Split_00029 - 8.53 (contd) b g b g 1 kJ kJ J 16.2 b301gb61g...

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8- 29 8.53 (cont’d) ΔΔ Δ ±± ± .. HH C H C vp l mp s =− + + + × 52 113 25 52 421 301 61 170 27 1 813 bg b g b g b g kJ mol 16.2 kJ mol J mol kJ 10 J kJ mol 3 Required heat transfer: QH n H == = ± . 2.05 mol kJ 1 kW s mol 1 kJ s kW 167 8.54 Basis: 100 kg wet film 95 kg dry film 5 kg acetone 0.5 kg acetone remain in film 4.5 kg acetone exit in gas phase 90% A evaporation a. = 35°C n 1 5 kg C H O( ) T f 95 kg DF 6 l 3 mol air , 1.01 atm n 1 0.5 kg C H O( ) 95 kg DF 6 l 3 mol air = 49°C, 1.0 atm T 1 T a 1 T f 2 4.5 kg C H O( ) (40% sat'd) 6 v 3 a 2 Antoine equation (Table B.4) mm Hg CHO 36 ⇒= p * . 59118 4.5 kg C H O 1 kmol 10 mol 58.08 kg kmol mol C H O in exit gas 3 = 77 5 . v y = 775 775 040 59118 760 405 1 1 . . . + =⇒ = = n n mm Hg 171.6 mol 22.4 L STP mol 95 kg DF LSTP kg DF b. R e f e r en c e s : A i r25C CHO 35C DF35C °° ° b g ,, , l Substance n in ± H in n out ± H out DF 95 0 95 1.33 T f 2 35 di n in kg ± H in kJ/kg CH O 61 4 l 4 v Air
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This note was uploaded on 01/07/2012 for the course CH E 210 at Pennsylvania State University, University Park.

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