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

Find a composition of vapor at bubble point for 30

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Unformatted text preview: m above = 7.84 ft/s Gas viscosity = 0.0091 cP and gas density = 0.171 lb/ft3. From Eq. (6-114), N ReV = uV DP ρV (7.84)(0.0049)(0.171) KW = (0.964) = 45, 000 (1 − ε ) µV (1 − 0.977 ) (0.0091)(0.000672) From Eq. (6-113), Ψ o = C p From Eq. (2), 64 1.8 64 1.8 + 0.08 = 0.421 + = 0.322 N ReV N ReV 45, 000 45, 0000.08 92.3(0.3048) 7.84 2 (0171) . 1 ∆Po = 0.322 = 162 lbf/ft3 or 0.0113 psi/ft . 3 0.977 2(32.2) 0.977 lT From Eq. (1), with hL = 0.0294 m3/m3 and liquid Froude number = 2.46 x 10-4 ∆P 0.977 − 0.0294 = 0.0113 0.977 lT 3/ 2 exp 1/ 2 13300 ( 2.46 ×10−4 ) = 0.0137 psi/ft 92.33/ 2 Stripping section of the column: Column diameter = DT = 3.6 ft From Eq. (6-102), 1 2 1 DP 2 1 0.0049 = 1+ = 1+ = 1039 . KW 3 1 − ε DT 3 1 − 0.977 3.6 Therefore, KW = 1/1.039 = 0.962 Superficial gas velocity at 70% of flooding from above = 5.9 ft/s Gas viscosity = 0.0092 cP and gas density = 0.179 lb/ft3. Exercise 7.53 (continued) Analysis: (e) Stripping section (continued) From Eq. (6-105), N ReV = uV DP ρV (5.9)(0.0049)(0.179) (0.962) = 35, 000 KW = (1 − ε ) µV (1 − 0.977 ) (0.0092)(0.000672) From Eq. (6-104), Ψ o = C p 64 1.8 64 1.8 + 0.08 = 0.421 + = 0.329 35, 000 35, 0000.08 N ReV N ReV From Eq. (2), ∆Po 92.3(0.3048) 5.9 2 (0.179) 1 = 0.329 = 1.0 lbf/ft3 or 0.0069 psi/ft 3 0.977 2(32.2) 0.962 lT From Eq. (1), with hL = 0.0412 m3/m3 and liquid Froude number = 7.80 x 10-4 ∆P 0.977 − 0.0412 = 0.0069 lT 0.977 3/ 2 exp 1/ 2 13300 7.80 × 10−4 ) = 0.0098 psi/ft 3/ 2 ( 92.3 From these results, the column diameter would be 4.0 ft without swaging. Take the packed height as 14 ft. The column height, allowing 4 ft for disengagement, 2 ft for feed entry, and 10 ft for sump at the bottom would be 30 ft. The total pressure drop is: (2.31/0.3048)(0.0069) + (1.67/0.3048)(0.0098) = 0.11 psi. For the plate column of Example 7.1, with 12.2 equilibrium stages, take tray efficiency = 80%. Therefore, need 16 trays, with a column height of 44 ft. Take a pressure drop of 0.08 psi/tray, giving a total pressure drop of 18(0.08) = 1.44 psi. The column diameter is estimated to be 4.5 ft. The following summarizes the comparison. with the plate column of Exercise 7.1: Plate column Column diameter, ft Column height, ft Pressure drop, psi 4.5 44 1.44 The packed column is very competitive with the plate column. Packed column 4.0 30 0.11 Exercise 7.54 Subject: Determining compositions and energy requirements for a mixture of nC6 and nC8 with an enthalpy-concentration diagram for 101 kPa. Given: Enthalpy-concentration diagram in Fig. 7.37. Find: (a) Composition of vapor at bubble-point for 30 mol% nC6. (b) Energy to vaporize 60 mol% of a mixture initially at 100oF with 20 mol% nC6 . (c) Vapor and liquid compositions resulting from part (b). Analysis: (a) In Fig. 7.37, below, the bubble point for 30 mol% nC6 in nC8 is shown as point B, which corresponds to a temperature of 211oF. An equilibrium tie line is drawn from point B to point M at the intersection with the sat...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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