Elementary Principles 144

Elementary Principles 144 - 5.51 n 4(kmol CO h n 5(kmol H 2...

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5- 32 5.51 ± ± n (kmo l CO/h ) l H /h ) 4 52 a. 5.0% XS H 2 : 22 2 3 2 100 kmol CO fed 2 kmol H reqd 1.05 kmol H fed kmol H 210 h 1 kmol CO fed 1 kmol H reqd h n == ± () 46 4 6 100 kmol CO 1 kmol C C balance: (1) 100 1 h 1 kmol CO nn n n =+⇒ = + ±± ± ± 56 5 6 H balance: 210(2) (2) (4) 210 2 (2) n n = + ± ± (O balance: 100 identical to C balance not independent) 5 6 456 6 6 6 6 100 , (2) 210 2 (100 ) (210 2 ) 310 2 tot n n n n n n ⇒= − =++= − + − += − ± ± ± ± ± ± 4 7- 2 p 2 -3 -8 9143.6 21.225+ 7.492ln 500K 500 K K T=500K 1.390 10 exp 9.11 10 kPa +4.076 10 500K -1.161 10 500K ⎛⎞ ⎜⎟ = × ×× ⎝⎠ K yP KP y yy n n n n n n kP a a p M CO H p 2 M CO H 6 6 6 6 6 6 p 2- 2 66 = = ====> = = −− di d i bg b g 13 ) 2 2 7 2 2 2 310 2 100 310 2 210 2 310 2 911 10 5000 22 775 310 2 100 210 2 () ( ± ± ± ± ± ± .. 3 4 6 2 16 1 2 Solving for 75.7 kmol CH OH/h , 100 24.3 kmol CO/h 210 2 58.6 kmol H / h Overall C balance: 75.7 kmol CO/h Overall H balance: (2) n n n n ⇒= = −= =−= =⇒ = ± ±
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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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