Split_00091 - 9.69 (cont'd) ge j CO: n = 0.988n +...

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9- 91 a. b. CH: mo l CH h 22 nn n 7 5 77 00155 5086 10 0006 7931 + = .. . bg ej CO: 0.00055 = 23311 mol CO h n 99 5 9 0988 5086 10 =+ × CO : 3458 mol CO h n 10 5 00068 5086 10 = H O: 0.0596 30313 mol H O h n 11 5 5086 10 = . Soot formation: 13 14 13 14 0 0567 00567 1 = ⇒= .) . (mol CH 1 mol C h1 m o l C H 4 4 Converter C balance: n n 14 13 14 13 5595 1 7931 2 23311 1 3458 1 48226 2 + + + ⇒=+ mol CH h mol C mol CH 44 b g b g b g b g b g b g b g Solve (1) & (2) simultaneously = 13 14 2899 51120 mol C s h mol CH h 4 , Converter H balance: 51120 mol CH 4 mol H m o l C h 4 4 CH C H H HO 42 2 2 2 =++ + 5595 4 7931 2 2 30313 2 8 b g bg b g b g n n 8 52816 = mol H h 2 Converter O balance: 0 96 2 3458 2 30313 1 15 . n b g b g g + 23311 mol CO 1 mol O m o l C O CO H O n 15 31531 mol h Converter N balance: 0.04 mol N h b g 31531 1261 12 12 Feed stream flow rates V CH 4 3 4 4 mol CH 0.0244 m STP m o l SCMH CH == 51120 1145 V O 2 3 2 2 O N 0.0244 m STP m o l N = + 31531 mol 706 SCMH O 2 2 b g Gas feed to absorber 5595 mol CH h 7931 mol C H h 23311 mol CO h
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This note was uploaded on 01/07/2012 for the course CH E 210H at Pennsylvania State University, University Park.

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