Solution Manual For Fundamentals of Heat and Mass Transfer, 8th Edition

Name: Solution Manual For Fundamentals of Heat and Mass Transfer, 8th Edition
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Author: Jack Murphy

Solution Manual For Fundamentals of Heat and Mass Transfer, 8th Edition is your textbook problem-solver, offering clear and concise solutions to difficult questions.

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PROBLEM 1.1
KNOWN: Temperature distribution in wall of Example 1.1.
FIND: Heat fluxes and heat rates at x = 0 and x = L.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through the wall, (2) constant thermal conductivity,
(3) no internal thermal energy generation within the wall.
PROPERTIES: Thermal conductivity of wall (given): k = 1.7 W/m·K.
ANALYSIS: The heat flux in the wall is by conduction and is described by Fourier’s law,
x
dT
q k dx
′′ = − (1)
Since the temperature distribution is T(x) = a + bx, the temperature gradient is
dT b
dx = (2)
Hence, the heat flux is constant throughout the wall, and is
( ) 2
1.7 W/m K 1000 K/m 1700 W/mx
dT
q k kb
dx
′′ = − = − = − ⋅ × − = <
Since the cross-sectional area through which heat is conducted is constant, the heat rate is constant and is
( )2
( ) 1700 W/m 1.2 m × 0.5 m 1020 Wx xq q W H′′= × × = × = <
Because the heat rate into the wall is equal to the heat rate out of the wall, steady-state conditions exist. <
COMMENTS: (1) If the heat rates were not equal, the internal energy of the wall would be changing
with time. (2) The temperatures of the wall surfaces are T1 = 1400 K and T2 = 1250 K.

PROBLEM 1.2
KNOWN: Thermal conductivity, thickness and temperature difference across a sheet of rigid
extruded insulation.
FIND: (a) The heat flux through a 3 m × 3 m sheet of the insulation, (b) the heat rate through
the sheet, and (c) the thermal conduction resistance of the sheet.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: (a) From Equation 1.2 the heat flux is
1 2
x 2
T - TdT W 12 K W
q = -k = k = 0.029 × = 13.9
dx L m K 0.025 m m
′′ ⋅ <
(b) The heat rate is
2
x x 2
W
q = q A = 13.9 × 9 m = 125 W
m
′′ ⋅ <
(c) From Eq. 1.11, the thermal resistance is
t,cond xR T / q = 12 K /125 W 0.096 K/W= ∆ = <
COMMENTS: (1) Be sure to keep in mind the important distinction between the heat flux
(W/m2) and the heat rate (W). (2) The direction of heat flow is from hot to cold. (3) Note that
a temperature difference may be expressed in kelvins or degrees Celsius. (4) The conduction
thermal resistance for a plane wall could equivalently be calculated from Rt,cond = L/kA.
qcond
A = 4 m 2
T 2T 1
k = 0.029
x
L = 20 mm
T 1 – T 2 = 10˚C
qcond
A = 4 m 2
T 2T 1
k = 0.029
x
L = 20 mm
T 1 – T 2 = 10˚C
9 m 2
12 °C
25 mm
qcond
A = 4 m 2
T 2T 1
k = 0.029
x
L = 20 mm
T 1 – T 2 = 10˚C
qcond
A = 4 m 2
T 2T 1
k = 0.029
x
L = 20 mm
T 1 – T 2 = 10˚C
9 m 2
12 °C
25 mm

PROBLEM 1.3
KNOWN: Thickness and thermal conductivity of a wall. Heat flux applied to one face and
temperatures of both surfaces.
FIND: Whether steady-state conditions exist.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties, (3) No internal energy
generation.
ANALYSIS: Under steady-state conditions an energy balance on the control volume shown is
2
in out cond 1 2( ) / 12 W/m K(50 C 30 C) / 0.01 m 24,000 W/mq q q k T T L′′ ′′ ′′= = = − = ⋅ ° − ° =
Since the heat flux in at the left face is only 20 W/m2, the conditions are not steady state. <
COMMENTS: If the same heat flux is maintained until steady-state conditions are reached, the
steady-state temperature difference across the wall will be
∆T = 2
/ 20 W/m 0.01 m /12 W/m K 0.0167 Kq L k′′ = × ⋅ =
which is much smaller than the specified temperature difference of 20°C.
q” = 20 W/m2
L = 10 mm
k = 12 W/m∙KT1 = 50°C
T2 = 30°C
q″cond

