Exercise 2.2
Subject:
Minimum work for separating a mixture of ethylbenzene and xylene isomers.
Given:
Component flow rates,
n
i
, of feed ,in lbmol/h. Component split fractions for
three products, Phase condition; temperature in
o
F; enthalpy,
h
, in Btu/lbmol; and
entropy,
s
, in Btu/lbmol
o
R for feed and three products. Infinite heat sink temperature =
T
0
= 560
o
R.
Find:
Minimum work of separation,
W
min
, in kJ/h
Analysis:
From Eq. (4), Table 2.1,
W
nb
nb
min
out
in
=

From Eq. 21,
b
h
T s
=

0
For the feed stream (in),
n
= 150 +190 + 430 + 230 = 1,000 lbmol/h
b
= 29,290  (560)(15.32) = 20,710 Btu/lbmol
For product 1 (out), using Eq. (12),
n
= 150(0.96) + 190 (0.005) + 430(0.004) = 146.7 lbmol/h
b
= 29,750  (560)(12.47) = 22,767 Btu/lbmol
For product 2 (out), using Eq. (12),
n
= 150(0.04) + 190(0.99) + 430(0.99) + 230(0.015) = 623.3
lbmol/h
b
= 29,550  (560)(13.60) = 21,934 Btu/lbmol
For product 3 (out), by total material balance,
n
= 1,000  146.7  623.3 = 230 lbmol/h
b =
28,320  (560)(14.68) = 20,099 Btu/lbmol
From Eq. (4), Table 2.1,
W
min
= 146.7(22,767) + 623.3(21,934) + 230(20,099)  1,000(20,710) =
924,200 Btu/h