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
, 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 q cond
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
dT
q = -k T1 - T2
x =k
dx L
Solving,
W
q"x = 0.029 10 K
×
m K 0.02 m
W
q′x′ = 14.5 <
2
m
The heat rate is
q =q
W
A= × 4 m2 = 58 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.
, 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,
q x
dT dx k , is a constant, and hence the temperature distribution is linear. The heat flux must be
q x
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 15 C
2
1W 133.3 W . (1)
q x k m K 0.30 m m
dx
k
L
qx q x A 133.3 2 20 2667 W . (2) <
W m m2
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
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient
iAmbient
Outside deair temperature, T2 (C)
surface
surfaceair
Wall thermal conductivity, k = 1.25 W/m.K k
= 1 W/m.K, concrete wall
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
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.
,
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
, 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 q cond
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
dT
q = -k T1 - T2
x =k
dx L
Solving,
W
q"x = 0.029 10 K
×
m K 0.02 m
W
q′x′ = 14.5 <
2
m
The heat rate is
q =q
W
A= × 4 m2 = 58 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.
, 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,
q x
dT dx k , is a constant, and hence the temperature distribution is linear. The heat flux must be
q x
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 15 C
2
1W 133.3 W . (1)
q x k m K 0.30 m m
dx
k
L
qx q x A 133.3 2 20 2667 W . (2) <
W m m2
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
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient
iAmbient
Outside deair temperature, T2 (C)
surface
surfaceair
Wall thermal conductivity, k = 1.25 W/m.K k
= 1 W/m.K, concrete wall
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
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.
