r r
Q ? ? ? ? ?
?
?
?
?
?
?
?
?
? ? ?
?
 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r r
r r r
T T T T
2
0
2 1
2 1
1 2 1
2
1
2 2 2
0 ? ?
?
?
? ? ? ?
? ? ?
ave
( )
( )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r r
r r r
T T T T ( )
( )
( )
( ) ? ? ? ?
?
?
? ? ?
1 1 2 2
2
0
2 1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 1 r 2 
T 1 k(T) 
r 
T 2 
Chapter 2 Heat Conduction Equation 
 2-61 
2-102   A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of 
heat transfer through the plate is to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Analysis The average thermal conductivity of the medium in this case 
is simply the conductivity value at the average temperature since the 
thermal conductivity varies linearly with temperature, and is 
determined to be 
 
K W/m 24 . 34
2
K 350) + (500
) K 10 (8.7 + 1 K) W/m 25 (
2
1 ) (
1 - 4 -
1 2
0 ave ave
? ?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
? ?
? ? ? ?
T T
k T k k
 
Then the rate of heat conduction through the plate becomes 
  W 30,820 ?
?
? ? ?
?
?
m 15 . 0
0)K 35 (500
m) 0.6  m K)(1.5 W/m 24 . 34 (
2 1
ave
L
T T
A k Q
?
 
Discussion We would obtain the same result if we substituted the given  k(T) relation into the second part of 
Eq, 2-76, and performed the indicated integration. 
 
 
T 2 
k(T) 
T 1 
 L 
Page 5


Chapter 2 Heat Conduction Equation 
 2-58 
 
Variable Thermal Conductivity 
 
2-94C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with 
constant thermal conductivity and no heat generation, the temperature in only the plane wall will vary 
linearly. 
 
2-95C  The thermal conductivity of a medium, in general, varies with temperature.  
 
2-96C During steady one-dimensional heat conduction in a plane wall in which the thermal conductivity 
varies linearly, the error involved in heat transfer calculation by assuming constant thermal conductivity at 
the average temperature is (a) none.  
 
2-97C No, the temperature variation in a plain wall will not be linear when the thermal conductivity varies 
with temperature. 
 
2-98C  Yes, when the thermal conductivity of a medium varies linearly with temperature, the average 
thermal conductivity is always equivalent to the conductivity value at the average temperature.  
 
2-99 A plate with variable conductivity is subjected to specified 
temperatures on both sides. The rate of heat transfer through the plate is to 
be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 
Thermal conductivity varies quadratically.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
2
1 ? . 
Analysis When the variation of thermal conductivity with temperature 
k(T) is known, the average value of the thermal conductivity in the 
temperature range between T T
1 2
  and   can be determined from 
? ? ? ?
? ?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
2
1 2 1
2
2 0
1 2
3
1
3
2 1 2 0
1 2
3
0
1 2
2
0
1 2
ave
3
1
3
3
) 1 ( ) (
2
1
2
1
2
1
T T T T k
T T
T T T T k
T T
T T k
T T
dT T k
T T
dT T k
k
T
T
T
T
T
T
?
?
?
?
 
This relation is based on the requirement that the rate of heat transfer through a medium with constant 
average thermal conductivity k
ave
  equals the rate of heat transfer through the same medium with variable 
conductivity k(T). Then the rate of heat conduction through the plate can be determined  to be 
 ? ?
L
T T
A T T T T k
L
T T
A k Q
2 1 2
1 2 1
2
2 0
2 1
ave
3
1
?
?
?
?
?
?
?
? ? ? ?
?
?
?
?
 
Discussion We would obtain the same result if we substituted the given k(T) relation into the second part of 
Eq. 2-76, and performed the indicated integration. 
T 2 
x 
k(T) 
 L 
T 1 
Chapter 2 Heat Conduction Equation 
 2-59 
2-100  A cylindrical shell with variable conductivity is subjected to specified temperatures on both sides. 
The variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is 
expressed as 
             
?
ln( / )
Q k L
T T
r r
cylinder ave
?
?
2
1 2
2 1
?    
  
where L is the length of the cylinder, r 1  is the inner radius, and 
r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 2 ?rL is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k L
r
dr
Q
1 1
) ( 2 ?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 
?
ln [( ) ( ) / ] Q
r
r
Lk T T T T
1
0 1
2
1
2
2 2 ? ? ? ? ? ? ? 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r
r r
T T T T
2
0
1
2 1
1 2 1
2
1
2 2 2
0 ? ? ? ? ? ?
? ? ?
ave
ln( / )
ln( / )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r
r r
T T T T ( )
ln( / )
ln( / )
( ) ? ? ? ? ? ? ?
1 1 2 2
2
0
1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 2 
T 2 
r 
r 1 
T 1 
k(T) 
Chapter 2 Heat Conduction Equation 
 2-60 
2-101  A spherical shell with variable conductivity is subjected to specified temperatures on both sides. The 
variation of temperature and the rate of heat transfer through the shell are to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Solution (a)  The rate of heat transfer through the shell is expressed as 
             
