College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 375
Heat Transfer
Spring 2007 Number 17629 Instructor: Larry Caretto
In-class Exercise Three
1. Steam at 320°C flows in a stainless steel pipe (k = 15 W/m·°C) whose inner and outer
diameters are 5 cm and 5.5 cm, respectively. The pipe is covered with 3-cm-thick glass
wool insulation (k = 0.038 W/m·°C). Heat is lost to the surroundings at 5°C by natural
convection and radiation, with a combined natural convection and radiation heat transfer
coefficient of 15 W/m2·°C. Taking the heat transfer coefficient inside the pipe to be 80
W/m2·°C, determine the rate of heat loss from the steam per unit length of the pipe. Also
determine the temperature drops across the pipe shell and the insulation. (Problem 3-72
in text.)
The problem can be modeled
as a thermal circuit where
there are four resistances:
inside convection, conduction
through the pipe wall and the
insulation and the combined convection plus radiation resistance on the outside. The overall heat
transfer is given by the equation.
Q 
T1  T 2

R1  R2  R3  R4
T1  T 2
r 
r
1
1
1

ln  o  
ln  ins
hi Di L 2k pipe L  ri  2k ins L  ro

1
 
 ho Do L
Dividing by L gives the heat transfer per unit length.
Q

L
1
1

hi Di 2k pipe
T1  T 2
r 
r 
1
1
ln  o  
ln  ins  
 ri  2k ins  ro  ho Do
We can calculate the individual resistances (multiplied by L) as follows.
0.0796 mo C
1
m 2 o C 1 1


hi Di
80 W  0.05 m
W
0.1845 mo C
1
m 2 o C 1
1


ho Do
15 W  0.115 m
W
r 
 D  1 mo C  5.5 cm  0.1011 mo C
1
1

ln  o  
ln  o  
ln 
2k pipe  ri  2k pipe  Di  2 15 W  5 cm 
15 W
r
1
ln  ins
2k ins  ro

D
1
 
ln  ins
 2kins  Do
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
 1 mo C
 11.5 cm  3.089 mo C
 
 
ln 
W
 2 0.038 W  5.5 cm 
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise three solutions
ME 375, L. S. Caretto, Spring 2007
Page 2
96.9 W
Q
320 o C 5o C
315o C



o
o
o
o
o
L 0.0796 m C 0.001011 m C 3.089 m C 0.1845 m C 3.354 m C
m



W
W
W
W
W
The temperature drops across individual resistances are simply found by the Ohm’s law analogy.
96.9 W 0.001011 mo C
Q
LR pipe 
 0.095 o C
L
m
W
96.9 W 3.089 mo C
Q
Tins  LRins  
 290 o C
L
m
W
T pipe 


Almost all of the 315oC temperature drop occurs across the insulation. The resistance of the
steel pipe is negligible.
2. The boiling temperature of nitrogen at atmospheric pressure at sea level (atmospheric
pressure) is –196°C. Therefore, nitrogen is commonly
used in low-temperature scientific studies since the
temperature of liquid nitrogen in a tank open to the
atmosphere will remain constant at –196°C until it is
depleted. Any heat transfer to the tank will result in the
evaporation of some liquid nitrogen, which has a heat of
vaporization of 198 kJ/kg and a density of 810 kg/m3 at
atmospheric pressure. Consider a 3-m-diameter spherical
tank that is initially filled with liquid nitrogen at
atmospheric pressure and –196°C. The tank is exposed to
ambient air at 15°C, with a combined convection and
radiation heat transfer coefficient of 35 W/m·°C. The
temperature of the thin-shelled spherical tank is observed
to be almost the same as the temperature of the nitrogen
inside. Determine the rate of evaporation of the liquid nitrogen in the tank as a result of
the heat transfer from the ambient air if the tank is (a) not insulated, (b) insulated with 5cm-thick fiberglass insulation (k = 0.035 W/m·°C), and (c) insulated with 2-cm-thick
superinsulation which has an effective thermal conductivity of 0.00005 W/m·°C. (Problem
3-85 in text.)
(a) If the tank is not insulated, and the outer tank temperature is almost the same as that of the
nitrogen, the only resistance is due to the external convection plus radiation.



35 W
Q  hAT  Ts   2 o 41.5 m 2 15o C  196 o C  208806 W
m  C
The heat transfer is the product of the evaporation rate and the latent heat of vaporization.
Q  m h fg
 m 
208806 W
1.055 kg
Q
kJ


198 kJ 1000 W  s
h fg
s
kg
If the tank is insulated, we have to consider two series
resistances: conduction through the insulation and
convection from the outer surface of the insulation. We
have the following equation for the sphere with heat
Exercise three solutions
ME 375, L. S. Caretto, Spring 2007
Page 3
transfer from the outside into the sphere.



