TEST BANK FOR Introduction to Heat 5th Edition By Incropera F.P. and Dewitt D.P.

KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal
conductivity k and inner temperature, T1.
FIND: The outer temperature of the wall, T2.
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,
q q q A=-k
dT
dx
A = kA
T T
cond x x L
= = ¢¢ × × 1 2
-
.
Solving for T2 gives
T T
q L
2 1 kA
= - cond .
Substituting numerical values, find
T C-
3000W 0.025m
2 0.2W/ m K 10m2 = ´
× ´
415$
T2 = 415 C- 37.5 C $ $
T2 C. = 378$ <
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1.
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.
ANALYSIS: From Fourier’s law, it is evident that the gradient, dT dx = -q¢ xk , is a constant, and
hence the temperature distribution is linear, if q¢ xand 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 T2 = -15°C
are
( ) 1 2 2
x
dT T T 25 C 15 C
q k k 1W m K 133.3W m
dx L 0.30m
- - - ¢¢ = - = = × =
$ $
. (1)
2 2
qx = q¢ x´A =133.3W m ´ 20m = 2667W. (2) <
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15
£ T2 £ 38°C, with different wall thermal conductivities, k.
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
-1500
-500
500
1500
2500
3500
Heat loss, qx (W)
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
( )T1 T2 ( ) 7 C
q k LW 1.4W/m K 11m 8m 4312 W
t 0.20m
= - = × ´ ° = <
The daily cost of natural gas that must be combusted to compensate for the heat loss is
g ( ) ( )
d 6 f
qC 4312W $0.01/MJ
C t 24h / d 3600s / h $4.14 / d
h 0.9 10 J /MJ
= D = ´ ´ =
´
<
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
T T m 40-20 C
x 1 2 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 1.5
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 k
L
W 15-5 C
q 1.4
m K 0.005m
q 2800 W/m .
1 2
x
x
2
x
′′ − =
′′ =
⋅
′′ =

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 1.6
KNOWN: Width, height, thickness and thermal conductivity of a single pane window and
the air space of a double pane window. Representative winter surface temperatures of single
pane and air space.
FIND: Heat loss through single and double pane windows.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state
conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced
motion).
ANALYSIS: From Fourier’s law, the heat losses are
Single Pane: qg kgA T1 LT2 1.4 W/m K(2m2 )03.050 5Cm 19,600 W
= - = × =
$
Double Pane: ( ) T1 T2 2 25 C qa kaA 0.024 2m 120 W L 0.010 m
= - = =
$
COMMENTS: Losses associated with a single pane are unacceptable and would remain
excessive, even if the thickness of the glass were doubled to match that of the air space. The
principal advantage of the double pane construction resides with the low thermal conductivity
of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use
of the double pane construction would also increase the surface temperature of the glass
exposed to the room (inside) air