Split_00027 - 8.51 n v ( mol / min) @ 65 o C, P0 (atm) y[...

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8- 27 8.51 ± (( [ ) nP y y v mol / min) @ 65 C, atm) mol P(v) / mol], sat'd (1- (mol H(v) / mol) o 0 100 mol 80 0 / s @ C, 5.0 atm .500 mol P(l) / mol 0.500 mol H(l) / mol o ± ( Q kJ / s) ± l mol / min) @ 65 C, atm) .41 mol P(l) / mol 0.59 mol H(l) / mol o 0 0 a. Degree of freedom analysis 5 unknowns – 2 material balances – 2 equilibrium relations (Raoult’s law) at outlet – 1 energy balance = 0 degrees of freedom Antoine equation (Table B.4) pp PH ** 65 65 oo C) = 1851 mm Hg, C) = 675 mm Hg Raoult's law for pentane and hexane 0.410 C) = 0.590 C) = (1 mol P(v) / mol mm Hg (1.52 atm) o o py P P y P P H * * ( () . 65 65 0656 1157 0 0 0 = = Total mole balance 100 mol = Pentane balance 50 mole P = 0.656 + 0.410 mol vapor / s mol liquid / s : ±± : ± . ± . nn n n vl v l + = = 36 6 634 Ideal gas equation of state mol 0.08206 L atm (65+ 273)K
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This note was uploaded on 01/07/2012 for the course CH E 210 at Pennsylvania State University, University Park.

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