T  Ts
T  Ts
4 15 o C   196 o C
Q  


 4233 W
ro  ri
1.55 m  1.5 m
1
1
Rins  Ro


0.035 W
4kri ro ho 4ro2
1.55 m 1.5 m  352 W
1.55 m 2
o
o
m C
m  C
As before, we compute the evaporation rate from the heat transfer and the latent heat.
m 
4233 W
0.0214 kg
Q
kJ


198 kJ 1000 W  s
h fg
s
kg
If the fiberglass is replaced with the superinsulator, we have the same analysis with different data.



T  Ts
T  Ts
4 15 o C   196 o C
Q  


 15.11 W
ro  ri
1.52 m  1.5 m
1
1
Rins  Ro


0.00005 W
4kri ro ho 4ro2
1.52 m 1.5 m  352 W
1.52 m 2
o
o
m C
m  C
m 
15.11 W
0.000076 kg
Q
kJ


198 kJ 1000 W  s
h fg
s
kg
3. A 10-in-thick, 30-ft-long, and 10-ft-high wall is
to be constructed using 9-in-long solid bricks
(k = 0.40 Btu/h·ft·°F) of cross section 7 in by 7
in, or identical size bricks with nine square air
holes (k = 0.015 Btu/h·ft·°F) that are 9 in long
and have a cross section of 1.5 in by 1.5 in.
There is a 0.5-in-thick plaster layer (k = 0.10
Btu/h·ft·°F) between two adjacent bricks on all
four sides and on both sides of the wall. The
house is maintained at 80°F and the ambient
temperature outside is 30°F. Taking the heat
transfer coefficients at the inner and outer
surfaces of the wall to be 1.5 Btu/h·ft2·°F and
4 Btu/h·ft2·°F, respectively, determine the rate
of heat transfer through the wall constructed of
(a) solid bricks and (b) bricks with air holes.
(Problem 3-58E in text.)
In this problem we have a combination of series and parallel resistances. The inner and outer
plaster walls are single resistances. Between these we have two or three resistances. For part
(a) we have resistance of plaster and resistance of the solid brick. For part (b) we have parallel
resistances of plaster, brick, and air. We also have inner and outer convection resistances for
each part. The equivalent thermal circuit when the bricks are solid is shown on the next page.
Here Ri and Ro represent the inside and outside convection resistances; R 1 and R5 are the
resistances of the plaster on the inner and outer sides of the wall. There are three components in
Exercise three solutions
ME 375, L. S. Caretto, Spring 2007
Page 4
the inner part of the wall: R2 is the resistance of a 7.5 in horizontal length of plaster that includes
the square of plaster surrounded by four bricks; R3 is the resistance of the 7 in vertical length of
plaster and R4 is the resistance of the brick. The typical inner section has an area of
(7.5 in)2/(ft2/144 in2) = 0.3906 ft2.
The inner and outer convection resistances for the area of the typical inner section with an area of
0.3906 ft2 are.
Ri 
1.707 ho F
1
h  ft 2 o F
1


hi A
15 Btu 0.3906 ft 2
Btu
R1  R5 
Ro 
0.640 ho F
1
h  ft 2 o F
1


ho A
4 Btu 0.3906 ft 2
Btu
L
L1
0.5 in
1.0607 h o F
h  ft o F
ft
 R5  5 

k1 A
k 5 A 0.10 Btu 0.3906 ft 2 12 in
Btu
For the inner, parallel, resistances, the bricks have an area of (7 in)2/(ft2/144 in2) = 0.34036 ft2; the
longer plaster strip has an area of (7.5 in)(0.57in)/(ft2/144 in2) = 0.02604 ft2; the shorter plastic
strip has an area of (7 in)(0.5 in)/(ft2/144 in2) = 0.02431 ft2; the sum of the three areas shown here
do not add to the total area of 0.3906 ft2 because of rounding. With these areas we can now
compute the values of the inner resistances as follows.
R2 
L2
9 in
288 h o F
h  ft o F
ft


k 2 A2 0.10 Btu 0.02604 ft 2 12 in
Btu
R3 
L3
9 in
309 h o F
h  ft o F
ft


k 3 A3 0.10 Btu 0.02431 ft 2 12 in
Btu
R4 
L4
9 in
5.51 h o F
h  ft o F
ft


k 4 A4 0.40 Btu 0.34036 ft 2 12 in
Btu
The total inner resistance is computed using the formula for parallel resistances.
Rinner 
5.31 h o F
1
1


1
1
1
Btu
Btu
Btu
Btu




R2 R3 R3 288 h o F 309 h o F 5.51 h o F
The total resistance is then found as the sum of this resistance plus the other resistances in
series.
Rtotal  Ri  R1  Rinner  R5  Ro 
1.707 ho F 1.0607 ho F 5.31 h o F 1.0607 h o F 0.640 h o F 9.794 h o F





Btu
Btu
Btu
Btu
Btu
Btu
Exercise three solutions
ME 375, L. S. Caretto, Spring 2007
Page 5
So the heat transfer for one section of the wall is
T  T 2 80 o F  30 o F 5.105 Btu
Q sec tion  1

