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

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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/...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online