Unformatted text preview: 530(26.6) 41.1 1/ 2 = 1.487 (1) For 24inch tray spacing, from Fig. 6.24, CF = 0.082 ft/s
From relations below Eq. (644), since FLV is < 0.1, Ad/A = 0.1
From relation below Eq. (642), FST = (σ/20)0.2 = (18.4/20)0.2 =0.98. FF = 0.75 , FHA = 1.0
From Eq. (642), C = FST FF FHA CF = (0.98)(0.75)(1.0)(0.082) = 0.0603 ft/s
From Eq. (640), U f = C ( ρ L − ρV ) / ρV 1/ 2 4VM V
From Eq. (644), DT =
fU f π (1 − Ad / A ) ρV = 0.0603 ( 41.1 − 1.924 ) /1.924
1/ 2 1/ 2 = 0.272 ft/s 4(530 / 3, 600)(26.6)
=
0.85(0.272)(3.14)(1 − 0.1)(1.924) 1/ 2 = 3. 53 ft Exercise 6.20
Subject: Hydraulic calculations for bottom tray of a sievetray column Given: Absorber of Exercise 6.16 with an absorbent flow rate of 40,000 gpm of nC8. Sieve
trays on 24inch spacing , with a weir height of 2.5 inches and 1/4inch holes. Foaming factor =
0.80 and fraction flooding = 0.70.
Find: (a)
(b)
(c)
(d)
(e)
(f) Column diameter at bottom.
Vapor pressure drop/tray.
Whether weeping will occur.
Entrainment rate.
Fractional decrease in Murphree efficiency due to entrainment.
Froth height in downcomer. Analysis: From Exercise 6.16, V = 800,000/359 = 2,230 lbmol/min of entering gas.
Average MW of gas = 0.725(2.016) + 0.25(16.04) + 0.025(30.05) = 6.22
Assuming ideal gas law for gas at 400 psia and 100oF,
ρV = PM/RT = (400)(6.22)/(10.73)(560) = 0.414 lb/ft3
For the liquid, neglect absorbed components. ρL = 43.9 lb/ft3 = 5.86 lb/gal
MW of nC8 = 114. Therefore, L = 40,000(5.86)/114 = 2,056 lbmol/min
(a) Use entrainment flooding correlation of Fig. 6.24, where the abscissa is,
FLV LM L ρV
=
VM V ρ L 1/ 2 = 2, 056(114) 0.414
2, 230(6.22) 43.9 1/ 2 = 1.64 (1) From Fig. 6.24, for 24inch tray spacing, CF = 0.08. Because FLV > 1, Ad /A = 0.2.
FHA = 1.0, FF = 0.8, and since σ = 20 dynes/cm, FST = 1.0.
From Eq. (624), C = FSTFFFHACF = (1)(0.8)(1)(0.08) = 0.064 ft/s
From Eq. (640), U f = C ρ L − ρV / ρV 1/ 2 = 0.064 43.9 − 0.414 / 0.414 4VM V
From Eq. (644), DT =
fU f π (1 − Ad / A ) ρV 1/ 2 1/ 2 = 0.66 ft / s 4(2, 230 / 60)(6.22)
=
0.70(0.66)(3.14)(1 − 0.2)(0.414) 1/ 2 = 44 ft This is a very large diameter. Would probably use two parallel columns of 31 ft each, but
continue with 44 ft diameter.
(b) Vapor pressure drop per tray is given by Eq. (649), ht = hd + hl + hσ
(2)
From the continuity equation, m = uAρ , hole velocity for 10% hole area =
uo = (2,230/60)(6.22)/(3.14/4)(44)2(0.414)(0.1) = 3.68 ft/s and superficial velocity = 0.368 ft/s
u 2 ρV
3.682 0.414
From Eq. (650), hd = 0186 o2
.
= 0186
.
= 0.045 in. of nC8
Co ρ L
0.732
43.9 Analysis: (continued) Exercise 6.20 (continued) Active bubbling area for Ad /A = 0.2 is Aa = A  2Ad = 0.6 A. So, Ua = 0.368/0.6 = 0.61 ft/s
From Eq. (653), Ks = U a ρV
ρ L − ρV 1/ 2 1/ 2 0.414
= 0.61
43.9 − 0.414 = 0.06 ft/s From Eq. (652), φe = exp(4.257Ks0.91) = exp[4.257(0.06)0.91] = 0.72
From Eq. (654), C = 0.362 + 0.317 exp( −35hw ) = 0.362 + 0.317 exp[ −3.5(2.5)] = 0.362
.
Take weir length, Lw = 0.73DT = 0.73(44) = 32.1 ft = 385 i...
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 Spring '11
 Levicky
 The Land

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