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

63 10 5 00126 psift 3 2 868 exercise 752 continued

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Unformatted text preview: ights are approximate because they are based on values of HOG that are assumed constant over the packed sections, which is not the case because m varies. (e) Compute pressure drops for the packed sections from the correlation of Billet and Schultes. The pressure drop per unit height of packed bed is given by Eq. (6-115), 3/ 2 1/ 2 ∆Po ε − hL ∆P 13300 = exp ( N FrL ) lT lT ε a 3/ 2 This equation will be used for each packing in each section. ∆Po a u2ρ 1 = Ψo 3 V V lT ε 2 g c KW The required packing parameters are as follows from Table 6.8 and above: NOR PAC Montz 23 FP , ft /ft 14 33 2 3 a, m /m 86.8 300 0.947 0.930 ε, m3/m3 Ch 0.651 0.482 0.350 0.295 Cp From Eq. (6-110), (1) (2) 1− ε 1 − 0.947 =6 = 0.00366 m or 0.012 ft for NOR PAC a 86.8 1− ε 1 − 0.930 DP = 6 =6 = 0.0014 m or 0.00459 ft for Montz a 300 From Eq. (6-112), DP = 6 Rectifying section of the column: For NOR PAC , column diameter = DT = 4.3 ft 1 2 1 DP 2 1 0.012 = 1+ = 1+ = 1035 . KW 3 1 − ε DT 3 1 − 0.947 4.3 Therefore, KW = 1/1.035 = 0.966 From Eq. (6-111), Analysis: (e) (continued) Exercise 7.52 (continued) For Montz, column diameter = DT = 5.3 ft 1 2 1 DP 2 1 0.00459 From Eq. (6-111), = 1+ = 1+ = 1008 . KW 3 1 − ε DT 3 1 − 0.930 5.3 Therefore, KW = 1/1.008 = 0.992 Superficial gas velocity at 70% of flooding from above = 13 ft/s for NOR PAC. Superficial gas velocity at 70% of flooding from above = 8.4 ft/s for Montz Gas viscosity = 0.011 cP and gas density = 0.0688 lb/ft3. From Eq. (6-114), for NOR PAC, u Dρ (13)(0.012)(0.0688) N ReV = V P V KW = (0.966) = 26,500 (1 − ε ) µV (1 − 0.947 ) (0.011)(0.000672) For Montz, N ReV = (8.4)(0.00459)(0.0688) (0.992) = 5, 090 (1 − 0.930 ) (0.011)(0.000672) From Eq. (6-113), for NOR PAC, Ψo = Cp 64 1.8 64 1.8 + 0.08 = 0.350 + = 0.280 N ReV N ReV 26, 500 26,5000.08 For Montz, Ψo = 0.295 64 18 . + = 0.272 5,090 5,0900.08 From Eq. (2), for NOR PAC, 2 ∆Po a uV ρV 1 86.8(0.3048) 132 (0.0688) 1 = Ψo 3 = 0.280 = 1.63 lbf/ft3 or 0.0113 psi/ft 3 lT ε 2 g c KW 0.947 2(32.2) 0.966 For Montz, ∆Po 300(0.3048) 8.4 2 (0.0688) 1 = 0.272 = 2.35 lbf/ft3 or 0.0163 psi/ft 3 lT 0.930 2(32.2) 0.992 From Eq. (1), for NOR PAC, with hL = 0.0162 m3/m3 and liquid Froude number = 6.63x10-5 ∆P 0.947 − 0.0162 = 0.0113 lT 0.947 1.5 exp 1/ 2 13300 6.63 ×10 −5 ) = 0.0126 psi/ft 3/ 2 ( 86.8 Exercise 7.52 (continued) Analysis: (e) Rectifying section (continued) From Eq. (1), for Montz, with hL = 0.0241 m3/m3 and liquid Froude number =9.91x10-5 ∆P 0.930 − 0.0241 = 0.0163 lT 0.930 1.5 exp 1/ 2 13300 9.91× 10−5 ) = 0.0161 psi/ft 3/ 2 ( 300 Stripping section of the column: For NOR PAC , column diameter = DT = 3.4 ft 1 2 1 DP 2 1 0.012 From Eq. (6-111), = 1+ = 1+ = 1044 . KW 3 1 − ε DT 3 1 − 0.947 3.4 Therefore, KW = 1/1.044 = 0.957 For Montz, column diameter = DT = 4.2 ft 1 2 1 DP 2 1 0.00459 From Eq. (6-102), = 1+ = 1+ = 1010 . KW 3 1 − ε DT 3 1 − 0.930 4.2 Therefore, KW = 1/1.010 = 0.990 Superficial gas velocity at 70% of flooding from above = 20 ft/s for NOR PAC. Superficial ga...
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