Chapter 4 Transient Heat Conduction 5 Notes | EduRev

: Chapter 4 Transient Heat Conduction 5 Notes | EduRev

 Page 1


Chapter 4 Transient Heat Conduction 
 4-84 
 
Review Problems 
 
4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. 
Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot 
air should be blown is to be determined. 
Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates 
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 steel plates are given to be  k = 43 W/m. ?C and  ? = 1.17 ?10
-5
 m
2
/s 
Analysis The characteristic length of the plates and the Biot number are 
 
1 . 0 019 . 0
) C W/m. 43 (
) m 02 . 0 )( C . W/m 40 (
m 02 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
      
min 8.0 s 482 ? ? ? ? ? ?
? ?
?
? ? ? ?
?
?
?
? ?
?
? ? ?
? ?
?
?
t e e
T T
T t T
L C
h
V C
hA
b
bt
i
c p p
s
)t s 000544 . 0 (
1 -
3 6
2
-1
50 15
50 0 ) (
s 000544 . 0
m) (0.02 ) C . J/m 10 675 . 3 (
C . W/m 40
? ?
 
where C . J/m 10 675 . 3
/s m 10 17 . 1
C W/m. 43
3 6
2 5
? ? ?
?
?
?
?
? ?
?
k
C
p
  
Alternative solution: This problem can also be solved using  the transient chart Fig. 4-13a, 
 2 . 0 15
769 . 0
50 15
50 0
6 . 52
019 . 0
1 1
2
? ? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
? ?
?
?
o
i
o r
t
T T
T T
Bi ?
? 
Then, 
 s 513 ?
?
?
?
?
?
?
/s) m 10 17 . 1 (
m) 02 . 0 )( 15 (
2 5
2 2
o
r
t 
The difference is due to the reading error of the chart.  
Steel plates 
T i = -15 ?C  
Hot gases 
T ? = 50 ?C 
 
Page 2


Chapter 4 Transient Heat Conduction 
 4-84 
 
Review Problems 
 
4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. 
Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot 
air should be blown is to be determined. 
Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates 
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 steel plates are given to be  k = 43 W/m. ?C and  ? = 1.17 ?10
-5
 m
2
/s 
Analysis The characteristic length of the plates and the Biot number are 
 
1 . 0 019 . 0
) C W/m. 43 (
) m 02 . 0 )( C . W/m 40 (
m 02 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
      
min 8.0 s 482 ? ? ? ? ? ?
? ?
?
? ? ? ?
?
?
?
? ?
?
? ? ?
? ?
?
?
t e e
T T
T t T
L C
h
V C
hA
b
bt
i
c p p
s
)t s 000544 . 0 (
1 -
3 6
2
-1
50 15
50 0 ) (
s 000544 . 0
m) (0.02 ) C . J/m 10 675 . 3 (
C . W/m 40
? ?
 
where C . J/m 10 675 . 3
/s m 10 17 . 1
C W/m. 43
3 6
2 5
? ? ?
?
?
?
?
? ?
?
k
C
p
  
Alternative solution: This problem can also be solved using  the transient chart Fig. 4-13a, 
 2 . 0 15
769 . 0
50 15
50 0
6 . 52
019 . 0
1 1
2
? ? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
? ?
?
?
o
i
o r
t
T T
T T
Bi ?
? 
Then, 
 s 513 ?
?
?
?
?
?
?
/s) m 10 17 . 1 (
m) 02 . 0 )( 15 (
2 5
2 2
o
r
t 
The difference is due to the reading error of the chart.  
Steel plates 
T i = -15 ?C  
Hot gases 
T ? = 50 ?C 
 
Chapter 4 Transient Heat Conduction 
 4-85 
4-106 A curing kiln is heated by injecting steam into it and raising its inner surface temperature to a 
specified value. It is to be determined whether the temperature at the outer surfaces of the kiln changes 
during the curing period. 
Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and 
the convection heat transfer coefficient inside is very large. Therefore, the wall can be considered to be a 
semi-infinite medium with a specified surface temperature of 45 ?C. 2 The thermal properties of the concrete 
wall are constant. 
Properties The thermal properties of the concrete wall are given to be k = 0.9 W/m. ?C and ? = 0.23 ?10
-5
 
m
2
/s. 
Analysis We determine the temperature at a depth of x = 0.3 m in 3 h using the analytical solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
       
C 9.1 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
) , (
1782 . 0 ) 952 . 0 (
) s/h 3600 h 3 )( /s m 10 23 . 0 ( 2
m 3 . 0
5 42
2 ) , (
2 5
t x T
erfc
erfc
t x T
 
which is greater than the initial temperature of 2 ?C. Therefore, heat will propagate through the 0.3 m thick 
wall in 3 h, and thus it may be desirable to insulate the outer surface of the wall to save energy. 
 
