Modern Engineering Thermodynamics

388 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.13 (Continued )
with
so that
v′ c = RT c
p c
0:0709 Btu/ lbm ⋅ R
= ½
ð ÞŠð240:
R Þ ð 778:16 ft ⋅ lbf/Btu Þ
ð507 lbf/in 2 Þð144 in 2 /ft 2
= 0:181 ft 3 /lbm
Þ
v′ R = v v′ c
= 0:122
0:181 = 0:67 ðnotice that we do not use the actual critical specific volume v c hereÞ
Using T R = T/T c = 1.60 and v′ R = v/v′ c = 0:67, we find from Figure 11.6 that
p R = p p c
= 2:10 and Z = 0:850
Then, we can calculate
p = p c p R = 507ð2:10Þ = 1070 psia
Exercises
37. Determine the pressure exerted by the carbon monoxide in Example 11.13 if it is at a temperature of 100.°F and all the
other variables remain unchanged. Answer: p = 1670 psia.
38. Suppose the tank containing the carbon monoxide in Example 11.13 was isothermally crushed by a giant winged
wombat, causing the pressure in the tank to increase to 2000. psia. Determine the final volume of the tank. Answer:
V final
= 0:516 ft 3 .
39. Management has just informed us that the tank in Example 11.13 really contains carbon dioxide rather than carbon
monoxide and that the tank was really at 178°F, not –78.0°F. Assuming all the remaining variables are unchanged,
determine the pressure of the CO 2 in the tank. Answer: p = 964 psia.
1.20
Compressibility factor, Z = pv/RT
1.10
1.00
1.20
1.10
1.00
0.90
0.80
0.70
0.60
3.00
1.60
1.20
2.00
1.40
1.00
0.90
0.80
0.70
0.60
0.50
0.45
0.40
0.35
0.30
0.25
1.20
1.30
1.40
v/v c
= 0.20
1.50
1.60
7.00
10.00
T R = T/T c = 3.50
2.50
2.00
1.80
5.00
T/T c = 15.0
0.50
1.15
0.40
0.30
0.20
0
1.05
1.10
T/T c = 1.00
1.0 2.0 3.0 4.0 5.0 6.0 7.0
Reduced pressure, p R = p/p c
8.0 9.0 10.0
FIGURE 11.6
The generalized (Nelson-Obert) compressibility chart—low-pressure range, 0 ≤ p R ≤ 10.0. Note that v′ R = v/v′ c = vp c /RT c .(Source:
“Nelson-Obert Compressibility Charts,” reprinted courtesy of Professor E. F. Obert, University of Wisconsin, Madison, from Obert, E. F.,
1948. Thermodynamics. McGraw-Hill, New York.)