Elementary - 5.26(cont'd Total mole balance n1 n 2 = n 3 n 2 = 7317 3722 = 3595 mol air s c n 2 = 13 n 2 V2 = b g min = 4674 mol s 4674 mol s 8.314

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5- 11 5.26 (cont’d) Total mole balance n n n n mol air / s 12 3 2 : ±± ± ± + = = = 7317 3722 3595 c. ± . ± nn m o l / s 22 min == 13 4674 bg ± .. / ± ./ ± ± . V 4674 mol / s m Pa mol K K 131,000 Pa ms V 3722 mol s mP a mol K K 110000 Pa V V m diluting air m fue l ga s 2 3 3 1 3 3 2 1 3 3 = = = = U V | | W | | = 8314 3982 118 8 314 2982 839 141 y mol/s mol / s mol/s+4674 mol/s 2 12 = + = 150 150 3722 100% 18% . d. The incoming propane mixture could be higher than 4.03%. If , min = fluctuations in the air flow rate would lead to temporary explosive conditions. 5.27 ( ) ( ) Basis: 12 breaths min 500 mL air inhaled breath 6000 mL inhaled min = 24 o C, 1 atm 6000 mL/min 37 o C, 1 atm ± . . n (m o l /m i n ) 0.206 O N HO in 2 2 2 0 774 0 020 blood ± . . . ( m o l i n ) 0.151 O CO N out 2 2 2 2 0 037 0 750 0 062 a. ± . n 6000 mL 1 L 273K 1 mol
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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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