Chapter 4 Transient Heat Conduction 4 Notes | EduRev

: Chapter 4 Transient Heat Conduction 4 Notes | EduRev

 Page 1


Chapter 4 Transient Heat Conduction 
 4-70 
4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are given to be  k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of 
thickness 2L = 5 cm.  
After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are 
     400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution can be written as 
 
? ?
? ?
C 323 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ? ?
?
?
) , 0 , 0 , 0 (
369 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 , 0 , 0 (
3
) 104 . 1 ( ) 5932 . 0 (
3
1
3
wall block
2
2
1
t T
e
t T
e A
T T
T t T
t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 ? ? C 445 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 115 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 208 . 2 ( ) 5932 . 0 (
2
t T e
t T
 
After 60 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 ? ? C 500 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 00109 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 624 . 6 ( ) 5932 . 0 (
2
t T e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Page 2


Chapter 4 Transient Heat Conduction 
 4-70 
4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are given to be  k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of 
thickness 2L = 5 cm.  
After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are 
     400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution can be written as 
 
? ?
? ?
C 323 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ? ?
?
?
) , 0 , 0 , 0 (
369 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 , 0 , 0 (
3
) 104 . 1 ( ) 5932 . 0 (
3
1
3
wall block
2
2
1
t T
e
t T
e A
T T
T t T
t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 ? ? C 445 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 115 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 208 . 2 ( ) 5932 . 0 (
2
t T e
t T
 
After 60 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 ? ? C 500 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 00109 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 624 . 6 ( ) 5932 . 0 (
2
t T e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Chapter 4 Transient Heat Conduction 
 4-71 
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r
o 
= D/2 = 2.5 cm and a plane wall of thickness 2L = 5 cm. 
After 10 minutes: The Biot number and the corresponding constants  for the long cylinder are 
  400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hr
Bi
o
? ? ? ? ? ?
1 1
08516 10931 . .   and   A 
To determine the center temperature, the product solution can be written as 
        
? ? ? ?
? ? ? ? C 331 ? ? ? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
) , 0 , 0 ( 352 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 (
) 104 . 1 ( ) 8516 . 0 ( ) 104 . 1 ( ) 5932 . 0 (
1 1
2 2
2
1
2
1
t T e e
t T
e A e A
T T
T t T
t t t
cyl wall
i
cyl wall block
? ? ? ?
? ? ?
 
After 20 minutes 
? ? ? ? C 449 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 107 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 208 . 2 ( ) 8516 . 0 ( ) 208 . 2 ( ) 5932 . 0 (
2 2
t T e e
t T
 
After 60 minutes 
      ? ? ? ? C 500 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 00092 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 624 . 6 ( ) 8516 . 0 ( ) 624 . 6 ( ) 5932 . 0 (
2 2
t T e e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Page 3


Chapter 4 Transient Heat Conduction 
 4-70 
4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are given to be  k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of 
thickness 2L = 5 cm.  
After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are 
     400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution can be written as 
 
? ?
? ?
C 323 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ? ?
?
?
) , 0 , 0 , 0 (
369 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 , 0 , 0 (
3
) 104 . 1 ( ) 5932 . 0 (
3
1
3
wall block
2
2
1
t T
e
t T
e A
T T
T t T
t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 ? ? C 445 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 115 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 208 . 2 ( ) 5932 . 0 (
2
t T e
t T
 
After 60 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 ? ? C 500 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 00109 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 624 . 6 ( ) 5932 . 0 (
2
t T e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Chapter 4 Transient Heat Conduction 
 4-71 
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r
o 
= D/2 = 2.5 cm and a plane wall of thickness 2L = 5 cm. 
After 10 minutes: The Biot number and the corresponding constants  for the long cylinder are 
  400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hr
Bi
o
? ? ? ? ? ?
1 1
08516 10931 . .   and   A 
To determine the center temperature, the product solution can be written as 
        
? ? ? ?
? ? ? ? C 331 ? ? ? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
) , 0 , 0 ( 352 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 (
) 104 . 1 ( ) 8516 . 0 ( ) 104 . 1 ( ) 5932 . 0 (
1 1
2 2
2
1
2
1
t T e e
t T
e A e A
T T
T t T
t t t
cyl wall
i
cyl wall block
? ? ? ?
? ? ?
 
