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Unformatted text preview: te. For the 3 ft diameter
column of Examples 6.1 and 6.4, liquid rate = 151.5 kmol/h of water. Assume the liquid rate is
proportional to the column crosssectional area. Thus, for the 11 ft diameter, the liquid rate =
151.5(11/3)2 = 2,037 kmol/h = 162 gpm. From Fig. 6.16, a singlepass tower is adequate.
Assume liquid flow path = ZL = 7 ft. Eq. (634) is used to compute EMV, which requires the
Peclet number, given by Eq. (636),
Zu
N Pe = L = 7(6) = 42
DE N
From Eq. (635), η = Pe
2 4 λEOV
1+
N Pe 1/ 2 −1 = 42
2 1+ 4(0.68)(0.55)
42 1/ 2 − 1 = 0.37 From Eq. (634), EMV/EOV = 1.20. Therefore, EMV = 1.20(0.55) = 0.66
From Eq. (637),
log [1 + EMV (λ − 1) ] log [1 + 0.66(0.68 − 1)]
Eo =
= 0.62
=
log λ
log 0.68
From these results, we see that the column is operating closely to plug flow for the liquid. Exercise 6.18
Subject: Estimation of column diameter based on conditions at bottom tray of a reboiled
stripper.
Given: Vapor and liquid conditions at bottom tray given in Fig. 6.48. Valve trays with 24inch
spacing.
Find: Column diameter for 80% of flooding.
Analysis: Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
1/ 2 LM L ρG
FLV =
VM G ρ L
From Fig. 6.48, V = 546.2 lbmol/h and L = 621.3 lbmol/h (1) Average molecular weights of the gas and liquid are computed from M = C
i =1 zi Mi , Component
M
y
My
x
Mx
Ethane
30.07 0.000006
0.0 0.000010
0.0
Propane
44.10 0.004817
0.2 0.001448
0.1
nButane
58.12 0.602573 35.0 0.391389 22.7
nPentane
72.15 0.325874 23.5 0.430599 31.1
nHexane
86.18 0.066730
5.8 0.176563 15.2
Total
64.5
69.1
Therefore, MV = 64.5 and ML = 69.1
Gas volumetric flow rate = QV = 6.192 ft3/s.
Therefore, ρV = VMV /QV = (546.2)(64.5)/(6.192)(3,600) = 1.58 lb/ft3
Liquid volumetric flow rate = 171.1 gpm = (171.1)(60)/7.48 = 1,372 ft3/h
Therefore, ρL = LML /QL = (621.3)(69.1)/1,372 = 31.3 lb/ft3
1/ 2
621.3(69.1) 158
.
From Eq. (1), FLV =
= 0.274
546.2(64.5) 31.3
Assume 24inch tray spacing. From Fig. 6.24, CF = 0.24 ft/s
From relations below Eq. (644), since FLV is between 0.1 and 1.0,
F − 0.1
0.274 − 01
.
Ad
= 0.1 + LV
= 01 +
.
= 0.119
A
9
9
From Perry's Handbook, surface tension = σ = 7 dyne/cm
From relation below Eq. (642), FST = (σ/20)0.2 = (7/20)0.2 = 0.8. FF = 1.0 , FHA = 1.0
From Eq. (642), C = FST FF FHA CF = (0.8)(1.0)(1.0)(0.24) = 0.192 ft/s
From Eq. (640), U f = C ( ρ L − ρV ) / ρV 1/ 2 = 0.192 ( 31.3 − 1.58 ) /1.58 1/ 2 = 0.833 ft/s Exercise 6.18 (continued)
From Eq. (644), 4VM V
DT =
fU f π (1 − Ad / A ) ρV 1/ 2 4(546.2 / 3, 600)(64.5)
=
0.80(0.833)(3.14)(1 − 0.119)(1.58) 1/ 2 = 3. 56 ft Exercise 6.19
Subject: Estimation of column diameter based on conditions at the top tray of an absorber. Given: Vapor and liquid conditions at the top tray. Valve trays with 24inch spacing.
Find: Flooding velocity and column diameter for 85% of flooding (f = 0.85) Analysis: Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
FLV LM L ρV
=
VM V ρ L 1/ 2 889(109) 1.924
=...
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 Spring '11
 Levicky
 The Land

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