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
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 =
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):
P = PW + xB PBs + xT PTs = 1.79 + 0.5(5.22) + 0.5(1.78) = 5.29 psia for the bubble
For the primary dew point, assume first that the water condenses first. Then, using Dalton's law:
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:
xi = 1 =
Therefore, solving Eq. (1) for P,
= 3.32 psia for the primary dew point.
HCs K i Exercise 4.71 (c...
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