College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
February 21 Homework Solutions
2-39C It is claimed that the temperature profile in a medium must be perpendicular to an
insulated surface. Is this a valid claim? Explain. (This problem was listed on the web site,
but not in the course outline.)
Yes. At an insulated surface the heat flux is zero. (Note that this is a typical definition of
“insulated” in the mathematics of heat transfer. In reality, an insulated surface will have a small,
but finite, heat transfer.). For a zero heat transfer, Fourier’s law tells us that q = –kT/n = 0,
where n is the direction normal to the surface. A line of zero gradient normal to surface means
that the line is perpendicular to the surface, so the claim is correct.
3-19
Consider a 1.2-m-high and 2-m-wide glass window whose thickness is 6 mm and thermal
conductivity is k = 0.78 W/m·oC. Determine the steady rate of heat transfer through this
glass window and the temperature of its inner surface for a day during which the room is
maintained at 24oC while the temperature of the outdoors is –5oC. Take the convection
heat transfer coefficients on the inner and outer surfaces of the window to be h1 = 10
W/m2·oC and h2 = 25 W/m2·oC, and disregard any heat transfer by radiation
We can solve this by an equivalent thermal circuit with three resistances: one for the inside
convection, one for conduction through the window and one for outside convection.
Rwindow
The convection resistances are 1/hA and the conduction resistance is L/kA. The combined
resistance is the sum of the three individual resistances and the heat transfer is the ratio of the
temperature difference to this combined resistance.
Q 
T1  T 2
24 o C  (5 o C )
 2o
 471 W
1
L
1
m  C 1
mo C 0.006 m m 2 o C 1




h1 A kA h2 A
10 W 2.4 m 2 0.78 W 2.4 m 2
25 W 2.4 m 2
The inner surface temperature is simply found from the heat flow, which is constant, and the initial
convection resistance.
Q
m 2 o C 1
Q  h1 AT1  T1   T1  T1 
 24 o C  471 W 
 4.4oC
2
h1 A
10 W 2.4 m
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
February 21 homework solutions
3-20
ME 375, L. S. Caretto, Spring 2007
Page 2
Consider a 1.2-m-high and 2-m-wide double-pane window consisting of two 3-mm-thick
layers of glass (k = 0.78 W/m·oC) separated by a 12-mm-wide stagnant air space (k = 0.026
W/m·oC). Determine the steady rate of heat transfer through this double-pane window and
the temperature of its inner surface for a day during which the room is maintained at 24 oC
while the temperature of the outdoors is –5 oC. T ake the convection heat transfer
coefficients on the inner and outer surfaces of the window to be h1 = 10 W/m2·oC and h2 =
25 W/m2·oC, and disregard any heat transfer by radiation
There are five separate resistances in the equivalent thermal circuit for this problem: two
convection resistances, one on each side of the double pane window, and three conduction
resistances in the two glass panes plus the air gap. A diagram of the double-pane window and
the equivalent thermal circuit are shown below.
The overall heat transfer coefficient is given by the following equation.
Q 
T1  T 2
T1  T 2

L
L
L
1
1
R1  R2  R3  R4  R5
 1  2  3 
h1 A k1 A k 2 A k 3 A h2 A
Using the data given we can find the resistances as follows.
0.0417 oC
1
m 2 o C 1


