All 14 Chapters Covered
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
Appendices
, 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
q = -k dT T -T
x =k 1 2
dx L
Solving,
W
q" = 0.029 ×
10 K
x
m K 0.02 m
W
q′x′ = 14.5 <
m2
The heat rate is
q = q A = 14.5
W × 4 m2 = 58 W <
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.
, PROBLEM 1.2
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:
L = 10 mm
T2 = 30 C
q” = 20 W/m2
qcond
T1 = 50 C k = 12 W/m·K
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
qin qout qcond k(T1 T2 ) / L 12 W/m K(50 C 30 C) / 0.01 m 24,000 W/m2
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
ΔT = q L / k 20 W/m 2 0.01 m /12 W/m K 0.0167 K
which is much smaller than the specified temperature difference of 20 C.
, PROBLEM 1.3
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 q x and k are each constant it is evident that the gradient,
dT dx q x 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
dT T1 T2
25 C 15 C
o 0.30 m o 2
1W m K 133.3 W m . (1)
qx k k
dx L
q x q x A 133.3 W m 2 20 m 2
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.
3500
2500
1500
Heat loss, qx (W)
500
-500
-1500 -20 -10 0 10 20 30 40
Outside surface temperature, T2 (C)
Wall thermal conductivity,
iAmbient
Ambient k d=eair
1.25 W/m.K k = 1
W/m.K, concrete wall
surface air
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.
, PROBLEM 1.4
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
T T
q k LW 1 2 1.4 W / m K 11m 8 m 7C 4312 W <
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 $0.02 / MJ 24 h / d 3600s / h $8.28 / d <
Δt
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.
, PROBLEM 1.5
KNOWN: Thermal conductivity and thickness of a wall. Heat flux through wall. Steady-state conditions.
FIND: Value of temperature gradient in K/m and in C/m.
SCHEMATIC:
k = 2.3 W/m·K
q”x = 10 W/m2
x L = 20 mm
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.
ANALYSIS: Under steady-state conditions,
dT q" 10 W/m2
x
4.35 K/m 4.35 C/m <
dx k 2.3 W/m K
Since the K units here represent a temperature difference, and since the temperature difference is the same in K
and C units, the temperature gradient value is the same in either units.
COMMENTS: A negative value of temperature gradient means that temperature is decreasing with increasing x,
corresponding to a positive heat flux in the x-direction.
, 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,
k=q L
40 W 0.05m
xT T
m2 40-20 oC
1 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.
, PROBLEM 1.7
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 T2 L
W
15-5 o C
q x 1.4 m K 0.005m
q x 2800 W/m 2 .
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W / m2 3m2
q = 8400 W. <
COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions.
, PROBLEM 1.8
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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
Appendices
, 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
q = -k dT T -T
x =k 1 2
dx L
Solving,
W
q" = 0.029 ×
10 K
x
m K 0.02 m
W
q′x′ = 14.5 <
m2
The heat rate is
q = q A = 14.5
W × 4 m2 = 58 W <
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.
, PROBLEM 1.2
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:
L = 10 mm
T2 = 30 C
q” = 20 W/m2
qcond
T1 = 50 C k = 12 W/m·K
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
qin qout qcond k(T1 T2 ) / L 12 W/m K(50 C 30 C) / 0.01 m 24,000 W/m2
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
ΔT = q L / k 20 W/m 2 0.01 m /12 W/m K 0.0167 K
which is much smaller than the specified temperature difference of 20 C.
, PROBLEM 1.3
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 q x and k are each constant it is evident that the gradient,
dT dx q x 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
dT T1 T2
25 C 15 C
o 0.30 m o 2
1W m K 133.3 W m . (1)
qx k k
dx L
q x q x A 133.3 W m 2 20 m 2
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.
3500
2500
1500
Heat loss, qx (W)
500
-500
-1500 -20 -10 0 10 20 30 40
Outside surface temperature, T2 (C)
Wall thermal conductivity,
iAmbient
Ambient k d=eair
1.25 W/m.K k = 1
W/m.K, concrete wall
surface air
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.
, PROBLEM 1.4
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
T T
q k LW 1 2 1.4 W / m K 11m 8 m 7C 4312 W <
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 $0.02 / MJ 24 h / d 3600s / h $8.28 / d <
Δt
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.
, PROBLEM 1.5
KNOWN: Thermal conductivity and thickness of a wall. Heat flux through wall. Steady-state conditions.
FIND: Value of temperature gradient in K/m and in C/m.
SCHEMATIC:
k = 2.3 W/m·K
q”x = 10 W/m2
x L = 20 mm
ASSUMPTIONS: (1) One-dimensional conduction, (2) Constant properties.
ANALYSIS: Under steady-state conditions,
dT q" 10 W/m2
x
4.35 K/m 4.35 C/m <
dx k 2.3 W/m K
Since the K units here represent a temperature difference, and since the temperature difference is the same in K
and C units, the temperature gradient value is the same in either units.
COMMENTS: A negative value of temperature gradient means that temperature is decreasing with increasing x,
corresponding to a positive heat flux in the x-direction.
, 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,
k=q L
40 W 0.05m
xT T
m2 40-20 oC
1 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.
, PROBLEM 1.7
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 T2 L
W
15-5 o C
q x 1.4 m K 0.005m
q x 2800 W/m 2 .
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W / m2 3m2
q = 8400 W. <
COMMENTS: A linear temperature distribution exists in the glass for the prescribed conditions.
, PROBLEM 1.8
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