Page 245 - Chemical process engineering design and economics

226 Chapter 5

known, the average compressibility factor for Equation 7.7.11 cannot be calcu-
lated until work done is calculated. To start the calculation, use the compressi-
bility factor at the inlet for one stage of compression, N = 1, in Equation 5.7.1.
Therefore,
n i)/n
T? T r ( P v ~ i
7 Z] K i i I I r 2 I I
Wpl = ——— —I -i I
I
(n-l)/n L P, ) J

0.97 8314.0 J 278.2K IY7.0 V' 1587 1
——
i
WPJ — ___ ————— ___ ———— ii I I __ I — i i i I
0.1587 1 kgmol-K 1 \\\A ) J
6
3
Wpi=4.114xl0 J/kgmol(1.770xl0 Btu/lbmol)

Now, calculate the discharge temperature from Equation 5.7.6.

(k-l)/k = (l.131-I)/1.131 =0.1158

0.1158 4.114xl0 6 J 1 kgmol-K
T 2 = ——— ————————— ————————— + 278.2 = 335.5 K (605 °R)
1 1 kgmol 8314.0 J

Therefore, T 2 is below 450 K (810 °R) given in Table 5.4. Thus, intercooling
is not required, and a single compression stage is adequate.
Now, find the compressibility factor at the compressor outlet, z, first calcu-
2
late the reduced temperature and pressure.
335.5
T 2
TR2 = — = —— = 0.9065
370.1
T c
P 2 7.0
P R2 = —= —— = 0.1645
PC 42.55

From Equation 5.7.12, z 2 = 0.93, and the average value of the compressibil-
ity factor,

z = (0.97 + 0.93) / 2 = 0.95

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