h1 A 10 W 2.4 m 2
W
o
L
L
m C 0.003 m 0.0016 oC
R2  R4  1  4 

k1 A k 4 A 0.078 W 2.4 m 2
W
o
o
L3
m C 0.012 m 0.1923 C
R3 


k 3 A 0.026 W 2.4 m 2
W
2 o
0.0164 oC
1
m  C 1
R5 


h2 A 10 W 2.4 m 2
W
R1 
We can now find the overall heat transfer.
Q 


T1  T 2
24 o C  5 o C

R1  R2  R3  R4  R5 0.0417 oC 0.0016 oC 0.1923 oC 0.0016 oC 0.0167 oC




W
W
W
W
W
Q  114 W
Comparing this answer with that of problem 3-19 we see that the double pane window has
decreased the transfer by 76% from 471 W to 114 W. This means that we have increased the
overall resistance since the temperature difference in the same. Computing the total resistance
February 21 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 3
for each problem shows that the resistance increased from 0.06155oC/W in problem 3-19 to
0.2539oC/W here. About 76% of the new total resistance comes from the air gap.
The temperature of the inner surface is computed from the convection resistance between the
inside air and the inside wall.
T T
Q  1 1
R1
0.0417 oC
= 4.4oC
 T1  T1  R1Q  24 o C  114 W 
W
3-29E A wall is constructed of two layers of 0.7-in-thick
sheetrock (k= 0.10 Btu/h·ft·oF), which is a plasterboard
made of two layers of heavy paper separated by a layer
of gypsum, placed 7 in apart. The space between the
sheetrocks is filled with fiberglass insulation (k = 0.020
Btu/h·ft·oF). Determine (a) the thermal resistance of the
wall and (b) its R-value of insulation in English units.
There are three separate thermal resistances in the
equivalent thermal circuit for this problem. A real wall
would have convection, but we are only considering th4e
equivalent resistance of the insulation. No area is given so
we will compute the resistance for a unit area (A = 1 ft2.):
The thickness of the sheetrock layers and their thermal
conductivities are the same so we have
L1 1
0.583 ft 2  ho F
h  fto F 0.7 in


k1 A h1 A 0.10 Btu 12 in ft
Btu
o
L1 1
7 in
29.17 ft 2  ho F
h  ft F
R2 


k1 A h1 A 0.020 Btu 12 in ft
Btu
R1  R3 
The total resistance is 30.34 ft2·h·oF/Btu, which corresponds to R-30 insulation.
3-36
The wall of a refrigerator is constructed of
fiberglass insulation (k = 0.035 W/m·oC)
sandwiched between two layers of 1-mm-thick
sheet metal (k = 15.1 W/m·oC). The refrigerated
space is maintained at 3oC, and the average heat
transfer coefficients at the inner and outer surfaces
of the wall are 4 W/m2·oC and 9 W/m2·oC,
respectively. The kitchen temperature averages
25oC. It is observed that condensation occurs on
the outer surfaces of the refrigerator when the
temperature of the outer surface drops to 20oC.
Determine the minimum thickness of fiberglass
insulation that needs to be used in the wall in order
to avoid condensation on the outer surfaces.
This problem is similar to problem 3-20 in that there are five thermal resistances: three for
conduction through the three materials in the refrigerator wall and two convection resistances,
one on each side of the wall. Unlike problem 3-20, however, we are not given all the data we
need to compute each resistance. The conduction resistance for the insulation, L/kA, will contain
the unknown length, L, that we are trying to find. If we try to find the heat transfer, we will get a
result that depends on L. However, we can find the heat transfer,
Q from the equation for
February 21 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 4
convection to the outer wall, q =houter(T,outer – Twall,outer), where we know all the other terms in the
equation. This will give us an equation that we can solve for L. (Twall,outer is the temperature of
20oC that we want to achieve to avoid condensation; note that this temperature is shown in the
diagram as 10oC as opposed to the problem statement where it is given as 20oC.) Since we are
not given the area of the wall we will use heat flux in our calculations to give the convection heat
transfer, which is the same as the heat transfer through all layers.
q 
9W
Q
W
 hT ,outer  Twall ,outer   2 o 25 o C  20 o C  45 2
A
m  C
m


We can now use this value in the overall heat transfer equation below. Note that when this
equation is written for heat flux, the area terms are not included in the thermal resistance..
q 
Q 45 W
T1  T 2
T1  T 2



2
A
AR1  R2  R3  R4  R5  1  L1  L2  L3  1
m
h1 k1 k2 k3 h2
Using the data given we can find the resistances as follows.
1 m2 o C 0.101 m2 o C


h1
9W
W
o
L1 L4
m C
6.62 x105 m2 o C
0.001 m 
AR2  AR4 


k1 A k4 15.1 W
W
o
o
L
m C
28.57 L m C
AR3  
L
k3 0.035 W
W
2o
1 m  C 0.25 m2 o C
RA5 


h2
4W
W
AR1 
Substituting these resistances into the overall heat flux equation gives an equation that we can
solve for L.
45 W
m2
T1  T 2
25o C 3o C


AR1  R2  R3  R4  R5  0.101 m2 o C 6.62 x105 m2 o C 28.57 L mo C 6.62 x105 m2 o C 0.25 m2 o C




W
W
W
W
W
45 W
m2
 0.101 m 2 o C 6.62 x10 5 m 2 o C 28.57 L mo C 6.62 x10 5 m 2 o C 0.25 m 2 o C 

  22 o C




W
W
W
W
W


22 o C 
L
45 W
m2
 0.101 m 2 o C 6.62 x10 5 m 2 o C 6.62 x10 5 m 2 o C 0.25 m 2 o C 





