College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Solutions to Exercise Six – More Unsteady Heat Transfer
1. The soil temperature in the upper layers of the earth varies with the variations in the
atmospheric conditions. Before a cold front moves in, the earth at a location is initially at
a uniform temperature of 10oC. Then the area is subjected to a temperature of –10oC and
high winds that resulted in a convection heat transfer coefficient of 40 W/m2·oC on the
earth’s surface for a period of 10 h. Taking the properties of the soil at that location to be k
= 0.9 W/m·oC and  = 1.6x10-5 m2/s, determine the soil temperature at distances 0, 10, 20,
and 50 cm from the earth’s surface at the end of this 10-h period.
We can assume that the solution for a semi-infinite medium applies here. In particular the
temperature, T, at any distance, x, and time, t, when there is a convection boundary condition
starting at t = 0 (when the semi-infinite region has an initial temperature, T i, is given by the
following equation.
 hx h 2t 

k 2 

T  Ti
 x   k 
 erfc
e
T  Ti
 4t 
 x
h t 

erfc


k
4

t


The terms in this equation using the heat transfer coefficient can be found as follows.
h t 40 W mo C 1.6 x10 5 m 2
3600 s
h 2t
10 h
 2o
 33.7
 33.7 2  1128
2
k
s
h
m  C 0.9 W
k
For x = 0, the parameter x/(4t)1/2 = 0; for x = 10 cm = 0,1 m, this parameter becomes
x

4t
0.1 m
5
1.6 x10 m 2
10 h 3600 s
4
s
h
 0.06588
For depths of 20 and 50 cm x/(4t)1/2 will be two and five times this value so that x/(4t)1/2 =
0.1318 for x = 20 cm and 0.3294 for x = 50 cm. Applying the parameters for x = 0 gives
T  Ti
 erfc0  e 01128erfc0  33.7 
T  Ti
I was not able to compute the second term. Both the argument of the exponential and the
argument of the complimentary error function gave me a numerical error. When I computed the
product for an increasing argument I got a limit of zero for this product. Because ot this I took the
product to be zero not only in the case of x = 0, but in the case of all other x values since they will
have even larger arguments than the ones for x = 0. Setting this term to zero gives the following
results.
 x
T  Ti  T  Ti erfc
 4t

 x
o
o
o
  10 C   10 C  10 C erfc

 4t



 x 
o
o
  10 C  20 C erfc


 4t 


For x = 0, the parameter x/(4t)1/2 = 0; and we get
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise five solutions
ME 375, L. S. Caretto, Spring 2007
 x
T  10 o C  20 o C erfc
 4t


Page 2

  10 o C  20 o C erfc0  10 o C  20 o C 1  –10oC .





For x = 10 cm, the parameter x/(4t)1/2 = 0.06588; and we get
 x
T  10 o C  20 o C erfc
 4t



  10 o C  20 o C erfc0.06588  10 o C  20 o C 0.9257   –8.5oC .





For x = 20 cm, the parameter x/(4t)1/2 = 0.1318; and we get
 x
T  10 o C  20 o C erfc
 4t



  10 o C  20 o C erfc0.1318  10 o C  20 o C 0.8519  –7.0oC .





For x = 50 cm, the parameter x/(4t)1/2 = 0.3294; and we get
 x
T  10 o C  20 o C erfc
 4t



  10 o C  20 o C erfc0.3294  10 o C  20 o C 0.6418  –2.8oC .





2. A short brass cylinder ( = 8530 kg/m3, cp = 0.389 kJ/kg·oC, k _ 110 W/m·oC, and  =
3.39x10-5 m2/s) of diameter D = 8 cm and height H = 15 cm is initially at a uniform
temperature of Ti = 150oC. The cylinder is now placed in atmospheric air at 20oC, where
heat transfer takes place by convection with a heat transfer coefficient of h = 40 W/m2·oC.
Calculate (a) the center temperature of the cylinder; (b) the center temperature of the top
surface of the cylinder; and (c) temperature of the top outer radius of the cylinder, and (d)
the total heat transfer from the cylinder 15 min after the start of the cooling.
The cylinder can be solved as the product of the solution for the infinite cylinder and that of the
infinite slab. The temperature at any point in the cylinder will be the product of two solutions. If
the Fourier number is greater than 0.2 we can use the approximate solutions shown below.
x, r , t  
2
2

 x  
 r 
T  T
 slab r , t cyl r , t    A1e1 cos 1 i   A1e1 J 0  1 
Ti  T
 Li  s 
 r0  c

