Elementary Principles 175

# Elementary Principles 175 - 6.17(cont'd n1 mol 15C 383.1 mm...

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6- 9 6.17 (cont’d) Saturation Condition: C mm Hg mm Hg mol H O mol HO 2 2 y p P 1 15 12 79 3831 0 03339 = ° == * . . . bg c . Dry air balance: 0.800 0.0258 mol b g =− = nn 11 1 003339 0 02135 .. Total mole balance: 0.0258 = 0.02135 + mol 22 0 00445 = . Mass of water condensed = 0.00445 mol 18.02 g mol g = 00802 . 6.18 Basis: 1 mol feed (mol), 15.6°C, 3 atm 1 mol, 90°C, 1 atm 0.10 mol H O (v)/mol 0.90 mol dry air/mol n 2 y 2 (mol H O (v)/mol)(sat'd) (1 – ) y 2 (mol DA/mol) (mol) H O( ), 15.6°C, 3 atm n 3 l 2 2 2 heat 100°C, 3 atm V (m ) 1 (mol) n 2 3 V (m ) 3 2 Saturation: C mm Hg atm 3 atm 760 mm Hg 2 y p P y 156 1329 0 00583 = ° * . . . Dry air balance: mol 0 90 1 1 0 00583 0 9053 . b g = H O mol balance: 0.10 mol 2 1 0 00583 09053 0 0947 33 b g =+ = . Fraction H O condensed: mol condensed mol fed mol condense mol fed 2 0 0947 0100 0 947 . . . = h yP p r = × ∗°
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