After 20 minutes 
? ? ? ? C 449 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 107 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 208 . 2 ( ) 8516 . 0 ( ) 208 . 2 ( ) 5932 . 0 (
2 2
t T e e
t T
 
After 60 minutes 
      ? ? ? ? C 500 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 00092 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 624 . 6 ( ) 8516 . 0 ( ) 624 . 6 ( ) 5932 . 0 (
2 2
t T e e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Chapter 4 Transient Heat Conduction 
 4-72 
4-82 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane wall of 
thickness 2L = 5 cm. Two infinite plane walls are exposed to the hot gases with a heat transfer coefficient of 
h ? ? 40 W/ m . C
2
 and one with h ? ? 80 W/ m . C
2
.  
After 10 minutes: The Biot number and the corresponding constants for h ? ? 40 W/ m . C
2
 are  
400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
The Biot number and the corresponding constants for h ? ? 80 W/ m . C
2
 are 
 800 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 80 (
2
?
?
?
? ?
k
hL
Bi 
? ? ? ? ? ?
1 1
0 7910 11016 . .   and   A 
The Fourier number is 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution method can 
be written as 
        
? ? ? ?
? ? ? ?
C 364 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
? ?
? ? ? ? ? ?
?
?
) , 0 , 0 , 0 (
284 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 , 0 (
) 104 . 1 ( ) 7910 . 0 (
2
) 104 . 1 ( ) 5932 . 0 (
1
2
1
wall
2
wall block
2 2
2
1
2
1
t T
e e
t T
e A e A
T T
T t T
t t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 
? ? ? ?
C 469 ? ? ? ? ?
? ?
?
?
? ?
) , 0 , 0 , 0 (
0654 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) 208 . 2 ( ) 7910 . 0 (
2
) 208 . 2 ( ) 5932 . 0 (
2 2
t T
e e
t T
 
After 60 minutes 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Page 4


Chapter 4 Transient Heat Conduction 
 4-70 
4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are given to be  k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of 
thickness 2L = 5 cm.  
After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are 
     400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution can be written as 
 
? ?
? ?
C 323 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ? ?
?
?
) , 0 , 0 , 0 (
369 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 , 0 , 0 (
3
) 104 . 1 ( ) 5932 . 0 (
3
1
3
wall block
2
2
1
t T
e
t T
e A
T T
T t T
t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 ? ? C 445 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 115 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 208 . 2 ( ) 5932 . 0 (
2
t T e
t T
 
After 60 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 ? ? C 500 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 00109 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 624 . 6 ( ) 5932 . 0 (
2
t T e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Chapter 4 Transient Heat Conduction 
 4-71 
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r
o 
= D/2 = 2.5 cm and a plane wall of thickness 2L = 5 cm. 
After 10 minutes: The Biot number and the corresponding constants  for the long cylinder are 
  400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hr
Bi
o
? ? ? ? ? ?
1 1
08516 10931 . .   and   A 
To determine the center temperature, the product solution can be written as 
        
? ? ? ?
? ? ? ? C 331 ? ? ? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
) , 0 , 0 ( 352 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 (
) 104 . 1 ( ) 8516 . 0 ( ) 104 . 1 ( ) 5932 . 0 (
1 1
2 2
2
1
2
1
t T e e
t T
e A e A
T T
T t T
t t t
cyl wall
i
cyl wall block
? ? ? ?
? ? ?
 
