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

1 x 10 2 cp and v u 00688 lbft3 where from eqs 6 134

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Unformatted text preview:  L (0.00898)(0.143)(48) = = 262 [(0.35)(0.000672)] ĀµL Take the surface tension of methanol at 154oF as Ļƒ = 17 dynes/cm = 0.001165 lbf/ft or 0.001165(32.2) = 0.0375 lbm/s2 2 u LĻ L d h ( 0.00898 ) (48)(0.143) = = = 0.0148 Ļƒ 0.0375 2 Weber number = N WeL ,h ( 0.00898) = 0.0000175 u2 = L= gd h 32.2(0.143) 2 Froude number = N FrL , h From (6-136), ( ) (N aPh āˆ’1/ 2 = 1.5 ( ad h ) N Re L ,h a āˆ’0.2 = 1.5 [ (86.8)(0.0436) ] āˆ’1/ 2 ) (N ) 0.75 We L , h āˆ’0.45 FrL , h ( 262 ) ( 0.0148) ( 0.0000175) āˆ’0.2 0.75 āˆ’0.45 = 1.485 a aPh Exercise 7.52 (continued) Analysis: (c) at the top of the column (continued) HL = 0.174 1 = 0.117 m 1.485 From Eq. (6-132), for Montz, using SI units, 11 HL = CL 12 = 0.0253 1/ 6 a aPh 4hL Īµ DL au L 1/ 2 uL a a 1 1 = 1.165 12 aPh 1/ 6 4(0.0241)(0.930) (3.0 Ɨ 10āˆ’9 )(300)(0.00180) 1/ 2 0.00180 300 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 = Īµ 0.930 =4 = 0.0124 m = 0.0407 ft a 300 uL d h Ļ L (0.00591)(0.0407)(48) = = 49.1 ĀµL [(0.35)(0.000672)] 2 u LĻ L d h ( 0.00898 ) (48)(0.407) = = 0.00421 Ļƒ 0.0375 2 Weber number = N WeL ,h = ( 0.00898) = 0.0000615 u2 = L= gd h 32.2(0.0407) 2 Froude number = N FrL ,h From (6-136), ( aPh āˆ’1/ 2 = 1.5 ( ad h ) N Re L ,h a = 1.5 [ (300)(0.0124)] ) (N āˆ’0.2 āˆ’1/ 2 HL = 0.0253 1 = 0.054 m 0.47 ) (N ) 0.75 We L , h āˆ’0.45 FrL , h ( 49.1) ( 0.00421) ( 0.0000615 ) āˆ’0.2 0.75 āˆ’0.45 = 0.47 a aPh Exercise 7.52 (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 = 131 for water, and V = 15.9 + 4( 2.31) + 611 = 312 for methanol and molecular weights are 32 for methanol and 18 for water. 0.00143T 1.75 DV = DAB = 1/ 2 1/ 3 1/ 3 2 2 P ( V ) A + ( V )B (1/ M A ) + (1/ M B ) 0.00143(350)1.75 = = 0.28 cm 2 /s 1/ 2 2 1/ 3 1/ 3 2 (13.1) + ( 31.2 ) (1/18) + (1/ 32) Use the following additional properties and parameters, together with ĀµĻ‚ = 1.1 x 10-2 cP and ĻV uĻ Āµ = 0.0688 lb/ft3 , where from Eqs. (6-134) and (6-135), N ReV = V V and N ScV = V aĀµV ĻV DV (1) CV uV , m/s (NSc)V (NRe)V NOR PAC 0.322 3.96 0.36 4,560 Montz 0.422 2.56 0.36 850 From Eq. (6-133), for NOR PAC, using SI units, 1 1/ 2 4Īµ HG = ( Īµ āˆ’ hL ) CV a4 1/ 2 (N ) (N ) āˆ’3/ 4 ReV āˆ’1/ 3 ScV 1 1/ 2 4(0.947) = ( 0.947 āˆ’ 0.0162 ) 0.322 86.84 1/ 2 uV a DV aPh ( 4, 560 ) āˆ’3/ 4 ( 0.36 ) āˆ’1/ 3 (3.96)(86.8) = 0.34 m (0.28 Ɨ 10āˆ’4 )(71.2) For Montz, 1 1/ 2 4(0.930) HG = ( 0.947 āˆ’ 0.0241) 0.422 3004 1/ 2 ( 850 ) āˆ’3/ 4 ( 0.36 ) āˆ’1/ 3 (2.56)(300) = 0.16 m (0.28 Ɨ 10āˆ’4 )(78) At the top of the column, vapor rate = V =1,495 lbmol/h and liquid rate = L = 727 lbmol/h. Therefore, mV/L = (0.42)(1,495)/(727) = 0.864 mV From Eq. (7.51), for NOR PAC, H OG = H G + H L = 0.34 + (...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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