College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
Midterm Exam Solutions
1. A flat wall with a temperature of 400oF loses heat by convection to ambient air with a
temperature of 60oF. The convection coefficient is 3 Btu/h·ft2·oF. Insulation with a thermal
conductivity of 0.06 Btu/h·ft·oF is to be added to cover the wall so that both of the
following conditions are satisfied:
a. The temperature of the outer surface of the insulation is less than or equal to 110oF.
b. The heat transfer is reduced by 80% or more compared to the value for no insulation.
What is the required insulation thickness?
The heat transfer with no added insulation is given by the usual equation for convective heat
transfer.
q  hTwall  T  
3 Btu
h  ft 
2 o
400
F
o

F  60o F 
1020 Btu
h  ft 2
The inner surface of the insulation will have a temperature of T wall and the outer surface of the
insulation will have a temperature of T s. The conduction heat flux through the insulation and the
convection heat flux from the outer insulation surface will be the same.
q  hTs  T  
k Twall  Ts 
L
 . We can find the solution required to satisfy
There are two unknowns in this equation, T s, and q
one of the two required conditions shown above and then see if that satisfies both conditions. If
we assume that Ts = 110oF, we can find the convection heat flux just as we did when there was
no insulation.
q  hTs  T  
3 Btu
h  ft 
2 o
150 Btu

110 F  60 F  
F
h  ft
o
o
2
The percent reduction in heat flux is found below.
1020 Btu 150 Btu

h  ft 2
h  ft 2
%Reduction 
 85.3%
1020 Btu
h  ft 2
So satisfying the condition that the surface temperature is 110oF satisfies the condition that the
new heat flux be an 80% or greater reduction of the original heat flux. We can now use this 150
Btu/h·ft2 heat flux as the conduction heat flux to compute the required insulation thickness.
k Twall  Ts 
q 

L
400
F
0.06 Btu
k Twall  Ts  h  ft 
L

q
o
o
F  110 o F

150 Btu
h  ft 2
L = 0.116 ft = 1.392 in
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Midterm solutions
2
ME 375, L. S. Caretto, Spring 2007
Page 2
A frying pan has a handle with a rectangular cross section that is 3 mm thick and 3 cm
wide. The point where the handle is attached to the pan has a temperature of 200 oC. The
desired temperature of the other side of the handle is 30oC. How long should the handle
be if there is a convection coefficient of 18 3 W/m2·oC to an ambient temperature of 20oC
and the handle is (a) duralumin (k = 180 W/m·oC) or stainless steel (k = 53 W/m·oC)?
The handle of the pan can be considered a fin. Since we are solving for length, we will use the
usual formula for no convection from the end with no end length correction. We can consider the
effect of this correction after we have solved the problem. According to this formula the
temperature at any point x, T, is given by the following equation
T  T coshmL  x 
hp

where m 
Tb  T
cosh mL
kAc
Since we want the temperature at the end of the handle (x = L) we can write this equation with
the given data T = 30oC, Tb = 200oC, and T = 20oC as follows.
cosh 0
1
T  T
30o C  20o C
1




 mL  cosh 1 18  3.5827
o
o
cosh mL cosh mL Tb  T 200 C  20 C 18
The value of m for the handle depends on the material. From the equation for m and the
condition that mL = 3.5827 we can write the following equation for L
L
3.5827
kAc
 3.5827  3.5827
m
hp
For the dimensions given, Ac = (3 mm)(3 cm) = (0.003 m)(.03 m) = 9x10-5 m2 and p = 2(0.003 m +
0.03 m) = 0.066 m. The required lengths for duralumin and stainless steel are


180 W
9 x10 5 m 2
o
kAc
m

C
L  3.5827
 3.5827
= 0.418 m for duralumin
18 W
hp


0
.
066
m
m 2 o C


53 W
9 x10 5 m 2
o
kAc
L  3.5827
 3.5827 m  C
= 0.227 m for stainless steel
18 W
hp


0
.
066
m
m 2 o C
If we considered the end correction, Lc = L + Ac/p = L + (9x10-5 m2)/(0.066 m) = L + 0.00136 m.
This would make a difference of 0.6% for the stainless steel and 0.3% for duralumin.
3
A rectangular prism that has sides of 0.3 m, 0.5 m, and 0.6 m is made of wood (k = 0.17
W/m·oC and  = 1.28x10-7 m2/s). It is initially at a uniform temperature of 15oC. It is exposed
to hot gases at 550oC a heat transfer coefficient of 35 W/m2·oC on the surface. How long
will it take for the center temperature to reach 200oC? What is the temperature of a corner
of the block at this time?
We can apply the lumped parameter model if the Biot number is less than 0.1. For a rectangular
prism, the characteristic length in the Biot number is found as follows.
Midterm solutions
Lc 
ME 375, L. S. Caretto, Spring 2007
Page 3
L1 L2 L3
V
0.5
0.5