4-107 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a 
particular location is to be determined. 
Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and 
thus the soil can be considered to be a semi-infinite medium with a specified surface temperature of          -
10 ?C. 2 The thermal properties of the soil are constant. 
Properties The thermal properties of the soil are given 
to be k = 0.7 W/m. ?C and ? = 1.4 ?10
-5
 m
2
/s. 
Analysis The depth at which the temperature drops to 
0 ?C in 75 days is determined using the analytical 
solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
m 7.05 ? ? ? ?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
?
x
x
erfc
                  
) s/h 3600 h/day 24 day 75 )( /s m 10 4 . 1 ( 2
15 10
15 0
2 5 
Therefore, the pipes must be buried at a depth of at least 7.05 m. 
2 ?C 42 ?C 
30 cm 
Kiln wall 
x 0 
Soil 
T i = 15 ?C 
Water pipe 
 T s =-10 ?C 
x 
Page 3


Chapter 4 Transient Heat Conduction 
 4-84 
 
Review Problems 
 
4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. 
Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot 
air should be blown is to be determined. 
Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates 
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 steel plates are given to be  k = 43 W/m. ?C and  ? = 1.17 ?10
-5
 m
2
/s 
Analysis The characteristic length of the plates and the Biot number are 
 
1 . 0 019 . 0
) C W/m. 43 (
) m 02 . 0 )( C . W/m 40 (
m 02 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
      
min 8.0 s 482 ? ? ? ? ? ?
? ?
?
? ? ? ?
?
?
?
? ?
?
? ? ?
? ?
?
?
t e e
T T
T t T
L C
h
V C
hA
b
bt
i
c p p
s
)t s 000544 . 0 (
1 -
3 6
2
-1
50 15
50 0 ) (
s 000544 . 0
m) (0.02 ) C . J/m 10 675 . 3 (
C . W/m 40
? ?
 
where C . J/m 10 675 . 3
/s m 10 17 . 1
C W/m. 43
3 6
2 5
? ? ?
?
?
?
?
? ?
?
k
C
p
  
Alternative solution: This problem can also be solved using  the transient chart Fig. 4-13a, 
 2 . 0 15
769 . 0
50 15
50 0
6 . 52
019 . 0
1 1
2
? ? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
? ?
?
?
o
i
o r
t
T T
T T
Bi ?
? 
Then, 
 s 513 ?
?
?
?
?
?
?
/s) m 10 17 . 1 (
m) 02 . 0 )( 15 (
2 5
2 2
o
r
t 
The difference is due to the reading error of the chart.  
Steel plates 
T i = -15 ?C  
Hot gases 
T ? = 50 ?C 
 
Chapter 4 Transient Heat Conduction 
 4-85 
4-106 A curing kiln is heated by injecting steam into it and raising its inner surface temperature to a 
specified value. It is to be determined whether the temperature at the outer surfaces of the kiln changes 
during the curing period. 
Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and 
the convection heat transfer coefficient inside is very large. Therefore, the wall can be considered to be a 
semi-infinite medium with a specified surface temperature of 45 ?C. 2 The thermal properties of the concrete 
wall are constant. 
Properties The thermal properties of the concrete wall are given to be k = 0.9 W/m. ?C and ? = 0.23 ?10
-5
 
m
2
/s. 
Analysis We determine the temperature at a depth of x = 0.3 m in 3 h using the analytical solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
       
C 9.1 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
) , (
1782 . 0 ) 952 . 0 (
) s/h 3600 h 3 )( /s m 10 23 . 0 ( 2
m 3 . 0
5 42
2 ) , (
2 5
t x T
erfc
erfc
t x T
 
which is greater than the initial temperature of 2 ?C. Therefore, heat will propagate through the 0.3 m thick 
wall in 3 h, and thus it may be desirable to insulate the outer surface of the wall to save energy. 
 
