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

# Analysis coefficient kla is obtained from a

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Unformatted text preview: (1.59)(0.0143)(0.148)(0.957) KW = = 3, 287 (1 − ε ) µV (1 − 0.919 ) [(0.018)(0.000672)] From Eq. (6-113), Ψ o = C p 64 1.8 64 1.8 + 0.08 = 0.698 + = 0.671 3, 287 3, 287 0.08 N ReV N ReV From Eq. (2), with the introduction of gc because of the use of American Engineering units, ∆Po a u2ρ 1 33.9 1.592 (0.148) 1 = Ψo 3 V V = 0.671 = 0.164 lbf/ft3 or 0.00114 psi/ft 3 lT ε 2 g c KW 0.919 2(32.2) 0.957 From Eq. (6-97), hL = 12 N FrL N Re L 1/ 3 ah a 2/3 Superficial liquid velocity = uL = m/SρL = (5,000/60)/(5.41)(62.4) = 0.247 ft/s From Perry's Handbook, liquid viscosity = 0.95 cP 2 u L a (0.247) 2 (33.9) From Eq. (6-99), N FrL = = = 0.064 g 32.2 uρ (0.247)(62.4) From Eq. (6-98), N Re L = L L = = 712 aµ L (33.9)(0.95)(0.000672) a 0. 0 From Eq. (6-101), h = 0.85Ch N Re25 N Fr.1 = 0.85(0.593)(712) 0.25 (0.247) 0.1 = 2.27 L L a (3) Exercise 6.28 (continued) Analysis: (d) (continued) 0.064 From Eq. (3), hL = 12 712 1/ 3 2.27 2 / 3 = 0.177 m3/m3 or ft3/ft3 From Eq. (1), ∆P ε = ∆Po ε − hL 3/ 2 exp 1/ 2 13300 N 3/ 2 ( FrL ) a ∆P 0.919 = ( 0.00114 ) lT 0.919 − 0.103 3/ 2 exp 13300 1/ 2 0.064 ) = 0.0241 psi/ft or 0.669 in. H2O/ft 3/ 2 ( 111.1 This is only 10% higher than the suggested value of 0.612 in. H2O/ft Exercise 6.29 Subject: Mass-transfer coefficients for conditions of Exercise 6.26. Given: Suggested mass-transfer coefficients in Exercise 6.26. Find: Mass-transfer coefficients kLa and kGa from Billet-Schultes correlation. Analysis: Coefficient kLa is obtained from a rearrangement of the left-most part of Eq. (6-131), uL kL a = (1) H L ( aPh / a ) with HL from Eq. (6-132) and (aPh /a) from (6-136). From Table 6.8, CL = 1.239 , ε = 0.919, and a = 111.1 m2/m3. From the results of Exercise 6.28, come the following parameters and properties: hL = 0.177 ft2/ft3, uL = 0.247 ft/s, ρL = 62.4 lb/ft3, µL = 0.95 cP Take the surface tension, σ = 0.005 lbf/ft or 0.005(32.2) = 0.161 lb/s2 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 From Exercise 6.28, Reynolds number = N ReL ,h = ε 0.919 =4 = 0.0331 m = 0.108 ft a 111.1 uL d h ρ L (0.247)(0.108)(62.4) = = 2, 607 [(0.95)(0.000672)] µL Weber number = N WeL ,h ( 0.247 ) (62.4)(0.108) = 2.55 u 2ρ d = L L h= σ 0.161 Froude number = N FrL , h ( 0.247 ) = 0.0175 u2 = L= gd h 32.2(0.108) 2 2 From (6-136), ( aPh −1/ 2 = 1.5 ( ad h ) N Re L ,h a ) (N −0.2 = 1.5 [ (111.1)(0.0331) ] −1/ 2 ) (N ) 0.75 We L , h −0.45 FrL , h ( 2607 ) ( 2.55 ) ( 0.0175 ) −0.2 0.75 −0.45 = 2.02 Exercise 6.29 (continued) Analysis: (continued) From Exercise 6.23, correcting for temperature, DL = 9.6 x 10-6(298/294) = 9.73 x 10-6 cm2/s or 1.05 x 10-8 ft2/s From (6-132), using American Engineering Units, 11 HL = CL 12 1/ 6 1 1 = 1.239 12 From (1), k L a = 4hL ε DL au L 1/ 6 1/ 2 uL a a a ph 4(0.177)(0.919) (1.05 × 10−8 )(33.9)(0...
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