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

# 15 k s psolid exp 26708 8712 0080 torr 29815 s pliquid

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Unformatted text preview: p 26.708 − 8,712 = 0.080 torr 298.15 s Pliquid = exp 16.1426 − 3992.01 = 0.234 torr 29815 − 71.29 . From Eq. (1), γL = 1 0.080 0.342 = x L 0.234 xL (2) Using the solubility data with Eq. (2), Solvent Benzene Cyclohexane Carbon tetrachloride n-Hexane Water Solubility, xL 0.2946 0.1487 0.2591 0.1168 0.18 x 10-5 γL 1.16 2.30 1.32 2.93 190,000 Exercise 2.9 Subject: Minimum isothermal work of separation for a binary gas mixture. Given: A feed gas mixture, F, of A and B to be separated at infinite surroundings temperature, T0, into two products, P1 and P2. Assumptions: Ideal gas law and ideal gas solution at temperature T0. Isobaric at P0. Find: Derive an equation for the Wmin in terms of the mole fractions of the feed and products. Plot Wmin/RT0nF versus the mole fraction of A in the feed for: (a) A perfect separation. (b) A separation with SFA = 0.98 and SFB = 0.02. (c) A separation with SRA = 9.0 and SRB = 1/9. (d) A separation with SFA = 0.95 and SRB = 361 Determine the sensitivity of Wmin to product purities. Does Wmin depend on the separation method? Prove that the largest value of Wmin occurs for an equimolar feed. Analysis: From Eq. (4), Table 2.1, Wmin = nb − out From Eq. 2-1, b = h − T0 s nb (1) in (2) Combining Eqs. (1) and (2) for one feed, F, in and two products, P1 and P2, out: Wmin = nP1 hP1 + nP2 hP2 − T0 nP1 sP1 + nP2 hP2 − nF hF + T0nF sF (3) However, for isothermal separation of an ideal gas mixture, the change in enthalpy = 0. Therefore, from Eq. (3), Wmin = T0nF sF − T0 nP1 sP1 + nP2 hP2 (4) From Eq. (3), Table 2.4, for an ideal gas mixture at T0 and P0 , s = −R yi ln yi i (5) Exercise 2.9 (continued) Analysis: (continued) Substituting Eq. (5) into Eq. (4), Wmin = − nF yA F ln yA F + 1 − yA F ln 1 − yA F RT0 + nP1 yA P ln yA P + 1 − yA P ln 1 − yA P 1 1 1 (6) 1 + nP2 yA P ln yA P + 1 − yA P ln 1 − yA P } 2 2 2 2 By combining a component material balance for A with a total material balance, nP1 = nF yA F − yA P (7) 2 yA P − yA P 1 2 nP2 = nF yA F − yA P (8) 1 yA P − yA P 2 1 Equations (6), (7), and (8) give a relationship for Wmin/RT0 in terms of the molar feed rate and the mole fractions of A in the feed and two products. (8), (a) Let product P1 be pure A and product P2 be pure B. Then, from Eqs. (7) and nP1 = yA F nF and nP2 = 1 − yA F nF (9) and (10) Combining Eqs. (6) through (10), noting that 1x ln(1) = 0 and 0x ln(0) = 0 Wmin = − yAF ln yA F + (1 − yAF ) ln (1 − yAF ) nF RT0 (11) (b) From Eq. (1-2), letting 1 be P1, yA P nP1 = 0.98 yA F nF or yA P = 0.98 yA F 1 1 nF nP1 (12) Exercise 2.9 (continued) Analysis: (b) (continued) and, nF nP1 yBP = 0.02 1 − yA F 1 (13) Combining (12 and (13) to give yA P + yBP = 1, we obtain, 1 1 nP1 = nF 0.96 yA F + 0.02 (14) By total material balance, nP2 = nF − nP1 or nP2 = nF 0.98 − 0.96 yA F (15) Also, from the SFA = 0.98 for the split fraction of A to P1 , we can write for the split fraction of A to P2 , yA P nP2 = 0.02 yA F nF or yA P = 0.02 yA...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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