4-107 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a 
particular location is to be determined. 
Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and 
thus the soil can be considered to be a semi-infinite medium with a specified surface temperature of          -
10 ?C. 2 The thermal properties of the soil are constant. 
Properties The thermal properties of the soil are given 
to be k = 0.7 W/m. ?C and ? = 1.4 ?10
-5
 m
2
/s. 
Analysis The depth at which the temperature drops to 
0 ?C in 75 days is determined using the analytical 
solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
m 7.05 ? ? ? ?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
?
x
x
erfc
                  
) s/h 3600 h/day 24 day 75 )( /s m 10 4 . 1 ( 2
15 10
15 0
2 5 
Therefore, the pipes must be buried at a depth of at least 7.05 m. 
2 ?C 42 ?C 
30 cm 
Kiln wall 
x 0 
Soil 
T i = 15 ?C 
Water pipe 
 T s =-10 ?C 
x 
Chapter 4 Transient Heat Conduction 
 4-86 
4-108 A hot dog is to be cooked by dropping it into boiling water. The time of cooking is to be determined. 
Assumptions 1 Heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both 
the axial x- and the radial r- directions. 2 The thermal properties of the hot dog 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 hot dog are given to be k = 0.76 W/m. ?C, ? = 980 kg/m
3
, C p = 3.9 
kJ/kg. ?C, and ? = 2 ?10
-7
 m
2
/s. 
Analysis This hot dog can physically be formed by the intersection of an infinite plane wall of  thickness 2L 
= 12 cm, and a long cylinder of radius r
o 
= D/2 = 1 cm. The Biot numbers and corresponding constants are 
first determined to be 
37 . 47
) C W/m. 76 . 0 (
) m 06 . 0 )( C . W/m 600 (
2
?
?
?
? ?
k
hL
Bi 2726 . 1   and   5381 . 1
1 1
? ? ? ? ? A ? 
895 . 7
) C W/m. 76 . 0 (
) m 01 . 0 )( C . W/m 600 (
2
?
?
?
? ?
k
hr
Bi
o
5515 . 1   and   1251 . 2
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 
 
2105 . 0
) 01 . 0 (
) 10 2 (
) 1251 . 2 ( exp ) 5515 . 1 (                  
) 06 . 0 (
) 10 2 (
) 5381 . 1 ( exp ) 2726 . 1 (
100 5
100 80
) , 0 ( ) , 0 ( ) , 0 , 0 (
2
7
2
2
7
2
1 1
2
1
2
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
? ?
t
t
e A e A t t t
cyl wall block
? ? ? ?
? ? ?
 
which gives 
 min 4.1 s 244 = ? t 
Therefore, it will take about 4.1 min for the hot dog to cook. Note that  
 2 . 0 49 . 0
m) 01 . 0 (
s) /s)(244 m 10 2 (
2
2 7
2
? ?
?
? ?
?
o
cyl
r
t ?
? 
and thus the assumption ? > 0.2 for the applicability of the one-term approximate solution is verified. 
Discussion This problem could also be solved by treating the hot dog as an infinite cylinder since heat 
transfer through the end surfaces will have little effect on the mid section temperature because of the large 
distance. 
Water 
100 ?C 
 
2 cm   Hot dog   Ti = 5 ?C 
Page 4


Chapter 4 Transient Heat Conduction 
 4-84 
 
Review Problems 
 
4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. 
Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot 
air should be blown is to be determined. 
Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates 
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 steel plates are given to be  k = 43 W/m. ?C and  ? = 1.17 ?10
-5
 m
2
/s 
Analysis The characteristic length of the plates and the Biot number are 
 
1 . 0 019 . 0
) C W/m. 43 (
) m 02 . 0 )( C . W/m 40 (
m 02 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
      
min 8.0 s 482 ? ? ? ? ? ?
? ?
?
? ? ? ?
?
?
?
? ?
?
? ? ?
? ?
?
?
t e e
T T
T t T
L C
h
V C
hA
b
bt
i
c p p
s
)t s 000544 . 0 (
1 -
3 6
2
-1
50 15
50 0 ) (
s 000544 . 0
m) (0.02 ) C . J/m 10 675 . 3 (
C . W/m 40
? ?
 
