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

From example 37 the diffusivity of methanol in water

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Unformatted text preview: NOR PAC, ah / a = 0.85(0.651)(69.2) 0.25 (6.63 × 10 −5 ) 0.1 = 0.61 For Montz, ah / a = 0.85(0.482)(13.2) 0.25 (9.91 × 10 −5 ) 0.1 = 0.31 From Eq. (6-97), the fractional liquid holdup is hL = 12 6.63 × 10 −5 For NOR PAC, hL = (12) 69.2 1/ 3 1/ 3 ( 0.61) 2/3 N FrL 1/ 3 N Re L = 0.0162 m3/m3 9.91× 10−5 2/3 For Montz, hL = (12) ( 0.31) = 0.0241 m3/m3 13.2 These liquid holdups at the top of the column are quite small. ah a 2/3 Exercise 7.52 (continued) Analysis: (a) (continued) Liquid holdup based on conditions at the bottom of the column. By the continuity equation, superficial liquid velocity is, uL = LML /ρLS = (2570/3600)(18.1)/ [62.4(3.14)(DT)2/4] = 0.264/(DT)2 ft/s 2 0.264 / DT 62.4 90800 u Lρ L From Eq. (6-98), N Re L = = = 2 µ L a (0.27)(0.000672)a DT a 2 2 u L a ( 0.264 / DT ) a a = = = 0.00216 4 g DT 32.2 2 From Eq. (6-99), N FrL Packing DT, ft uL, ft/s a, ft2/ft3 N Re L NOR PAC Montz 3.4 4.2 0.0228 0.0150 26.5 91.4 296 56.3 N FrL 4.28 x 10-4 6.39 x 10-4 0. 0 Since both values of N Re L are > 5, use Eq. (6-101), ah / a = 0.85Ch N Re25 N Fr.1 L L . For NOR PAC, ah / a = 0.85(0.651)(296) 0.25 (4.28 × 10 −4 ) 0.1 = 1057 For Montz, ah / a = 0.85(0.482)(56.3) 0.25 (6.39 × 10 −4 ) 0.1 = 0.54 From Eq. (6-97), the fractional liquid holdup is hL = 12 4.28 × 10 −4 For NOR PAC, hL = (12) 69.2 1/ 3 1/ 3 (1.057 ) 2/3 N FrL N Re L 1/ 3 ah a 2/3 = 0.0269 m3/m3 6.39 × 10 −4 2/3 For Montz, hL = (12) ( 0.54 ) = 0.0341 m3/m3 56.3 These liquid holdups at the bottom of the column are quite small. mV (c) From Eq. (7.52), HOG = HG + H L , where we must use the slope, m, of the equilibrium L curve instead of a K-value because the equilibrium curve is curved. At the top of the column: Near a mole fraction, xD , of 0.9, m = (0.9780 - 0.9359)/0.10 = 0.42 Estimate HL from Eq. (6-132), using the following properties and parameters: NOR PAC Montz CL 1.080 1.165 3 3 hL , m /m 0.0162 0.0241 0.947 0.930 ε , m3/m3 uL, ft/s 0.00898 0.00591 2 3 a, m /m 86.8 300 ah, m2/m3 52.9 93 Exercise 7.52 (continued) Analysis: (c) at the top of the column (continued) Need an estimate of the diffusivity of methanol in water at high concentrations of methanol. From Example 3.7, the diffusivity of methanol in water at a mole fraction of 0.8 and 25oC is 1.5 x 10-5 cm2/s. Use Eq. (3-39) to correct this for temperature to 170oF or 350 K and viscosity. For methanol, viscosity at 25oC (298 K) = 0.55 cP and at 350 K, viscosity = 0.32 cP. Therefore, 350 0.55 DMeOH = 150 × 10−5 . = 3.0 × 10−5 cm2/s 298 0.32 From Eq. (6-132), for NOR PACK, using SI (instead of American Engineering) units, 11 HL = CL 12 1/ 6 4hL ε DL au L 1/ 2 uL a 1/ 6 a 1 1 = 1.08 12 aPh 4(0.0162)(0.947) (3.0 × 10−9 )(86.8)(0.00274) 1/ 2 0.00274 86.8 a m aPh 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). = 0.174 d h = packing hydraulic diameter = 4 Reynolds number = N ReL , h = ε 0.947 =4 = 0.0436 m = 0.143 ft 86.8 a u L d h...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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