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

# Mol a test 1 test 2 stage vapor liquid

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Unformatted text preview: c) Number of equilibrium stages required. Analysis: (a) Take a basis of F = 100 mol/h. Therefore, 79.1 mol/h of N2 and 20.9 mol/h of O2 in the feed. Bottoms product vapor contains 0.6(20.9) = 12.54 mol/h of O2 and (0.2/99.8)(12.54) = 0.025 mol/h N2. By material balance, the overhead vapor contains 20.9 - 12.54 = 8.36 mol/h O2 and 79.1 - 0.025 = 79.075 mol/h of N2. The mol% N2 in the overhead vapor = 79.075/(8.36 + 79.075) x 100% = 90.4 %. (b) Assume constant molar overflow. Then because the feed is assumed to be a saturated liquid, the moles of vapor generated in the reboiler per 100 moles of feed = mol/h of overhead vapor = 8.36 + 79.075 = 87.435 moles per 100 moles of feed. (c) Use a y-x diagram for N2 because it is the more volatile. The slope of the operating line is L/V = 100/87.435 = 1.14. At the top of the column, the operating line terminates at a (y-x) of (0.904, 0.791). At the bottom of the column, with a total reboiler, the operating line terminates at a (y-x) of (0.002, 0.002). To determine the number of equilibrium stages, it is convenient to use two diagrams, the usual one and a second one for just the very low mole fraction region so as to gain accuracy in the region of the lower end of the operating line. From the two diagrams, it is seen that just less than 8 equilibrium stages are required. Analysis (d) (continued): Exercise 7.9 (continued) Exercise 7.10 Subject: Stage composition data for distillation of an A-B mixture. Given: Saturated liquid feed of 40 mol% A. Test results for vapor and liquid compositioins for 3 successive stages between the feed stage and a total condenser. ____________Mol % A_____________ Test 1 Test 2 Stage Vapor Liquid Vapor Liquid M+2 79.5 68.0 75.0 68.0 M+1 74.0 60.0 68.0 60.5 M 67.9 51.0 60.5 53.0 Assumption: Constant molar overflow. Find: Reflux ratio and overhead composition for each test. Analysis: Test 1: By material balance for component A around stage M + 1, Vy M + Lx M + 2 = Vy M +1 + Lx M +1 (1) Solving Eq. (1) for L/V and substituting for y and x values from above table, L y − yM 0.740 − 0.679 = M +1 = = 0.763 V x M + 2 − x M +1 0.680 − 0.600 From Eq. (7-7), reflux ratio = R = L/D = (L/V)/(1 - L/V) = 0.763/(1 - 0.763) = 3.22 D/V = (L/V)/(L/D) = 0.763/3.22 = 0.237. Noting that stages in the rectifying section are counted here from the bottom up instead of the top down, Eq. (7-5) becomes, L D y M +1 = x M + 2 + x D (2) V V Solving Eq. (2) for xD , L y M +1 − x M +2 0.740 − (0.763)(0.680) V xD = = = 0.933 D /V 0.237 Therefore the composition of the distillate is 93.3 mol% A and 6.7 mol% B. Test 2: Because y M +1 = x M + 2 and y M = x M +1 , operation is at total reflux, i.e. reflux ratio = infinity. In this case, Eq. (2) can not be solved for xD because D/V = 0. We can not determine the composition of the distillate. We do know that it must at least 75 mol% A. Exercise 7.11 Subject: Five procedures for continuous distillation of a mixture of benzene (A) and toluene (B). Given: Operation at 1 atm to produce a distillate of 80 mol% benzene (i.e. yD = 0.8) from a saturated liquid feed of 70...
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