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Unformatted text preview: 0.37)2.24
=
= 0.028 s
φ eU a
0.37(137) 0
From Eq. (667), k L a = 78.8 DL.5 ( F + 0.425) = 78.8(181 × 10−5 )(190 + 0.425) = 0.78 s1
.
. From Eq. (666),
kG a = 0
1,030 DV .5 f − 0.842 f 2
0.5
l h = 1,030(7.86 × 10 −2 ) 0.5 0.44 − 0.842 0.44
2.24 0.5 2 = 53.4 s1 From Eq. (663), N L = k L at L = 0.78(28.6) = 22.3
From Eq. (662), N G = k G at G = 53.4(0.28) = 15
.
From Example 6.7, KV/L = 0.662
1
1
1
From Eq. (661), N OG =
=
=
= 144
.
1
KV / L
1 0.662 0.667 + 0.030
+
+
NG
NL
15 22.3
.
Mass transfer is controlled by the vapor phase.
From a rearrangement of Eq. (652), EOV = 1 − exp(− N OG ) = 1 − exp(−1.44) = 0.76 or 76%. Exercise 6.22
Subject: Sizing, hydraulics, and mass transfer for an acetone absorber Given: Acetone absorber of Fig. 6.1, using sieve trays with 10% hole area, 3/16inch holes, and
an 18inch tray spacing. Foaming factor = 0.85.
Find: (a) Column diameter for 75% of flooding.
(b) Vapor pressure drop per tray.
(c) Number of transfer units.
(d) Number of overall transfer units.
(e) Controlling resistance to mass transfer.
(f) Murphree point vapor efficiency
If 30 trays are adequate.
Analysis: (a) Base column diameter on the top tray, where pressure is the lowest and gas flow
rate is the highest. Use data in Fig. 6.1. L = 1,943 kmol/h, ML = 18, ρL = 1,000 kg/m3
V = 6.9 + 144.3 + 536.0 + 22.0 + 0.1 = 709.3 kmol/h, MV = 28.8, P = 90 kPa, T = 25oC
From the ideal gas law, ρV = PM/RT = (90)(28.8)/(8.314)(298) = 1.05 kg/m3
Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
1/ 2 1/ 2 LM L ρV
1,943(18) 105
.
FLV =
=
= 0.056
VMV ρ L
709.2(28.8) 1,000
From Fig. 6.24, for 18inch tray spacing, CF = 0.28 ft/s. Because FLV < 1, Ad /A = 0.1.
FHA = 1.0, FF = 0.85, and since σ = 70 dynes/cm, FST = (70/20)0.2 = 1.285
From Eq. (624), C = FSTFFFHACF = (1.285)(0.85)(1)(0.28) = 0.306 ft/s
From Eq. (640), U f = C ρ L − ρV / ρV
From Eq. (644),
4VM V
DT =
fU f π (1 − Ad / A ) ρV 1/ 2 1/ 2 = 0.306 1,000 − 105 / 105
.
. 1/ 2 4(709.2 / 3, 600)(28.8)
=
0.75(9.44 / 3.28)(3.14)(1 − 0.1)(1.05) = 9.44 ft / s 1/ 2 = 1.89 m = 6.2 ft (b) Vapor pressure drop per tray is given by Eq. (649), ht = hd + hl + hσ
(1)
From the continuity equation, m = uAρ , hole velocity for 10% hole area =
uo = (709.2/3,600)(28.8)/(3.14/4)(1.9)2(1.05)(0.1) = 19.0 m/s =62.3 ft/s
Superficial velocity = (0.1)(19.0) = 1.90 m/s = 6.23 ft/s
u 2 ρV
62.32 105
.
From Eq. (650), hd = 0186 o2
.
= 0186
.
= 142 in. of liquid
.
2
Co ρ L
0.73 1,000
Active bubbling area for Ad /A = 0.1 is Aa = A  2Ad = 0.8 A. So, Ua = 6.23/0.8 = 7.8 ft/s
From Eq. (653), Ks = U a ρV
ρ L − ρV 1/ 2 1.05
= 7.8
1,000 − 105
. 1/ 2 = 0.253 ft/s From Eq. (652), φe = exp(4.257Ks0.91) = exp[4.257(0.253)0.91] = 0.30
Assume a 2inch weir height = hw Exercise 6.22 (continued) Analysis: (b) (continued)
.
.
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(6.2) = 4.52 ft = 54.3 i...
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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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