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Split_00027

# Split_00027 - 8.51 n v mol min 65 o C P0(atm y mol P(v mol...

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8- 27 8.51 ± ( ( [ ) n P 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) ± ( ( n P 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) p p P H * * ( ( 65 65 o o 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 p yP p y P y P P H * * ( ( ) . 65 65 0 656 1157 0 0 0 = = Total mole balance 100 mol = Pentane balance 50 mole P = 0.656 + 0.410 mol vapor / s mol liquid / s : ± ± : ± ± ± . ± . n n n n n n v l v l v l + = = 36 6 634 Ideal gas equation of state mol 0.08206 L atm (65+ 273)K s mol K atm
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