ch03
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ch03


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COMMENTS: (1) The temperature at the inner surface can be found from an energy balance on the
i-node using the value found for L.
,i i ,o i
rad i
cv,o cd cv,i
T T T T
q 0 T 298.3 C
R R R
\u221e \u221e\u2212 \u2212
\u2032\u2032+ + = = °
\u2032\u2032 \u2032\u2032 \u2032\u2032+
It follows that Ti is close to T\u221e,i since the wall represents the dominant resistance of the system.
(2) Verify that 2iq 50 W / m\u2032\u2032 = and 2oq 150 W / m .\u2032\u2032 = Is the overall energy balance on the system
satisfied?
PROBLEM 3.12
KNOWN: Configurations of exterior wall. Inner and outer surface conditions.
FIND: Heating load for each of the three cases.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction, (3) Constant properties, (4)
Negligible radiation effects.
PROPERTIES: (T = 300 K): Table A.3: plaster board, kp = 0.17 W/m\u22c5K; urethane, kf = 0.026 W/m\u22c5K;
wood, kw = 0.12 W/m\u22c5K; glass, kg = 1.4 W/m\u22c5K. Table A.4: air, ka = 0.0263 W/m\u22c5K.
ANALYSIS: (a) The heat loss may be obtained by dividing the overall temperature difference by the
total thermal resistance. For the composite wall of unit surface area, A = 1 m2,
( ) ( ) ( ) ( ) ( )
,i ,o
i p p f f w w o
T T
q
1 h L k L k L k 1 h A
\u221e \u221e
\u2212
=
+ + + +\uf8ee \uf8f9\uf8f0 \uf8fb
( )
( ) 2 2
20 C 15 C
q
0.2 0.059 1.92 0.083 0.067 m K W 1m
\u2212 \u2212
=
+ + + + \u22c5\uf8ee \uf8f9\uf8f0 \uf8fb
$ $
35 C
q 15.0 W
2.33 K W
= =
$
<
(b) For the single pane of glass,
( ) ( ) ( )
,i ,o
i g g o
T T
q
1 h L k 1 h A
\u221e \u221e
\u2212
=
+ +\uf8ee \uf8f9\uf8f0 \uf8fb
( ) 2 2
35 C 35 C
q 130.3 W
0.269 K W0.2 0.002 0.067 m K W 1m
= = =
+ + \u22c5\uf8ee \uf8f9\uf8f0 \uf8fb
$ $
<
(c) For the double pane window,
( ) ( ) ( ) ( )
,i ,o
i g g a a o
T T
q
1 h 2 L k L k 1 h A
\u221e \u221e
\u2212
=
+ + +\uf8ee \uf8f9\uf8f0 \uf8fb
( ) 2 2
35 C 35 C
q 75.9 W
0.461K W0.2 0.004 0.190 0.067 m K W 1m
= = =
+ + + \u22c5\uf8ee \uf8f9\uf8f0 \uf8fb
$ $
<
COMMENTS: The composite wall is clearly superior from the standpoint of reducing heat loss, and the
dominant contribution to its total thermal resistance (82%) is associated with the foam insulation. Even
with double pane construction, heat loss through the window is significantly larger than that for the
composite wall.
PROBLEM 3.13
KNOWN: Composite wall of a house with prescribed convection processes at inner and
outer surfaces.
FIND: (a) Expression for thermal resistance of house wall, Rtot; (b) Total heat loss, q(W); (c)
Effect on heat loss due to increase in outside heat transfer convection coefficient, ho; and (d)
Controlling resistance for heat loss from house.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction, (2) Steady-state conditions, (3)
Negligible contact resistance.
