Elementary Principles 146

# Elementary Principles 146 - 5.52(cont'd nA = 1 1 nB = 1 1 1...

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5- 34 5.52 (cont’d) n n n n n n yn n y y y y py P A B C D E tot AA t o t B C D E ii =− = =+ = = + U V | | | | | W | | | | | == + =+− + + = 1 1 1 2 1 2 1 1 2 3 1 2 6 2 21 6 26 1 1 12 2 2 1 1 11 1 ξ ξξ bg b g b g K pp p yy y p (1) 1 CO O CO BC A 2 2 = +− −+ = ⇒− + = + 1 2 22 6 5 03272 0 3272 1 6 2 236 2 2 2 b g . .. K p y p (2) 2 NO ON E C D 11212 = = ⇒+ = −− di b g b g 2 01222 01222 2 2 2 2 2 2 2 . . Solve (1) and (2) simultaneously with E-Z Solve = = 0 20167 012081 ., y mol CO mol y mol N mol y mol CO mol y mol NO mol y m o l O m o l A2 D 2 BE C2 += = = 2 1 6 0 2574 03030 0 0650 0 0390 0 3355 . 5.53 a. 81 0 864 PX=C H , TPA=C H O , S=Solvent ± ± ± ± V (m / h) @105 C, 5.5 atm n ( k m o l O/ h ) ( k m o l N/ h ) n (kmol H O(v) / h) 3 3O 3N 3o 2 2 3W 2 ± ± . V (m / h) at 25 C, 6.0 atm ( kmo l/h ) 0.21 O N 2 2 2 2 079 ± . . (kmo ) 4 O N 2 2 004 096 ± ± n (kmol H O(v) / h) V ( m/ h ) 3W 3W 2 3 condenser 100 mol TPA / s ± () ± nk m o l P X / h 100 kmol TPA / h 3p s m (kg S/h ) reactor
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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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