Elementary Principles 178 - 6.22 (cont'd) c. 6.53 kmol/h...

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6- 12 6.22 (cont’d) c. 80% condensation: nl L = 080 015 653 .( .) ( . ( ) / kmol / h) = 0.7836 kmol C H h 61 4 Mole balance: 653 07836 5746 .. . = + = nn VV kmol / h Hexane balance: 015 653 0 7836 0 03409 .) ( . ) . . = + = yy kmol C H / kmol 4 Raoult’s law: py P p T Hex Hex == × (. ) ( ( ) * 0 03409 2 760 mmHg) = 51.82 mmHg = Antoine equation : o 10 1175.817 log 51.82 6.88555 2.52 C 224.867 T T =− = + 6.23 Let H=n-hexane a. 50% relative saturation at inlet: yP p oH = 0500 80 *o C) y o = (. ) ( . 0500 1068 0703 mmHg) 760 mmHg = kmolH / kmol Saturation at outlet : 0 05 0 05 760 11 ) ( ) . ( ** PpT pT HH =⇒= mmHg) = 38 mmHg Antoine equation: o 10 1 1 1175.817 log 38 6.88555 3.26 C 224.867 T T = + Mole balance: N balance: kmol / min kmol / min 2 ±± . ) ± . ± ± . ± . n n 01 0 1 150 1 0 703 0 95 218
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This note was uploaded on 02/27/2012 for the course CHEMICAL E 312 taught by Professor Cheung during the Fall '11 term at The University of Akron.

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