PROPERTIES: Table A-3, ( ) ( )( )i oT T T / 2 20 15 C/2=2.5 C 300K := + = \u2212 \u2248$ $ Fiberglass
blanket, 28 kg/m3, kb = 0.038 W/m\u22c5K; Plywood siding, ks = 0.12 W/m\u22c5K; Plasterboard, kp =
0.17 W/m\u22c5K.
ANALYSIS: (a) The expression for the total thermal resistance of the house wall follows
from Eq. 3.18.
p b s
tot
i p b s o
L L L1 1R .
h A k A k A k A h A
= + + + + <
(b) The total heat loss through the house wall is
( )tot i o totq T/R T T / R .= \u2206 = \u2212
Substituting numerical values, find
[ ]
tot 2 2 2 2
2 2 2
5 5
tot
1 0.01m 0.10mR
30W/m K 350m 0.17W/m K 350m 0.038W/m K 350m
0.02m 1
 
0.12W/m K 350m 60W/m K 350m
R 9.52 16.8 752 47.6 4.76 10 C/W 831 10 C/W\u2212 \u2212
= + +
\u22c5 × \u22c5 × \u22c5 ×
+ +
\u22c5 × \u22c5 ×
= + + + + × = ×$ $
The heat loss is then,
( ) -5q= 20- -15 C/831 10 C/W=4.21 kW.\uf8ee \uf8f9 ×\uf8f0 \uf8fb$ $ <
(c) If ho changes from 60 to 300 W/m2\u22c5K, Ro = 1/hoA changes from 4.76 × 10-5 °C/W to 0.95
× 10-5 °C/W. This reduces Rtot to 826 × 10
-5
 °C/W, which is a 0.5% decrease and hence a
0.5% increase in q.
(d) From the expression for Rtot in part (b), note that the insulation resistance, Lb/kbA, is
752/830 \u2248 90% of the total resistance. Hence, this material layer controls the resistance of the
wall. From part (c) note that a 5-fold decrease in the outer convection resistance due to an
increase in the wind velocity has a negligible effect on the heat loss.
PROBLEM 3.14
KNOWN: Composite wall of a house with prescribed convection processes at inner and
outer surfaces.
FIND: Daily heat loss for prescribed diurnal variation in ambient air temperature.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional, steady-state conduction (negligible change in wall
thermal energy storage over 24h period), (2) Negligible contact resistance.
PROPERTIES: Table A-3, T \u2248 300 K: Fiberglass blanket (28 kg/m3), kb = 0.038 W/m\u22c5K;
Plywood, ks = 0.12 W/m\u22c5K; Plasterboard, kp = 0.17 W/m\u22c5K.
ANALYSIS: The heat loss may be approximated as 
24h
,i ,o
tot0
T TQ dt where
R
\u221e \u221e\u2212
= \u222b
p b s
tot
i p b s o
tot 2 2 2
tot
L L L1 1 1
R
A h k k k h
1 1 0.01m 0.1m 0.02m 1
R
0.17 W/m K 0.038 W/m K 0.12 W/m K200m 30 W/m K 60 W/m K
R 0.01454 K/W.
= + + + +
= + + + +
\u22c5 \u22c5 \u22c5
\u22c5 \u22c5
=
\uf8ee \uf8f9\uf8ef \uf8fa\uf8ef \uf8fa\uf8f0 \uf8fb\uf8ee \uf8f9\uf8ef \uf8fa\uf8f0 \uf8fb
Hence the heat rate is
12h 24h
tot 0 12
1 2 2Q 293 273 5 sin t dt 293 273 11 sin t dt
R 24 24
\u3c0 \u3c0\uf8f1 \uf8fc\uf8ee \uf8f9 \uf8ee \uf8f9\uf8f4 \uf8f4\uf8ee \uf8f9 \uf8ee \uf8f9
= \u2212 + + \u2212 +\uf8f2 \uf8fd\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f4 \uf8f4\uf8f3 \uf8fe
\u222b \u222b
12 24
0 12
W 24 2 t 24 2 tQ 68.8 20t+5 cos 20t+11 cos K h
K 2 24 2 24
\u3c0 \u3c0
\u3c0 \u3c0
\uf8f1 \uf8fc\uf8ee \uf8f9 \uf8ee \uf8f9\uf8f4 \uf8f4\uf8ee \uf8f9 \uf8ee \uf8f9
= + \u22c5\uf8f2 \uf8fd\uf8ef \uf8fa \uf8ef \uf8fa\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f4 \uf8f4\uf8f3 \uf8fe
( ) ( )60 132Q 68.8 240 1 1 480 240 1 1 W h
\u3c0 \u3c0
\uf8f1 \uf8fc\uf8ee \uf8f9 \uf8ee \uf8f9
= + \u2212 \u2212 + \u2212 + + \u22c5\uf8f2 \uf8fd\uf8ef \uf8fa \uf8ef \uf8fa\uf8f0 \uf8fb \uf8f0 \uf8fb\uf8f3 \uf8fe
{ }Q 68.8 480-38.2+84.03 W h= \u22c5
8Q=36.18 kW h=1.302 10 J.\u22c5 × <
COMMENTS: From knowledge of the fuel cost, the total daily heating bill could be
determined. For example, at a cost of 0.10$/kW\u22c5h, the heating bill would be $3.62/day.
