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Unformatted text preview: 7 inches liquid = 0.175 psi/tray.
This is a very high pressure drop/tray.
Consider next, the conditions at the bottom tray of the column.
From the continuity equation, uo = m/AhρV
uo = (1,789/3,600)(18.1)/[(2.26)(0.0408)] = 97.9 ft/s and Co = 0.73
97.9 2 0.0408
From Eq. (3), hd = 0186
.
= 2.28 inches of liquid
0.732
59.5 (4) Superficial velocity based on bubbling area = Ua = 97.9(2.26/22.6) = 9.8 ft/s
From Eq. (653), Ks = U a ρV
ρ L − ρV 1/ 2 = 9.8 0.0408
59.5 − 0.0408 1/ 2 = 0.255 ft/s From Eq. (652), φe = exp(4.257Ks0.91) = exp[4.257(0.255)0.91] = 0.293 Analysis: (b) (continued) Exercise 7.49 (continued) From Eq. (654), Cl = 0.362 + 0.317 exp(−3.5hw ) = 0.362 + 0.317 exp[−3.5(2.0)] = 0.362
Weir length = Lw = 42.5 in.
Volumetric liquid rate = qL = 3,351(18.1)/[60(8.33)(59.5/62.4)] =127 gpm
From Eq. (651), hl = φe hw + Cl qL
Lw φe 2/3 127
= 0.293 2 + 0.362
( 42.5 ) 0.293 2/3 = 1.08 in. liq From Eq. (655), with maximum bubble size = DH = 3/16 inch = 0.00476 m and liquid density =
59.5 lb/ft3 or 954 kg/m3, and σ = 58 dyne/cm = 0.058 kg/s2,
hσ = 6σ / gρ L DBmax = 6(0.058) / (9.8)(954)(0.00476) = 0.0078 m = 0.31 in. liquid
From Eq. (2), ht = hd + hl + hσ =2.28 + 1.08 + 0.31 = 3.67 inches liquid = 0.125 psi/tray.
This pressure drop is a slightly on the high side.
(c) From Fig. 6.28 for the top tray, with FLV = 0.0192 and % flood = 95%, fractional
entrainment = ψ = 0.50, which is fairly high.
From Fig. 6.28 for the bottom tray, with FLV = 0.0473 and % flood = 51%, fractional
entrainment = ψ = 0.017, which is low.
The fractional entrainment is defined as, ψ = e/(L + e). Rearranging, entrainment rate = e =
ψL/(1  ψ).
At the top tray, e = 0.50(945)/(1  0.50) = 945 lbmol/h, which is very excessive.
At the bottom tray, e = 0.017(3,351)/(1  0.017) = 58 lbmol/h
(d) From Eqs. (670) and (672), the froth height in the downcomer = hdf =(ht + hl + hda) / 0.5
Estimate the head loss for flow under the downcomer from Eq. (671). Area for flow under the
downcomer apron = Ada = Lwha = 42.5(2  0.5) = 63.8 in2 = 0.433 ft2 77.2
For the top tray, qL = 77.2 gpm. hda = 0.03
100(0.433)
hdf = (6.47 + 0.61 + 0.095) / 0.5 = 14.4 in. liquid 2 = 0.095 in. liquid 128
For the bottom tray, qL = 128 gpm. hda = 0.03
100(0.433)
hdf = (3.67 + 1.08 + 0.262) / 0.5 = 10.0 in. liquid 2 = 0.262 in. liquid The new operation is not acceptable in the rectifying section because % flooding is 95%, causing
very high entrainment. Should retrofit that section with structured packing. Bottom section is
operable with the trays if the higher pressure drop is acceptable. Exercise 7.50
Subject:
Determination of HETP from experimental data for a packed column separating
benzene and dichloroethane at total reflux.
Given: 10 feet of packing, with operation at 1 atm. Vaporliquid equilibrium data in terms of
benzene mole fractions. Measured benzene mole fractions are xD = 0.653 and xB = 0.298.
Find: HETP in inches. Limitations of the use of the HETP for design.
Analysis: In the plot below, the yx equilibrium data and the operating line...
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 Spring '11
 Levicky
 The Land

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