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

# 29 assumptions negligible liquid holdup on trays and

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Unformatted text preview: rom Exercise 7.29 Assumptions: Perfect mixing in the still. No holdup on the stages or in the condenser. No pressure drop. Find: (a) (b) (c) (d) (e) Number of theoretical plates in the column. Reflux ratio when the residue contains 25 mol% ethanol. Instantaneous distillate rate in lbmol/h when the residue contains 15 mol% ethanol. Lbmol of distillate product. Lbmol of residual. Analysis: First, compute compositions in moles. Density of charge = 8.33(0.871) = 7.26 lb/gal. Amount of charge = 7.26(2,000) = 14,500 lb. Molecular weights are 46.07 for ethanol and 18.02 for water. Lbmoles of ethanol in the charge = 14,500(0.70)/46.07 = 220 lbmoles. Lbmoles of water in the charge = 14,500(0.30)/18.02 = 241 lbmoles. Therefore, the total charge = 220 + 241 = 461 lbmoles. The mole fraction of ethanol in the charge = 220/461 = 0.477. Take a basis of 100 lb for the final residue. The lbmoles of ethanol in the final residue = 3/46.07 = 0.0651. The lbmoles of water in the final residue = 97/18.02 = 5.3829. The total lbmoles of final residue = (0.0651 + 5.3829) = 5.4480 lbmoles/100 lb. The ethanol mole fraction in the final residue = 0.0651/5.4480 = 0.0119. (a) One approach to determine a reasonable number of theoretical stages is to use the McCabe-Thiele method for estimating the minimum number of stages at total reflux and then multiplying that value by two to obtain the number of theoretical stages at actual reflux. The minimum number of stages can be stepped between the constant xD = 0.85 and the final xW = 0.0119. The construction is shown on the following page, where approximately 10 stages are stepped off. Therefore, the number of theoretical stages at actual reflux is estimated at 2(10) = 20 (19 plus a reboiler). (b) To estimate the reflux ratio, use the McCabe-Thiele method for estimating the minimum reflux ratio, Rmin and then assume the actual reflux ratio = R = 1.2Rmin. The construction is shown on the following page, where the position of the operating line is controlled not by a line from the constant xD = 0.85 to xW = 0.25, but by the equilibrium line, as in Fig. 7.12b. The slope is 0.62 = (L/V)min . From Eq. (7-27), Rmin = (L/V)min/[1 - (L/V)min] = 0.62/(1 - 0.62) = 1.63. Therefore, take the actual reflux ratio = R = 1.2Rmin = 1.2(1.63) = 1.98. Exercise 13.14 (continued) Analysis: (a) and (b) (continued) Exercise 13.14 (continued) Analysis: (continued) (d) and (e) Compute these before Part (c). A total molar material balance is: W0 = 461 = W + D. (1) An ethanol molar material balance is: 220 = 0.0119W + 0.85D Solving Eqs. (1) and (2), D = 256 lbmoles and W = 205 lbmoles (2) (c) When the residue contains 15 mol% ethanol, the minimum reflux is still controlled by the equilibrium curve. Therefore, take a reflux ratio of 1.98. Assume that the distillate rate is constant over the 20 hour period. Therefore, the distillate rate = 256/20 = 12.8 lbmol/h. For a reflux ratio of 1.98, the boilup (vapor rate) = 2.98D = 2.98(12.8) = 38.1 lbmol/h Alternatively, this exercise...
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