After 20 minutes 
? ? ? ? C 449 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 107 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 208 . 2 ( ) 8516 . 0 ( ) 208 . 2 ( ) 5932 . 0 (
2 2
t T e e
t T
 
After 60 minutes 
      ? ? ? ? C 500 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 00092 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 624 . 6 ( ) 8516 . 0 ( ) 624 . 6 ( ) 5932 . 0 (
2 2
t T e e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Chapter 4 Transient Heat Conduction 
 4-72 
4-82 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane wall of 
thickness 2L = 5 cm. Two infinite plane walls are exposed to the hot gases with a heat transfer coefficient of 
h ? ? 40 W/ m . C
2
 and one with h ? ? 80 W/ m . C
2
.  
After 10 minutes: The Biot number and the corresponding constants for h ? ? 40 W/ m . C
2
 are  
400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
The Biot number and the corresponding constants for h ? ? 80 W/ m . C
2
 are 
 800 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 80 (
2
?
?
?
? ?
k
hL
Bi 
? ? ? ? ? ?
1 1
0 7910 11016 . .   and   A 
The Fourier number is 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution method can 
be written as 
        
? ? ? ?
? ? ? ?
C 364 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
? ?
? ? ? ? ? ?
?
?
) , 0 , 0 , 0 (
284 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 , 0 (
) 104 . 1 ( ) 7910 . 0 (
2
) 104 . 1 ( ) 5932 . 0 (
1
2
1
wall
2
wall block
2 2
2
1
2
1
t T
e e
t T
e A e A
T T
T t T
t t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 
? ? ? ?
C 469 ? ? ? ? ?
? ?
?
?
? ?
) , 0 , 0 , 0 (
0654 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) 208 . 2 ( ) 7910 . 0 (
2
) 208 . 2 ( ) 5932 . 0 (
2 2
t T
e e
t T
 
After 60 minutes 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Chapter 4 Transient Heat Conduction 
 4-73 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 
? ? ? ?
C 500 ? ? ? ? ?
? ?
?
?
? ?
) , 0 , 0 , 0 (
000186 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) 624 . 6 ( ) 7910 . 0 (
2
) 624 . 6 ( ) 5932 . 0 (
2 2
t T
e e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r
o 
= D/2 = 2.5 cm exposed to the hot gases with a heat transfer coefficient of h ? ? 40 W/ m . C
2
 and a plane 
wall of thickness 2L = 5 cm exposed to the hot gases with h ? ? 80 W/ m . C
2
. 
After 10 minutes: The Biot number and the corresponding constants  for the long cylinder are 
  400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hr
Bi
o
? ? ? ? ? ?
1 1
08516 10931 . .   and   A 
To determine the center temperature, the product solution method can be written as 
 
? ? ? ?
? ? ? ?
C 370 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
) , 0 , 0 (
271 . 0 ) 0931 . 1 ( ) 1016 . 1 (
500 20
500 ) , 0 , 0 (
) , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 (
) 104 . 1 ( ) 8516 . 0 ( ) 104 . 1 ( ) 7910 . 0 (
1
wall
1
wall block
2 2
2
1
2
1
t T
e e
t T
e A e A
T T
T t T
t t t
cyl
i
cyl
? ? ? ?
? ? ?
 
After 20 minutes 
? ? ? ? C 471 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 06094 . 0 ) 0931 . 1 ( ) 1016 . 1 (
500 20
500 ) , 0 , 0 (
) 208 . 2 ( ) 8516 . 0 ( ) 208 . 2 ( ) 7910 . 0 (
2 2
t T e e
t T
After 60 minutes 
? ? ? ? C 500 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 0001568 . 0 ) 0931 . 1 ( ) 1016 . 1 (
500 20
500 ) , 0 , 0 (
) 624 . 6 ( ) 8516 . 0 ( ) 624 . 6 ( ) 7910 . 0 (
2 2
t T e e
t T
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Page 5


Chapter 4 Transient Heat Conduction 
 4-70 
4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are given to be  k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane walls of 
thickness 2L = 5 cm.  
After 10 minutes: The Biot number, the corresponding constants, and the Fourier number are 
     400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution can be written as 
 
