Elementary Principles 223

Elementary Principles 223 - 6.77 (cont'd) m0 = 2150 kg feed...

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6- 57 6.77 (cont’d) Mass balance kg / h MgSO balance: 0.35 kg MgSO h kg feed / h kg soln / h The crystals would yield 0.488 1000 kg / h = 488 kg anhydrous MgSO h 4 4 4 : ±± . ± .( ) / ± ± mm m m 01 0 1 1000 0 23 0 488 1000 2150 1150 =+ = = × 6.78 Basis : 1 lb m feed solution. Figure 6.5-1 a saturated KNO 3 solution at 25 o C contains 40 g KNO 3 /100 g H 2 O x KNO 3 3m 3 m 3 g KNO (40 +100) g solution g K N Og = 0 . 2 8 6 l b K N Ol b == 40 0 286 ./ / x 1 lb m solution @ 80 o C 0.50 lb m KNO 3 /lb m m 1 (lb m ) sat’d solution @ 25 o C 0.50 lb m H 2 O/lb m 0.286 lb m KNO 3 /lb m soln 0.714 lb m H 2 O/lb m soln m 2 [lb m KNO 3 (s)] Mass balance 1 lb KNO balance 0.50 lb KNO = 0.700 lb solution / lb feed 0.300 lb crystals / lb feed Solid / liquid mass ratio = 0.300 lb crystals / lb feed 0.700 lb solution / lb feed = 0.429 lb crystals / lb solution m 3 : :. = m m 12 1 2 0286 6.79 a. Basis : 1000 kg NaCl(s)/h. Figure 6.5-1 a saturated NaCl solution at 80 o C contains 39 g NaCl/100 g H 2 O xg NaCl (39 +100) g solution
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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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