Elementary Principles 24 - 3.15(a m = 175 m3 h 1000 L m 3...

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3- 8 3.15 (a) ± m == 175 2526 m 1000 L 0.866 kg 1 h h m L 60 min kg / min 3 3 (b) ± n 2526 457 kg 1000 mol 1 min min 92.13 kg 60 s mol / s (c) Assumed density (SG) at T, P of stream is the same as the density at 20 o C and 1 atm 3.16 (a) 200 0 0150 936 .. kg mix kg CH OH kmol CH OH 1000 mol kg mix 32.04 kg CH OH 1 kmol mol CH OH 33 3 3 = (b) ± m mix 100.0 lb - mole MA 74.08 lb MA 1 lb mix h 1 lb - mole MA 0.850 lb MA /h mm m m 8715 lb 3.17 M =+ = 025 2802 075 202 852 . . . mol N g N mol N mol H g H mol H gmo l 22 2 2 ± m N 2 3000 2470 kg kmol kmol N kg N h 8.52 kg kmol feed kmol N kg N h 2 2 3.18 M suspension g g g =− = 565 65 500 , M CaCO 3 g g = = 215 65 150 (a) ± V = 455 mL min , ± m = 500 g min (b) ρ = ± / ± . mV 500 110 g / 455 mL g mL (c) 150 500 0 300 g CaCO g suspension g CaCO g suspension /. = 3.19 Assume 100 mol mix. m CHOH 25 10.0 mol C H OH 46.07 g C H OH mol C H OH
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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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