Split_00028 - 8.52 a. B=benzene; T=toluene n 2 mol/s 95o C...

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8- 28 8.52 a. B=benzene; T=toluene Q 1320 mo l/s 25 o C 0.500 mol B/ mo l 0.500 mol T/ mol n 2 mol/s 95 o C 0.735 mol B/ mo l 0.265 mol T/ mol n 3 mol/s 95 o C 0.425 mol B/ mo l 0.575 mol T/ mol Total mole balance: Benzene balance: 1320(0.500) = mol / s mol / s 1320 0 735 0 425 319 1001 23 2 3 =+ + U V W = = R S T nn n n (. ) ) References : B(l, 25 o C), T(l, 25 o C) Substance ± n in (mol/s) ² ( H in kJ / mol) ± n out (mol/s) ² ( H out kJ / mol) B(l) 660 0 425 9.838 B(v) -- -- 234 39.91 T(l) 660 0 576 11.78 T(v) -- -- 85 46.06 4 ˆˆ 2.42 10 kW ii out in Qn H n H =−= × ∑∑ b. Antoine equation (Table B.4) C torr , C torr Raoult's law Benzene: 0.425 torr Toluene: 0.575 torr Analyses are inconsistent. oo ⇒= = =⇒ = = U V | W | ⇒≠ pp PP BT ** . . ..' ' ' 95 1176 95 476 9 1176 0 735 680 476 9 0 265 1035 ej bg b g Possible reasons: The analyses are wrong; the evaporator had not reached steady state when the samples were taken; the vapor and liquid product streams are not in equilibrium; Raoult’s law is
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This note was uploaded on 01/07/2012 for the course CH E 210 at Penn State.

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