11.3 Temperature Difference and Heat Transfer 321
Example 11.6 Determine the thermal resistanceRand theR-value for the glass window of Example 11.5.
The thermal resistanceRand theR-value of the window can be determined from Equa-
tions (11.16) and (11.17), respectively.And theR-value or theR-factor for the given glass pane is
Example 11.7 For Example 11.6, convert the thermal resistanceRandR-factor results from SI units to
U.S. Customary units.Or, if theR-value is expressed in terms of in
2
,Example 11.8 A double-pane glass window consists of two pieces of glass, each having a thickness of 8 mm,
with a thermal conductivity ofk1.4 W/mK. The two glass panes are separated by an
air gap of 10 mm, as shown in Figure 11.10. Assuming the thermal conductivity of air to be
k0.025 W/mK, determine the totalR-value for this window.
The total thermal resistance of the window is obtained by adding the resistance offered by
each pane of glass and the air gap in the following manner:R (^) totalR (^) glassR (^) airR (^) glass
Lglass
kglass
Lair
kair
Lglass
kglass
R
L
k
0.0057a
m
(^2) #
K
W
b°
1 W
3.4123
Btu
h
¢°
5
9
(^) °R
K
¢a
39.37 in.
1 m
b
2
1.43
°R#in
2
Btu
h
R
L
k
0.0057a
m
(^2) #
K
W
b°
1 W
3.4123
Btu
h
¢°
5
9
(^) °R
K
¢a
3.28 ft
1 m
b
2
0.01
°R#ft
2
Btu
h
R¿0.00317a
K
W
b°
1 W
3.4123
Btu
h
¢°
5
9
(^) °R
K
¢5.161 10
4
°R
Btu
h
R
L
k
0.008 m
1.4a
W
m#K
b
0.0057
m
(^2) #
K
W
R¿
L
kA
0.008 m
1.4a
W
m#K
b11.8 m
2
2
0.00317
K
W
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