PROBLEM 1.4
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall.
FIND: Heat loss by conduction through the wall as a function of outer surface temperatures ranging
from -15 to 38°C.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)
Constant properties.
ANALYSIS: From Fourier’s law, if xq′′ and k are each constant it is evident that the gradient,
xdT dx q k′′= − , is a constant, and hence the temperature distribution is linear. The heat flux must be
constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends
only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C
are
( ) 21 2
x
25 C 15 CdT T T
q k k 1W m K 133.3 W m
dx L 0.30 m
− −−
′′ = − = = ⋅ =
 
. (1)
2 2
x xq q A 133.3 W m 20 m 2667 W′′= × = × = . (2) <
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of outer surface temperature,
-15 ≤ T2 ≤ 38°C, with different wall thermal conductivities, k.
For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearly from +2667 W to -867 W and is zero
when the inside and outer surface temperatures are the same. The magnitude of the heat rate increases
with increasing thermal conductivity.
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane
wall would not be linear.
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
-1500
-500
500
1500
2500
3500
Heat loss, qx (W)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
Outside surface
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
-1500
-500
500
1500
2500
3500
Heat loss, qx (W)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
Outside surface

PROBLEM 1.5
KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency
of gas furnace and cost of natural gas.
FIND: Daily cost of heat loss.
SCHEMATIC:
Furnace, = 0.90ηf
Natural gas,
C = $0.01/MJg Concrete, k = 1.4 W/m-K
t = 0.2 m
L = 11 m
W = 8 m
T = 17 C1 o
T = 10 C2 o
q
Warm air
ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.
ANALYSIS: The rate of heat loss by conduction through the slab is
( ) ( )1 2T T 7 C
q k LW 1.4 W / m K 11m 8 m 4312 W
t 0.20 m
− °
= = ⋅ × = <
The daily cost of natural gas that must be combusted to compensate for the heat loss is
( ) ( )g
d 6
f
q C 4312 W $0.02 / MJ
C t 24 h / d 3600s / h $8.28 / d
0.9 10 J / MJ
η
×
= ∆ = × =
×
<
COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation
between it and the concrete.

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PROBLEM 1.6
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed
thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be
determined from Fourier’s law, Eq. 1.2. Rearranging,
( )
L W 0.05m
k=q 40
T T m 40-20 C
x 21 2
′′ =
- 
k = 0.10 W / m K.⋅ <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.

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PROBLEM 1.7
KNOWN: Inner and outer surface temperatures and thermal resistance of a glass window of
prescribed dimensions.
FIND: Heat loss through window. Thermal conductivity of glass.
SCHEMATIC:
A =
1m x 3m = 3m2,
R t,cond =
1.19 x 10 -3 K/W
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: From Eq. 1.11,
( )3
t,cond
15-5 CT T
q 8400 W
R 1.19 10 K/W
1 2
x -
-
= = =
×

<
The thermal resistance due to conduction for a plane wall is related to the thermal
conductivity and dimensions according to
t,condR = L/kA
Therefore
3 2
t,condk L/(R A) 0.005 m / (1.19 10 K/W 3 m ) 1.40 W/m K-
= = × × = ⋅ <
COMMENTS: The thermal conductivity value agrees with the value for glass in Table A.3.

