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Unformatted text preview: s = 169 psia, and yA b = 0.001609. At the top, PAs = 1995 psia, yA b = 0.0882, T = 126oF,
.
.
and P = 14.3 psia.
Assumptions: All masstransfer resistance resides in the gas phase because pure water is used.
Assume air rate pertains to moisturefree air. Bulkflow effect may be appreciable.
Find: Overall masstransfer coefficient, Ky
Analysis: From a rearrangement of Eq. (3234) and including the bulkflow effect,
nA 10 − yA LM
.
Ky =
(1)
A yA i − yA b
LM However, no data are given on the packing to enable the determination of the area, A.
Therefore, let a = area, A, for mass transfer per unit volume, V, of tower. Then Eq. (1) becomes, Kya = nA 10 − yA
. LM V yA i − yA b LM (2) Therefore, Kya will be determined instead of Ky .
The rate of mass transfer of water into the air is obtained by material balance, with the water
mole fraction data and the air flow rate (waterfree basis), where, nA = n yA top
1.0  yA top − yA bottom
1.0  yA bottom = 0.401 0.0882
0.001609
−
= 0.0381 lbmol/h
1.0  0.0882 1.0  0.001609 The volume over which the data were taken = V = SH = 0.5(0.7) = 0.35 ft2
Interface mole fractions of water vapor are, yA itop = PAstop / Ptop = 1995 / 14.3 = 0.1395
. yA i bottom = PAsbottom / Pbottom = 1.69 / 14.1 = 01199
. Exercise 3.40 (continued)
Analysis: (continued) y A i − yA b LM = yA itop − yA btop − yA ibottom − yAbbottom
ln = yA itop − yA btop
yA ibottom − yA bbottom 01395 − 0.0882 − 01199 − 0.001609
.
.
= 0.0802
01395 − 0.0882
.
ln
01199 − 0.001609
. For the bulkflow effect, (1.0  yA)LM at the top = 1 − yA i top ln − 1 − yA b top = bottom = 1 − yA i top 1 − yA b .
1 − 01395 − 1 − 0.0882
= 0.886
1 − 0.1395
ln
1 − 0.0882 top (1.0  yA)LM at the bottom = 1 − yA i bottom ln − 1 − yA b 1 − yA i bottom 1 − yA b 1 − 01199 − 1 − 0.001609
.
= 0.938
1 − 0.1199
ln
1 0.001609 bottom The average value of (1.0  yA)LM is 0.912.
From Eq. (2),
K ya = 0.0381(0.912)
= 1.24 lbmol/hft3unit mole fraction
(0.35)(0.0802) For the same packing and gas and liquid velocities, this value can be used to design a large unit. Exercise 4.1
Subject: Degrees of freedom analysis for a threephase equilibrium stage.
Given: Equilibrium stage of Figure 4.35, with two feeds (one vapor, one liquid), vapor entering
from stage below, liquid entering from stage above, three exiting streams (one vapor, two liquid),
and heat transfer.
Assumptions: Equilibrium stage
Find: (a)
(b)
(c)
(d)
Analysis: List and count of variables.
List and count of equations.
Number of degrees of freedom.
List of reasonable set of design variables.
(a) With 4 streams in and 3 streams out, and heat transfer,
Number of variables = NV = 7(C+3) + 1 = 7C + 22 The variables are 7 total flow rates, 7 temperatures, 7 pressures, 1 heat transfer rate and C mole
fractions for each of the 7 streams.
(b) The equations are:
C
Component material balances
1
Energy balance
2
Pressure identity equations for the 3 exiting streams
2
Temperature identity equation...
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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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