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

The equilibrium liquid has a mole fraction of 035 for

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Unformatted text preview: urated vapor line. The vapor composition at point M is 69 mol% nC6 and 31 mol% nC8. (b) From Fig. 7.37, reproduced below, the enthalpy of the initial mixture of 20 mol% nC6 at 100oF, shown at point G, is 6,500 Btu/lbmol = HG . A vertical line is drawn upward from point G until it reaches the point E, located on the dashed tie line between the saturated liquid and saturated vapor curves, at a point that is 60% of the way from the saturated liquid curve to the saturated vapor curve. The enthalpy of the two-phase mixture at point E is 23,000 Btu/lbmol = HE. Therefore, the energy required = Q = HE - HG = 23,000 - 6,500 = 16,500 Btu/lbmol. (c) From the reproduction of Fig. 7.37 below, the equilibrium liquid phase composition is at point D, which is 7.5 mol% nC6 , while the equilibrium vapor composition is at point F, which is 28.5 mol% nC6 . Checking this by material balance on nC6 , for a basis of 100 lbmoles of mixture: 0.20(100) = 20 lbmoles 0.075(40) + 0.285(60) = 20.1 lbmoles This is a reasonably good check. Analysis: (continued) Exercise 7.54 (continued) Exercise 7.55 Subject: Determining compositions and energy requirements for a mixture of nC6 and nC8 with an enthalpy-concentration diagram for 101 kPa. Given: Enthalpy-concentration diagram in Fig. 7.37. Find: (a) Temperature and compositions of liquid and vapor, where F1 = 950 lb/h of 30 mol% nC6 at 180oF is adiabatically mixed with F2 = 1,125 lb/h of 80 mol% nC6 at 240oF. (b) Energy and resulting phase compositions when a mixture of 60 mol% nC6 at 260oF is cooled and partially condensed to 200oF. (c) Compositions and relative amounts of the two phases when the equilibrium vapor from part (b) is further cooled to 180oF. Analysis: (a) In Fig. 7.37, reproduced on the next page, the two feeds before mixing are shown at F1 and F2. When these two feeds are adiabatically mixed, the resulting two-phase mixture contains the following, where it is necessary to convert mass to moles. Component nC6 nC8 Total MW 86.11 114.14 Feed F1 : lb 232 718 950 lbmol 2.69 6.29 8.98 Feed F2 : lb 845 280 1,125 lbmol 9.81 2.45 12.26 Total feeds: lbmol 12.5 8.74 21.24 mole fraction 0.5885 0.4115 100 Based on the mole fractions of the combined feed, point M, the adiabatic mixing point is located in Fig.7.37 on the next page. By coincidence the mixing line connecting the two feeds is also a tie line for a temperature of 204oF. The equilibrium vapor is located at point V, while the equilibrium liquid is located at point L. The equilibrium vapor has a mole fraction of 0.75 for nC6 and 0.25 for nC8. The equilibrium liquid has a mole fraction of 0.35 for nC6 and 0.65 for nC8. Although not requested, the moles of equilibrium vapor and liquid can be determined by material balances. By total material balance, F1 + F2 = 21.24 = V + L (1) By material balance on n-hexane, 12.5 = 0.75V + 0.35L (2) Solving Eqs. (1) and (2), V = 12.68 lbmol and L = 8.56 lbmol The molar % vaporized = 12.69/21.24 x 100% = 59.7%. This is consistent with the value obtained by the inverse lever arm rule on the diagra...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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