PROBLEM 3.15
KNOWN: Dimensions and materials associated with a composite wall (2.5m × 6.5m, 10 studs each
2.5m high).
FIND: Wall thermal resistance.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Temperature of composite depends only on x
(surfaces normal to x are isothermal), (3) Constant properties, (4) Negligible contact resistance.
PROPERTIES: Table A-3 (T \u2248 300K): Hardwood siding, kA = 0.094 W/m\u22c5K; Hardwood,
kB = 0.16 W/m\u22c5K; Gypsum, kC = 0.17 W/m\u22c5K; Insulation (glass fiber paper faced, 28 kg/m3),
kD = 0.038 W/m\u22c5K.
ANALYSIS: Using the isothermal surface assumption, the thermal circuit associated with a single
unit (enclosed by dashed lines) of the wall is
( ) ( )A A A
0.008mL / k A 0.0524 K/W
0.094 W/m K 0.65m 2.5m
= =
\u22c5 ×
( ) ( )B B B
0.13mL / k A 8.125 K/W
0.16 W/m K 0.04m 2.5m
= =
\u22c5 ×
( ) ( )D D D
0.13mL /k A 2.243 K/W
0.038 W/m K 0.61m 2.5m
= =
\u22c5 ×
( ) ( )C C C
0.012mL / k A 0.0434 K/W.
0.17 W/m K 0.65m 2.5m
= =
\u22c5 ×
The equivalent resistance of the core is
( ) ( )1 1eq B DR 1/ R 1/ R 1/ 8.125 1/ 2.243 1.758 K/W\u2212 \u2212= + = + =
and the total unit resistance is
tot,1 A eq CR R R R 1.854 K/W.= + + =
With 10 such units in parallel, the total wall resistance is
( ) 1tot tot,1R 10 1/ R 0.1854 K/W.\u2212= × = <
COMMENTS: If surfaces parallel to the heat flow direction are assumed adiabatic, the thermal
circuit and the value of Rtot will differ.
PROBLEM 3.16
KNOWN: Conditions associated with maintaining heated and cooled conditions within a refrigerator
compartment.
FIND: Coefficient of performance (COP).
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state operating conditions, (2) Negligible radiation, (3) Compartment
completely sealed from ambient air.
ANALYSIS: The Case (a) experiment is performed to determine the overall thermal resistance to heat
transfer between the interior of the refrigerator and the ambient air. Applying an energy balance to a
control surface about the refrigerator, it follows from Eq. 1.11a that, at any instant,
g outE E 0\u2212 =\ufffd \ufffd
Hence,
elec outq q 0\u2212 =
where ( )out ,i ,o tq T T R\u221e \u221e= \u2212 . It follows that
( ),i ,o
t
elec
T T 90 25 C
R 3.25 C/W
q 20 W
\u221e \u221e
\u2212
\u2212
= = =
$
$
For Case (b), heat transfer from the ambient air to