Solution Manual for Fundamentals of Heat and Mass Transfer 8th Edition by Theodore L. Bergman||ISBN:9781119722489

Solutions Manual for Fundamentals of Heat and Mass Transfer, 8e
Bergman, Lavine, Incropera, DeWitt (All Chapters Download link at th
end of this file)
PROBLEM 1.1


KNOWN: 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): k = 1.7 W/m·K.

ANALYSIS: The heat flux in the wall is by conduction and is described by Fourier’s law,
dT
q  k (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  k  kb  1.7 W/m K 1000 K/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  qx (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 K and T2 = 1250 K.




MEDSTUDY.COM

, PROBLEM 1.2

KNOWN: Thermal conductivity, thickness 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



k = 0.029
qcond

T1 – T2 = 1
10˚C
12 C
C
102˚C


T1 T2


L=2205m
25
20 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 K W
q = -k =k = 0.029 × = 13.9 <
x
dx L mK 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 K / 125 W  0.096 K/W <
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. (4) The conduction
thermal resistance for a plane wall could equivalently be calculated from Rt,cond = L/kA.

, PROBLEM 1.3

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

qin  qout  qcond  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

T = qL / k  20 W/m2  0.01 m /12 W/m  K  0.0167 K

which is much smaller than the specified temperature difference of 20C.




MEDSTUDY.COM

, PROBLEM 1.4
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 T T 25∘C  15∘C
qx  k  k 1 2  1W m  K  133.3 W m2 . (1)
dx L 0.30 m

qx  qx  A  133.3 W m2  20 m2  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

OAmbient
m
sidbeie
utsi dsn aair
eurtsurface
facire temperature, T2 (C)

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.