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

# 2 for ng simplifies to ng 000162 dy 000162 ln

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Unformatted text preview: uperficial velocity is, m = uoSρ. Therefore, the superficial velocity is, uo = m/Sρ (1) 3 From the ideal gas law, ρ = PM/RT = (1)(29.7)/(0.7302)(460+77) = 0.076 lb/ft Therefore, from Eq. (1), uo = (2189)/(3.14)(0.076) = 9,170 ft/h or 2.55 ft/s. From Eq. (6-102), noting that for the dilute caustic solution at 25oF, f{ρL} = 1 and f{µL} = 0.98 for a liquid viscosity of 0.95 cP at 25oC, Y= 2 uo FP ρV g ρH 2O f {ρ L }f {µ L }= (2.55) 2 (580) 0.076 (1.0)(0.98)=0.14 32.2 62.4 (uo)flood = uo/0.75 =2.55/0.75 = 3.40 ft/s. Therefore, Yflood = 0.14/(0.75)2 = 0.25 LM L From Fig. 6.36, VM V ρ LM L = X (GM G ) L ρG 1/ 2 62.4 = 0.009(2,191) 0.076 ρV ρL 1/ 2 = 0.009 1/ 2 = 565 lb/h or 0.071 kg/s of entering liquid Exercise 6.27 (continued) Analysis: (b) Assume that GeCl4 is the controlling species, with mass transfer controlled in the gas phase. From Table 6.7, column height = lT = HGNG where, HG = V/kyaS (1) (1 − y ) LM dy (1 − y )( y − yI ) and, N G = (2) At the interface, yI = 0 because it is given that dissolved GeCl4 has no vapor pressure In entering gas, y = (288/214.6)/803 = 0.00162. Therefore (1 - y)bottom = (1 - 0.00162) = 0.9984 In exiting gas, y = 0.01 of y in entering gas = 0.0000162. Therefore (1 - y)top = 1.0000 Therefore, (1 - y)LM = 0.9992. Also, on the average, (1 - y) is approximately 0.9992. Therefore, Eq. (2) for NG simplifies to, NG = 0.00162 dy 0.00162 = ln = 4.61 y 0.0000162 0.0000162 V DPV ′ Equation given for ky is, k y = 1.195 S µ (1 − ε o ) −0.36 ( NSc ) −2 / 3 (3) V = 803/24(3,600) = 0.0093 kmol/s S = 3.14 ft2 or 0.292 m2 V' = (V/S)MG = (0.0093/0.292)(29.7) = 0.944 kg/s-m2 DP = 0.01774 m At 25oC, from Perry's Handbook, for air, µ = 0.018 cP or 1.8 x 10-5 kg/m-s εο = ε - hL ε = 0.63 hL = 0.03591(L')0.331 (4) L' =(LML) /S= 0.071/0.292 = 0.243 kg/s-m2 From Eq. (4), hL = 0.0359(0.243)0.331 = 0.0225 Therefore, εo = 0.63 - 0.0225 = 0.608 NSc = µ/ρDGeCl4 (5) 3 3 From above, ρ of the gas = 0.076 lb/ft = 1.21 kg/m DGeCl4 = 6 x 10-6 m2/s From Eq. (5), NSc = (1.8 x 10-5)/(1.21)(6 x 10-6) = 2.48 0.0093 (0.01774)(0.944) From Eq. (3), k y = 1195 . 0.292 (18 × 10 −5 ) 1 − 0.608 . Equation given for interfacial area is, Equation given for exponent, Therefore, from Eq. (6), a = −0.36 2.48 −2 / 3 14.69(808 V ′ / ρ1/ 2 ) n a= 0.111 ( L′ ) = 126 × 10 −3 kmol/s-m2 . (6) n = 0.01114 L' + 0.148 = 0.01114(0.243) + 0.148 = 0.151 (808)(0.944) 14.69 (121)1/ 2 . 0.243 0.111 0.151 = 46.2 m2/m3 Exercise 6.27 (continued) Analysis: (b) (continued) Kya = kya = (1.26 x 10-3)(46.2) = 0.058 kmol/s-m3 From Eq. (1), HG = HOG = (V/S)/kya = (0.0093/0.292)/0.058 = 0.55 m or 1.8 ft Packed height = HGNG = 0.55(4.61) = 2.53 m or 8.3 ft. Therefore, a 10 foot height is sufficient. Now check the assumption that GeCl4 controls. Because gas is dilute in Cl2 and 99% of it is absorbed, NG is the same as for GeCl4. Note that in Eq. (3), k y is proportional to ( N Sc ) −2 / 3 or Di2 / 3 . Since the given gas diffusivity of Cl2 is about twice that of GeCl4, the...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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