{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# 433 hdf 647 061 0095 05 144 in liquid 2

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. (6-53), 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. (6-52), φe = exp(-4.257Ks0.91) = exp[-4.257(0.255)0.91] = 0.293 Analysis: (b) (continued) Exercise 7.49 (continued) From Eq. (6-54), 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. (6-51), 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. (6-55), 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. (6-70) and (6-72), the froth height in the downcomer = hdf =(ht + hl + hda) / 0.5 Estimate the head loss for flow under the downcomer from Eq. (6-71). 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. Vapor-liquid 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 y-x equilibrium data and the operating line...
View Full Document

{[ snackBarMessage ]}