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

# 01 the stages are stepped off with the optimal feed

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Unformatted text preview: (45o line for total reflux condition) are plotted and equilibrium stages are stepped off between x = 0.298 and 0.653. The number of stages = Nt = 14.4. From Eq. (6-73), HETP = 10/14.4 = 0.7 ft. This value must be used with caution for design, because the HETP is a function of the L/V ratio. The value of 0.7 ft applies only to L/V =1. Exercise 7.51 Subject: atm. Design of a packed column for separation of a mixture of ethanol and water at 1 Given: Bubble-point feed of 10 mol% ethanol. Bottoms of 1 mol% ethanol and distillate of 80 mol% ethanol. Reflux ratio at 1.5 times minimum. Equilibrium data from Exercise 7.29. Assumptions: Constant molar overflow. Find: (a) (b) (c) (d) (e) Equilibrium stages above and below feed. Number of transfer units above and below the feed. Height of plate column for 18-inch tray spacing and 80% tray efficiency. Height of packing for HOG = 1.2 ft. HTU from HTU for benzene-toluene system. Analysis: By material balance, with a basis of 100 moles of feed, Total material balance, F = 100 = D + B Benzene material balance, 0.10F = 10 = 0.80D + 0.01B Solving these two equations, D = 11.39 moles and B = 88.61 moles. (a) Below is the McCabe-Thiele diagram showing the equilibrium curve from the data of Exercise 7.29, a vertical q-line for the saturated liquid feed at xF = 0.10, and the construction of the rectification section operating line for determining the minimum reflux ratio. The slope of that line is, (0.80 - 0.44)/(0.80 - 0.10) = 0.514 = Lmin/V. From a rearrangement of Eq. (7-7), Rmin = Lmin / V 0.514 = = 1.058 and R = 1.5 Rmin = 1.5(1.058) = 1.59 1 − Lmin / V 1 − 0.514 Slope of the rectification section operating line is, from Eq. (7-7), L/V = 1.59/(1 + 1.59) = 0.614. In the McCabe-Thiele diagrams below, one for the higher concentration region and one for the lower concentration region, a rectification section operating line of this slope is drawn that passes through the 45o line at x = xD = 0.8. A stripping section operating line is drawn that extends from the intersection of the rectifying operating line and the q-line to the 45o line at x = xB = 0.01. The stages are stepped off, with the optimal feed stage location. The result is 3 equilibrium stages in the stripping section, including one for the partial reboiler, and 13.8 equilibrium stages in the rectifying section. Analysis: (a) (continued) Exercise 7.51 (continued) Exercise 7.51 (continued) Analysis: (a) (continued) Exercise 7.51 (continued) Analysis: (a) (continued) Exercise 7.51 (continued) Analysis: (continued) (b) Because the value of HOG is given in part (d), calculate NOG for each section, using from Table 6.7 for EM diffusion, N OG = dy y −y * For the stripping section, the first stage is the partial reboiler. Therefore, from the McCabeThiele diagram, starting from Stage 2, the limits on y are y = 0.09 at x = 0.03 to y = 0.37 at x = 0.10. Using the trapezoidal method to solve the integral, ( NOG )SS = ∆y (0.23 − 0.09) (0.37 − 0.23) = + = 2.4 0.10 0.14 ( y *−...
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