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Split_00085 - 9.66(cont'd H b = H v H 2 O(25 C(C p H 2 O(v...

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9- 85 a. b. ± ( ± ) ( ) H H C dT b p T = + z Δ v H O(25 C) H O(v) 25 2 2 D 4 o c CH out out in in ˆ Table B.2 for ( ), ( ) 44.01 kJ/mol v H O 2 5 2 2 8 3 3 12 4 Energy Balance ˆ ˆ ˆ ( ) 0 0.247( 890.36) 0.494(44.01) 0.0963( 25) 1.02 10 ( 25 ) 0.305 10 ( 25 ) 1.61 10 ( 25 pi C T H H H n H n H T T T T ξ Δ = Δ = Δ + = + + + × + × × ² ² ² ² 4 ) 0.6175(3.78) 2.32(3.66) 0 = 5 2 8 3 12 4 o 211.4 0.0963 1.02 10 0.305 10 1.61 10 0 1832 C ad ad ad ad T T T T T ⇒ − + + × + × × = = 2 2 2 2 o H O 2 Table B.3 * * H O H O H O In product gas, 1832 C, 1.05 760 798 mmHg 0.494 mol/s y 0.155 mol H O/mol (0.124 2.32 0.247 0.494) mol/s Raoult's law : y ( ) (0.155)(798) 124 mmHg 56 C Degr. superheat dp dp T P P p T p T = = × = = = + + + = = = = D = 1832 C 56 C = 1776 C D D D 9.67 a. CH l O g CO (g) + 2H O(v) 4 2 2 2 ( ) ( ) + 2 Basis 1 mol CH 4 : Theoretical oxygen 1 mol CH 2 mol O 1 mol CH mol O 4 2 4 2 = = 2 00 . 30 130 2 00 2 60 376 2 60 9 78 excess air mol O mol N 2 2 % . ( . ) . , . . . = × = 1 mol CH 4 n 2 (mol CO 2 ) 2.60 mol O 2 n 3 (mol H 2 O) 9.78 mol N 2 n 4 (mol O 2 ) 25
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