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PROBLEM 1.8
KNOWN: Net power output, average compressor and turbine temperatures, shaft dimensions and
thermal conductivity.
FIND: (a) Comparison of the conduction rate through the shaft to the predicted net power output of
the device, (b) Plot of the ratio of the shaft conduction heat rate to the anticipated net power output of
the device over the range 0.005 m ≤ L ≤ 1 m and feasibility of a L = 0.005 m device.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Net power output is
proportional to the volume of the gas turbine.
PROPERTIES: Shaft (given): k = 40 W/m⋅K.
ANALYSIS: (a) The conduction through the shaft may be evaluated using Fourier’s law, yielding
( ) ( )2 3 2( ) 40W/m K(1000 400) C
" / 4 (70 10 m) / 4 92.4W
1m
h c
c
k T T
q q A d
L
π
π −− ⋅ − °
= = = × =
The ratio of the conduction heat rate to the net power output is
6
6
92.4W 18.5 10
5 10 W
q
r P
−
= = = ×
× <
(b) The volume of the turbine is proportional to L3. Designating La = 1 m, da = 70 mm and Pa as the
shaft length, shaft diameter, and net power output, respectively, in part (a),
d = da × (L/La); P = Pa × (L/La)3
and the ratio of the conduction heat rate to the net power output is
( ) ( )( )22 2
3 2
3 2 6 6 2
2 2
( ) ( ) ( )
/ 4 / / 4 /" 4
( / )
40W/m K(1000 400) C (70 10 m) 1m / 5 10 W 18.5 10 m4=
h c h c h c
a a a a a
c
a a
k T T k T T k T T
d d L L d L Pq A L Lr P P P L L L
L L
π
π
π
π − −
− − −
= = = =
⋅ − ° × × × ×
=
Continued…

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PROBLEM 1.8 (Cont.)
The ratio of the shaft conduction to net power is shown below. At L = 0.005 m = 5 mm, the shaft
conduction to net power output ratio is 0.74. The concept of the very small turbine is not feasible since
it will be unlikely that the large temperature difference between the compressor and turbine can be
maintained. <
COMMENTS: (1) The thermodynamics analysis does not account for heat transfer effects and is
therefore meaningful only when heat transfer can be safely ignored, as is the case for the shaft in part
(a). (2) Successful miniaturization of thermal devices is often hindered by heat transfer effects that
must be overcome with innovative design.
Ratio of shaft conduction to net power
0 0.2 0.4 0.6 0.8 1
L (m)
0.0001
0.001
0.01
0.1
1
r

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PROBLEM 1.9
KNOWN: Heat flux at one face and air temperature and convection coefficient at other face of plane
wall. Temperature of surface exposed to convection.
FIND: If steady-state conditions exist. If not, whether the temperature is increasing or decreasing.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) No internal energy generation.
ANALYSIS: Conservation of energy for a control volume around the wall gives
st
in out g
dE E E E
dt = − +  
[ ]st
in in
2 2 2
( ) ( )
20 W/m 20 W/m K(50 C 30 C) 380 W/m
s s
dE q A hA T T q h T T A
dt
A A
∞ ∞
′′ ′′= − − = − −
= − ⋅ ° − ° = −  
Since dEst/dt ≠ 0, the system is not at steady-state. <
Since dEst/dt < 0, the stored energy is decreasing, therefore the wall temperature is decreasing. <
COMMENTS: When the surface temperature of the face exposed to convection cools to 31°C, qin =
qout and dEst/dt = 0 and the wall will have reached steady-state conditions.
q” = 20 W/m2
Ts = 50°C
h = 20 W/m2∙K
T∞ = 30°C
Air
q” conv

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PROBLEM 1.10
KNOWN: Expression for variable thermal conductivity of a wall. Constant heat flux.
Temperature at x = 0.
FIND: Expression for temperature gradient and temperature distribution.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction.
ANALYSIS: The heat flux is given by Fourier’s law, and is known to be constant, therefore
x
dT
q k constant
dx
′′ = − =
Solving for the temperature gradient and substituting the expression for k yields
x xq qdT
dx k ax b
′′ ′′
= − = − + <
This expression can be integrated to find the temperature distribution, as follows:
xqdT dx dx
dx ax b
′′
= − +∫ ∫
Since xq constant′′ = , we can integrate the right hand side to find
( )xq
T ln ax b c
a
′′
= − + +
where c is a constant of integration. Applying the known condition that T = T1 at x = 0,
we can solve for c.
Continued…
q”
x
k = ax + b
T1