The subscripts s for slab and c for cylinder indicate that the values of A1 and 1 will be different for
the different solutions. In addition, the Fourier numbers will be different for the slab (with a half
width, L = (15 cm)/2 = 7.5 cm = 0.075 m, and the cylinder with a radius of (8 cm)/2 = 4 cm = 0.04
m. Using the total time of 15 minutes = 900 s, the Fourier number for the cylinder is
 cyl 
t 3.39 x10 5 m 2
r02

s
900 s 
19.069 0.2
0.04 m 2
The Fourier number for the slab is

t 3.39 x10 5 m 2
L2

s
900 s 
 5.424 0.2
0.04 m 2
Since both Fourier numbers are greater than 0.2 the use of the approximate solution is justified.
Next we compute the Biot numbers to determine the values of A1 and 1 for each solution.
 mo C 
hL 40 W
  0.02727
Bi slab 
 2 o 0.075 m
k
mC
 110 W 
Exercise five solutions
ME 375, L. S. Caretto, Spring 2007
Bicyl 
Page 3
 mo C 
hro 40 W
  0.01455
 2 o 0.04 m
k
mC
 110 W 
For Bi = 0.4 Table 4-2 shows that that A1 = 1.0045 and 1 = 0.1620 for the slab and A1 = 1.0036
and 1 = 0.1667 for the cylinder. We can now compute the center temperature of the cylinder
where x = r = 0. (Note that cos(0) = J0(0) = 1.)

2
2
 r 
T  T   12
 x  
  A1e cos 1   A1e  1 J 0  1   A1e  1
Ti  T 
 L  s 
 r0  c
 A e 
 12
s
1
Substituting numerical values for, A1 and 1 for each geometry gives.

2
T  T
 A1e  1
Ti  T
 A e   1.0045e
 12
s
1
 0.1620 2 5.424 
c
1.00361e
 0.1667 2 19.069 
c
  0.8710.587  0.511
We can then find the temperature at the center of the cylinder.


T  T  Ti  T 0.511  20o C  150o C  20o C 0.511  –86.4oC .
To find the temperature at the top center of the cylinder, the radial solution will still have the value
of 0.587, but the rectangular solution has to be evaluated at x = L (equivalent to  = x/L = 1). This
solution is

2
2
 x 
 slab L, t    A1e  1 cos 1 i   1.0045e  0.1620 5.424 cos0.1620  0.860
 Li  s

So the product  at the top center of the cylinder =s (0.860)(0.587) = 0.505. With this  we find
the desired temperature as


T  T  Ti  T 0.505  20o C  150o C  20o C 0.505  –85.6oC .
To compute the temperature at the top outer radius we can use the axial solution that we just
obtained, but we not have to determine the radial solution at the point where r/r o = 1. This
requires that we multiply the exponential solution for r = 0 by the Bessel function J o(1r/ro) = Jo(1)
= Jo(0.1667) = 0.9931. This gives a radial solution at the outer radius of (0.9931)(0.587) = 0.583.
Multiplying this by the solution slab = 0.860 for the rectangular problem found above gives a
product solution  = 0.501. Solving for the temperature gives


T  T  Ti  T 0.501  20o C  150o C  20o C 0.501  –85.2oC .
Note that this problem has a low Biot number and the temperature throughout the solid is close to
uniform.
The heat transfer can be computed from the equations for Q/Q max for the one-dimensional
solutions given in equations (4-33) and (4-34) on page 235 of the text.
 Q

 Qmax

sin 1
  1   0, wall
1
 slab
 Q

 Qmax

J  

 1  2 0,cyl 1 1
1
 cylinder
The product solution for a two-dimensional case is given by the following modified product
solution in equation 94-53) on page 251. Here the 2D solution is written in terms of the first and
second dimension.
Exercise five solutions
ME 375, L. S. Caretto, Spring 2007
 Q

 Qmax

 Q
  
 2 D  Qmax

 Q
  
 first  Qmax
  Q


1  
 sec ond   Qmax
Page 4
 
 
 first 
We can compute the values for the cylinder and the slab as the first and second dimensions
directly. We know 0,slab =0.871 for the slab and 0,cyl =0.587 for the cylinder from the solution for
the center temperatures, so we can substitute these values as well as the appropriate values for
1 to determine the Q/Qmax ratio for each dimension.
 Q

 Qmax

sin 1
sin 0.1620
  1   0, wall
 1  0.871
 0.133
1
0.1620
 slab
Similarly, for the infinite cylinder solution we get
 Q

 Qmax

J 0.1677 
0.0835

 1  20.587  1
 1  20.587 
 0.415
0
.
1677
0
.
1677
 cylinder
Now we can compute the value of Q/Qmax for the short clinder.
 Q

 Qmax
  Q

 Q 
 Q 
  
  

1  
 2 D  Qmax  first  Qmax  sec ond   Qmax
 0.133  0.4151  0.133  0.493
 
 
 first 
To compute Q, we have to first find Qmax. This is the heat transfer if the entire solid changes
temperature from Ti to T∞.
Qmax  mc p T  Ti   D(2 L)c p T  Ti  
8350 kg
0.389 kJ
 .08 m.15 m
150o C  20o C  325 kJ
3
m
kg o C

We thus find Q = Qmax(Q/Qmax) = (325 kJ)(0.493) or Q = 160 kJ .
