Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Analysis the rachford rice flash equations can be

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Unformatted text preview: 5.8)/2 = 103.7oC. In this case, the values of Ψ, xW, and xA are computed from Eqs. (1) and (3), where the vapor pressures are computed from Eqs. (5) and (6) to be 867 torr for W and 473 for A. Values of yW and yA are obtained from Eq. (2). Using, again, a spreadsheet with a trial and error procedure, the following result is quickly obtained: V/F = 0.49 xW = 0.4100 xA = 0.5900 yW = 0.5937 yA = 0.4063 Exercise 4.34 Subject: Bubble point, dew point, and flash of a toluene (1) - n-butanol (2) mixture. Given: Feed of z1 =0.4 and z2 =0.6 at 1 atm. Liquid-phase activity coefficients for 1 and 2 as a function of liquid-phase mole fractions from the van Laar equations. Assumptions: Modified Raoult's law, Eq. (2-72) applies. Find: Dew point, bubble point, and equilibrium vapor and liquid at a temperature halfway between the bubble and dew points. Analysis: The Rachford-Rice flash equations can be used from Table 4.4: f {Ψ} = zi 1 − Ki =0 i =1 1 + Ψ Ki − 1 C (1) zi Ki (2) 1 + Ψ Ki − 1 zi xi = (3) 1 + Ψ Ki − 1 The modified Raoult's law is: γ iL Pi s Ki = (4) P Antoine vapor pressure (in torr) equations are obtained by fitting the vapor pressure data for toluene that are given in Exercise 4.8 and from Perry's Handbook for n-butanol: yi = 3896.3 T ( C ) + 255.67 (5) 1305198 . T ( C ) + 173.427 (6) P1s = exp 17.2741 − log P2s = 7.36366 − o o The van Laar equations, Table 2.9, with the given constants are: ln γ 1 = ln γ 2 = 0.855 0.855x1 1+ 1306 x2 . 1306 . 1306 x2 . 1+ 0.855x1 (7) (8) Exercise 4.34 (continued) Analysis: (continued) Since P = 1 atm and the normal boiling points of toluene and n-butanol are 110.8oC and 117oC, respectively, it might be expected that the dew and bubble point of the mixture would be in the vicinity of 110oC, unless the liquid-phase activity coefficients are much different from 1. To check this, activity coefficients are computed from Eqs. (7) and (8), with a spread sheet, with the following result as a plot. It is seen that in the vicinity of mole fractions equal to 0.5, the coefficients are not large, but are about 1.3. At the bubble point, Ψ = V/F = 0, and Eq. (1), combined with (4), becomes: f {T} = z1 1 − γ 1 P s {T} γ P s {T} 1 + z2 1 − 2 2 =0 P P (9) Exercise 4.34 (continued) Analysis: (continued) Also, at the bubble point, x1 = z1 = 0.4 and x2 = z2 = 0.6. Then, the only unknown in Eq. (9) is T. Solving nonlinear Eq. (9), by trial and error with a spreadsheet, starting from a guess of T = 100oC, quickly leads to a bubble-point temperature of 106.9oC. The composition of the vapor bubble is obtained from Eq. (2), which at the bubble point reduces to yi = xiKi = ziKi.. The K-values at the bubble point are computed to be K1 = 1.363 and K2 = 0.758, giving y1 = 0.545 and y2 = 0.455. At the dew point, Ψ = V/F =1, and Eq. (1), combined with (4), becomes: f{x1, x2, T} = z1 P P − 1 + z2 −1 = 0 s γ 1 {x1 , x2 } P {T} γ 2 {x1 , x2 } P2s {T} 1 where because y1 = z1 = 0.4 and y2 = z2 = 0.6, x1 , and x2 =(1 - x1) are left as unknowns. The liquid phase mole fractions are from Eq. (3), xi...
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