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constants for the system.
Find: Minimum work for the separation in kJ/kmol of feed.
Analysis: Material balance for 1 kmol of feed.
Let the two products be R and S, where the former is acetonerich.
Total mole balance: 1 = nR + nS
Acetone balance:
0.35 = 0.99 nR + 0.02 nS
Solving, nR = 0.3469 kmol and nS = 0.6531 kmol
From the problem statement,
Wmin
= nR
RT0 xi ln γ i xi + nS iR xi ln γ i xi − nF iS xi ln γ i xi (1) iS The activity coefficients are given by the van Laar equations, (3) in Table 2.9, which with
the given constants, A12 = 2.0 and A21 = 1.7, become,
ln γ 1 = 2.0
2.0 x1
1+
1.7 x2 2 (2) ln γ 2 = .
17
1.7 x2
1+
2.0 x1 2 (3) Applying Eqs. (2) and (3) to the given molefraction compositions, Component
Acetone (1)
Water (2) x
0.35
0.65 Feed γ
2.116
1.291 Substituting the above values into Eq. (1), Product R x
0.99
0.01 γ
1.000
5.318 Product S_____
x
γ
0.02
6.735
0.98
1.000 Exercise 2.21 (continued)
Analysis: (continued)
Wmin
= 0.3469 {0.99 ln [ (0.99)(1.00)] + 0.01ln [ (0.01)(5.318) ]}
RT0
+0.6531{0.02 ln [ (0.02)(6.735) ] + 0.98ln [ (0.98)(1.00) ]} − 1.0000 {0.35 ln [ (0.35)(2.116)] + 0.65 ln [ (0.65)(1.291)]}
= 0.1664 kmol Wmin = 0.1664 RT0 = 0.1664(8.314)(298) = 412 kJ/kmol feed
For ideal liquid solutions, all values of γ are equal to 1.0. Therefore,
Wmin
= 0.3469 {0.99 ln [ (0.99)(1.00)] + 0.01ln [ (0.01)(1.00]}
RT0 +0.6531{0.02 ln [ (0.02)(1.00)] + 0.98 ln [ (0.98)(1.00) ]} − 1.0000 {0.35 ln [ (0.35)(1.00)] + 0.65 ln [ (0.65)(1.00)]}
= 0.5639 kmol Wmin = 0.5639 RT0 = 0.5639(8.314)(298) = 1,397 kJ/kmol feed
Thus, the minimum work of separation for the ideal solution is 3.4 times that of the
nonideal solution. Exercise 2.22
Subject: Relative volatility and activity coefficients of the benzene (B)  cyclohexane
(CH) azeotropic system at 1 atm
Given: Experimental vaporliquid equilibrium data, including liquidphase activity
coefficients, and Antoine vapor pressure constants.
Assumptions: Ideal gas and gas solutions
Find: (a) Relative volatility of benzene with respect to cyclohexane as a function of
benzene mole fraction in the liquid phase.
(b) Van Laar constants from the azeotropic point and comparison of van Laar
predictions with experimental data.
Analysis: (a) From Eqs. (221), (219), and (3) in Table 2.3,
α B,CH = KB
y /x
yB / xB
γ Ps
= B B=
= B Bs
KCH yCH / xCH
γ CH PCH
1 − yB / 1 − xB (1) Using the yx data for benzene, the following values of relative volatility are computed
from (1): Temperature, oC
79.7
79.1
78.5
78.0
77.7
77.6
77.6
77.6
77.8
78.0
78.3
78.9
79.5 xB
0.088
0.156
0.231
0.308
0.400
0.470
0.545
0.625
0.701
0.757
0.822
0.891
0.953 yB
0.113
0.190
0.268
0.343
0.422
0.482
0.544
0.612
0.678
0.727
0.791
0.863
0.938 αB,CH
1.317
1.269
1.218
1.173
1.095
1.049
0.996
0.946
0.898
0.854
0.821
0.771
0.746 Exercise 2.22 (continued)
Analysis: (a) (continued) (b) From the data, take the azeotrope at xB = yB = 0.545 and xCH = yCH = 1  0.545 =
0.455, and with γΒ = 1.079 and γCH = 1.102.
To determine the van La...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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