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

596 x1 0371 benzene material balance around the

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Unformatted text preview: L (3) Solving for yB L 3 y B = y1 + ( x1 − x D ) = 0.750 + (0.545 − 0.750) = 0.596 V 4 Analysis: (b) (continued) Exercise 7.12 (continued) The mole fraction of benzene in the bottoms product is in equilibrium with yB =0.596. Therefore, the form of Eq. (2) applies, yB 0.596 xB = = = 0.371 y B + α (1 − y B ) 0.596 + 2.5(1 − 0.596) Overall total material balance, F = 100 = D + B (4) Overall benzene material balance, xFF = xDD + xBB or 50 = 0.75D + 0.371B (5) Solving Eqs. (4) and (5), D = 34.2 moles or 34.2 mol/100 mol feed, and B = 65.8 moles. (c) With the feed to the theoretical plate, the following results apply from part (b), y1 = 0.75 xD = 0.75 x1 = 0.545 Benzene material balance around Stage 1, which now includes the feed, xFF + y BV + x D L = y1V + x1 L (6) Solving for yB, V L L F V L L 100 y B = y1 + x1 − xD − xF = 0.750 + 0.545 − 0.750 − 0.50 (7) V V V V V V V V Because the feed is a saturated liquid, , V =V and L = L + 100 From above, V = 4D and L/V = 3/4. Also, L / V = L / V + 100 / V = 3 / 4 + 100 / V Therefore, Eq. (7) becomes, 3 100 3 100 4.5 1125 . (8) y B = 0.750 + 0.545 + − 0.750 − 0.50 = 0.596 − = 0.596 − 4V 4 V V D The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yB xB = (9) y B + 2.5(1 − y B ) Overall total material balance, F = 100 = D + B (10) Overall benzene material balance, xFF = xDD + xBB or 50 = 0.75D + xBB (11) Solving Eqs. (8), (9), (10), and (11), yB = 0.647, xB = 0.423, D = 23.5 moles or 23.5 mol/100 mol feed, B = 76.5 moles (d) With a partial condenser, the mole fraction of the liquid reflux is that in equilibrium with the vapor distillate. Therefore, from the above results, yD = 0.75 xR = 0.545 y1 = 0.596 x1 = 0.371 Benzene material balance around the theoretical plate, which includes the feed, xFF + y BV + x R L = y1V + x1 L (12) Solving for yB, V L L F V L L 100 y B = y1 + x1 xR − xF = 0.596 + 0.371 − 0.545 − 0.50 (13) V V V V V V V V Exercise 7.12 (continued) Analysis: (d) (continued) Because the feed is a saturated liquid, From above, V = 4D and L/V = 3/4. Therefore, Eq. (13) becomes, 3 100 y B = 0.596 + 0.371 + − 0.545 4V , V =V and L = L + 100 Also, L / V = L / V + 100 / V = 3 / 4 + 100 / V 3 100 12.9 3.23 − 0.50 = 0.466 − = 0.466 − 4 V V D (14) The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yB xB = (15) y B + 2.5(1 − y B ) Overall total material balance, F = 100 = D + B Overall benzene material balance, xFF = yDD + xBB or 50 = 0.75D + xBB Solving Eqs. (14), (15), (16), and (17), yB = 0.405, xB = 0.214, D = 53.4 moles or 53.4/100 mol feed, (16) (17) B = 46.6 moles (e) At minimum reflux, with the feed sent to the still pot (partial reboiler), an infinite number of theoretical plates is needed between the condenser and reboiler. This part is not completely specified. In order to compute the distillate, we must assume a bottoms benzene mole fraction less than that in the feed. Suppose we choose t...
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