where C . J/m 10 675 . 3
/s m 10 17 . 1
C W/m. 43
3 6
2 5
? ? ?
?
?
?
?
? ?
?
k
C
p
  
Alternative solution: This problem can also be solved using  the transient chart Fig. 4-13a, 
 2 . 0 15
769 . 0
50 15
50 0
6 . 52
019 . 0
1 1
2
? ? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
? ?
?
?
o
i
o r
t
T T
T T
Bi ?
? 
Then, 
 s 513 ?
?
?
?
?
?
?
/s) m 10 17 . 1 (
m) 02 . 0 )( 15 (
2 5
2 2
o
r
t 
The difference is due to the reading error of the chart.  
Steel plates 
T i = -15 ?C  
Hot gases 
T ? = 50 ?C 
 
Chapter 4 Transient Heat Conduction 
 4-85 
4-106 A curing kiln is heated by injecting steam into it and raising its inner surface temperature to a 
specified value. It is to be determined whether the temperature at the outer surfaces of the kiln changes 
during the curing period. 
Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and 
the convection heat transfer coefficient inside is very large. Therefore, the wall can be considered to be a 
semi-infinite medium with a specified surface temperature of 45 ?C. 2 The thermal properties of the concrete 
wall are constant. 
Properties The thermal properties of the concrete wall are given to be k = 0.9 W/m. ?C and ? = 0.23 ?10
-5
 
m
2
/s. 
Analysis We determine the temperature at a depth of x = 0.3 m in 3 h using the analytical solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
       
C 9.1 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
) , (
1782 . 0 ) 952 . 0 (
) s/h 3600 h 3 )( /s m 10 23 . 0 ( 2
m 3 . 0
5 42
2 ) , (
2 5
t x T
erfc
erfc
t x T
 
which is greater than the initial temperature of 2 ?C. Therefore, heat will propagate through the 0.3 m thick 
wall in 3 h, and thus it may be desirable to insulate the outer surface of the wall to save energy. 
 
4-107 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a 
particular location is to be determined. 
Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and 
thus the soil can be considered to be a semi-infinite medium with a specified surface temperature of          -
10 ?C. 2 The thermal properties of the soil are constant. 
Properties The thermal properties of the soil are given 
to be k = 0.7 W/m. ?C and ? = 1.4 ?10
-5
 m
2
/s. 
Analysis The depth at which the temperature drops to 
0 ?C in 75 days is determined using the analytical 
solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
m 7.05 ? ? ? ?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
?
x
x
erfc
                  
) s/h 3600 h/day 24 day 75 )( /s m 10 4 . 1 ( 2
15 10
15 0
2 5 
Therefore, the pipes must be buried at a depth of at least 7.05 m. 
2 ?C 42 ?C 
30 cm 
Kiln wall 
x 0 
Soil 
T i = 15 ?C 
Water pipe 
 T s =-10 ?C 
x 
Chapter 4 Transient Heat Conduction 
 4-86 
4-108 A hot dog is to be cooked by dropping it into boiling water. The time of cooking is to be determined. 
Assumptions 1 Heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both 
the axial x- and the radial r- directions. 2 The thermal properties of the hot dog 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 hot dog are given to be k = 0.76 W/m. ?C, ? = 980 kg/m
3
, C p = 3.9 
kJ/kg. ?C, and ? = 2 ?10
-7
 m
2
/s. 
Analysis This hot dog can physically be formed by the intersection of an infinite plane wall of  thickness 2L 
= 12 cm, and a long cylinder of radius r
o 
= D/2 = 1 cm. The Biot numbers and corresponding constants are 
first determined to be 
37 . 47
) C W/m. 76 . 0 (
) m 06 . 0 )( C . W/m 600 (
2
?
?
?
? ?
k
hL
Bi 2726 . 1   and   5381 . 1
1 1
? ? ? ? ? A ? 
895 . 7
) C W/m. 76 . 0 (
) m 01 . 0 )( C . W/m 600 (
2
?
?
?
? ?
k
hr
Bi
o
5515 . 1   and   1251 . 2
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 
 
