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
m ⋅K 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
dT T -T
q′′x = -k =k 1 2
dx L
Solving,
W 10 K
q"x = 0.029 ×
m⋅K 0.02 m
W
q′′x = 14.5 <
m2
The heat rate is
W
q x = q′′x ⋅ A = 14.5 2
× 4 m 2 = 58 W <
m
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
q″cond
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
′′ = qout
qin ′′ = qcond
′′ = k (T1 − T2 ) / L = 12 W/m ⋅ K(50°C − 30°C) / 0.01 m = 24,000 W/m 2
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
q′′x = − k
dT
=k
T1 − T2
= 1W m ⋅ K
25o C − −15o C
= 133.3 W m 2 .
( )
(1)
dx L 0.30 m
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
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient
Outside air temperature, T2 (C)
surface
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
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 7°C
q = k ( LW ) 1 2 = 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
q Cg 4312 W × $0.02 / MJ
Cd = ( Δt ) = ( 24 h / d × 3600s / h ) = $8.28 / d <
ηf 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/m 2
dx
=− x =−
k 2.3 W/m ⋅ K
= −4.35 K/m = −4.35 °C/m <
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,
L W 0.05m
k=q′′x = 40
T1 − T2 m2 ( 40-20 )o 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.
, 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.
T −T
q′′x = k 1 2
L
o
W (15-5 ) 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 = q ′′x × 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
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:
Combustion
Compressor chamber Turbine
d = 70 mm
Tc = 400°C
Th = 1000°C
Shaft P = 5 MW
k = 40 W/m·K
L = 1m
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
q = q " Ac =
L
(
k (Th − Tc )
πd2 /4 = )
40W/m ⋅ K(1000 − 400)°C
1m
(
π (70 × 10−3 m) = 92.4W )
The ratio of the conduction heat rate to the net power output is
q 92.4W
r= = = 18.5 × 10−6 <
P 5 × 10 W
6
(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
r=
q " Ac
=
k (Th − Tc )
L
(
πd2 /4
=
)
k (Th − Tc )
L
(
π ( d a L / La ) / 4
2
=
)
k (Th − Tc )π 2
4
d a La / Pa
P P Pa ( L / La )3 L2
40W/m ⋅ K(1000 − 400)°Cπ
(70 × 10−3 m)2 × 1m / 5 × 106 W
4 18.5 × 10−6 m2
= =
L2 L2
Continued…
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
m ⋅K 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
dT T -T
q′′x = -k =k 1 2
dx L
Solving,
W 10 K
q"x = 0.029 ×
m⋅K 0.02 m
W
q′′x = 14.5 <
m2
The heat rate is
W
q x = q′′x ⋅ A = 14.5 2
× 4 m 2 = 58 W <
m
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
q″cond
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
′′ = qout
qin ′′ = qcond
′′ = k (T1 − T2 ) / L = 12 W/m ⋅ K(50°C − 30°C) / 0.01 m = 24,000 W/m 2
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
q′′x = − k
dT
=k
T1 − T2
= 1W m ⋅ K
25o C − −15o C
= 133.3 W m 2 .
( )
(1)
dx L 0.30 m
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
Heat loss, qx (W)
1500
500
-500
-1500
-20 -10 0 10 20 30 40
Ambient
Outside air temperature, T2 (C)
surface
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
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 7°C
q = k ( LW ) 1 2 = 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
q Cg 4312 W × $0.02 / MJ
Cd = ( Δt ) = ( 24 h / d × 3600s / h ) = $8.28 / d <
ηf 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/m 2
dx
=− x =−
k 2.3 W/m ⋅ K
= −4.35 K/m = −4.35 °C/m <
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,
L W 0.05m
k=q′′x = 40
T1 − T2 m2 ( 40-20 )o 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.
, 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.
T −T
q′′x = k 1 2
L
o
W (15-5 ) 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 = q ′′x × 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
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:
Combustion
Compressor chamber Turbine
d = 70 mm
Tc = 400°C
Th = 1000°C
Shaft P = 5 MW
k = 40 W/m·K
L = 1m
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
q = q " Ac =
L
(
k (Th − Tc )
πd2 /4 = )
40W/m ⋅ K(1000 − 400)°C
1m
(
π (70 × 10−3 m) = 92.4W )
The ratio of the conduction heat rate to the net power output is
q 92.4W
r= = = 18.5 × 10−6 <
P 5 × 10 W
6
(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
r=
q " Ac
=
k (Th − Tc )
L
(
πd2 /4
=
)
k (Th − Tc )
L
(
π ( d a L / La ) / 4
2
=
)
k (Th − Tc )π 2
4
d a La / Pa
P P Pa ( L / La )3 L2
40W/m ⋅ K(1000 − 400)°Cπ
(70 × 10−3 m)2 × 1m / 5 × 106 W
4 18.5 × 10−6 m2
= =
L2 L2
Continued…
