Unformatted text preview: nt than the liquidphase resistance.
From Eq. (656), EOV = 1 − exp(− N OG ) = 1 − exp(−1.488) = 0.774 or 77.4%
This is in very good agreement with the performance data. Exercise 7.42
Subject:
Estimation of efficiencies, EMV and Eo from EOV for methanolwater mixture, as
measured with a small Oldershaw column.
Given: Column conditions from Exercise 7.41.
Find: EMV and Eo
Analysis: Case 1: Assume complete mixing on the trays.
EMV = EOV = 0.65 or 65% Case 2: Assume plug flow of liquid with no longitudinal diffusion. Take
conditions at the top of the column.
From Eq. (77), L/V = R/(R + 1) = 0.947/(1 + 0.947) = 0.486
From Eq. (633), λ = m/(L/V)
From the vaporliquid equilibrium data given in Exercise 7.41,
m = dy/dx = (1  0.915)/(1  0.793) = 0.41
λ = 0.41/0.486 = 0.844
From Eq. (632),
EMV = 1
1
[ exp(λEOV ) − 1] =
{exp [0.844(0.65)] − 1} = 0.754 or 75.4%
λ
0.844 The actual value of EMV probably lies inbetween 65% and 75.4%, or say 70%.
From Eq. (637), assuming that the equilibrium and operating lines are straight, Eo = log [1 + EMV (λ − 1) ] log [1 + 0.70(0.844 − 1) ]
=
= 0.68 or 68%
log λ
log(0.844) Exercise 7.43
Subject:
Estimation of column diameter at top tray of a distillation column separating
benzene from monochlorobenzene.
Given: Conditions at the top tray in Fig. 7.46.
Find: Column diameter for a valve tray at 85% of flooding.
Analysis: Use Fig. 6.24, where the value of FLV is needed.
V = 336.5 lbmol/h, L = 274.7 lbmol/h
Because both the vapor and liquid are almost pure benzene, take MV and ML = 78.1
From Perry's Handbook, the liquid density of benzene at 204oF and 23 psia = ρL = 52 lb/ft3
From the ideal gas law, ρV = PMV/RT = (23)(78.1)/[(10.73)(664)] = 0.252 lb/ft3
FLV 1/ 2 LM L ρV
=
VMV ρ L = (274.7)(78.1) 0.252
(336.5)(781) 52
. 0.5 = 0.057 Assume 24inch tray spacing. From Fig. 6.24, CF = 0.37
From below Eq. (644), Ad/A = 0.1
From the Handbook of Chemistry and Physics, for benzene under its own vapor, σ = 30 dyne/cm.
From below Eq. (642), FST = (30/20)0.2 = 1.084
Assume that the foaming factor, FF = 1.0 and take FHA = 1.0
.
From Eq. (642),
C = 1.084(1.0)(1.0)(0.37) = 0.40 ft/s
From Eq. (640),
ρ − ρV
Uf =C L
ρV 1/ 2 52 − 0.252
= 0.40
0.252 0.5 = 5.73 ft / s From Eq. (644), 4VM V
DT =
fU f π(1 − Ad / A)ρV 1/ 2 4(336.5 / 3600)(78.1)
=
0.85(5.73)(3.14)(1 − 0.1)(0.252) 0.5 = 2.9 ft Exercise 7.44
Subject: Diameters and heights for separation of propylene from propane in two columns. Given: Feed and product conditions and numbers of trays for sievetray columns in Fig. 7.47.
Assumptions: Constant molar overflow. 85% of flooding.
Find: Column diameters, tray efficiencies, numbers of actual trays, and column heights.
Analysis:
Column diameters: Assume that the diameter of the second column is controlled at the top tray,
and than the diameter of the first column is controlled by the bottom tray.
Top tray of second column: Use the properties of pure propylene.
Distillate rate = 3.5 + 360  12.51 = 350.99 lbmol/h
With an R = 15.9, L =...
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 Spring '11
 Levicky
 The Land

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