33 kmolh each for benzene and toluene next calculate

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Unformatted text preview: lance: F = nF = 10.66 = D + B (1) overall benzene mole balance: FxF = 5.33 = 0.95D + 0.05B (2) Solving Eqs. (1) and (2), D = 5.33 kmol/h and B = 5.33 kmol/h Therefore in terms of mass flow rates, total total distillate rate is, mD = 0.95(5.33)(78.11) + 0.05(5.33)(92.14) = 420.1 kg/h Therefore the bottoms rate = mB = 907.3 - 420.1 = 487.2 kg/h (e) First compute the kmol/h of vapor leaving the reboiler, using the assumption of constant molar overflow. From part (c), the reflux ratio = 1.57. Therefore, the reflux rate = 1.57(5.33) = 8.37 kmol/h. Below the feed stage, the liquid rate = 8.37 + 10.66 = 19.03 kmol/h. The vapor rate leaving the reboiler = 19.03 - 5.33 = 13.70 kmol/h. From the plot above, the composition of the reboiler vapor = 12 mol% benzene. Neglecting the sensible heat and using the enthalpy data given, after converting from Btu/lbmol to kJ/kmol, the reboiler heat duty is, QR = 2.324[0.12(13.7)(18,130 - 4,900) + 0.88(13.7)(21,830 - 8,080)] = 436,000 kJ/kmol. From Perry's Handbook, latent heat of v...
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This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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