Elementary Principles 111

# Elementary Principles 111 - 4.78(cont'd SO 2 0.03 100 g S b...

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4- 70 4.78 (cont’d) SO 2 : 0.03 100 g S 1 mol S 1 mol SO g S 1 mol S n m o l S O .0956 mol O consumed 2 52 2 bg 32 06 0 0936 0 . . ⇒= F H G I K J 25% excess O 2 n 12 mol O =+ + 125 7 244 2 475 0 0936 12 27 .. . . . O balance: 12.27 mol O fed mol O consumed mol O 23 2 2 2 n =− + + = 7 244 2 475 0 0936 246 ... . N balance: 2 n mol mol N 22 == 3761227 4614 . SO in stack SO balance around mixing point xx x n 0 0936 010 1 0 0936 0 00936 0 0842 5 F H I K +− = + b g b g mol SO 2 Total dry gas in stack (Assume no CO 2 , O 2 , or N 2 is absorbed in the scrubber) 7 244 2 46 4614 0 00936 0 0842 5585 0 0842 . . . . CO O N SO 2 2 mol dry gas bgb gb g b g ++ + + = + 612 5 . ppm SO dry basis in stack gas 2 000936 00842 5585 0 0842 612 5 10 10 0 295 30% 6 . . . + + = × x x x bypassed 4.79 Basis: 100 mol stack gas C + O 2 CO 2 2H + O 2 1 2 H O 2 S + O 2 SO 2 (mol C) n 1 (mol H) n 2 (mol S) n 3 (mol O ) n 42 (mol O ) n 3.76 100 mol 0.7566 N 2 0.1024 CO 2 0.0827 H O 2 0.0575 O 2 0.000825 SO 2 a. b. C balance: mol C H balance:
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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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