?
Q k r r
T T
r r
sphere ave
?
?
?
4
1 2
1 2
2 1
?    
  
where  r 1 is the inner radius, r 2 is the outer radius, and 
             ?
?
?
?
?
? ?
? ? ?
2
1 ) (
1 2
0 ave ave
T T
k T k k ?    
  
is the average thermal conductivity. 
(b) To determine the temperature distribution in the shell, we begin with the Fourier’s law of heat 
conduction expressed as 
            
?
( ) Q k T A
dT
dr
? ?  
  
where the rate of conduction heat transfer 
?
Q is constant and the heat conduction area A = 4 ?r
2
 is variable. 
Separating the variables in the above equation and integrating from r = r 1 where  T r T ( )
1 1
?  to any r where 
T r T ( ) ? , we get 
             
? ?
? ?
T
T
r
r
dT T k
r
dr
Q
1 1
) ( 4
2
?
?
 
 
Substituting k T k T ( ) ( ) ? ?
0
1 ? and performing the integrations gives 
 ] 2 / ) ( ) [( 4
1 1
2
1
2
1 0
1
T T T T k
r r
Q ? ? ? ? ?
?
?
?
?
?
?
?
?
? ? ?
?
 
Substituting the 
?
Q expression from part (a) and rearranging give 
 T T
k
k
r r r
r r r
T T T T
2
0
2 1
2 1
1 2 1
2
1
2 2 2
0 ? ?
?
?
? ? ? ?
? ? ?
ave
( )
( )
( ) 
which is a quadratic equation in the unknown temperature T. Using the quadratic formula, the temperature 
distribution T(r) in the cylindrical shell is determined to be 
 T r
k
k
r r r
r r r
T T T T ( )
( )
( )
( ) ? ? ? ?
?
?
? ? ?
1 1 2 2
2
0
2 1
2 1
1 2 1
2
1
? ? ? ?
ave
 
Discussion The proper sign of the square root term (+ or -) is determined from the requirement that the 
temperature at any point within the medium must remain between T T
1 2
  and  .  
r 1 r 2 
T 1 k(T) 
r 
T 2 
Chapter 2 Heat Conduction Equation 
 2-61 
2-102   A plate with variable conductivity is subjected to specified temperatures on both sides. The rate of 
heat transfer through the plate is to be determined. 
Assumptions 1 Heat transfer is given to be steady and one-dimensional. 2 Thermal conductivity varies 
linearly.  3 There is no heat generation. 
Properties The thermal conductivity is given to be k T k T ( ) ( ) ? ?
0
1 ? . 
Analysis The average thermal conductivity of the medium in this case 
is simply the conductivity value at the average temperature since the 
thermal conductivity varies linearly with temperature, and is 
determined to be 
 
K W/m 24 . 34
2
K 350) + (500
) K 10 (8.7 + 1 K) W/m 25 (
2
1 ) (
1 - 4 -
1 2
0 ave ave
? ?
?
?
?
?
?
?
? ? ?
?
?
?
?
?
?
?
? ?
? ? ? ?
T T
k T k k
 
Then the rate of heat conduction through the plate becomes 
  W 30,820 ?
?
? ? ?
?
?
m 15 . 0
0)K 35 (500
m) 0.6  m K)(1.5 W/m 24 . 34 (
2 1
ave
L
T T
A k Q
?
 
Discussion We would obtain the same result if we substituted the given  k(T) relation into the second part of 
Eq, 2-76, and performed the indicated integration. 
 
 
T 2 
k(T) 
T 1 
 L 
Chapter 2 Heat Conduction Equation 
 2-62 
2-103  
"GIVEN" 
A=1.5*0.6 "[m^2]" 
L=0.15 "[m]" 
"T_1=500 [K], parameter to be varied" 
T_2=350 "[K]" 
k_0=25 "[W/m-K]" 
beta=8.7E-4 "[1/K]" 
 
"ANALYSIS" 
k=k_0*(1+beta*T) 
T=1/2*(T_1+T_2) 
Q_dot=k*A*(T_1-T_2)/L 
 
T1 [W] Q [W] 
400 9947 
425 15043 
450 20220 
475 25479 
500 30819 
525 36241 
550 41745 
575 47330 
600 52997 
625 58745 
650 64575 
675 70486 
700 76479 
 
400 450 500 550 600 650 700
0
10000
20000
30000
40000
50000
60000
70000
80000
T
1
  [K]
Q  [W]
 
 
 
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