Rtotal
h
9.794 h o F
Btu
If the heat flux through this typical section applies to the entire wall we have the following result.
Q wall  q wall Awall  q sec tion Awall
5.105 Btu
Q sec tion
h
30 ft 10 ft   3921 Btu

Awall 
2
Asec tion
h
0.3906 ft
When the bricks have air spaces, the equivalent circuit has an additional resistance for the air.
All the resistances in this case are the same as they were in the previous problem except for the
resistance of the bricks which now has a lower area, and the new resistance of the air gaps. The
area of the air gaps is 9(1.5 in)2/(ft2/144 in2) = 0.140625 ft2. The remaining area of brick is
0.34036 ft2 – 0.150625 ft2= 0.19965 ft2. Thus, the two new resistances are
R4 
R5 
L4
9 in
9.39 h o F
h  ft o F
ft


k 4 A4 0.40 Btu 0.19965 ft 2 12 in
Btu
L5
9 in
355.6 h o F
h  ft o F
ft


k 5 A5 0.015 Btu 0.140625 ft 2 12 in
Btu
As before, we compute the total inner resistance and then use that to compute the total
resistance and the heat transfer over the typical section.
Rinner 
1
1
1
1
1



R2 R3 R4 R5
Rtotal  Ri  R1  Rinner  R5  Ro 

1
Btu
288 h F
o

Btu
309 h F
o

Btu
9.39 h F
o

Btu

8.62 h o F
Btu
355.6 h o F
1.707 h o F 1.0607 h o F 8.62 h o F 1.0607 h o F 0.640 h o F 13.10 h o F





Btu
Btu
Btu
Btu
Btu
Btu
T  T 2 80 o F  30 o F 3.817 Btu
Q sec tion  1

Rtotal
h
13.10 h o F
Btu
Exercise three solutions
ME 375, L. S. Caretto, Spring 2007
Page 6
We have the previous equation for the assumption that the heat flux through this typical section
applies to the entire wall.
Q wall
3.817 Btu
Q sec tion
h
30 ft 10 ft   2932 Btu

Awall 
2
Asec tion
h
0.3906 ft
4. The roof of a house consists of a 15-cmthick concrete slab (k = 2 W/m·°C) that is
15 m wide and 20 m long. The
convection heat transfer coefficients on
the inner and outer surfaces of the roof
are 5 and 12 W/m2·°C, respectively. On a
clear winter night, the ambient air is
reported to be at 10°C, while the night
sky temperature is 100 K. The air inside
the house and the interior surfaces
(walls and floors) of the house are
maintained at a constant temperature of
20°C. The emissivity of both the upper
and lower surfaces of the concrete roof
is 0.9. Considering both radiation and
convection heat transfers, determine the
rate of heat transfer through the roof, and the inner surface temperature of the roof.
(Problem 3-30 in text.)
In this problem we have convection from the inside air at 20oC (293.15 K)and radiation from the
inside walls, also at 20oC to the roof. This combined inside heat transfer is than transferred
through the roof by conduction and the heat leaving the roof is then transferred by convection to
air at 10oC (283.15 K) and radiation to 100 K. The unknowns in this problem are the inner and
outer surface temperatures of the roof, Ts,i, and Ts,o. Because of the radiation equations we will
use the temperatures in kelvins.
For these calculations we will assume that the radiation heat transfer for both the inside and the
outside of the house can be modeled as a small object in a large enclosure. The area of the roof
is actually (15 m)(20 m) = 300 m 2, but it is certainly smaller than the night sky and it is reasonable
to assume that is small compared to the area of all the walls and floors in the house.
The insider heat transfer is



Q  A hT ,i  Ts ,i    T4,i  Ts4,i 
8
300 m 2  52Wo 293.15 K  Ts ,i   0.9 5.6702 x10 4 W 293.15 K 4  Ts4,i
m K
m  C


This same heat transfer is maintained by conduction through the roof.
Q 
kATs ,i  Ts ,o 
L

2 W 300 m 2
mo C 0.15 m
Finally, this same heat is transferred to the air and sky by convection and radiation.



4
Q  A hTs,o  T,i    Ts4,oi  Tsky

Ts,o  283.15 K   0.9 5.6702x10 4 W Ts4,i  100 K 4 
300 m2  122 W
o
m  C
8
m K
Exercise three solutions
ME 375, L. S. Caretto, Spring 2007
Page 7
We now have three equations with two unknown temperatures and an unknown heat flux. We
have to solve by iteration or calculator/computer approach. Using the goal seek tool of Excel, I
found the following results.
Q  3.754 x10 4 W
Ts,i  280.40 K  7.25o C
Ts,o  271.05 K  2.10o C
Note that the very cold sky temperature leaves the outer wall temperature less than the air
temperature. The convection heat transfer on the outside is actually going into the roof.