EXAMPLES Fundamentals Of Heat And Mass Transfer, Heat and mass transfer
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PROBLEM 1.1
KNOWN:
Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T
1
.
FIND:
The outer temperature of the wall, T
2
.
SCHEMATIC:
ASSUMPTIONS:
(1) One-dimensional conduction in the x-direction, (2) Steady-state conditions,
(3) Constant properties.
ANALYSIS:
The rate equation for conduction through the wall is given by Fourier’s law,
A=kA
TT
L
-
= -k
dT
dx
12
q
q
q
.
==¢¢ ×
×
x
x
cond
Solving for T
2
gives
TT
qL
kA
cond
=-
.
2
1
Substituting numerical values, find
3000W 0.025m
0.2W / m K
´
×
415
$
T
=
C -
2
2
´
10m
415
$
$
T
=
C - 37.5 C
2
<
378
$
T
=
C.
2
COMMENTS:
Note direction of heat flow and fact that T
2
must be less than T
1
.
 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 ambient air temperatures ranging from
-15 to 38
°
C.
SCHEMATIC:
ASSUMPTIONS:
(1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3)
Constant properties, (4) Outside wall temperature is that of the ambient air.
dT dx
=-
q
¢¢
k
ANALYSIS:
From Fourier’s law, it is evident that the gradient,
, is a constant, and
x
q
¢¢
hence the temperature distribution is linear, if
and k are each constant. 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 T
2
= -15
x
°
C
are
(
)
$
$
25 C
--
15 C
dT
T
-
T
2
12
¢¢
q
=-
k
=
k
=
1W m K
×
=
133.3 W m
.
(1)
x
dx
L
0.30 m
(2)
<
2
2
q
¢=´=
q
A
133.3 W m
´
20 m
=
2667 W
.
xx
Combining Eqs. (1) and (2), the heat rate q
x
can be determined for the range of ambient temperature, -15
£
T
2
£
38
°
C, with different wall thermal conductivities, k.
3500
2500
1500
500
-500
-1500
-20
-10
0
10
20
30
40
Ambient air 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 linearily from +2667 W to -867 W and is zero
when the inside and ambient 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 1.3
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
TT
-
7C
°
<
12
(
)
(
)
q
=
k
LW
=
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
qC
4312 W
´
$0.01/ MJ
g
<
()
(
)
C
=
D
t
=
24 h / d
´
3600 s / h
=
$4.14 / d
d
6
h
0.9
´
10 J / MJ
f
COMMENTS:
The loss could be reduced by installing a floor covering with a layer of insulation
between it and the concrete.
 PROBLEM 1.4
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
40
x
2
TT
12
m
40-20
C
<
k = 0.10 W / m K.
°
C or K temperature units may be used interchangeably when
evaluating a temperature difference.
COMMENTS:
Note that the
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