2105 . 0
) 01 . 0 (
) 10 2 (
) 1251 . 2 ( exp ) 5515 . 1 (                  
) 06 . 0 (
) 10 2 (
) 5381 . 1 ( exp ) 2726 . 1 (
100 5
100 80
) , 0 ( ) , 0 ( ) , 0 , 0 (
2
7
2
2
7
2
1 1
2
1
2
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
? ?
t
t
e A e A t t t
cyl wall block
? ? ? ?
? ? ?
 
which gives 
 min 4.1 s 244 = ? t 
Therefore, it will take about 4.1 min for the hot dog to cook. Note that  
 2 . 0 49 . 0
m) 01 . 0 (
s) /s)(244 m 10 2 (
2
2 7
2
? ?
?
? ?
?
o
cyl
r
t ?
? 
and thus the assumption ? > 0.2 for the applicability of the one-term approximate solution is verified. 
Discussion This problem could also be solved by treating the hot dog as an infinite cylinder since heat 
transfer through the end surfaces will have little effect on the mid section temperature because of the large 
distance. 
Water 
100 ?C 
 
2 cm   Hot dog   Ti = 5 ?C 
Chapter 4 Transient Heat Conduction 
 4-87 
4-109 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. 
The temperature of the sheet metal after quenching and the rate at which heat needs to be removed from the 
oil in order to keep its temperature constant are to be determined.  
Assumptions 1 The thermal properties of the balls are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface.  3 The Biot number is Bi < 0.1 so that the lumped system analysis is 
applicable (this assumption will be checked). 
Properties The properties of the steel plate are given to be  k = 60.5 W/m. ?C, ? = 7854 kg/m
3
, and C p = 434 
J/kg. ?C (Table A-3). 
Analysis The characteristic length of the steel 
plate and the Biot number are 
         
1 . 0 036 . 0
C W/m. 5 . 60
) m 0025 . 0 )( C . W/m 860 (
m 0025 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
s 30 min 5 . 0
m/min 10
m 5
velocity
length
time
s 10092 . 0
m) C)(0.0025 J/kg. 434 )( kg/m (7854
C . W/m 860
1 -
3
2
? ? ? ?
?
?
?
? ? ?
c p p
s
L C
h
V C
hA
b
? ?
 
Then the temperature of the sheet metal when it leaves the oil bath is determined to be 
 C 82.53 ? ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
) (
45 820
45 ) ( ) (
s)  30 )( s 10092 . 0 (
-1
t T e
t T
e
T T
T t T
bt
i
 
The mass flow rate of the sheet metal through the oil bath is 
 kg/min 4 . 785 m/min) 10 ( m) 005 . 0 ( m) 2 )( kg/m 7854 (
3
? ? ? ? ? ? wtV V m
?
? 
Then the rate of heat transfer from the sheet metal to the oil bath and thus the rate at which heat needs to be 
removed from the oil in order to keep its temperature constant at 45 ?C becomes 
 kW 213.2 = J/min  10 279 . 1 C ) 45 53 . 82 )( C J/kg. 434 )( kg/min 4 . 785 ( ] ) ( [
7
? ? ? ? ? ? ? ?
?
T t T C m Q
p
?
?
 
Steel plate 
10 m/min 
 Oil bath 
45 ?C 
Page 5


Chapter 4 Transient Heat Conduction 
 4-84 
 
Review Problems 
 
4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. 
Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length of time the hot 
air should be blown is to be determined. 
Assumptions 1 Heat conduction in the plates is one-dimensional since the plate is large relative to its 
thickness and there is thermal symmetry about the center plane. 3 The thermal properties of the steel plates 
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 steel plates are given to be  k = 43 W/m. ?C and  ? = 1.17 ?10
-5
 m
2
/s 
Analysis The characteristic length of the plates and the Biot number are 
 
1 . 0 019 . 0
) C W/m. 43 (
) m 02 . 0 )( C . W/m 40 (
m 02 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
      
min 8.0 s 482 ? ? ? ? ? ?
? ?
?
? ? ? ?
?
?
?
? ?
?
? ? ?
? ?
?
?
t e e
T T
T t T
L C
h
V C
hA
b
bt
i
c p p
s
)t s 000544 . 0 (
1 -
3 6
2
-1
50 15
50 0 ) (
s 000544 . 0
m) (0.02 ) C . J/m 10 675 . 3 (
C . W/m 40
? ?
 