W
W
W
W


o
45 W  28.57 m C 


W
m 2 

L = 0.45 m
February 21 homework solutions
ME 375, L. S. Caretto, Spring 2007
Page 5
Note that the resistances of the metal sheets are very small and we would get the same answer
(to two significant figures) if we ignored them. Thin meal sheets, with both a small thickness and
a high thermal conductivity usually pose a negligible resistance in problems where there are a
series of thermal resistances.
3-47
Six identical power transistors with aluminum
casing are attached on one side of a 1.2-cmthick 20-cm by 30-cm copper plate (k = 386
W/m·oC) by screws that exert an average
pressure of 10 MPa. The base area of each
transistor is 9 cm2, and each transistor is
placed at the center of a 10-cm by 10-cm
section of the plate. The interface roughness
is estimated to be about 1.4 mm. All
transistors are covered by a thick Plexiglas
layer, which is a poor conductor of heat, and
thus all the heat generated at the junction of
the transistor must be dissipated to the
ambient at 23oC through the back surface of
the copper plate. The combined
convection/radiation heat transfer coefficient
at the back surface can be taken to be 30
W/m2·oC. If the case temperature of the
transistor is not to exceed 75oC, determine
the maximum power each transistor can
dissipate safely, and the temperature jump at
the case-plate interface.
In this problem we really should do a two-dimensional analysis since the heat supplied by the
transistors is localized. However, we can assume that the copper plate, with its high thermal
conductivity will quickly even out differences in temperature in the cross sectional area, and, to a
first approximation we can consider this a one-dimensional problem. Since the Plexiglas is an
insulator, we can start the analysis with the transistors. The data about the interface roughness
and contact pressure suggest that we should consider the contact resistance. Thus we will have
a thermal circuit that has three components between the transistors and the air: (1) the contact
resistance, (2) the conduction resistance
through the copper, and (3) the convection
resistance from the copper to the air, as shown
at the left.
We find the contact resistance as the inverse of the contact conductance which is given in Table
3-2 on page 145. That table has data for copper-aluminum contact resistance with roughness
levels of 1.17 to 1.4 m, which is the level in this problem at pressure of 5 and 15 MPa.
Interpolating between these two pressures gives the approximate contact conductance at the
pressure of 10 MPa specified here as 49,000 W/m2·oC. We use this contact conductance with
the contact area between the plate and transistors of 6 X (9 cm 2) = 54x10-4 m2. The plate
conduction and convection resistances use the full area of 100 cm 2 = 0.01 m2. Applying the value
for contact conductance, the areas just stated, and the other data given in the problem in the
expression for series heat transfer, along with the usual equations for conductive and convective
resistances gives the following result.
February 21 homework solutions
Q 
ME 375, L. S. Caretto, Spring 2007
Page 6
Tcase  Tair
75 o C  23o C

1
L
1
m 2 o C
1
mo C 0.012 m m 2 o C
1




4
2
2
hcontact A kA hconvection A 49,000 W 54 x10 m
386 W 0.01 m
30 W 0.01 m 2
Q = 15.5 W
3-133E A row of 3-ft-long and 1-in-diameter used
uranium fuel rods that are still radioactive are
buried in the ground parallel to each other
with a center-to-center distance of 8 in at a
depth 15 ft from the ground surface at a
location where the thermal conductivity of the
soil is 0.6 Btu/h·ft·oF. If the surface
temperature of the rods and the ground are
350oF and 60oF, respectively, determine the
rate of heat transfer from the fuel rods to the
atmosphere through the soil.
  Sk T  T  . Shape factors
This problem is analyzed using the shape factor analysis where Q
1
2
for various geometries are found in Table 3-7. For the
2L
system of buried cylinders considered here, this table gives
S per cylinder 
the equation shown at the right for S per cylinder where L
2z 
 2w
ln 
sinh

and D are the length and diameter of the cylinders, w is the
w 
 D
distance between cylinders and z is the perpendicular
distance from the ground to the center of each cylinder. This formula is valid L >> D, z, and w >
1.5 D. All these constraints appear to be satisfied except the one that L >> D, z. I assumed that
this means that L >> D and L >> z. However we have L < z!!! Ignoring these constraints and
applying the equation we get the total S, by multiplying the per-cylinder S by 4.
Stotal  4
2L
2(3 ft )
4
 0.5298 ft
2z 
 2w
 2(8 in )
2(15 ft ) 
ln 
sinh


ln 
sinh
w 
 D
(8 12 ft ) 
 (8 in )
With this value we can compute the heat transfer.


0.6 Btu
Q  Sk T1  T2   0.5298 ft 
350oF  60oF = 92.2 Btu/h
o
h  ft F