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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. (242), 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 threecomponent 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 dewpoint 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 5050
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: Bubblepoint 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. (242) 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. (434):
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. (413) 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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 Spring '11
 Levicky
 The Land

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