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cooled to obtain crystals of naphthalene.
Assumptions: Equilibrium upon cooling, using phase diagram of Fig. 4.23.
Find: Temperature necessary to crystallize 80% of the naphthalene. Flow rates and
compositions of crystals and mother liquor.
Analyze: The feed is (0.60)(6,000) = 3,600 lb/h of naphthalene and 2,400 lb/h of benzene. If
80% of the naphthalene is crystallized, the crystals flow rate will be (0.8)(3,600) = 2,880 lb/h.
This leaves (3,600 - 2,880) = 720 lb/h of naphthalene in the mother liquor, with 2,400 lb/h of
benzene. The total flow rate of mother liquor is 720 + 2,400 = 3,120 lb/h. Therefore, the
naphthalene concentration in the mother liquor is 720/3,120 x 100% = 23 wt%. In the figure
below, A is the feed solution at 50oC. It lies in the homogeneous solution region
C is the feed solution in the two-phase region, which separates into naphthalene crystals
at D and mother liquor at B, with the 23 wt% solubility.
It is seen that a temperature of 3oC is needed to achieve a naphthalene solubility of 23 wt%. Exercise 4.57
Subject: Crystallization from a mixture of benzene and naphthalene.
Given: 10,000 kg/h of 90 wt% benzene and 10 wt% naphthalene in liquid state at 30oC to be
cooled to 0oC to obtain crystals.
Assumptions: Equilibrium upon cooling to 0oC, using phase diagram of Fig. 4.23.
Find: Flow rates and compositions of crystals and mother liquor.
Analyze: The feed is (0.10)(10,000) = 1,000 kg/h of naphthalene and 9,000 kg/h of benzene.
In the figure below:
A is the feed solution at 30oC. It lies in the homogeneous solution region.
C is the feed solution at 0oC. It now lies in the two-phase region and separates
into benzene crystals at B and mother liquor at D.
From the diagram, the mother liquor at 0oC contains 15 wt% naphthalene and, therefore, 85 wt%
benzene. Because all of the naphthalene is in the mother liquor, the mother liquor flow rate is
1,000/(0.15) = 6,667 kg/h. Therefore, the benzene crystals flow rate is 10,000 - 6,667 = 3,333
kg/h. The % crystallization of benzene is 3,333/9,000 x 100% = 37.0%. Exercise 4.58
Subject: Crystallization of Na2SO4 from an aqueous solution by cooling.
Given: 1,000 lb/h of Na2SO4 dissolved in 4,000 lb/h of H2O at 50oC.
Assumptions: Equilibrium according to phase diagram of Fig. 4.24.
Find: Temperature at which crystallization begins. Temperature to obtain 50% crystallization
of the sulfate. Hydrate form of the crystals.
Analysis: In the diagram below, the feed of 20 wt% sulfate, at point A, is in the homogeneous
solution region at 50oC. Crystallization begins at point B, corresponding to a temperature of
24oC. As the temperature is lowered further, crystals of the decahydrate, Na 2SO 4 ⋅ 10H 2 O form.
For 50% crystallization of Na2SO4 , must crystallize 0.5(1,000) = 500 lb/h. For the decahydrate,
with molecular weights of 18 for water and 142 for the sulfate, the crystals contain:
= 634 lb/h or water of crystallization
Therefore, the total flow rate of crystals = 500 + 634 = 1,134 lb/...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
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