Modern Engineering Thermodynamics

11.10 Compressibility Factor and Generalized Charts 395
and since this is an isothermal process, T 2 = T 1 and the ideal gas portion of this equation for a constant specific heat
gas is h 2
* − h 1
* = c p ðT 2 − T 1 Þ = 0: The properties of ethylene at its critical state and its molecular mass are found in
Table C.12a as T c = 508.3 R, p c = 742 psia, and M = 28.05 lbm/lbmoles. Then,
p R1 = 150:
742 = 0:202 and T 80:0 + 459:67
R1 = = 1:06
508:3
p R2 =
15:0 × 103
742
= 20:2 and T R2 = T R1 = 1:06
Using p R1 = 0.202 and T R1 = 1.06, Figure 11.9 gives the enthalpy correction for state 1 as
h* − h
kJ
= 1:50
T c 1
kgmole.K = 1:50 kJ 1 Btu/ðlbmole.RÞ
= 0:360
Btu
kgmole.K 4:1865 kJ/ðkgmole.KÞ lbmole.R
and using p R2 = 20.2 and T R2 = 1.06, Figure 11.9 gives the enthalpy correction for state 2 as
h* − h
T c 2
Then, Eq. (11.41) gives
kJ
= 31:5
kgmole.K = 31:5 kJ
kgmole.K
1 Btu/ðlbmole.RÞ
4:1865 kJ/ðkgmole.KÞ
h 2 − h 1 = ðh*− 2 h
h* − h
*Þ 1
−
−
h* − h
T c 2
T c 1
= 0 − ½7:52 − 0:360 Btu/ðlbm.RÞŠ
= −130: Btu
lbm
Tc
M
Btu
= 7:52
lbmole.R
508:3R
28:05 lbm/lbmole
Note that, since the compressibility charts cannot be read to more than two or three significant figures, our final
calculation results are limited to this accuracy as well.
b. The compressibility charts do not give values for the specific internal energy, so it must be calculated from the definition
of enthalpy as u = h – pv, oru 2 – u 1 = h 2 – h 1 – (p 2 v 2 – p 1 v 1 ), where v 1 = Z 1 RT 1 /p 1 and v 2 = Z 2 RT 2 /p 2 . For p R1 = 0.202
and T R1 = 1.06, Figure 11.5 gives Z 1 = 0.940, and for p R2 = 20.2 and T R2 = T R1 = 1.06, Figure 11.7 gives Z 2 = 2.15. The
gas constant R for ethylene can be found in Table C.13a as R = 55.1 ft ·lbf/(lbm·R). Then,
v 1 = Z 1RT 1
p 1
= 0:940½55:1ft .lbf/ðlbm.RÞŠð80:0 + 459:67 RÞ
ð150: lbf/in 2 Þð144 in 2 /ft 2 Þ
= 1:29 ft3
lbm
and
v 2 = Z 2RT 2
p 2
= 2:15½55:1ft .lbf/ðlbm.RÞŠð80 + 459:67 RÞ
ð15:0 × 10 3 lbf/in 2 Þð144 in 2 /ft 2 Þ
= 0:030 ft3
lbm
Then,
u 2 − u 1 = h 2 − h 1 − ðp 2 v 2 − p 1 v 1 Þ
= 130: Btu
− 15:0 ×
lbf
103
lbm in 2 144 in2
ft 2 0:0300 ft3 1 Btu
lbm 778:16 ft .lbf
− 150: lbf
in 2 144 in2
ft 2 1:29 ft3 1 Btu
lbm 778:16 ft .lbf
= − 180: Btu
lbm
c. Finally, from Eq. (11.42), we have the change in specific entropy as
s 2 − s 1 = ðs 2
* − s*Þ 1
− ðs* − sÞ 2
− ðs* − sÞ 1
1
M
(Continued )