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PROBLEM 1.10 (Cont.)
1
x
1
x
1
T(x 0) T
q ln b c T
a
q
c T ln b
a
= =
′′
− + =
′′
= +
Therefore, the temperature distribution is given by
( )x x
1
q q
T ln ax b T ln b
a a
′′ ′′
= − + + + <
x
1
q b
T ln
a ax b
′′
= + + <
COMMENTS: Temperature distributions are not linear in many situations, such as when the
thermal conductivity varies spatially or is a function of temperature. Non-linear temperature
distributions may also evolve if internal energy generation occurs or non-steady conditions exist.

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PROBLEM 1.11
KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil
water. Rate of heat transfer to the pan.
FIND: Outer surface temperature of pan for an aluminum and a copper bottom.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan.
ANALYSIS: From Fourier’s law, the rate of heat transfer by conduction through the bottom
of the pan is
T T1 2q kA L
−
=
Hence,
qL
T T1 2 kA
= +
where ( )22 2A D / 4 0.22m / 4 0.038 m .
π
π= = =
Aluminum: ( )
( )
600W 0.008 m
T 110 C 110.5 C1 2240 W/m K 0.038 m
= + =
⋅
  <
Copper: ( )
( )
600W 0.008 m
T 110 C 110.3 C1 2390 W/m K 0.038 m
= + =
⋅
  <
COMMENTS: Although the temperature drop across the bottom is slightly larger for
aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for
both materials. To a good approximation, the bottom may be considered isothermal at T ≈
110 °C, which is a desirable feature of pots and pans.
Aluminum
(k=240 W/m-K)
or
Copper
(k=390 W/m-K)
L = 5 mm
D = 200 mm
q = 600 W
T 1
T = 110 C2 o
L = 8 mm
D = 220 mm

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PROBLEM 1.12
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
( )sq h T T∞′′ = −
For the air stream:
( )2 2
airq 40 W m K 30 8 K 1,520 W m′′ = ⋅ − − =   <
For the water stream:
( )2 2
waterq 900 W m K 30 10 K 18, 000 W m′′ = ⋅ − = <
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 contrast, the heat
loss in a normal room environment is only 30 W/m2 which is a factor of 50 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.

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PROBLEM 1.13
KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder
with an imbedded electrical heater for different air velocities.
FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display
the results graphically, and (b) Assuming that the convection coefficient depends upon air velocity as
h = CVn, determine the parameters C and n.
SCHEMATIC:
V(m/s) 1 2 4 8 12
¢Pe (W/m) 450 658 983 1507 1963
h (W/m2⋅K) 22.0 32.2 48.1 73.8 96.1
ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation
exchange between the cylinder surface and the surroundings, (3) Steady-state conditions.
ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by the
electrical heater is transferred by convection to the air stream. Using Newton’s law of cooling on a
per unit length basis,
( )( )e sP h D T T
π ∞′ = −
where eP′ is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s
condition, using the data from the table above, find
( ) 2h 450 W m 0.025 m 300 40 C 22.0 W m K
π= × − = ⋅
 <
Repeating the calculations, find the convection coefficients for the remaining conditions which are
tabulated above and plotted below. Note that h is not linear with respect to the air velocity.
(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C =
22.12 W/m2⋅K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope of
the h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable
choice. Hence, C = 22.12 and n = 0.6. <
0 2 4 6 8 10 12
Air velocity, V (m/s)
20
40
60
80
100
Coefficient, h (W/m^2.K)
Data, smooth curve, 5-points
1 2 4 6 8 10
Air velocity, V (m/s)
10
20
40
60
80
100
Coefficient, h (W/m^2.K)
Data , smooth curve, 5 points
h = C * V^n, C = 22.1, n = 0.5
n = 0.6
n = 0.8
COMMENTS: Radiation may not be negligible, depending on surface emissivity.

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