1. (Welty, Rorrer, Foster, 6th Edition International Student Version 14.12)
Determine the heat transfer rate per square meter of wall area for the case of a
furnace with inside air at 1340K. The furnace wall is composed of a 0.106-m
layer of fireclay brick and a 0.635-cm thickness of mild steel on its outside
surface. Heat transfer coefficients of the inside and outside wall surfaces are 5110
and 45 W/m2.K respectively; outside air is at 295K. What will be the temperatures
are each surface and at the brick-steel interface?
2. (WWWR #15.26) Given that the furnace wall and other conditions are as
specified by in Problem 1, what thickness of celotex (k=0.065 W/m.K) must be
added to the furnace wall in order that the surface temperature of the insulation do
not exceed 340K.
3. (Modified from Welty, Rorrer, Foster, 6th Edition International Student
Version 14.1) An asbestos pad is square in cross section, measuring 5 cm on a
side on its small end increasing linearly to 10 cm in a side at the large end. The
pad is 15 cm high. If the small end is held at 600K and the large end at 300K,
what heat-flow rate will be obtained if the four sides are insulated? Assume onedimensional heat conduction. The thermal conductivity of asbestos may be taken
as 0.173 W/m.K.
4. (WWWR#15.2) Solve Problem 3 for the case of the larger surface area exposed
to the higher temperature and the smaller end at 300K.
5. (WWWR#15.3) Solve Problem 3 if, in addition to a varying cross-sectional area,
the thermal conductivity varies according to
, where ko = 0.138,
, T is temperature in Kelvin, and k is in W/m.K. Compare this
result to that using a k value evaluated at the arithmetic mean temperature.
6. (Welty, Rorrer, Foster, 6th Edition International Student Version 15.1)
The Fourier Field equation in cylindrical coordinates is
a. What form does the equation reduce to for the case of a steady-state, radial heat
transfer?
b. Given the boundary conditions
T = Ti
at r = ri
T = To
at r = ro
c. Generate an expression for the heat flow rate, qr, using the result from part (b).
7. (WWWR # 16.2) Perform the same operations as in parts (a), (b) and (c) of
Problem 6 with respect to a spherical system.
(
)
8. ID 1.16 (Modified)
A square silicon chip (k = 150 W/m.K) is of width w = 5 mm on a side and of thickness t
= 1mm. The chip is mounted in a substrate such that its side and back surfaces are
insulated, while the front surface is exposed to a coolant.
If 4 W are being dissipated in circuits mounted to the back surface of the chip, what is the
steady-state temperature difference between the front and back surfaces?
9. ID 1.18 (Modified)
You’ve experience convection cooling if you’ve ever extended your hand out the window of
a moving vehicle or into a flowing stream of water. With the surface of your hand at a
temperature of 30oC, determine the convection heat flux for (a) a vehicle speed of 35 km/h in
air at with a convection coefficient of 40 W/m2.K and (b) a velocity of 0.2 m/s in a water
stream at with a convection coefficient of 900 W/m2.K. Which condition would feel colder?
Contrast these results with a heat loss of approximately 30 W/m 2 under normal room
conditions.
(1)
1340K
h=5110

fireclay
brick
Ri =
1
=
AiHi
Ro =
1
1
=
= 0.022
A0 H 0
(45)(1)
R1 =
x
0.106

 0.0938
kA (1.13)(1)
R2 =
0.00635
= 1.48x10-4
(42.9)(1)

mild
steel
1
=1.97 x 10 –4
(5110) 1
 R = Ri + Ro + R1 + R2 = 0.1161
q=
(1340  295)
= 9000W
0.1161
9000=
1340  T 1 T 1  T 2 T 2  T 3 T 3  295



Ri
R1
R2
R0
T1 = 1338K
T2 = 494K
T3 = 493K
295K
h=45
0.635cm
0.106m

(2)
From Question (1)
R  Ri  Ro  R1  R2  RCELOTEX
 0.1611  L / 0.069  0.1611  14.49 L
340  295
 2027W
0.022
(1340  295)
2027 
 L  0.0276m
0.1611  14.49 L
S=Ax+B
X = 0, S = 5
X = 15, S = 15
S = 5 + x/3 (cm)
q
10cm
(3)
S
x
S  5  cm
3
x
A  S 2  (5  )2 cm2
3
q  kA
dT
dx
qdx  kAdT
q
15
0
5cm
300K
y
z
600K
x
x
15cm
300
dx
 k  dT
2
600
(5  x / 3)

