College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Solutions to Exercise Five – Unsteady Heat Transfer
1. Consider a sphere of diameter 5 cm, a cube of side length 5 cm, and a rectangular prism of
dimension 4 cm by 5 cm by 6 cm, all initially at 0oC and all made of silver (k = 429 W/m2·oC,
 = 10,500 kg/m3, cp = 0.235 kJ/kg·oC). Now all three of these geometries are exposed to
ambient air at 33oC on all of their surfaces with a heat transfer coefficient of 12 W/m2·oC.
Determine how long it will take for the temperature of each geometry to rise to 25 oC.
Each geometry follows the same analysis; we will do the calculations for the sphere first, followed
by the cube and the rectangular prism. First we compute the characteristic length and the Biot
number to see if the lumped parameter analysis is applicable.
 3
D
V 6
D 5 cm 0.05 m
Lc  



 0.008333 m
A D 2
6
6
6
 mo C
hLc
12 W
 2 o 0.008333 m 
 429 W
k
m  C

Bi 

  0.00023  0.1


Since the Biot number is less than 0.1, we can use the lumped parameter analysis. In such an
analysis, the time to reach a certain temperature is given by the following equation.
1  T  T 

t   ln 
b  Ti  T 
b
hA
h

c p V c p Lc
From the data in the problem we can compute the parameter, b, and then compute the time for
the ratio (T – T)/(Ti – T) to reach the desired value.
b
h
c p Lc

m3
kgo C
1
1J
0.0005386

s
m 2 o C 10,500 kg 235 J 0.00833 m W  s
12 W
For this problem we have Ti =0oC, T = 33oC, and T = 25oC. The required time for the sphere is
thus found as follows.
 25o C  33o C 
1  T  T 
s
  2428 s for sphere
  
t   ln 
ln  o
b  Ti  T 
0.0005386  0 C  33o C 
Repeating the same calculations for the cube, without intermediate comments gives.
Lc 
V L3
L 5 cm 0.05 m
 2  

 0.08333 m
A 6L
6
6
6
Since this is the same value as Lc for the sphere, the results will be exactly the same. Lumped
parameter analysis is applicable and the time to reach 25oC is 2428 s for cube.
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
Page 2
Finally, repeating the calculations for the rectangular prism gives
Lc 
L1L2 L3
4 cm5 cm6 cm
V


 0.8108 cm  0.008108 m
A 2L1L2  L2 L3  L1L3  24 cm5 cm  5 cm6 cm  4 cm6 cm
Bi 
 mo C
hLc
12 W
 2 o 0.008108 m 
 429 W
k
m  C


  0.00023  0.1


Since the Biot number is less than 0.1, we can use the lumped parameter analysis.
m3
kgo C
1
1J
0.0005998
b
 2o

c p Lc m  C 10,500 kg 235 J 0.008108 m W  s
s
h
12 W
For this problem we have Ti =0oC, T = 33oC, and T = 25oC. The required time for the sphere is
thus found as follows.
1  T  T
t   ln 
b  Ti  T
 25o C  33o C 

s

  
ln  o
 0 C  33o C  2363 s for prism.
0
.
0005998



2. Cylindrical brass pellets (k = 64.1 Btu/h ft·oF,  = 532 lbm/ft3, and cp = 0.092 Btu/lbm·oF) that
are 2-in in diameter and 1 in long initially at 250oF are quenched in a water bath at 120oF
for a period of 2 minutes. If the convection heat transfer coefficient is 42 Btu/h ft2·oF,
determine the temperature of the balls after quenching
First we compute the characteristic length and the Biot number to see if the lumped parameter
analysis is applicable.
V
Lc  
A

LD 2
1
1
ft
4


 0.25 in
 0.02083 ft
2
2 4
2
4
12 in
D


2
 DL
L D 1 in 2 in
4
 h  fto F 
hLc
42 Btu
  0.014  0.1
0.02083 ft 
Bi 


k
64
.
1
Btu
h  ft 2 o F


Since the Biot number is less than 0.1, we can use the lumped parameter analysis. The
temperature after a time, t, is given by the following equation.
T  T   Ti  T e  bt
b
hA
h

c p V c p Lc
From the data in the problem we can compute the parameter, b, and then compute the
temperature after 2 minutes.
b
h
c p Lc

lbm o F
ft 3
1
41.18 1 h
0.6865


h 60 min 60 min
h  ft 2 o F 532 lbm 0.092 Btu 0.02083 ft
42 Btu
From the problem data we have T = 120oF and Ti = 250oF. The temperature after a quenching
time of 2 minutes is found as follows.
Exercise five solutions
ME 375, L. S. Caretto, Spring 2007
T  T  Ti  T e
 bt


 120 F  250 F  120 F e
o
o
o

0.6865
2 min 
min
=
Page 3
153oF
3. A 35-cm diameter cylindrical shaft made of stainless steel 304 (k = 14.9 W/m·oC,  = 7900
kg/m3, cp = 477 kJ/kg·oC) comes out of an oven at a uniform temperature of 400oC. The
shaft is then allowed to cool slowly in a chamber at 150oC with an average heat transfer
coefficient of 60 W/m2·oC. Determine the temperature center of the shaft 20 min after the
start of the cooling process. (Problem 4-41 in text.)
We can use the equation below from the class notes to compute the characteristic length of the
cylinder. Since we do not have any data for the length of the “long” cylinder we will assume that
is the ratio D/L << 2 and can be neglected in computing the characteristic length.
V
Lc  
A
 2
D L
4
D
D 35 cm

 4  
 8.75 cm  0.0875 m
 2
2 4
D
4
4
2 D  DL

1
4
L D
2L
Bi 
1
 mo C
hLc
60 W
 2 o 0.0875 m 
 14.9 W
k
m  C

The Biot number is greater than 0.1 so we have to use the
analysis with charts. The chart for long cylinders is Figure 4-16
on page 233 of the text. To use this chart we have to know two
of the following three parameters: (T 0 – T)/(Ti – T), k/hro, and
t/ro2. In this problem, the unknown parameter is (T 0 – T)/(Ti –
T); we can find this parameter from the chart and then use the
known values of Ti and T to find the temperature T0. The two
known parameters are computed below.
k
14.9 W

hro
mo C
 m 2 o C 
1


 1.42
 60 W  0.175 m


t
k t

c p ro2
60 W m 3 kg C 20 min 60 s 1 J
 0.155
mo C 7900 kg 377 J 0.175 m 2 min W  s
ro2
o

An extract from the chart is shown at the right. The value of
t/r02 = 0.155 is between the vertical axis (t/r02 = 0) and the
first grid line (t/r02 = 0.2). The extension of the line for k/hr0 =
1.4, shown in blue, intersects a line with t/r02 = 0.155 at a
dimensionless temperature of approximately 0 = 0.92. Using
this value we can find the center temperature from the known
values of Ti = 400oC and T =150oC as follows.


T  T  Ti  T 0  150o C  400o C  150o C 0.92
T = 380oC

  0.352

