Modern Engineering Thermodynamics

11.10 Compressibility Factor and Generalized Charts 387
Compressibility factor, Z = pv/RT
1.00
0.90
0.80
0.70
0.60
0.50
0.60
0.65
0.75
0.70
8.00
0.80
6.00
5.00
4.00
3.50
3.00
0.85
2.60
2.40
2.20
2.00
1.80
0.90
1.60
1.50
1.40
1.30
0.95
1.20
T R =T/T c 3.00
2.00
1.60
1.40
1.10
1.00
0.70
0.90
0.80
U R
= 0.70
1.30
1.20
1.15
1.10
1.05
T R = 1.00
0.40
0.30 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Reduced pressure, p R = p/p c
FIGURE 11.5
The generalized (Nelson-Obert) compressibility chart—low-pressure range, 0 ≤ p R ≤ 1.0. Note that v′ R = v/v′ c = vp c /RT c .(Source:
“Nelson-Obert Compressibility Charts,” adapted courtesy of Professor E. F. Obert, University of Wisconsin, Madison, from Obert, E. F., 1960.
Concepts of Thermodynamics. McGraw-Hill, New York.)
Though the van der Waals equation is not very accurate as a universal equation of state, Su’s modified compressibility
factor formulation as an approximate approach to the law of corresponding states has found universal
acceptance within the engineering community.
EXAMPLE 11.13
Using the compressibility charts, find the pressure exerted by 8.20 lbm of carbon monoxide in a 1.00 ft 3 rigid tank at –78.0°F.
Solution
From Table C.12a, we find that
T c = 240: R
and
p c = 507 psia
v c =
1:49 ft3 /lbmole
28:011 lbm/lbmole = 0:053 ft3 /lbm
Also, from Table C.13a, we find that R = 0.0709 Btu/lbm· R. Then, we have
T R = T T c
=
− 78:0 + 460:
240:
= 1:60
and
v =
1:00 ft3
8:20 lbm = 0:122 ft3 /lbm
(Continued )