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

Composition of exit gas accounting for stripping of

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Unformatted text preview: om O'Connell correlation, Eq. (6-23), with ML = 250, ρL = 57.9 lb/ft3 from Example 6.3, log E o = 1587 − 0.199 log C − 0.0896 log C . where, C = 2 KM L µ L (2.0)(250)(1.4) = = 12.1 ρL 57.9 Therefore, from Eq. (2), 2 log E o = 1587 − 0.199 log 12.1 − 0.0896 log 12.1 = 1277 . . which gives Eo = 18.9% The propane tray efficiency is lower than predicted. (2) Exercise 6.16 Subject: Oil absorption of a gas to produce 95% hydrogen Given: 800,000 scfm (0oC, 1 atm) of fuel gas containing 72.5% H2, 25% CH4, and 2.5% C2H6. Absorbent of nC8. Absorber operates at 400 psia and 100oF. Exiting gas is to contain at least 80% of the entering H2 at a H2 purity of 95 mol%. Assumptions: Ideal gas law. Find: (a) (b) (c) (d) (e) (f) (g) Minimum absorbent rate in gpm. Actual absorbent rate at 1.5 times minimum. Number of theoretical stages. Stage efficiency of each component from O'Connell correlation. Number of actual trays needed. Composition of exit gas, accounting for stripping of octane. Annual cost of lost octane. Analysis: At 0oC and 1 atm, have 359 scf/lbmol. Therefore, entering V = 800,000/359 = 2,230 lbmol/min, with 0.725(2,230) = 1,616 lbmol/min H2, 558 lbmol/min CH4, 56 lbmol/min C2H6 . From Fig. 2.8, at 400 psia, 100oF, K-values are: 30 for H2 , 7 for CH4 , and 1.75 for C2H6 . (a) Assume the key component is CH4. If we neglect stripping of octane and assume complete absorption of C2H6, with 20% absorption of H2, then exit gas will be 0.8(1,616) = 1,293 lbmol/min H2 and the exit liquid will contain 1,616 - 1,293 = 323 lbmol/min H2. For a purity of 95 mol% H2, then will have 5 mol% CH4 or 1,293(5/95) = 68 lbmol/min of CH4 in the exit gas and 558 - 68 = 490 lbmol/min of CH4 in exit liquid. Can not use Eq. (6-11) to compute the minimum absorbent rate because we do not have a dilute system. Instead, for infinite stages, assume the exiting rich oil is in equilibrium with the inlet gas. Then, for CH4, K = 7 = y/x. But y for the inlet gas = 0.25. Therefore, x in the exiting oil = 0.25/7= 0.0357. Therefore, with 490 lbmol/min of CH4 in this oil, the total exiting oil = 490/0.0357 = 13,700 lbmol/min. Therefore, the entering oil is 13,700 - 323 - 490 - 56 = 12,831 lbmol/min of nC8 = Lmin. For nC8 , MW = 114 and from Perry's Handbook, SG = 0.703. Therefore, liquid density = 0.703(8.33) = 5.86 lb/gal. Therefore, Lmin = 12,831(114)/5.86 = 250,000 gpm. (b) For 1.5 times minimum, inlet absorbent rate = 19,200 lbmol/min = 375,000 gpm. (c) Calculate the number of equilibrium stages from the Kremser equation based on CH4 absorption, inlet vapor and liquid flow rates. From Eq. (6-15), ACH 4 = L / KV = 19,200 / (7)(2,230) = 123 . From Eq. (6-13), for CH4, with a fraction absorbed = 490/558 = 0.878, fraction of CH4 absorbed = 0.878 = N ACH+41 − ACH 4 N ACH+41 − 1 123 N +1 − 123 . . = N +1 1.23 − 1 (1) Analysis: (c) (continued) Exercise 6.16 (continued) Solving Eq. (1), N =4.1 stages. Check absorption of H2. Absorption factor = A = 19,200/(30)(2,230) =0.287 N AH 2+1 − AH...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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