Split_00042 - 8.65 (contd) (b) T = 75 C pC6 H 6 = 648 mm Hg...

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8- 42 8.65 (cont’d) (b) Tp =°⇒ = 75 648 C m m H g CH 66 , p 78 mm Hg = 244 (from Table 6.1-1) Raoult's law ⇒= + = + ∗∗ px p x p t ank 0 439 648 0561 244 .. bg b g b g =+ 284 137 mm Hg = 421 mmHg P 0.554 atm tank yv 6 6 284 mm Hg 421 mm Hg mol C H mol == 0675 . 0.541 C H ( ) n v l 6 93.19 mol/s 6 0.459 C H ( ) l 90°C, P 0 atm (mol/s), 75°C 0.675 C H ( ) v 0.325 C H ( ) v n L (mol/s), 75°C 0.439 C 6 H 6 ( l ) 0.541 C 7 H 8 ( l ) 0.554 atm Mole balance: = C H balance: 0.541 40.27 mol vapor s 52.92 mol liquid s 9319 9319 0 675 0 439 . . nn n n vL v L + U V W = = (c) Reference states : C H , C H at 75 C ll ° Substance in mo ls k Jmo l in in out out ± ² ± ² ± . ² . nH vn lH v l −− 2718 310 5041 216 2323 0 1309 353 42 78 2 64 29 69 0 C H , 90 C kJ mol °= = b g b g : ² 0144 90 75 216 C H kJ mol = b g b g : ² 0176 90 75 2 64 CH , 75C kJ mol C vH T d T H v ++ + × = A ° z b g b g : ² . . ² . 0144 801 75 30 77 0 074 0330 10 80 1 3 80.1 75 Δ kJ mol T d T + + + × = z b g b g : ² . . 0176 1106 75 3347 0 0942 0 380 10 3 110.6 75 Energy balance: ±± ± ² ± ² Q H ii = =−= = ∑∑ Δ out in 1 kW s 1 kJ s 1082 kW 1082 kJ (d) The feed composition changed; the chromatographic analysis is wrong; the heating rate
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