? ?
? ?
C 323 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ? ?
?
?
) , 0 , 0 , 0 (
369 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 , 0 , 0 (
3
) 104 . 1 ( ) 5932 . 0 (
3
1
3
wall block
2
2
1
t T
e
t T
e A
T T
T t T
t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 ? ? C 445 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 115 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 208 . 2 ( ) 5932 . 0 (
2
t T e
t T
 
After 60 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 ? ? C 500 ? ? ? ? ? ? ?
?
?
?
) , 0 , 0 , 0 ( 00109 . 0 ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
3
) 624 . 6 ( ) 5932 . 0 (
2
t T e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Chapter 4 Transient Heat Conduction 
 4-71 
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r
o 
= D/2 = 2.5 cm and a plane wall of thickness 2L = 5 cm. 
After 10 minutes: The Biot number and the corresponding constants  for the long cylinder are 
  400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hr
Bi
o
? ? ? ? ? ?
1 1
08516 10931 . .   and   A 
To determine the center temperature, the product solution can be written as 
        
? ? ? ?
? ? ? ? C 331 ? ? ? ? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
) , 0 , 0 ( 352 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 (
) 104 . 1 ( ) 8516 . 0 ( ) 104 . 1 ( ) 5932 . 0 (
1 1
2 2
2
1
2
1
t T e e
t T
e A e A
T T
T t T
t t t
cyl wall
i
cyl wall block
? ? ? ?
? ? ?
 
After 20 minutes 
? ? ? ? C 449 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 107 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 208 . 2 ( ) 8516 . 0 ( ) 208 . 2 ( ) 5932 . 0 (
2 2
t T e e
t T
 
After 60 minutes 
      ? ? ? ? C 500 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 00092 . 0 ) 0931 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 (
) 624 . 6 ( ) 8516 . 0 ( ) 624 . 6 ( ) 5932 . 0 (
2 2
t T e e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Chapter 4 Transient Heat Conduction 
 4-72 
4-82 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center 
temperatures of each geometry in 10, 20, and 60 min are to be determined. 
Assumptions 1 Heat conduction in the cubic block is three-dimensional, and thus the temperature varies in 
all x-, y, and z- directions. 2 Heat conduction in the cylindrical block is two-dimensional, and thus the 
temperature varies in both axial  x- and radial r- directions. 3 The thermal properties of the granite are 
constant. 4 The heat transfer coefficient is constant and uniform over the entire surface. 5 The Fourier 
number is ? > 0.2 so that the one-term approximate solutions (or the transient temperature charts) are 
applicable (this assumption will be verified). 
Properties The thermal properties of the granite are k = 2.5 W/m. ?C and ? = 1.15 ?10
-6
 m
2
/s. 
Analysis:  
Cubic block: This cubic block can physically be formed by the intersection of three infinite plane wall of 
thickness 2L = 5 cm. Two infinite plane walls are exposed to the hot gases with a heat transfer coefficient of 
h ? ? 40 W/ m . C
2
 and one with h ? ? 80 W/ m . C
2
.  
After 10 minutes: The Biot number and the corresponding constants for h ? ? 40 W/ m . C
2
 are  
400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hL
Bi ? ? ? ? ? ?
1 1
05932 10580 . .   and   A 
The Biot number and the corresponding constants for h ? ? 80 W/ m . C
2
 are 
 800 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 80 (
2
?
?
?
? ?
k
hL
Bi 
? ? ? ? ? ?
1 1
0 7910 11016 . .   and   A 
The Fourier number is 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
1104 0 2
( .
( .
. .
 m / s)(10 min 60 s / min)
 m)
2
2
 
To determine the center temperature, the product solution method can 
be written as 
        
? ? ? ?
? ? ? ?
C 364 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ?
? ?
? ? ? ? ? ?
?
?
) , 0 , 0 , 0 (
284 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) , 0 , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 , 0 (
) 104 . 1 ( ) 7910 . 0 (
2
) 104 . 1 ( ) 5932 . 0 (
1
2
1
wall
2
wall block
2 2
2
1
2
1
t T
e e
t T
e A e A
T T
T t T
t t t
i
 
