SOLUTION MANUAL
,TABLE OF CONTENTS
1. Introduction
2. Introduction to Conduction
3. One-Dimensional, Steady-State Conduction
4. Two-Dimensional, Steady-State Conduction
5. Transient Conduction
6. Introduction to Convection
7. External Flow
8. Internal Flow
9. Free Convection
10. Boiling and Condensation
11. Heat Exchangers
12. Radiation: Processes and Properties
13. Radiation Exchange Between Surfaces
14. Diffusion Mass Transfer
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PROBLEM 1.1
KNOWN: Thermal conductivity, thickness and temperature difference across a sheet
of rigid extruded insulation.
FIND: (a) The heat flux through a 2 m 2 m sheet of the insulation, and (b) The heat
rate through the sheet.
SCHEMATIC:
A = 4 m2
W
k = 0.029 mK
qcond
T1 – T2 = 10˚C
T1 T2
L = 20 mm
x
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: From Equation 1.2 the heat flux is
q = -k T1 - T2
dT =k
x
dx L
Solving,
q" =
W
0.029
×
10 K
x
m K 0.02 m
W
q′x′ = 14.5 <
m2
The heat rate is
q =q A=
× 4 m2 = 58 W <
W
14.5
x x
m2
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.
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PROBLEM 1.2
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 and k are each constant it is evident that the gradient,
qx
dT dx k , is a constant, and hence the temperature distribution is linear. The heat flux must be
qx
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
dT T1 T2
25 C C
15
2
1W 0.30 133.3 W . (1)
qx k k
dx L m K m m
q x q x A 133.3 W 2 20
m 2
2667 W . (2) <
m
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.
3500
2500
1500
500
Heat loss, qx (W)
-500
-1500 -20 -10 0 10 20 30 40
Outside surface temperature, T2 (C)
Wall thermal conductivity,
Ambientdeairk = 1.25 W/m.K
iAmbient
k = 1 W/m.K, concreteairwall
surface
k = 0.75 W/m.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.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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PROBLEM 1.3
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:
ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.
ANALYSIS: The rate of heat loss by conduction through the slab is
T1 T2
q k LW 1.4 W / m K 7 4312 W <
C
11m 8 m
t 0.20 m
The daily cost of natural gas that must be combusted to compensate for the heat loss is
Cd q Cg 4312 W 24 h / d 3600 s / h $4.14 / d <
Δt
$0.01/ MJ
f 0.9 106 J / MJ
COMMENTS: The loss could be reduced by installing a floor covering with a layer of
insulation between it and the concrete.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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PROBLEM 1.4
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,
k=q L
W 0.05m
4
0
xT T
1 2 m2 40-20 C
k = 0.10 W / m K. <
COMMENTS: Note that the C or K temperature units may be used interchangeably
when evaluating a temperature difference.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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PROBLEM 1.5
KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.
FIND: Heat loss through window.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from
Fourier’s law, Eq. 1.2.
q′x′ k T1
T
15-5 C
2L
W
qx 1.4 m K 0.005m
qx 2800 W/m 2 .
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W /
3m2
m2 q = 8400 <
W.
COMMENTS: A linear temperature distribution exists in the glass for the prescribed
conditions.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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PROBLEM 1.6
KNOWN: Width, height, thickness and thermal conductivity of a single pane window
and the air space of a double pane window. Representative winter surface
temperatures of single pane and air space.
FIND: Heat loss through single and double pane windows.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-
state conditions, (3) Enclosed air of double pane window is stagnant (negligible
buoyancy induced motion).
ANALYSIS: From Fourier’s law, the heat losses are
Single
Pane:
qg
T
kgA 1
T
2 1.4 W/m
K 2m
2
35
C
19, 600 W <
L 0.005m
Double
Pane:
qa
T
ka A 1
T
2 0.024 2m
2
C
25 120 W <
L 0.010 m
COMMENTS: Losses associated with a single pane are unacceptable and would
remain excessive, even if the thickness of the glass were doubled to match that of
the air space. The principal advantage of the double pane construction resides with
the low thermal conductivity of air (~ 60 times smaller than that of glass). For a fixed
ambient outside air temperature, use of the double pane construction would also
increase the surface temperature of the glass exposed to the room (inside) air.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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PROBLEM 1.7
KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.
FIND: Thickness of styrofoam insulation needed to maintain heat load below
prescribed value.
SCHEMATIC:
ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5
2
walls of area A = , (3) Steady-state conditions, (4) Constant properties.
4m
ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is
q = q A ΔT
= k L total
A
Solving for L and recognizing that Atotal = 5 , find
W
2
5 k Δ T2 W
L=
q
⎡⎣35 - -10 ⎤⎦ C 4m2
5 0.03
W/m K
L=
500 W
L = 0.054m = 54mm. <
COMMENTS: The corners will cause local departures from one-dimensional
conduction and a slightly larger heat loss.
