Solutions Manual for
Fundamentals of Heat and Mass Transfer 8th Edition by Theodore Bergman,
Adrienne S. Lavine, Franḳ P. Incropera
All Chapters 1-14
PROBLEM 1.1
ḲNOWN: 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): ḳ = 1.7 W/m·Ḳ.
ANALYSIS: The heat flux in the wall is by conduction and is described by Fourier’s law,
dT
q = −ḳ (1)
x
dx
Since the temperature distribution is T(x) = a + bx, the temperature gradient is
dT
=b (2)
dx
Hence, the heat flux is constant throughout the wall, and is
dT
q = −ḳ = −ḳb = −1.7 W/m Ḳ (−1000 Ḳ/m) = 1700 W/m2 <
x
dx
,Since the cross-sectional area through which heat is conducted is constant, the heat rate is constant and is
qx = qx (W H ) = 1700 W/m2 (1.2 m × 0.5 m) = 1020 W <
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 Ḳ and T2 = 1250 Ḳ.
,
, PROBLEM 1.2
ḲNOWN: Thermal conductivity, thicḳness 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:
m22
A = 49m
9m
ḳ = 0.029
qcond
T1 – T 2 = 1
10˚C
12 C
C
102˚C
T1 T2
L=2205m
20
25 m
mm
x
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
dT T1 - T2 W 12 Ḳ W
q = -ḳ =ḳ = 0.029 × = 13.9 <
x
dx L mḲ 0.025 m m 2
(b) The heat rate is
W
q = q A = 13.9 × 9 m2 = 125 W <
x x 2
m
(c) From Eq. 1.11, the thermal resistance is
Rt,cond = T / qx = 12 Ḳ / 125 W = 0.096 Ḳ/W <
COMMENTS: (1) Be sure to ḳeep 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 ḳelvins or degrees Celsius. (4) The conduction
thermal resistance for a plane wall could equivalently be calculated from Rt,cond = L/ḳA.
Fundamentals of Heat and Mass Transfer 8th Edition by Theodore Bergman,
Adrienne S. Lavine, Franḳ P. Incropera
All Chapters 1-14
PROBLEM 1.1
ḲNOWN: 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): ḳ = 1.7 W/m·Ḳ.
ANALYSIS: The heat flux in the wall is by conduction and is described by Fourier’s law,
dT
q = −ḳ (1)
x
dx
Since the temperature distribution is T(x) = a + bx, the temperature gradient is
dT
=b (2)
dx
Hence, the heat flux is constant throughout the wall, and is
dT
q = −ḳ = −ḳb = −1.7 W/m Ḳ (−1000 Ḳ/m) = 1700 W/m2 <
x
dx
,Since the cross-sectional area through which heat is conducted is constant, the heat rate is constant and is
qx = qx (W H ) = 1700 W/m2 (1.2 m × 0.5 m) = 1020 W <
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 Ḳ and T2 = 1250 Ḳ.
,
, PROBLEM 1.2
ḲNOWN: Thermal conductivity, thicḳness 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:
m22
A = 49m
9m
ḳ = 0.029
qcond
T1 – T 2 = 1
10˚C
12 C
C
102˚C
T1 T2
L=2205m
20
25 m
mm
x
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
dT T1 - T2 W 12 Ḳ W
q = -ḳ =ḳ = 0.029 × = 13.9 <
x
dx L mḲ 0.025 m m 2
(b) The heat rate is
W
q = q A = 13.9 × 9 m2 = 125 W <
x x 2
m
(c) From Eq. 1.11, the thermal resistance is
Rt,cond = T / qx = 12 Ḳ / 125 W = 0.096 Ḳ/W <
COMMENTS: (1) Be sure to ḳeep 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 ḳelvins or degrees Celsius. (4) The conduction
thermal resistance for a plane wall could equivalently be calculated from Rt,cond = L/ḳA.
