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

# 4030 12 097d 002b d 121 kgh b 179 kgh solving

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Unformatted text preview: hat mole fraction to be 0.45. Then the operating line will intersect the equilibrium line at x = 0.45, creating the pinch zone of infinite stages. The value of y at the intersection is given by Eq. (1): 2.5(0.45) = 0.672 1 + 15(0.45) . Therefore the operating line passes through the two points, as {y, x}, of {0.75, 0.75} and {0.672, 0.450}. Therefore, the slope = L/V = (0.75 - 0.672)/(0.75 - 0.45) = 0.260. Now compute the overall material balances: Overall total material balance, F = 100 = D + B (18) Overall benzene material balance, xFF = yDD + xBB or 50 = 0.75D + 0.45B (19) Solving Eqs. (18) and (19), D = 16.67 moles or 16.67 mol/100 mol feed, and B = 83.33 moles y= Calculations for other values of the benzene mole fraction in the bottoms can be made in the same manner. (f) At total reflux, there is no distillate, but there is a boilup. The moles of distillate per 100 moles of feed = 0. Exercise 7.13 Subject: Distillation of a mixture of benzene and toluene at 101 kPa for specified reflux ratio and product compositions. Given: Feed of 30 kg/h of saturated liquid feed containing 40 mass% benzene and 60 mass% toluene. Distillate to contain 97 mass% benzene and bottoms to contain 98 mass% toluene. Reflux ratio = 3.5 and feed is to optimal stage. Table of vapor-liquid equilibrium data in mole fractions. At 101 kPa. Assumptions: Total condenser and partial reboiler. Saturated liquid reflux. Constant molar overflow. Find: (a) Flow rates of distillate and bottoms. (b) Number of equilibrium stages needed. Analysis: First solve the material balance in mass units. Then convert to moles and mole fractions so that the McCabe-Thiele method can be used for part (b). Overall total mass balance: 30 = D + B (1) (2) Overall benzene mass balance: 0.40(30) = 12 = 0.97D + 0.02B D = 12.1 kg/h B = 17.9 kg/h Solving Eqs. (1) and (2): Converting to moles with molecular weights of 78.11 for benzene and 92.13 for toluene, Benzene Toluene Product kmol/h Mass fraction Mole fraction Mass fraction Mole fraction Distillate 0.154 0.97 0.974 0.03 0.0235 Bottoms 0.196 0.02 0.026 0.98 0.9765 1.00 1.000 1.00 1.0000 Total: 0.350 (b) Because benzene is the more volatile component of the feed, the x and y coordinates will be those of benzene in the diagram on the next page. . In moles, the feed consists of: Component kmol/h Mole fraction Benzene 0.154 0.44 Toluene 0.196 0.56 Total: 0.350 1.00 For a saturated liquid feed, the q-line is vertical and passes through x = 0.44. The slope of the rectifying operating line, L/V, is obtained from Eq. (7-7), using the specified reflux ratio = 3.5, L/V = R/(1 + R) = 3.5/4.5 = 0.778 For saturated liquid reflux, the rectifying operating line passes through the point {0.974, 0.974}. See the McCabe-Thiele construction on the next page, where it is seen that slightly more than 10 stages + a partial reboiler acting as an equilibrium stage are required. The top 5 stages are in the rectifying section. Analysis: (b) (continued) Exercise 7.13 (continued) McCabe-Thiele Diagram Exercise 7.14 Subject: Distillation of a mixture of benzene and...
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