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

In example 147 the feed side mass transfer

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Unformatted text preview: ol/h and by overall total balance, P1 = 60 lbmol/h: A M in m 2 = Therefore, −5 Total balance around the mixing point: 0 = 100 + P2 - F1 (5) Total balance around Stage 1: 0 = 60 + R1 - F1 (6) Eq. (1) for Stage 1: 0 = 3.05*x1 - y1 - 2.05*y1*x1 (7) Eq. (1) for Stage 2: 0 = 3.05*x2 - y2 - 2.05*y2*x2 (8) Eq. (3) for Stage 1: 0 = xF1 - x1 - T1*y1 + T1*x1 (9) Eq. (3) for Stage 2: 0 = x1 - x2 - T2*y2 + T2*x2 (10) Eq. (4) for equal AM: y1*T1*F1*(1551*x2 - 76.6* y2) = y2*T2*R1*(1551*x1 - 76.6*y1) (11) C3= balance around the mixing point: 0 = xF1*F1-60-y2*P2 (12) C3= balance around Stage 1: 0 = 60*y1 + x1*R1 - xF1*F1 (13) C3= balance around Stage 2: 0 = y2*P2 + 40*x2 -x1*R1 (14) Eqs. (5) to (14) are a set of 10 equations (8 of which are nonlinear) in 10 unknowns. Solving with a nonlinear solver, such as Polymath, the following results are obtained, based on the following guesses to start the iterative process: Guess Result F1 133 150.2 R1 73 90.2 P2 33 50.2 T1 0.45 0.399 T2 0.45 0.556 xF1 0.55 0.610 x1 0.5 0.510 y1 0.8 0.761 x2 0.4 0.359 y2 0.7 0.631 Thus, the product obtained is a permeate, P1, of 60 lbmol/h with a propylene mole fraction of 0.761. The membrane area for each stage is computed from Eq. (4), which for Stage 1 is: A M 1 in m 2 = 0.761(0.399)(150.2)(2820) = AM 2 = 195 m2 9.0 × 10 0.510(1,551) − 0.761(76.6) (10,000) −5 Exercise 14.11 (continued) Analysis: (continued) Case 2: Configuration of Fig. 14.12b Again take α = 3.05. Assume this value is constant for both stages. Again assume that the membrane areas for Stages 1 and 2 are equal. The governing equations then are as follows where R1 = 40 lbmol/h and by overall total balance, P2 = 60 lbmol/h: Total balance around the mixing point: 0 = 100 + R2 - F1 (15) Total balance around Stage 1: 0 = 40 + P1 - F1 (16) Eq. (1) for Stage 1: 0 = 3.05*x1 - y1 - 2.05*y1*x1 (17) Eq. (1) for Stage 2: 0 = 3.05*x2 - y2 - 2.05*y2*x2 (18) Eq. (3) for Stage 1: 0 = xF1 - x1 - T1*y1 + T1*x1 (19) Eq. (3) for Stage 2: 0 = x1 - x2 - T2*y2 + T2*x2 (20) Eq. (4) for equal AM: y1*T1*F1*(1551*x2 - 76.6* y2) = y2*T2*P1*(1551*x1 - 76.6*y1) (21) C3= balance around the mixing point: 0 = xF1*F1-60-x2*R2 (22) C3= balance around Stage 1: 0 = P1*y1 + x1*40 - xF1*F1 (23) C3= balance around Stage 2: 0 = y2*60 + R2*x2 -y1*P1 (24) Eqs. (15) to (24) are a set of 10 equations (8 of which are nonlinear) in 10 unknowns. Solving with a nonlinear solver, such as Polymath, the following results are obtained, based on the following guesses to start the iterative process: Guess Result F1 160 100 R2 60 0 P1 120 60 T1 0.75 0.60 T2 0.50 1.00 xF1 0.60 0.60 x1 0.3 0.4407 y1 0.7 0.7062 x2 0.6 0.4407 y2 0.8 0.7062 Unfortunately, this is not a useful solution because all of the feed to Stage 2 passes through the membrane. Thus, nothing is achieved by Stage 2. Therefore, other solutions were sought in an attempt to achieve a higher mole fraction of propylene in the final permeate. This was achieved by eliminating the equal area condition of Eq. (21) and replacing it by...
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