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

# The slope of the operating line lv 19455835 0333

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Unformatted text preview: he bottoms contains 290.9 kmol/h of water. But, the flow rate of water entering in the two feeds = 125 + 83.3 - 100 = 108.3 kmol/h. Therefore, the open steam flow rate = 290.9 + 5 - 108.3 = 187.6 kmol/h In the McCabe-Thiele diagram on the next page, the three operating lines are drawn and the equilibrium stages are stepped off so as to place the two feeds at their optimal locations. As seen, the number of equilibrium stages required = Nt = 14. From Eq. (6-21), for a plate efficiency of 70%, the actual number of trays = Na = Nt /Eo = 14/0.7 = 20 plates. Analysis: (continued) Exercise 7.30 (continued) Exercise 7.31 Subject: intercooler. Distillation of a mixture of n-hexane and n-octane in a column with an Given: Saturated liquid feed of 40 mol% hexane in octane. Intercooler at second stage from the top removes heat so as to condense 50 mol% of the vapor rising from the third. Distillate is to contain 95 mol% of hexane and bottoms is to contain 5 mol% of hexane. Reflux ratio, L/D, at the top, is equal to 0.5. Vapor-liquid equilibrium data for 1 atm is plotted in Fig. 4.4. Assumptions: Constant molar overflow. Total condenser and partial reboiler. Operating pressure of 1 atm. Find: (a) Equations to locate operating lines. (b) Number of equilibrium stages if optimal feed stage location is used. Analysis: First compute overall material balance. Take a basis of F = 100 kmol/h. Overall total mole balance: F = 100 = D + B (1) Overall hexane mole balance: FxF = 40 = DxD + BxB = 0.95D + 0.05B (2) Solving Eqs. (1) and (2), D = 38.9 kmol/h and B = 61.1 kmol/h (a) For a reflux ratio of 0.5, in the section above the intercooler, L = 0.5D = 19.45 kmol/h. The overhead vapor rate is V = L + D = 19.45 + 38.9 = 58.35 kmol/h. The slope of the operating line = L/V = 19.45/58.35 = 0.333. Using Eq. (7-6), the equation for the operating line y = 0.333x + DxD/V = 0.333x + (38.9)(0.95)/(58.35) = 0.333x + 0.633 (3) is, Now consider the section of stages between the intercooler at stage 2 from the top and the feed stage. Because 50 mol% of the vapor from this section is condensed at stage 2 by the intercooler, the vapor rate in this section = V' = 2V = 2(58.35) = 116.7 kmol/h. The liquid rate in this section is L' = V' - D = 116.7 - 38.9 = 77.8 kmol/h. The slope of the operating line = L'/V' = 77.8/116.7 = 0.667. In this section, by hexane material balance, yV' = xL' + xDD or, y = (L'/V')x + DxD/V' = 0.667x + (38.9)(0.95)/116.7 = 0.667x + 0.317 (4) In the section below the feed stage, for a saturated liquid feed, L"= L' + F = 77.8 + 100 = 177.8 kmol/h. The vapor rate = V" = V' = 116.7 kmol/h. The slope of the operating line = L"/V" = 177.8/116.7 = 1.524. From Eq. (7-11), y = (L"/V")x - BxB/V”= 1.524x - (61.1)(0.05)/116.7 = 1.524x - 0.026 (5) (b) A McCabe-Thiele diagram in terms of hexane, the more volatile component, is shown on the next page, where the equilibrium curve is obtained from Fig. 4.4 and the operating lines for the three sections are draw...
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