where C . J/m 10 675 . 3
/s m 10 17 . 1
C W/m. 43
3 6
2 5
? ? ?
?
?
?
?
? ?
?
k
C
p
  
Alternative solution: This problem can also be solved using  the transient chart Fig. 4-13a, 
 2 . 0 15
769 . 0
50 15
50 0
6 . 52
019 . 0
1 1
2
? ? ?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
? ?
?
?
o
i
o r
t
T T
T T
Bi ?
? 
Then, 
 s 513 ?
?
?
?
?
?
?
/s) m 10 17 . 1 (
m) 02 . 0 )( 15 (
2 5
2 2
o
r
t 
The difference is due to the reading error of the chart.  
Steel plates 
T i = -15 ?C  
Hot gases 
T ? = 50 ?C 
 
Chapter 4 Transient Heat Conduction 
 4-85 
4-106 A curing kiln is heated by injecting steam into it and raising its inner surface temperature to a 
specified value. It is to be determined whether the temperature at the outer surfaces of the kiln changes 
during the curing period. 
Assumptions 1 The temperature in the wall is affected by the thermal conditions at inner surfaces only and 
the convection heat transfer coefficient inside is very large. Therefore, the wall can be considered to be a 
semi-infinite medium with a specified surface temperature of 45 ?C. 2 The thermal properties of the concrete 
wall are constant. 
Properties The thermal properties of the concrete wall are given to be k = 0.9 W/m. ?C and ? = 0.23 ?10
-5
 
m
2
/s. 
Analysis We determine the temperature at a depth of x = 0.3 m in 3 h using the analytical solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
       
C 9.1 ? ?
? ?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
) , (
1782 . 0 ) 952 . 0 (
) s/h 3600 h 3 )( /s m 10 23 . 0 ( 2
m 3 . 0
5 42
2 ) , (
2 5
t x T
erfc
erfc
t x T
 
which is greater than the initial temperature of 2 ?C. Therefore, heat will propagate through the 0.3 m thick 
wall in 3 h, and thus it may be desirable to insulate the outer surface of the wall to save energy. 
 
4-107 The water pipes are buried in the ground to prevent freezing. The minimum burial depth at a 
particular location is to be determined. 
Assumptions 1 The temperature in the soil is affected by the thermal conditions at one surface only, and 
thus the soil can be considered to be a semi-infinite medium with a specified surface temperature of          -
10 ?C. 2 The thermal properties of the soil are constant. 
Properties The thermal properties of the soil are given 
to be k = 0.7 W/m. ?C and ? = 1.4 ?10
-5
 m
2
/s. 
Analysis The depth at which the temperature drops to 
0 ?C in 75 days is determined using the analytical 
solution, 
 
?
?
?
?
?
?
?
?
?
?
?
t
x
erfc
T T
T t x T
i s
i
? 2
) , (
 
Substituting,  
m 7.05 ? ? ? ?
?
?
?
?
?
?
?
?
?
?
? ? ?
?
? ?
?
?
x
x
erfc
                  
) s/h 3600 h/day 24 day 75 )( /s m 10 4 . 1 ( 2
15 10
15 0
2 5 
Therefore, the pipes must be buried at a depth of at least 7.05 m. 
2 ?C 42 ?C 
30 cm 
Kiln wall 
x 0 
Soil 
T i = 15 ?C 
Water pipe 
 T s =-10 ?C 
x 
Chapter 4 Transient Heat Conduction 
 4-86 
4-108 A hot dog is to be cooked by dropping it into boiling water. The time of cooking is to be determined. 
Assumptions 1 Heat conduction in the hot dog is two-dimensional, and thus the temperature varies in both 
the axial x- and the radial r- directions. 2 The thermal properties of the hot dog 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 hot dog are given to be k = 0.76 W/m. ?C, ? = 980 kg/m
3
, C p = 3.9 
kJ/kg. ?C, and ? = 2 ?10
-7
 m
2
/s. 
Analysis This hot dog can physically be formed by the intersection of an infinite plane wall of  thickness 2L 
= 12 cm, and a long cylinder of radius r
o 
= D/2 = 1 cm. The Biot numbers and corresponding constants are 
first determined to be 
37 . 47
) C W/m. 76 . 0 (
) m 06 . 0 )( C . W/m 600 (
2
?
?
?
? ?
k
hL
Bi 2726 . 1   and   5381 . 1
1 1
? ? ? ? ? A ? 
895 . 7
) C W/m. 76 . 0 (
) m 01 . 0 )( C . W/m 600 (
2
?
?
?
? ?
k
hr
Bi
o
5515 . 1   and   1251 . 2
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 
 
