Unformatted text preview: n.
Volumetric liquid rate = qL = m/ρL = (1,943/60)(18)/(1,000)(0.003785 m3/gal) = 154 gpm
2/3 qL
From Eq. (651), hl = φ e hw + C
Lw φ e 154
= 0.30 2 + 0.362
54.3 0.30 2/3 = 1.09 in. of liquid From Eq. (655), with maximum bubble size of 3/16 inch = 0.00476 m,
hσ = 6σ / gρ L DBmax = 6(70 / 1000) / (9.8)(1,000)(0.00476) = 0.009 m = 0.35 in. of liquid
From Eq. (2), ht = hd + hl + hσ =1.42 + 1.09 + 0.35 = 2.86 in. liquid = 0.103 psi/tray =
0.7 kPa/tray
(c) DT = 1.89 m, A = 2.804 m2 = 28,040 cm2, Aa = 0.8(2.804) = 2.24 m2 = 22,400 cm2
φe = 0.30, hl = 1.09 in. = 2.77 cm
Ua = 7.8 ft/s = 238 cm/s, Uf =9.44 ft/s, f = Ua/Uf = 7.8/9.44 = 0.826
F = UaρV0..5 = 2.38(1.05)0.5 = 2.44 (kg/m)0.5/s , qL = 154 gpm = 9,716 cm3/s
hA
2.77(22,400)
From Eq. (664), t L = l a =
= 6.39 s
qL
9,716
1 − φ e hl (1 − 0.30)2.77
=
= 0.027 s
φ eU a
0.30(238)
From the WilkeChang Eq. (339), DL = 1.12 x 105 cm2/s
0
.
From Eq. (667), k L a = 78.8 DL.5 ( F + 0.425) = 78.8(112 × 10−5 )(2.44 + 0.425) = 0.756 s1
From Perry's Handbook, with a temperature correction based on Eq. (336), DV = 0.127 cm2/s
From Eq. (666), From Eq. (665), t G = kG a = 0
1,030 DV .5 f − 0.842 f 2 hl0.5 = 1,030(0127) 0.5 0.826 − 0.842 0.826
.
2.77 0.5 2 = 55.4 s1 From Eq. (663), N L = k L at L = 0.756(6.39) = 4.83
From Eq. (662), N G = kG atG = 55.4(0.027) = 1.5
(d) From p. 271, for acetone, A = L/KV = 1.38. Therefore, KV/L = 1/1.38 = 0.725
1
1
1
From Eq. (661), N OG =
=
=
= 122
.
1
KV / L
1 0.725 0.667 + 0150
.
+
+
NG
NL
15 4.83
.
(e) Mass transfer is controlled by the vapor phase.
(f) From a rearrangement of Eq. (656), E OV = 1 − exp( − N OG ) = 1 − exp( −122) = 0.705 or 70 %
.
Now, the separation requires 10 equilibrium stages. Assume the liquid on a tray is well mixed.
Then, from Eq. (631), EMV = EOV = 0.70. From below Eq. (633), take λ=KV/L=0.725.
log 1 + E MV ( λ − 1) log 1 + 0.70(0.725 − 1)
From Eq. (637), Eo =
=
= 0.66 (worst case)
log λ
log 0.725
Therefore, at most, we need from Eq. (621), Na =Nt /Eo = 10/0.66 = 15. Therefore, 30 okay. Exercise 6.23
Subject: Design of a column to strip VOCs from water by air at 15 psia and 70oF.
Given: Conditions in Example 6.2 except that entering flow rates are twice as much.
Assumptions: Dilute system such that changes in vapor and liquid rates in the column are
negligible. Sieve trays on 24inch spacing, with 10 % hole area of 3/16inch holes, FF = 0.9,
FHA = 1.0 and f = 0.80.
Find: (a)
(b)
(c)
(d)
(e) Number of equilibrium stages required.
Column diameter for sieve trays.
Vapor pressure drop per tray.
Murphree vaporpoint efficiency from ChanFair method.
Number of actual trays. Analysis: (a) Need 3 equilibrium stages, as determined in Example 6.2 and which is unchanged
when flow rates are doubled.
(b) liquid rate = 2(500) = 1,000 gpm or 1,000(60)(8.33)/18.02 = 27,800 lbmol/h = L
vapor rate = 2(3,400) = 6,800 scfm or 6,800(60)/379 = 1,077 lbmol/h = V
ρL = 62.4 lb/ft3. From ideal gas law, ρV = PM/RT = (15)(29)/(10.73)(530)...
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 Spring '11
 Levicky
 The Land

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