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

148 from fig 636 x 319 vm v l 765 623 from fig

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Unformatted text preview: mass transfer coefficient for Cl2 is higher and the corresponding HG is lower. Therefore, the assumption that GeCl4 controls is correct. (c) If a packed height of 10 ft rather than the computed 8.3 ft is used, then for GeCl4 , NG = lT /HG = 10/1.8 = 5.55 rather than 4.61. y From above, N G = ln in = 5.55 yout y Solving, in = 258 Therefore, yout = 0.00162/258 = 0.0000063 yout % absorption of GeCl4 = (1 - 0.0000063/0.00162) = 0.996 or 99.6% compared to 99% specified For Cl2, from above, HG = HG of GeCl4 (DGeCl4/DCl2)2/3 = 1.8(0.000006/0.000013)2/3 = 1.08 Therefore, NG = 10/1.08 = 9.26 = = ln yin yout Solving, yin = 10,500 yout % absorption of Cl2 = (1 - 1/10,500) = 0.9999 or 99.99% compared to 99% specified Exercise 6.28 Subject: Packed tower diameter and pressure drop for conditions of Exercise 6.26. Given: Suggested tower diameter of 0.80 m and pressure drop of 500 N/m2-m of packed height (0.612 in. H2O/ft). Find: (a) (b) (c) (d) Fraction of flooding if FP = 24 ft2/ft3. Pressure drop at flooding. Pressure drop at operating conditions using GPDC chart. Pressure drop at operating conditions using Billet-Schultes correlation. Analysis: (a) Using data from Exercise 6.26 and Fig. 6.36, LML = 600(8.33) = 5,000 lb/min VMV = (1,000/379)(29) = 76.5 lb/min From ideal gas law, ρV = PM/RT = (2)(29)/(0.7302)(460+77) = 0.148 lb/ft3 ρL =62.3 lb/ft3 1/ 2 1/ 2 LM L ρV 5, 000 0.148 From Fig. 6.36, X = = = 3.19 VM V ρ L 76.5 62.3 From Fig. 6.36, at flooding, Y = 0.0053 For the actual operation, by the continuity equation, m = uoSρ Area for flow = S = πD2/4 = (3.14)(0.8/0.3048)2/4 = 5.41 ft2 Superficial velocity = uo = m/Sρ = (76.5/60)/(5.41)(0.148) = 1.59 ft/s From Eq. (6-102), noting that for the dilute aqueous solution at 25oF, f{ρL} = 1 and f{µL} = 0.98 for a liquid viscosity of 0.95 cP at 25oC, 2 uo FP ρV Y= g ρH 2O (1.59) 2 (24) 0.148 (1.0)(0.98)=0.0044 f {ρ L }f {µ L }= 32.2 62.3 1/ 2 u Y 0.0044 Fraction of flooding = o = = uflood Yflood 0.0053 This is too high. Should increase tower diameter. 1/ 2 = 0.91 or 91% (b) From Eq. (6-104), ∆Pflood = 0.115 FP0.7 = 0.115(24)0.7 = 1.06 in. H2O/ft of packed height (c) From Fig. 6.36, ∆P at design = 1.50 in. H2O/ ft of packed height (seems high) (d) From Table 6.8, for 2-inch (50-mm) plastic Pall rings, the packing characteristics are: a = 111.1 m2/m3 or 33.9 ft2/ft3 , ε = 0.919 m3/m3 , Ch = 0.593 , Cp = 0.698, Cs = 2.816 The pressure drop per unit height of packed bed is given by Eq. (6-115), ∆P ε = ∆Po ε − hL 3/ 2 exp 1/ 2 13300 N 3/ 2 ( FrL ) a (1) Exercise 6.28 (continued) Analysis: (d) (continued) where, hL is given by (6-97) and is in m2/m3 and N FrL is given by (6-99). From Eq. (6-110), ∆Po a u2ρ 1 = Ψo 3 V V lT ε 2 KW (2) 1− ε 1 − 0.919 =6 = 0.0143 ft a 33.9 1 2 1 DP 2 1 0.0143 From Eq. (6-111), = 1+ = 1+ = 1045 . KW 3 1 − ε DT 3 1 − 0.919 2.62 Therefore, KW = 0.957 . From Eq. (6-112), DP = 6 From Perry's Handbook, µ for air at 25oC = 0.018 cP, and uo = uV = 1.59 ft/s From Eq. (6-114), N ReV = uV DP ρV...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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