 15
3
q 
 |  300 K
 (5  x / 3)  0
q 0.6  0.3  300(0.173W / m.K )
Are the units consistent on the two
sides of this equation?
q  1.73W
(4)
Same value as in previous problem expect heat flowing in opposite direction
q  1.73W
(5)
15
q
0
600
dx
 5  x / 3
2
  k0
 (1   T )dT
300
3 

 

q 
 k0 T 1  (T1  T2 ) 

 5  x / 3 0
 2
 300
15
600
q[0.3cm1 ]  [0.138W /  m  K ][300 K ]  [1  1.95 104 ](450)
q  1.50W
The q value is the same as that obtained by using the thermal conductivity at the
arithmetic mean temperature.
(6)
In Cylindrical Coordinates
2
d T 1 dT

 0 or
dr 2 r dr
 dT 
d r
1  dr 
0
r
dr
dT
 c1
dr
T  C1 ln r  C2
r
B.C.
Ti  c1 ln ri  c 2  (i )
To  c1 ln ro  c 2  (ii)
by (i) & (ii)
(Ti  T 0)
ln(r 0 / ri )
C 2  Ti  C1 ln ri
C1  
ln(r / ri )
 (b)
ln(r 0 / ri )
dT
dT
q  kA
 k (2 rL)
dr
dr
2 kL
 2 kLC1 
(Ti  T 0) (c)
ln(r 0 / ri )
T  Ti  (Ti  T 0)
 (a)
(7)
In Spherical Coordinates:
1 d  2 dT
r
r 2 dr  dr
dT
r2
 C1
dr

  0    (a)

C
T   1  C2
r
C
B.C. Ti   1  C2
ri
C
To   1  C2
ro
T T
C1  i o
C2  Ti  C1 / ri
1/ ro  1/ ri
(1/ r  1/ ri )
(Ti  To )    (b)
(1/ ro  1/ ri )
dT
q  k (4 r 2 )
 4 kc1
dr
4 k

(Ti  To )    (c)
(1/ r0  1/ ri )
T  Ti 
(8) PROBLEM 1.16 (ID) (Modified)
A surface silicon chip (k=150 W/m•K) is of width w=5mm on a side and of thickness
t=1mm. The chip is mounted in a substrate such that its side and back surface are
insulated, while the front surface is exposed to a coolant.
Solution:
KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one
surface.
FIND: Temperature drop across the chip.
SCHEMATIC:
coolant
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat
dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction
in chip.
ANALYSIS: All of the electrical power dissipated at the back surface of the chip is
transferred by conduction through the chip. Hence, from Fourier’s law,
COMMENTS: For fixed P, the temperature drop across the chip decreases with
increasing k and W, as well as with decreasing t.
(9) PROBLEM 1.18 (ID)
You’ve experienced convection cooling if you’ve ever extended your hand out of a
moving vehicle or into a flow water stream. With the surface of your hand at a
temperature of 30℃, determine the convection heat flux for (a) a vehicle of speed of
35km/h in air at -5℃ with a convection coefficient of 40 W/m•K and (b) a velocity of 0.2
m/s in a water stream at 10℃ with a convection coefficient of 900 W/m•K. Which
condition will feel colder? Contrast these results with a heat loss of approximately 30
W/m2 under normal room conditions.
Solution:
KNOWN: Hand experiencing convection heat transfer with moving air and water.
FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30
W/m2 under normal room conditions.
SCHEMATIC:
ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection
coefficient is uniform over the hand, and (3) Negligible radiation exchange between hand and
surroundings in the case of air flow.
ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The
heat loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as
COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger
than when in the air stream for the given temperature and convection coefficient conditions. In
2
contrast, the heat loss in a normal room environment is only 30 W/m which is a factor of 400
times less than the loss in the air stream. In the room environment, the hand would feel
comfortable; in the air and water streams, as you probably know from experience, the hand would
feel uncomfortably cold since the heat loss is excessively high.