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Unformatted text preview: 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 diagram on the next page. Analysis: (a) (continued) Exercise 7.55 (continued) Exercise 7.55 (continued)
Analysis: (continued) (b) From the diagram below, the enthalpy of the feed, F , of 60 mol% nC6 at 260oF is equal to HF = 27,700 Btu/lbmol. Upon cooling, the point P, in the two-phase region, is reached at 200oF. From the tie line through point P, the equilibrium vapor and liquid phases are shown at points V and L. The enthalpy of the two-phase mixture at point P = HP = 17,700 Btu/lbmol. Therefore, the energy that must be removed = Q = 27,700 - 17,700 = 10,000 Btu/lbmol mixture. The composition of the equilibrium vapor at point V is 77 mol% nC6. The composition of equilibrium liquid at point L is 38 mol% nC6 . The amounts of vapor and liquid can be obtained by material balances or by the in...
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