Elementary Principles 130 - 5.35 Basis 100 kmol dry product...

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5- 18 5.35 Basis: 100 kmol dry product gas n (kmo l m (kg CH ) 1 1x y CH) xy a. N balance: 0.79n n =106.6 kmol air 22 2 =⇒ 0842 100 .() O balance: 2 0.21n n n kmol H O 23 3 2 bg b g b g =+ + = 100 2 0105 2 0 053 1317 .. . C balance: n kmo l CH x kmol C kmol C H n x =10.5 1 y 1 di b g 100 0105 . H balance: n y = 2n n y 2 13 n 1 3 ====> = = 13 17 26 34 . . Divide 2 by 1 y x mol H / mol C ⇒= = 26 34 105 251 . . O fed: 0.21 106.6 kmol air kmol 2 = 22 4 O in excess = 5.3 kmol Theoretical O = 22.4 -5.3 kmol =17.1 kmol % excess = 5.3 kmol O 17.1 kmol O excess air 2 2 ×= 100% 31% b. V N m ( S T P ) kmol kPa kPa K 273 K m 2 2 3 3 == 106.6 kmol 22 4 1013 98 303 2740 m= nx kmo l C 12.0 kg kmol n y kmol H kg kmol m k g 1 11 ny=26 .34 nx=10 .5 1 1 1 + =====> = 101 152 6 . V m = 2740 m air kg fuel m a i r kg fuel 2 1 33 152 6 18 0 . . = 5.36
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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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