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courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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,TABLE OF CONTENTS
1. Introduction
2. Introduction to Conduction
3. One-Dimensional, Steady-State Conduction
4. Two-Dimensional, Steady-State Conduction
5. Transient Conduction
6. Introduction to Convection
7. External Flow
8. Internal Flow
9. Free Convection
10. Boiling and Condensation
11. Heat Exchangers
12. Radiation: Processes and Properties
13. Radiation Exchange Between Surfaces
14. Diffusion Mass Transfer
, www.konkur.in
PROBLEM 1.1
KNOWN: Thermal conductivity, thickness and temperature difference across a sheet
of rigid extruded insulation.
FIND: (a) The heat flux through a 2 m 2 m sheet of the insulation, and (b) The heat
rate through the sheet.
SCHEMATIC:
A = 4 m2
W
k = 0.029 mK
qcond
T1 – T2 = 10˚C
T1 T2
L = 20 mm
x
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: From Equation 1.2 the heat flux is
q = -k T1 - T2
dT =k
x
dx L
Solving,
q" =
W
0.029
×
10 K
x
m K 0.02 m
W
q′x′ = 14.5 <
m2
The heat rate is
q =q A=
× 4 m2 = 58 W <
W
14.5
x x
m2
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.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
, www.konkur.in
PROBLEM 1.2
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 and k are each constant it is evident that the gradient,
qx
dT dx k , is a constant, and hence the temperature distribution is linear. The heat flux must be
qx
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
dT T1 T2
25 C C
15
2
1W 0.30 133.3 W . (1)
qx k k
dx L m K m m
q x q x A 133.3 W 2 20
m 2
2667 W . (2) <
m
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.
3500
2500
1500
500
Heat loss, qx (W)
-500
-1500 -20 -10 0 10 20 30 40
Outside surface temperature, T2 (C)
Wall thermal conductivity,
Ambientdeairk = 1.25 W/m.K
iAmbient
k = 1 W/m.K, concreteairwall
surface
k = 0.75 W/m.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.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
, www.konkur.in
PROBLEM 1.3
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:
ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties.
ANALYSIS: The rate of heat loss by conduction through the slab is
T1 T2
q k LW 1.4 W / m K 7 4312 W <
C
11m 8 m
t 0.20 m
The daily cost of natural gas that must be combusted to compensate for the heat loss is
Cd q Cg 4312 W 24 h / d 3600 s / h $4.14 / d <
Δt
$0.01/ MJ
f 0.9 106 J / MJ
COMMENTS: The loss could be reduced by installing a floor covering with a layer of
insulation between it and the concrete.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
, www.konkur.in
PROBLEM 1.4
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,
k=q L
W 0.05m
4
0
xT T
1 2 m2 40-20 C
k = 0.10 W / m K. <
COMMENTS: Note that the C or K temperature units may be used interchangeably
when evaluating a temperature difference.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
, www.konkur.in
PROBLEM 1.5
KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.
FIND: Heat loss through window.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from
Fourier’s law, Eq. 1.2.
q′x′ k T1
T
15-5 C
2L
W
qx 1.4 m K 0.005m
qx 2800 W/m 2 .
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W /
3m2
m2 q = 8400 <
W.
COMMENTS: A linear temperature distribution exists in the glass for the prescribed
conditions.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
, www.konkur.in
PROBLEM 1.6
KNOWN: Width, height, thickness and thermal conductivity of a single pane window
and the air space of a double pane window. Representative winter surface
temperatures of single pane and air space.
FIND: Heat loss through single and double pane windows.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-
state conditions, (3) Enclosed air of double pane window is stagnant (negligible
buoyancy induced motion).
ANALYSIS: From Fourier’s law, the heat losses are
Single
Pane:
qg
T
kgA 1
T
2 1.4 W/m
K 2m
2
35
C
19, 600 W <
L 0.005m
Double
Pane:
qa
T
ka A 1
T
2 0.024 2m
2
C
25 120 W <
L 0.010 m
COMMENTS: Losses associated with a single pane are unacceptable and would
remain excessive, even if the thickness of the glass were doubled to match that of
the air space. The principal advantage of the double pane construction resides with
the low thermal conductivity of air (~ 60 times smaller than that of glass). For a fixed
ambient outside air temperature, use of the double pane construction would also
increase the surface temperature of the glass exposed to the room (inside) air.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
, www.konkur.in
PROBLEM 1.7
KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.
FIND: Thickness of styrofoam insulation needed to maintain heat load below
prescribed value.
SCHEMATIC:
ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5
2
walls of area A = , (3) Steady-state conditions, (4) Constant properties.
4m
ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is
q = q A ΔT
= k L total
A
Solving for L and recognizing that Atotal = 5 , find
W
2
5 k Δ T2 W
L=
q
⎡⎣35 - -10 ⎤⎦ C 4m2
5 0.03
W/m K
L=
500 W
L = 0.054m = 54mm. <
COMMENTS: The corners will cause local departures from one-dimensional
conduction and a slightly larger heat loss.
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in
courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976
United States Copyright Act without the permission of the copyright owner is unlawful.
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CLICK ON THE L.I.N.K
ON THE NEXT PAGE
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