Split_00024 - 8.49 Let A denote acetone. Q( kW) Ws = -25.2...

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8- 24 8.49 Let A denote acetone. ± ( ± QW s kW) 25.2 kW =− 142 L / s @ C, 1.3 atm mol / s) [mol A(v) / mol], sat'd (1 mol air / mol) o 150 0 0 0 ± ( )( n y y ± )( n y y 1 1 1 (mol / s) @ 18 C, 5 atm [mol A(v) / mol], sat'd (1 mol air / mol) o ± [ n 2 18 mol A(l) / s]@ C, 5 atm o a. Degree of freedom analysis : 6 unknowns ( ± , ± , ± ,,, ± ) nnnyyQ 01201 –2 material balances –1 equation of state for feed gas –1 sampling result for feed gas –1 saturation condition at outlet –1 energy balance 0 degrees of freedom b. Ideal gas equation of state Raoult’s law (1) ± ± n PV RT 0 00 0 = (2) y p p A A 1 18 = * * ( ) o C) 5 atm (Antoine equation for Feed stream analysis (3) y PR T 0 4 017 300 mol A mol [(4.973 g A][1 mol A / 58.05 g] L) mol feed gas F H G I K J = .) [( . / ] Air balance : ± ±
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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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