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

# It is seen that a temperature of 3oc is needed to

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Unformatted text preview: o be 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: (10)(18) = 634 lb/h or water of crystallization 142 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.

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