Elementary Principles 29 - 3.28 (cont'd) 55 gal 3.7854 L...

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3- 13 3.28 (cont’d) (b) t V V == = ± min 55 60 144 gal 3.7854 L s 87 L s 236 10 mL 1.03 g 0.0500 g H SO 1 lbm 1 L mL g 453.59 g lb H SO 3 24 m2 4 = . (c) u V A × = ± (/ ) . 87 4 0513 L m 1 min min 1000 L 60 s 0.06 m m/s 3 22 π t L u = 45 88 m 0.513 m / s s 3.29 (a) ± . . n 3 150 1147 L 0.659 kg 1000 mol min L 86.17 kg mol / min Hexane balance: 0 (mol C H / min) Nitrogen balance: 0.820 (mol N 61 4 2 . ± . ± . ± . ± /m in) 180 0050 1147 0950 12 nn =+ = U V W = R S | T | solve mol / min = 72.3 mol / min ± . ± n n 1 2 838 (b) Hexane recovery = × = ± ± . .. n n 3 1 100% 0180 838
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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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