After 20 minutes 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
2 208 0 2
( .
( .
. .
 m / s)(20 min 60 s / min)
 m)
2
2
 
 
? ? ? ?
C 469 ? ? ? ? ?
? ?
?
?
? ?
) , 0 , 0 , 0 (
0654 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) 208 . 2 ( ) 7910 . 0 (
2
) 208 . 2 ( ) 5932 . 0 (
2 2
t T
e e
t T
 
After 60 minutes 
Ti = 20 ?C 
 
Hot  
gases 
500 ?C 
 
5 cm ? 5 cm ? 5 cm 
Ti = 20 ?C 
 
5 cm ? 5 cm 
Chapter 4 Transient Heat Conduction 
 4-73 
 ?
?
? ?
? ?
? ?
?
t
L
2
6
115 10
0 025
6 624 0 2
( .
( .
. .
 m / s)(60 min 60 s / min)
 m)
2
2
 
 
? ? ? ?
C 500 ? ? ? ? ?
? ?
?
?
? ?
) , 0 , 0 , 0 (
000186 . 0 ) 1016 . 1 ( ) 0580 . 1 (
500 20
500 ) , 0 , 0 , 0 (
) 624 . 6 ( ) 7910 . 0 (
2
) 624 . 6 ( ) 5932 . 0 (
2 2
t T
e e
t T
 
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Cylinder: This cylindrical block can physically be formed by the intersection of a long cylinder of radius r
o 
= D/2 = 2.5 cm exposed to the hot gases with a heat transfer coefficient of h ? ? 40 W/ m . C
2
 and a plane 
wall of thickness 2L = 5 cm exposed to the hot gases with h ? ? 80 W/ m . C
2
. 
After 10 minutes: The Biot number and the corresponding constants  for the long cylinder are 
  400 . 0
) C W/m. 5 . 2 (
) m 025 . 0 )( C . W/m 40 (
2
?
?
?
? ?
k
hr
Bi
o
? ? ? ? ? ?
1 1
08516 10931 . .   and   A 
To determine the center temperature, the product solution method can be written as 
 
? ? ? ?
? ? ? ?
C 370 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
? ?
?
?
) , 0 , 0 (
271 . 0 ) 0931 . 1 ( ) 1016 . 1 (
500 20
500 ) , 0 , 0 (
) , 0 , 0 (
) , 0 ( ) , 0 ( ) , 0 , 0 (
) 104 . 1 ( ) 8516 . 0 ( ) 104 . 1 ( ) 7910 . 0 (
1
wall
1
wall block
2 2
2
1
2
1
t T
e e
t T
e A e A
T T
T t T
t t t
cyl
i
cyl
? ? ? ?
? ? ?
 
