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

Separation Process Principles 2n Seader& Henley Solutions Manual

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Unformatted text preview: .0000631) HL = 0.24 −0.2 0.75 1 = 0.154 m 1.56 −0.45 = 1.56 a aPh Exercise 7.53 (continued) Analysis: (c) at the top of the column (continued) Now estimate the value of HG from Eq. (6-133). Estimate the gas diffusivity from Eq. (3-36), where from Table 3.1, . V = 6(15.9 + 2.31) − 18.3 = 90.96 for benzene, and V = 7(15.9 + 2.31) + 2.31 − 18.3 = 1115 for toluene and molecular weights are 787 for benzene and 92 for toluene. 0.00143T 1.75 DG = DAB = 1/ 2 1/ 3 2 P VA+ (1 / M A ) + (1 / M B ) 1/ 3 2 V B 0.00143(355)1.75 = 1/ 2 = 0.052 cm2 / s 2 1/ 3 90.96 + 1115 . (1 / 78) + (1 / 92) Use the following additional properties and parameters, together with µV = 0.0091 cP and ρV = uρ (7.84)(0.171) 0.171 lb/ft3 , where from Eq. (6-134), N ReV = V V = = 7,800 aµV (28.1)(0.0091)(0.000672) µ (0.0091)(0.000672) From (6-135) NScV = V = = 0.639 ρV DV 0.052 (0.171) (2.54) 2 (12) 2 From Eq. (6-133), with CV = 0.408 from Table 6.8, 1/ 3 2 (1) 1 1/ 2 4ε HG = ( ε − hL ) CV a4 1/ 2 (N ) (N ) −3/ 4 ReV −1/ 3 ScV 1 1/ 2 4(0.977) = ( 0.977 − 0.0273) 0.408 92.34 1/ 2 uV a DV aPh ( 7,800 ) −3/ 4 ( 0.639 ) −1/ 3 (2.39)(92.3) = 0.23 m (0.052 × 10−4 )(144) At the top of the column, vapor rate = V =712 lbmol/h and liquid rate = L = 437 lbmol/h. Therefore, V/L = (712)/(437) = 1.63 At the bottom of the column: Near a mole fraction, xB , of 0.05, m = 2.1 Need an estimate of the liquid diffusivity of benzene in toluene at low concentrations of benzene. At the bottom, temperature = 227oF = 382 K and liquid viscosity = 0.25 cP. This time, benzene is component A and toluene is B. From Eq. (3-42), T 1.29 (PB0.5 / PA0.5 ) 3821.29 (24550.5 / 205.30.42 ) . DL = DAB = 155 × 10−8 . = 155 × 10−8 . = 6.64 × 10−5 0.92 0.23 0.92 0.23 µB υ B 0.25 118.2 Exercise 7.53 (continued) Analysis: (c) bottom of the column (continued) From Eq. (6-132), using SI units 11 HL = CL 12 1/ 6 1 1 = 1.168 12 4hL ε DL au L 1/ 6 1/ 2 uL a a aPh 4(0.0312)(0.977) (6.64 ×10 −9 )(92.3)(0.00911) 1/ 2 0.00911 92.3 a a = 0.26 aPh aPh m To compute (aPh /a), we need for the liquid phase, the Reynolds, Weber, and Froude numbers from equations (6-138), (6-139), and (6-140), respectively, based on the packing hydraulic diameter from (6-137). d h = packing hydraulic diameter = 4 Reynolds number = N ReL ,h = ε 0.977 =4 = 0.0423 m = 0.139 ft a 92.3 uL d h ρ L (0.0299)(0.139)(48.7) = = 1205 [(0.25)(0.000672)] µL Take the surface tension of toluene at 227oF as σ = 18 dynes/cm = 0.00123 lbf/ft or 0.00123(32.2) = 0.0397 lbm/s2 2 uL ρ L d h ( 0.0299 ) (48.7)(0.139) = = 0.152 0.0397 σ 2 Weber number = N WeL ,h = ( 0.0299 ) = 0.000200 u2 = L= gd h 32.2(0.139) 2 Froude number = N FrL ,h From (6-136), ( ) (N aPh −1/ 2 = 1.5 ( ad h ) N Re L ,h a −0.2 = 1.5 [ (92.3)(0.0423) ] −1/ 2 ) (N ) 0.75 We L , h −0.45 FrL , h (1205 ) ( 0.152 ) ( 0.000200 ) HL = 0.26 −0.2 0.75 1 = 0.126 m 2.07 −0.45 = 2.07 Exercise 7.53 (continued) Analysis: (c) bottom of the column (continued) Now estimate the value of HG from Eq. (6-133). Estimate the gas diffusivity from Eq. (3-36), by ratio...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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