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

# 3 15 1663 kmolh and v v 15575 kmolh thus lv

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Unformatted text preview: hanol boils at 78.4oC. From Fig. 2.4, the vapor pressure of ethanol at 64.7oC = 8.2 psia. The vapor pressure of methanol at 78.4oC = 25 psia. For the Raoult's law K-value, Eq. (2-44) applies. Combining this equation with the definition of the relative volatility of Eq. (2-21), gives, for methanol with respect to the less volatile ethanol, αΜ,Ε = (Ps)M/ (Ps)E . At 64.7oC, αΜ,Ε = 14.696/8.2 = 1.79. At 78.4oC, αΜ,Ε = 25/14.696 = 1.70. Since these values are close (within about 5%), use an equilibrium curve based on a constant α = (1.70 + 1.79)/2 = 1.745. From Eq. (7-3), the curve is computed from, y = αx/[1 + x(α-1)] = 1.745x/(1+0.745x) (3) The McCabe-Thiele diagram on the next page shows the operating lines for determining the minimum reflux ratio. It assumes that the pinch region occurs at the feed stage and not at the sidestream stage. From Eq. (7-18), q for 25 mol% vaporized = 0.75. From Eq. (7-26), the slope of the q-line = q/(q-1) = 0.75/(0.75-1) = -3. The upper section operating line intersects the q-line and equilibrium curve at y=0.823 and x=0.727. Thus, slope of the upper section operating line = L/V = (0.96-0.823)/(0.96-0.727) = 0.588. From Eq. (7-27), Rmin = (L/V)/[1 - (L/V)] = 0.588/(10.588) = 1.427. Therefore, L = 1.427(74.45) = 106.3 kmol/h and V = L + D = 106.3 + 74.45 = 180.75 kmol/h. In the middle section between the feed stage and sidestream stage, for 25 mol% vaporization, L' = L + 0.75(100) = 106.3 + 75 = 181.3 kmol/h and V' = V - 0.25(100) = 180.75 25 = 155.75 kmol/h. Thus, L'/V' = 181.3/155.75 = 1.164. The middle section operating line has this slope and passes through y=0.823 and x=0.727. In the lower section operating line below the sidestream, for a liquid sidestream flow rate of 15 kmol/h, L" = L' - S = 181.3 - 15 = 166.3 kmol/h and V" = V' = 155.75 kmol/h. Thus, L"/V" = 166.3/155.75 = 1.068. This line passes through the point y=0.05 and x = 0.05 with the slope of 1.068. It also intersects the middle section operating line at the sidestream composition, xs = 0.20. These three operating lines are drawn in the McCabe-Thiele diagram, showing that the middle section operating line lies below Analysis: (continued) Exercise 7.36 (continued) the equilibrium curve. Therefore, the assumption that the minimum reflux is controlled by the feed stage is verified and the minimum reflux ratio is 1.427. For actual operation, reflux ratio = 1.2Rmin = 1.2(1.427) = 1.712. Reflux rate in upper section = L = 1.2Lmin = 1.2(106.3) = 127.6 kmol/h. Vapor rate = V = L + D = 127.6 + 74.45 = 202.05 kmol/h. Slope of upper section operating line = L/V = 127.6/202.05 = 0.632. For constant molar overflow, the flow rates in the other sections are: Middle section: L' = 202.6 kmol/h V' = 177.05 kmol/h L'/V' = 1.144 Lower section: L" = 187.6 kmol/h V" = 177.05 kmol/h L"/V" = 1.060 Analysis: (continued) Exercise 7.36 (continued) In the McCabe-Thiele diagrams below, upper, middle, and lower operating lines are based on these values start...
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