After 20 minutes 
? ? ? ? C 471 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 06094 . 0 ) 0931 . 1 ( ) 1016 . 1 (
500 20
500 ) , 0 , 0 (
) 208 . 2 ( ) 8516 . 0 ( ) 208 . 2 ( ) 7910 . 0 (
2 2
t T e e
t T
After 60 minutes 
? ? ? ? C 500 ? ? ? ? ? ? ?
?
?
? ?
) , 0 , 0 ( 0001568 . 0 ) 0931 . 1 ( ) 1016 . 1 (
500 20
500 ) , 0 , 0 (
) 624 . 6 ( ) 8516 . 0 ( ) 624 . 6 ( ) 7910 . 0 (
2 2
t T e e
t T
Note that ? > 0.2 in all dimensions and thus the one-term approximate solution for transient heat conduction 
is applicable. 
Chapter 4 Transient Heat Conduction 
 4-74 
4-83 A cylindrical aluminum block is heated in a furnace. The length of time the block should be kept in the 
furnace and the amount of heat transfer to the block are to be determined. 
Assumptions 1 Heat conduction in the cylindrical block is two-dimensional, and thus the temperature varies 
in both axial  x- and radial r- directions. 2 The thermal properties of the aluminum are constant. 3 The heat 
transfer coefficient is constant and uniform over the entire surface. 4 The Fourier number is ? > 0.2 so that 
the one-term approximate solutions (or the transient temperature charts) are applicable (it will be verified). 
Properties The thermal properties of the aluminum block are given to be k = 236 W/m. ?C, ? = 2702 kg/m
3
, 
C p = 0.896 kJ/kg. ?C, and ? = 9.75 ?10
-5
 m
2
/s. 
Analysis This cylindrical aluminum block can physically be formed by the intersection of an infinite plane 
wall of thickness 2L = 20 cm, and a long cylinder of radius r
o 
= D/2 = 7.5 cm. The Biot numbers and the 
corresponding constants are first determined to be 
0339 . 0
) C W/m. 236 (
) m 1 . 0 )( C . W/m 80 (
2
?
?
?
? ?
k
hL
Bi       0056 . 1 and      1811 . 0
1 1
? ? ? ? ? A ? 
0254 . 0
C W/m. 236
) m 075 . 0 )( C . W/m 80 (
2
0
?
?
?
? ?
k
hr
Bi 0063 . 1 and      2217 . 0
1 1
? ? ? ? ? A ? 
Noting that ? ? ? t L /
2
 and assuming ? > 0.2 in all dimensions and thus the one-term approximate solution 
for transient heat conduction is applicable, the product solution for this problem can be written as 
7627 . 0
) 075 . 0 (
) 10 75 . 9 (
) 2217 . 0 ( exp ) 0063 . 1 (
) 1 . 0 (
) 10 75 . 9 (
) 1811 . 0 ( exp ) 0056 . 1 (
1200 20
1200 300
) , 0 ( ) , 0 ( ) , 0 , 0 (
2
5
2
2
5
2
cy l
1
wall
1 cy l wall block
2
1
2
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ? ? ? ?
? ?
? ? ? ? ? ?
t t
e A e A t t t
Solving for the time t gives t = 241 s = 4.0 min.  We note that  
 2 . 0 35 . 2
m) 1 . 0 (
s) /s)(241 m 10 75 . 9 (
2
2 5
2
? ?
?
? ?
?
L
t ?
? 
and thus the assumption of  ? > 0.2 for the applicability of the one-term approximate solution is verified.  
 The maximum amount of heat transfer is 
   
? ?
kJ 100 , 10 C ) 1200 20 )( C kJ/kg. 896 . 0 )( kg 550 . 9 ( ) (
kg 550 . 9 m) 2 . 0 ( m) 075 . 0 ( ) kg/m 2702 (
max
2. 3 2
0
? ? ? ? ? ? ?
? ? ? ?
?
T T mC Q
L r V m
i p
? ?? ?
 
Then we determine the dimensionless heat transfer ratios for both geometries as 
2415 . 0
1811 . 0
) 1811 . 0 sin(
) 7627 . 0 ( 1
) sin(
1
1
1
,
max
? ? ?
?
?
? ? ?
?
?
?
?
?
?
?
?
wall o
wall
Q
Q
 
2425 . 0
2217 . 0
1101 . 0
) 7627 . 0 ( 2 1
) (
2 1
1
1 1
,
max
? ? ?
?
?
? ? ?
?
?
?
?
?
?
?
? J
Q
Q
cyl o
cyl
 
The heat transfer ratio for the short cylinder is 
4254 . 0 ) 2415 . 0 1 )( 2425 . 0 ( 2415 . 0 1
max max max max
? ? ? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
wall
plane
cylinder
long
wall
plane
cylinder
short Q
Q
Q
Q
Q
Q
Q
Q
 
Then the total heat transfer from the short cylinder as it is cooled from 300 ?C at the center to 20 ?C becomes 
 kJ 4297 ? ? ? kJ) 100 , 10 )( 4254 . 0 ( 4236 . 0
max
Q Q 
which is identical to the heat transfer to the cylinder as the cylinder at 20 ?C is heated to 300 ?C at the center. 
L 
z 
Cylinder 
Ti = 20 ?C 
r 0 
Furnace 
T ? = 1200 ?C 
 
L 
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