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when taking into account entrainment, occlusion, backmixing, weeping, and tray hydraulics.
The percentage reduction in the number of variables and equations for the ratebased
model is not nearly as great as for the equilibrium model because, at least not for many
components, that number is dominated by the 5C term. Exercise 12.4
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 acetone (1), methanol (2), and water (3) at 14.7 psia
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,
z1 = (0.4913 + 0.5291) / 2 = 0.5102
z2 = (0.4203 + 0.4070) / 2 = 0.4136
z3 = (0.0884 + 0.0639) / 2 = 0.0762
z
z
z
0.5102 0.4136 0.0762
V
R11 = 1 + 2 + 3 =
+
+
= 0.000509 h / lbmol
k13 k12 k13
2154
1750
2154
z
z
z
0.4136 0.5102 0.0762
V
R22 = 2 + 1 + 3 =
+
+
= 0.000487 h / lbmol
k23 k 21 k 23
2503
1750
2503
V
R12 = − z1 1
1
1
1
= −0.5102
−
= −0.000055 h / lbmol
−
1750 2154
k12 k13 V
R21 = − z2 1
1
1
1
= −0.4136
−
= −0.000071 h / lbmol
−
1750 2503
k 21 k 23 In matrix form: RV = 0.000509 − 0.000055 −0.000071 0.000487 From Eq. (1229), by matrix inversion, κ V = RV −1 = 1996 225.4 291 2086 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%. Exercise 12.4 (continued)
Analysis: (continued)
From Eq. (1227) with units of JV in lbmol/h
V
J1
V
J2 = κV
11 κV
12 y1 − y1I κV
21 κV
22 I
y2 − y2 V
I
J1V = κ11 ( y1 − y1I ) + κV2 ( y2 − y2 )
1 = 1996 ( 0.4913 − 0.5291) + 225.4(0.4203 − 0.4070) = −72.45 lbmol/h V
I
J 2 = κV ( y1 − y1I ) + κV2 ( y2 − y2 )
21
2 = 291(0.4913 − 0.5291) + 2086(0.4203 − 0.4070) = 16.74 lbmol/h
From Eq. (1223):
V
V
J 3 = − J1V − J 2 = 72.45 − 16.74 = 55.71 lbmol/h (b) NT = Vn+1  Vn = 1164  1200 = 36 lbmol/h.
From Eq. (1219), but with diffusion and masstransfer rates instead of fluxes,
V
N1V = J1V + z1 NT = −72.45 + 0.5102(36) = 90.8 lbmol/h
V
V
V
N 2 = J 2 + z2 NT = 16.74 + 0.4136(−36) = 1.9 lbmol/h
V
V
V
N 3 = J 3 + z3 NT = 55.71 + 0.0762(−36) = 53.0 lbmol/h (c)
Approximate values of the Murphree vapor tray efficiency are obtained from (123), with
Kvalues at phase interface conditions:
E MVi = yi ,n − yi ,n +1 / KiI,n xi ,n − yi ,n +1 Accordingly: EMV1 = ( 0.4913 − 0.4106 )
(1.507 )( 0.3683) − 0.4106 = 0.559 = 55. 9% EMV2 = ( 0.4203 − 0.4389 )
( 0.900 )( 0.4487 ) − 0.4389 = 0.530 = 53. 0% EMV3 = ( 0.0884 − 0.1505 )
( 0.3247 )...
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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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