Elementary Principles 188 - 6.33 n0 (kmol/min wet air) @...

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6- 22 6.33 Dry pulp balance: 1500 1 10 7 5 1 0 0015 858 11 × + =− = . ± (.) ± mm kg / min 50% rel. sat’n at inlet : 2 *o 1H O 1 2 0.50 (28 C) 0.50(28.349 mm Hg)/(760 mm Hg) = 0.0187 mol H O/mol yP p y =⇒ = 40 o C dew point at outlet: yP p y 22 40 324 770 = HO 2 2 C) 55 mm Hg) / ( mm Hg) = 0.0718 mol H O / mol (( . Mass balance on dry air : ± ± ( ) nn 01 1 0 0187 1 0 0718 1 −= Mass balance on water: ± (. ) ( . ) (. /. ) ± ) ( ) )() 0 1 0 0187 18 0 1500 0 75 175 0 0718 18 858 0 0015 2 kg/kmol + = + Solve (1) and (2) ± . ± . 622 8 658 4 = = kmol / min, kmol / min Mass of water removed from pulp : [1500(0.75/1.75)–858(.0015)]kg H 2 O = 642 kg / min Air feed rate
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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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