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Unformatted text preview: ( 0.1830 ) âˆ’ 0.1505 = 0.682 = 68.2% Exercise 12.5
Subject: Reciprocal rate functions for a fivecomponent system.
Given:. Eqs. (1231) and (1232) for a general Ccomponent system.
Find: Expanded forms of Eqs. (1231) and Eq. (1232).
Analysis: From Eq. (1231):
P
R11 = z1
z
z
z
z
+ 2+ 3+ 4+ 5
P
P
P
P
P
k15 k12 k13 k14 k15 P
R22 = z2
z
z
z
z
+ 1+ 3+ 4+ 5
P
P
P
P
P
k25 k 21 k 23 k 24 k 25 P
R33 = z
z
z
z
z3
+ 1+ 2+ 4+ 5
P
P
P
P
P
k 35 k 31 k 32 k 34 k 35 P
R44 = z4
z
z
z
z
+ 1+ 2+ 3+ 5
P
P
P
P
P
k45 k 41 k 42 k 43 k 45 P
R12 = âˆ’ z1 1
1
âˆ’P
P
k12 k15 P
R13 = âˆ’ z1 1
1
âˆ’P
P
k13 k15 1
1
âˆ’P
P
k14 k15 P
R21 = âˆ’ z2 1
1
âˆ’P
P
k 21 k 25 P
R23 = âˆ’ z2 1
1
âˆ’P
P
k 23 k 25 P
R24 = âˆ’ z2 1
1
âˆ’P
P
k 24 k 25 P
R31 = âˆ’ z3 1
1
âˆ’P
P
k 31 k 35 P
R32 = âˆ’ z3 1
1
âˆ’P
P
k 32 k 35 P
R34 = âˆ’ z3 1
1
âˆ’P
P
k 34 k 35 P
R41 = âˆ’ z4 1
1
âˆ’P
P
k 41 k 45 P
R42 = âˆ’ z4 1
1
âˆ’P
P
k 42 k 45 P
R43 = âˆ’ z4 1
1
âˆ’P
P
k 43 k 45 P
R14 = âˆ’ z1 Exercise 12.6
Subject: Masstransfer rates and tray efficiencies from data for a perfectly mixed tray.
Given: Vaporliquid traffic, component mole fractions, and masstransfer coefficients for a tray
in a column separating methanol (1), water (2), and acetone (3) at 14.7 psia from Example 12.1,
but with the designations 1, 2, 3 changed to see if different results are obtained.
Find: (a) Component molar diffusion rates.
(b) Component masstransfer rates.
(c) Murphree vapor tray efficiencies.
Analysis:
Because masstransfer coefficients are given for the gas phase, work with the values of
yi ,n and yiI,n . Because rates instead of fluxes are given, the equations developed in this
section are used with rates rather than fluxes.
(a) Compute the reciprocal rate functions, R, from Eqs. (1231) and (1232), assuming
linear mole fraction gradients such that zi can be replaced by ( yi + yiI ) / 2 . Thus,
with the new designations for 1, 2, 3, the values from Example 12.1 become:
z1 = 0.4654
z2 = 0.2100
z3 = 0.3246
z
z
z
0.4654 0.2100 0.3246
V
R11 = 1 + 2 + 3 =
+
+
= 0.000479 h / lbmol
k13 k12 k13
1955
2797
1955
z
z
z
0.2100 0.4654 0.3246
V
R22 = 2 + 1 + 3 =
+
+
= 0.000388 h / lbmol
k23 k 21 k 23
2407
2797
2407
V
R12 = âˆ’ z1 1
1
1
1
= âˆ’0.4654
âˆ’
= âˆ’0.0000717 h / lbmol
âˆ’
2797 1955
k12 k13 V
R21 = âˆ’ z2 1
1
1
1
= âˆ’0.2100
âˆ’
= âˆ’0.0000122 h / lbmol
âˆ’
2797 2407
k 21 k 23 In matrix form: RV = 0.000497 0.0000717 0.0000122 0.000388 From Eq. (1229), by matrix inversion, Îº V = RV âˆ’1 = 2021 âˆ’ 3735
. âˆ’63.6 2589 Exercise 12.6 (continued)
Analysis: (continued)
Because the offdiagonal terms in the above 2 x 2 matrix are much smaller that the
diagonal terms, the effect of coupling in this example is small, approximately 10%.
From Eq. (1227) with units of JV in lbmol/h
V
J1
V
J2 = ÎºV
11 ÎºV
12 y1 âˆ’ y1I ÎºV
21 ÎºV
22 I
y2 âˆ’ y2 V
V
I
J1V = Îº11 ( y1 âˆ’ y1I ) + Îº12 ( y2 âˆ’ y2 ) = 2021( 0.4631 âˆ’ 0.4677 ) âˆ’ 373.5(0.2398 âˆ’ 0.1802) = âˆ’31.6 lbmol/h I
J = ÎºV ( y1 âˆ’ y1I ) + ÎºV2 ( y2 âˆ’ y2 )
21
2 V
2 = âˆ’63.6(0.4631 âˆ’ 0.4677) + 2589(0.2398 â...
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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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