2105 . 0
) 01 . 0 (
) 10 2 (
) 1251 . 2 ( exp ) 5515 . 1 (                  
) 06 . 0 (
) 10 2 (
) 5381 . 1 ( exp ) 2726 . 1 (
100 5
100 80
) , 0 ( ) , 0 ( ) , 0 , 0 (
2
7
2
2
7
2
1 1
2
1
2
1
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
? ?
?
?
? ?
t
t
e A e A t t t
cyl wall block
? ? ? ?
? ? ?
 
which gives 
 min 4.1 s 244 = ? t 
Therefore, it will take about 4.1 min for the hot dog to cook. Note that  
 2 . 0 49 . 0
m) 01 . 0 (
s) /s)(244 m 10 2 (
2
2 7
2
? ?
?
? ?
?
o
cyl
r
t ?
? 
and thus the assumption ? > 0.2 for the applicability of the one-term approximate solution is verified. 
Discussion This problem could also be solved by treating the hot dog as an infinite cylinder since heat 
transfer through the end surfaces will have little effect on the mid section temperature because of the large 
distance. 
Water 
100 ?C 
 
2 cm   Hot dog   Ti = 5 ?C 
Chapter 4 Transient Heat Conduction 
 4-87 
4-109 A long roll of large 1-Mn manganese steel plate is to be quenched in an oil bath at a specified rate. 
The temperature of the sheet metal after quenching and the rate at which heat needs to be removed from the 
oil in order to keep its temperature constant are to be determined.  
Assumptions 1 The thermal properties of the balls are constant. 2 The heat transfer coefficient is constant 
and uniform over the entire surface.  3 The Biot number is Bi < 0.1 so that the lumped system analysis is 
applicable (this assumption will be checked). 
Properties The properties of the steel plate are given to be  k = 60.5 W/m. ?C, ? = 7854 kg/m
3
, and C p = 434 
J/kg. ?C (Table A-3). 
Analysis The characteristic length of the steel 
plate and the Biot number are 
         
1 . 0 036 . 0
C W/m. 5 . 60
) m 0025 . 0 )( C . W/m 860 (
m 0025 . 0
2
? ?
?
?
? ?
? ? ?
k
hL
Bi
L
A
V
L
c
s
c
 
Since Bi < 0.1 , the lumped system analysis is applicable. Therefore, 
s 30 min 5 . 0
m/min 10
m 5
velocity
length
time
s 10092 . 0
m) C)(0.0025 J/kg. 434 )( kg/m (7854
C . W/m 860
1 -
3
2
? ? ? ?
?
?
?
? ? ?
c p p
s
L C
h
V C
hA
b
? ?
 
Then the temperature of the sheet metal when it leaves the oil bath is determined to be 
 C 82.53 ? ? ? ? ? ?
?
?
? ? ? ?
?
?
? ?
?
?
) (
45 820
45 ) ( ) (
s)  30 )( s 10092 . 0 (
-1
t T e
t T
e
T T
T t T
bt
i
 
The mass flow rate of the sheet metal through the oil bath is 
 kg/min 4 . 785 m/min) 10 ( m) 005 . 0 ( m) 2 )( kg/m 7854 (
3
? ? ? ? ? ? wtV V m
?
? 
Then the rate of heat transfer from the sheet metal to the oil bath and thus the rate at which heat needs to be 
removed from the oil in order to keep its temperature constant at 45 ?C becomes 
 kW 213.2 = J/min  10 279 . 1 C ) 45 53 . 82 )( C J/kg. 434 )( kg/min 4 . 785 ( ] ) ( [
7
? ? ? ? ? ? ? ?
?
T t T C m Q
p
?
?
 
