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

Analysis material balance for 1 kmol of feed let the

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Unformatted text preview: r constants for the system. Find: Minimum work for the separation in kJ/kmol of feed. Analysis: Material balance for 1 kmol of feed. Let the two products be R and S, where the former is acetone-rich. Total mole balance: 1 = nR + nS Acetone balance: 0.35 = 0.99 nR + 0.02 nS Solving, nR = 0.3469 kmol and nS = 0.6531 kmol From the problem statement, Wmin = nR RT0 xi ln γ i xi + nS iR xi ln γ i xi − nF iS xi ln γ i xi (1) iS The activity coefficients are given by the van Laar equations, (3) in Table 2.9, which with the given constants, A12 = 2.0 and A21 = 1.7, become, ln γ 1 = 2.0 2.0 x1 1+ 1.7 x2 2 (2) ln γ 2 = . 17 1.7 x2 1+ 2.0 x1 2 (3) Applying Eqs. (2) and (3) to the given mole-fraction compositions, Component Acetone (1) Water (2) x 0.35 0.65 Feed γ 2.116 1.291 Substituting the above values into Eq. (1), Product R x 0.99 0.01 γ 1.000 5.318 Product S_____ x γ 0.02 6.735 0.98 1.000 Exercise 2.21 (continued) Analysis: (continued) Wmin = 0.3469 {0.99 ln [ (0.99)(1.00)] + 0.01ln [ (0.01)(5.318) ]} RT0 +0.6531{0.02 ln [ (0.02)(6.735) ] + 0.98ln [ (0.98)(1.00) ]} − 1.0000 {0.35 ln [ (0.35)(2.116)] + 0.65 ln [ (0.65)(1.291)]} = 0.1664 kmol Wmin = 0.1664 RT0 = 0.1664(8.314)(298) = 412 kJ/kmol feed For ideal liquid solutions, all values of γ are equal to 1.0. Therefore, Wmin = 0.3469 {0.99 ln [ (0.99)(1.00)] + 0.01ln [ (0.01)(1.00]} RT0 +0.6531{0.02 ln [ (0.02)(1.00)] + 0.98 ln [ (0.98)(1.00) ]} − 1.0000 {0.35 ln [ (0.35)(1.00)] + 0.65 ln [ (0.65)(1.00)]} = 0.5639 kmol Wmin = 0.5639 RT0 = 0.5639(8.314)(298) = 1,397 kJ/kmol feed Thus, the minimum work of separation for the ideal solution is 3.4 times that of the nonideal solution. Exercise 2.22 Subject: Relative volatility and activity coefficients of the benzene (B) - cyclohexane (CH) azeotropic system at 1 atm Given: Experimental vapor-liquid equilibrium data, including liquid-phase activity coefficients, and Antoine vapor pressure constants. Assumptions: Ideal gas and gas solutions Find: (a) Relative volatility of benzene with respect to cyclohexane as a function of benzene mole fraction in the liquid phase. (b) Van Laar constants from the azeotropic point and comparison of van Laar predictions with experimental data. Analysis: (a) From Eqs. (2-21), (2-19), and (3) in Table 2.3, α B,CH = KB y /x yB / xB γ Ps = B B= = B Bs KCH yCH / xCH γ CH PCH 1 − yB / 1 − xB (1) Using the y-x data for benzene, the following values of relative volatility are computed from (1): Temperature, oC 79.7 79.1 78.5 78.0 77.7 77.6 77.6 77.6 77.8 78.0 78.3 78.9 79.5 xB 0.088 0.156 0.231 0.308 0.400 0.470 0.545 0.625 0.701 0.757 0.822 0.891 0.953 yB 0.113 0.190 0.268 0.343 0.422 0.482 0.544 0.612 0.678 0.727 0.791 0.863 0.938 αB,CH 1.317 1.269 1.218 1.173 1.095 1.049 0.996 0.946 0.898 0.854 0.821 0.771 0.746 Exercise 2.22 (continued) Analysis: (a) (continued) (b) From the data, take the azeotrope at xB = yB = 0.545 and xCH = yCH = 1 - 0.545 = 0.455, and with γΒ = 1.079 and γCH = 1.102. To determine the van La...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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