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

# Therefore the holdup is independent of the gas rate

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Unformatted text preview: NOR PAC and FP = 33 for Montz. Analysis: (b) (continued) Exercise 7.52 (continued) 2 From a rearrangement of the ordinate of Fig. 6.36a, uo = Y g ρH2 O (L) FP ρV 1 f {ρ L } f {µ L } 32.2 62.4 1 = 341 (ft/s)2 and uo = 18.5 ft/s 14 0.0689 16(0.8) . 32.2 62.4 1 2 For Montz, uo = 0.21 = 145 (ft/s)2 and uo = 12.0 ft/s 33 0.0689 16(0.8) . For fraction of flooding = f = 0.7, uV = uo f = 18.5(0.7) = 13.0 ft/s for NOR PAC, uV = uo f = 12.0(0.7) = 8.4 ft/s for Montz. 2 For NOR PAC, uo = 0.21 4VM V From Eq. (6-103), column diameter = DT = fu0 πρV 4(1495 / 3600)(30.9) For NORPAC, DT = 13(3.14)(0.0689) 1/ 2 4VM V = uV πρV 1/ 2 1/ 2 = 4.3 ft. 1/ 2 4(1495 / 3600)(30.9) For Montz, DT = = 5.3 ft 8.4(3.14)(0.0689) Column diameter based on conditions at the bottom of the column, where T = 207oF and P = 15.5 psia = 1.05 atm. Use Figs. 6.36a, b, and c to compute flooding velocity, following Example 6.13. From the ideal gas law, ρV = PMV/RT = (1.05)(18.1)/[0.730(207 + 460)] = 0.039 lb/ft3 For the liquid, which is mainly water, use ρL = 1.00 g/cm3 = 62.4 lb/ft3 and µ L = 0.27 cP. LM L The abscissa in Fig. 6.36a = X = VM V ρV ρL 1/ 2 2569.7(18.1) 0.039 = 1367.8(18.1) 62.4 1/ 2 = 0.047 From Fig. 6.36a, Y at flooding = 0.17; from Fig. 6.36b, for ρwater/ ρL = 1/1 = 1, f{ρL }= 1; and from Fig. 6.36c, for µL = 0.27 cP, f{µL } = 0.77. From Table 6.8, FP = 14 for NOR PAC and FP = 33 for Montz. g ρH2 O (L) 1 2 From a rearrangement of the ordinate of Fig. 6.36a, uo = Y FP ρV f {ρ L } f {µ L } 32.2 62.4 1 = 812 (ft/s)2 and uo = 28.5 ft/s 14 0.039 1(0.77) 32.2 62.4 1 2 For Montz, uo = 0.17 = 345 (ft/s)2 and uo = 18.6 ft/s 33 0.039 1(0.77) For fraction of flooding = f = 0.7, uV = uo f = 28.5(0.7) = 20.0 ft/s for NOR PAC, uV = uo f = 18.6(0.7) = 13.0 ft/s for Montz. 2 For NOR PAC, uo = 0.17 4VM V From Eq. (6-103), column diameter = DT = fu0 πρV 1/ 2 4VM V = uV πρV 1/ 2 Analysis: (b) (continued) Exercise 7.52 (continued) 4(1367.8 / 3600)(18.1) For NOR PAC, DT = 20(3.14)(0.039) 1/ 2 = 3.4 ft. 1/ 2 4(1367.8 / 3600)(18.1) For Montz, DT = = 4.2 ft 13.0(3.14)(0.039) (a) For liquid holdup estimates, assume the column operates in the preloading region. Therefore, the holdup is independent of the gas rate. Follow Example 6.12. Pertinent packing characteristics from Table 6.8: Ch Packing a, m2/m3 a, ft2/ft3 ε NOR PAC 86.8 26.5 0.947 0.651 Montz 300 91.4 0.930 0.482 Use Eqs. (6-97) to (6-101), which requires calculating the liquid Reynolds and Froude numbers. Liquid holdup based on conditions at the top of the column. By the continuity equation, superficial liquid velocity is, uL = LML /ρLS = (727/3600)(30.9)/ [48(3.14)(DT)2/4] = 0.166/(DT)2 ft/s 2 0166 / DT 48 . 33900 u Lρ L From Eq. (6-98), N Re L = = = 2 µ L a (0.35)(0.000672)a DT a 2 2 u L a ( 0.166 / DT ) a a = = = 0.000856 4 g 32.2 DT 2 From Eq. (6-99), N FrL Packing DT, ft a, ft2/ft3 uL, ft/s N Re L N FrL NOR PAC 4.3 0.00898 26.5 69.2 6.63 x 10-5 Montz 5.3 0.00591 91.4 13.2 9.91 x 10-5 0. 0 Since both values of N Re L > 5, use Eq. (6-101), ah / a = 0.85Ch N Re25 N Fr.1 L L For...
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