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

# Given a 50 mol benzene b 50 mol water w b 50 mol

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Unformatted text preview: 315 wt. Fraction. C = rate of crystal formation = 0.8(1,000) = 800 kg/h Therefore, sulfate dissolved in the mother liquor = 1,000 - 800 = 200 kg/h. From the solubility of the sulfate, water in the mother liquor = Therefore, we must evaporate, 4,000 - 435 = 3,565 kg/h (1 − 0.315) (200) = 435 kg/h 0.315 (b) MW of Na2SO4 = 142.05 Therefore, have 200/142.05 = 1.41 kmol/h of Na2SO4 in the mother liquor. Also, we have 435/18.02 = 24.14 kmol/h of water in the mother liquor. The mole fraction of water in the mother liquor = 24.14 = 0.945 24.14 + 1.41 At 60oC, the vapor pressure of water = 2.89 paia. From Raoult's law, given by Eq. (2-42), the partial pressure of water = s xW PW = 0.945(2.89) = 2.73 psia Assuming that Na2SO4 has no vapor pressure in solution, crystallizer pressure = 2.73 psia. Exercise 4.71 Subject: Bubble, secondary dew, and primary dew points of a three-component mixture assuming immiscibility of water and hydrocarbons. Given: (a) 50 mol% benzene (B), 50 mol% water (W). (b) 50 mol% toluene (T), 50 mol% water (W). (c) 40 mol% benzene (B), 40 mol% toluene (T), and 20 mol% water (W) Assumptions: Ideal hydrocarbon solutions and immiscibility with water. Find: Bubble, secondary dew, and primary dew point pressures at 50oC. Analysis: From the CHEMCAD program, vapor pressures at 50oC are: benzene 5.22 psia, toluene 1.78 psia, water 1.79 psia (a) At the bubble point, both pure liquid phases are present. Therefore, P = sum of the vapor pressures = 5.22 + 1.79 = 7.01 psia. This is also the secondary dew-point pressure. For a binary mixture, the primary dew point occurs when the pressure is increased to the point where the partial pressure of one component equals it vapor pressure. Since we have a 50-50 molar mixture, this occurs for the component with the lowest vapor pressure, water. Thus, water will condense first, with P = pW + pB = 1.79 + 1.79 = 3.58 psia. (b) Calculations for toluene and water are made in an identical manner to part (a). The results are: Bubble-point pressure = 1.79 + 1.78 = 3.57 psia. This is also the secondary dew point. The primary dew point is controlled by the toluene. Thus, the primary dew point pressure = 1.78 +1.78 = 3.56 psia. (c) Assuming Raoult's law from Eq. (2-42) to hold for the hydrocarbon phase, which is equimolar in B and T, and the water phase exerting its full vapor pressure, we have for the bubble point pressure from Eq. (4-34): s P = PW + xB PBs + xT PTs = 1.79 + 0.5(5.22) + 0.5(1.78) = 5.29 psia for the bubble point. For the primary dew point, assume first that the water condenses first. Then, using Dalton's law: p P s 179 . P= W = W = = 8.95 psia yW yW 0.2 But this higher than the bubble point pressure. So, instead, assume that the hydrocarbon phase condenses first. Apply the dew point equation, Eq. (4-13) to just the hydrocarbon phase: zi zi P xi = 1 = = (1) s HCs HCs Ki HCs P i Therefore, solving Eq. (1) for P, 1 1 P= = = 3.32 psia for the primary dew point. zi 0.4 0.4 + 5.22 1.78 HCs K i Exercise 4.71 (c...
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