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Unformatted text preview: = 0.0765 lb/ft3
Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
FLV LM L ρV
=
VMV ρ L 1/ 2 27,800(18) 0.0765
=
1,077(29) 62.4 1/ 2 = 0.56 From Fig. 6.24, for 24inch tray spacing, CF = 0.18 ft/s.
Because 0.1 < FLV > 1, from below Eq. (644), Ad /A = 0.1 + (FLV  0.1)/9 = 0.15
FHA = 1.0, FF = 0.9, and since σ = 80 dynes/cm, FST = (80/20)0.2 = 1.32
From Eq. (624), C = FSTFFFHACF = (1.32)(0.9)(1)(0.18) = 0.21 ft/s
From Eq. (640), U f = C ρ L − ρV / ρV
From Eq. (644), for f = 0.80,
4VM V
DT =
fU f π (1 − Ad / A ) ρV 1/ 2 1/ 2 = 0.21 62.4 − 0.0765 / 0.0765 1/ 2 = 6.0 ft / s 4(1, 077 / 3, 600)(29)
=
0.80(6.0)(3.14)(1 − 0.15)(0.0765) (c) Vapor pressure drop per tray is given by Eq. (649), ht = hd + hl + hσ
From the continuity equation, m = uAρ . Hole velocity for 10% hole area =
uo = (1,077/3,600)(29)/(3.14/4)(6)2(0.0765)(0.1) = 40 ft/s
Superficial velocity = (0.1)(40) = 4 ft/s 1/ 2 = 6.0 ft
(1) Exercise 6.23 (continued) Analysis: (c) (continued) 2
uo ρV
402
0.0765
= 0186
.
= 0.68 in. of liquid
2
2
Co ρ L
0.73
62.4
Active bubbling area for Ad /A = 0.15 is Aa = A  2Ad = 0.7 A. Therefore, Ua = 4/0.7 = 5.7 ft/s From Eq. (650), hd = 0186
. ρV
ρ L − ρV From Eq. (653), Ks = U a 1/ 2 0.0765
= 5.7
62.4 − 0.0765 1/ 2 = 0.20 ft/s From Eq. (652), φe = exp(4.257Ks0.91) = exp[4.257(0.20)0.91] = 0.37
Assume a 2inch weir height = hw
.
.
From Eq. (654), C = 0.362 + 0.317 exp( −35hw ) = 0.362 + 0.317 exp[ −35(2.0)] = 0.362
Take weir length, Lw = 0.73DT = 0.73(6)(12) = 53 in.
qL
From Eq. (651), hl = φe hw + C
Lw φe 2/3 1, 000
= 0.37 2 + 0.362
( 53) 0.37 2/3 = 2. 58 in. of liquid From Eq. (655), with maximum bubble size of 3/16 inch = 0.00476 m,
hσ = 6σ / gρ L DBmax = 6(80 / 1,000) / (9.8)(1,000)(0.00476) = 0.0103 m = 0.40 in. of liquid
From Eq. (2), ht = hd + hl + hσ =0.68 + 2.58 + 0.40 = 3.66 in. liquid = 0.13 psi/tray
(d) DT = 6 ft, A = 28.3 ft2, Aa = 0.7(28.3) = 19.8 ft2 = 18,390 cm2
φe = 0.37, hl = 2.58 in. = 6.55 cm, ρG = 0.0765 lb/ft3 = 1.23 kg/m3
Ua = 5.7 ft/s = 1.74 m/s , Uf =6.0 ft/s, f = Ua/Uf = 5.7/6.0 = 0.95
F = UaρV0..5 =1.74(1.23)0.5 = 1.93 (kg/m)0.5/s , qL = 1,000 gpm = 63,100 cm3/s
hA
6.55(18,390)
From Eq. (664), t L = l a =
= 19 s
.
qL
63,100
1 − φ e hl (1 − 0.37)6.55
=
= 0.064 s
φ eU a
0.37(174)
From the WilkeChang Eq. (339), DL = 0.96 x 105 cm2/s
0
From Eq. (667), k L a = 78.8 DL.5 ( F + 0.425) = 78.8(0.96 × 10−5 )(193 + 0.425) = 0.58 s1
.
From Perry's Handbook, with T and P corrections based on Eq. (336), DV = 0.086 cm2/s
From Eq. (665), t G = From Eq. (666),
kG a = 0.5
1, 030 DV ( f − 0.842 f 2 )
0.5
l h = 1, 030(0.086)0.5 0.95 − 0.842 ( 0.95 ) From Eq. (663), N L = k L at L = 0.58(1.9) = 1.1 6.55 0.5 2 = 2. 56 s1 Analysis: (continued) Exercise 6.23 (continued) From Eq. (662), N G = kG atG = 2.56(0.064) = 0.16
(d) From Example 6.2, for benzene, S = KV/L = 9.89
1
1
1
From Eq. (661), N OG =
=
=
= 0.066
1
KV / L
1
9.89 6.25 + 8.99
+
+
NG
NL
016 11
.
.
From a rearrangement of...
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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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