Elementary Principles 344

# Elementary Principles 344 - 8.90 Basis: 200 kg/h feed...

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8-65 8.90 Basis: 200 kg/h feed solution . A = NaC H O 232 (kmol A-3H O( )/h) 200 kg/h @ 60°C (kmol/h) n 0 0.20 A 0.80 H O 2 (kmol H O( )/h) n 12 v 50°C, 16.9% of H O 2 in feed Product slurry @ 50°C n 22 v (kmol solution/h) n 3 0.154 A 0.896 H O 2 (kJ/hr) Q . . . . a. Average molecular weight of feed solution: MM M A =+ 0 200 0800 .. HO 2 =+= 0 200 82 0 0800 18 0 308 . bg b g b g kg k Molar flow rate of feed: n 0 649 == 200 kg 1 kmol h 30.8 kg kmol h . b. 16.9% evaporation ⇒= = nv 1 0169 080 649 0877 . . b g b g b g kmol h kmol H O h 2 A balance: 020 649 3 0154 2 3 . b g kmol h kmol H O 1 mole h1 m o l e 3 H O 2 2 = + E nA A A n + = nn 23 130 (1) H O balance: kmol h kmol H O 3 moles H O m o l e 3 H O 2 2 080 649 3 0846 3 4 315 2 32 3 . . b g +⇒ + = A n 2 Solve 1 and simultaneously kmol 3H O s h kmol solution h 2 b g 21 1 3 1095 2 3 = n . . Mass flow rate of crystals 1.13 kmol 3H O 136 kg 3H O k m
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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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