Steel plate 
10 m/min 
 Oil bath 
45 ?C 
Chapter 4 Transient Heat Conduction 
 4-88 
4-110E A stuffed turkey is cooked in an oven. The average heat transfer coefficient at the surface of the 
turkey, the temperature of the skin of the turkey in the oven and the total amount of heat transferred to the 
turkey in the oven are to be determined. 
Assumptions 1 The turkey is a homogeneous spherical object. 2 Heat conduction in the turkey is one-
dimensional because of symmetry about the midpoint. 3 The thermal properties of the turkey 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 are applicable (this assumption will be verified). 
Properties The properties of the turkey are given to be k = 0.26 Btu/h.ft. ?F, ? = 75 lbm/ft
3
, C p = 0.98 
Btu/lbm. ?F, and ? = 0.0035 ft
2
/h. 
Analysis (a) Assuming the turkey to be spherical in shape, its radius is 
determined to be 
 
ft 3545 . 0
4
) ft 1867 . 0 ( 3
4
3
3
4
ft 1867 . 0
lbm/ft 75
lbm 14
3
3
3
3
3
3
? ? ? ? ? ? ?
? ? ? ? ? ? ?
? ?
?
?
?
V
r r V
m
V V m
o o
 
The Fourier number is 1392 . 0
ft) 3545 . 0 (
h) /h)(5 ft 10 5 . 3 (
2
2 3
2
?
?
? ?
?
o
r
t ?
? 
which is close to 0.2 but a little below it. Therefore, assuming the one-term approximate solution for 
transient heat conduction to be applicable, the one-term solution formulation at one-third the radius from the 
center of the turkey can be expressed as 
 
1
1 ) 14 . 0 (
1
1
1
1
333 . 0
) 333 . 0 sin(
491 . 0
325 40
325 185
/
) / sin( ) , (
) , (
2
1
2
1
?
?
?
?
?
?
? ?
?
?
?
?
? ?
?
?
?
?
?
?
e A
r r
r r
e A
T T
T t x T
t x
o
o
i
sph
 
By trial and error, it is determined from Table 4-1 that the equation above is satisfied when Bi = 20 
corresponding to ?
1 1
2 9857 19781 ? ? . .   and   A . Then the heat transfer coefficient can be determined from 
  F . Btu/h.ft 14.7
2
? ?
?
? ? ? ? ? ?
) ft 3545 . 0 (
) 20 )( F Btu/h.ft. 26 . 0 (
o
o
r
kBi
h
k
hr
Bi 
 (b) The temperature at the surface of the turkey is 
 
F 317 ? ? ? ? ?
? ? ?
?
?
? ?
) , (
02953 . 0
9857 . 2
) 9857 . 2 sin(
) 9781 . 1 (
/
) / sin(
325 40
325 ) , (
) 14 . 0 ( ) 9857 . 2 (
1
1
1
2 2
1
t r T
e
r r
r r
e A
t r T
o
o o
o o o
?
?
? ?
 
(c) The maximum possible heat transfer is 
 Btu 3910 = F ) 40 325 )( F Btu/lbm. 98 . 0 )( lbm 14 ( ) (
max
? ? ? ? ? ?
? i p
T T mC Q 
Then the actual amount of heat transfer becomes 
 
Btu 3238 ? ? ?
?
?
? ?
?
? ? ? ?
? ? ?
Btu) 3910 )( 828 . 0 ( 828 . 0
828 . 0
) 9857 . 2 (
) 9857 . 2 cos( ) 9857 . 2 ( ) 9857 . 2 sin(
) 491 . 0 ( 3 1
) cos( ) sin(
3 1
max
3 3
1
1 1 1
,
max
Q Q
Q
Q
sph o
 
Discussion The temperature of the outer parts of the turkey will be greater than that of the inner parts when 
the turkey is taken out of the oven. Then heat will continue to be transferred from the outer parts of the 
turkey to the inner as a result of temperature difference. Therefore, after 5 minutes, the thermometer reading 
will probably be more than 185 ? F . 
Oven 
T ? = 325 ?F 
Turkey 
T i = 40 ?F 
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