 0.07143 m
1
1
1
A 2L1 L2  L1 L3  L2 L3  1  1  1


L1 L2 L3 0.3 m 0.5 m 0.6 m
The Biot number for this problem is
Bi 
 mo C
hLc
35 W
 2 o 0.07143 m 
 0.17 W
k
m  C


  14.7


This is definitely not less than 0.1 so we have to use the product solution.
We can find the solution to this problem as the solution to three separate one-dimensional infinite
slabs. Each slab will have a separate length parameter, L, the half length, which will be equal to
0.15 m, 0.25 m, and 0.3 m for the three sides in our prism. The product solution can be written as
follows
x, y, z, t   1x, t 1 y, t 1z, t 
We will assume that the Fourier number, , is greater than 0.2 so that we can use the
approximate solution and check this assumption after we solve for . If we denote the coordinate
distance as xi and the half-length as Li for each of the coordinate directions, the one-dimensional
solution for an infinite slab is.
1 xi , t   A1e  1 cos 1  A1e
2

21t
L2i
 x 
cos 1 i 
 Li 
We first have to find the Biot numbers to determine the values of A1 and 1. The three Biot
numbers for this problem are.
Bi1 
 mo C
hL1
35 W
 2 o 0.15 m 
 0.17 W
k
m  C


  30.9


 mo C 
35 W
  51.5
 2 o 0.25 m 


0
.
17
W
m  C


o
 m C 
35 W
  61.8
 2 o 0.3 m
m  C
 0.17 W 
Bi 2 
hL2
k
Bi1 
hL1
k
Interpolating in Table 4-2 we find the following values of A1 and 1: For Bi = 30.9, A1 = 1.2718
and 1 = 1.5213; for Bi = 51.5, A1 = 1.2723 and 1 = 1.5249; for Bi = 61.8, A1 = 1.2725 and 1 =
1.5342.
The value of  = t/Li2, will be different for each dimension because the length parameter will be
different for each dimension. However, the value of time, t, will be the same. So we have to write
the equation in terms of the time, t. Substituting the one dimensional solutions, with the
appropriate values of A1, 1, , and the length parameter for each dimension and setting the
dimensionless distance in the cosine term to 0 for to get the solution at the center gives.
1.5213 1.28 x107 m 2



t
T  T
200 o C  550 o C
s
0.15 m 2


 0.6542  1.2718e
cos0 


Ti  T
15 o C  550 o C


2
2
7 2
7 2




1
.
5249
1
.
28
x
10
m
1
.
5342
1
.
28
x
10
m

 


t

t
2
2
s
s
1.2723e 0.25 m 
cos0  1.2725e 0.3 m 
cos0

 


 

2
Midterm solutions
ME 375, L. S. Caretto, Spring 2007
Page 4
The cosine of zero is one and we can rearrange this equation by writing the product of three
exponentials as a single exponential.
0.6542
e
1.27181.27231.2725

1.28 x10 7 m 2  1.52132 1.52492 1.53422
t


 0.15 m 2 0.25 m 2 0.3 m 2
s





Taking the natural logarithm of each side gives us an equation that we can solve for the time, t.


0.6542
ln 

 1.27181.27231.2725
= 5.39x104 s = 15.0 h
t
2
2
2
7 2
1.28 x10 m  1.5213 1.5249 1.5342 


 0.15 m 2  0.25 m 2  0.3 m 2 
s


We have to check the value of  to make sure that it is greater than 0.2. We only have to check
the value for a length parameter of 0.15 m. If  > 2 for this length, then it will also be > 0.2 for the
larger length parameters of 0.25 m and 0.3 m.


 5.39 x10 4 s 

t 1.28 x10  7 m 2 



 0.307  0.2
2
2
s


0.15 m
L
Thus, our initial assumption that we could use the approximate solutions was correct.
To get the corner temperatures, where all the dimensionless coordinates are one, we have to
solve the following equation, with t = 5.39x104 s..
1.5213 1.28 x107 m 2



t
T  T 
s
0.15 m 2

 1.2718e
cos1.5213 

Ti  T 


2
2
7 2
7 2




1
.
5249
1
.
28
x
10
m
1
.
5342
1
.
28
x
10
m

 


t

t
2
2
s
s
1.2723e 0.25 m 
cos1.5249  1.2725e 0.3 m 
cos1.5342

 


 

2
We can solve this equation as written or we can use the following short cut: the prefactors and
exponential terms (multiplied together) are simply the dimensionless temperature value for the
center point, 0.6452; multiplying this by the three cosine terms gives our answer:  = 5.44x10-5.
So the corner temperature = T + (Ti – T) = 550oC + (15oC – 550oC)(5.44x10-5) = 549.97oF .
In this problem, the low thermal conductivity of the wood allows a large temperature difference
between the corner, which will have the highest temperature and the center of the block that will
have the lowest temperature.