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

Let v kgh of overflow l kgh of liquid in the underflow

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Unformatted text preview: zed as follows: Component Insoluble oxide Na2CO3 Water Total Lo , Solids, kg/h 2.400 1,350 0 3,750 So , Solvent, kg/h 0 0 4,000 4,000 L1 , Overflow, kg/h 0 810 2,400 3,210 S1 , Underflow, kg/h 2,400 540 1,600 4,540 The % extraction of Na2CO3 = 810/1,350 x 100% = 60% Exercise 4.54 Subject: Leaching of Na2CO3 from a solid by water. Given: Lo = 3,750 kg/h of a solid containing 1,350 kg/h Na2CO3, contacted with So = 4,000 kg/h of water. Assumptions: Ideal leaching stage so that compositions of overflow and underflow liquid are equal. Underflow contains 40 wt% water on a solute-free (dissolved Na2CO3-free) basis. Only 80% of the Na2CO3 is dissolved. Find: Compositions and flow rates, L1 and S1, respectively, of overflow and underflow. Analysis: Because only 80% of the Na2CO3 is dissolved, the underflow will contain (3,750 1,350) = 2,400 kg/h of insoluble oxide plus (0.20)(1,350) = 270 kg/h of solid Na2CO3 or a total of 2,670 kg/h of total solids. Let: V = kg/h of overflow L = kg/h of liquid in the underflow. Total underflow = L + 2,670 kg/h y = mass fraction of dissolved Na2CO3 in overflow or underflow liquid (equilibrium) Total mass balance: Na2CO3 mass balance: Lo + So = 4,000 + 3,750 = 7,750 = V + L + 2,670 1,350 = 270 + y(V + L) (1) (2) Solving Eqs. (1) and (2), V + L = 5,080 kg/h and y = 0.2126 Flow rate of underflow on a solute-free basis = (1 - y)L + 2,670 = 0.7874 L + 2,670 kg/h Therefore the flow rate of water in the underflow = 0.40(0.7874 L + 2,670) Water mass balance: 4,000 = (1 - 0.2126)V + 0.40(0.7874 L + 2,670) Solving linear Eqs. (1) and (3), L = 2,260 kg/h and V = 2,820 kg/h The material balance may be summarized as follows: Component Insoluble oxide Insoluble Na2CO3 Soluble Na2CO3 Water Total Lo , Solids, kg/h 2.400 1,350 0 3,750 So , Solvent, kg/h 0 0 0 4,000 4,000 The % extraction of Na2CO3 = 600/1,350 x 100% = 44.4% Overflow, kg/h 0 0 600 2,220 2,820 Underflow, kg/h 2,400 270 480 1,780 4,930 (3) Exercise 4.55 Subject: Crystallization from a mixture of benzene and naphthalene. Given: 6,000 lb/h of 40 wt% benzene and 60 wt% naphthalene in liquid state at 50oC, which is cooled to 15oC. Assumptions: Equilibrium at 15oC. Equilibrium phase diagram of Fig. 4.23. Find: Flow rates and compositions of crystals and mother liquor. Analysis: The feed is (0.60)(6,000) = 3,600 lb/h of naphthalene and 2,400 lb/h of benzene. In the figure below: A is the feed solution at 50oC. It lies in the homogeneous solution region. C is the feed solution at 15oC. It now lies in the two-phase region and separates into naphthalene crystals at D and mother liquor at B. From the diagram, the mother liquor at 15oC contains 31.5wt% naphthalene and, therefore, 68.5 wt% benzene. Because all of the benzene is in the mother liquor, the mother liquor flow rate is 2,400/(0.685) = 3,500 lb/h. Therefore, the naphthalene crystals flow rate is 6,000 - 3,500 = 2,500 lb/h. The % crystallization of naphthalene is 2,500/3,600 x 100% = 69.4%. Exercise 4.56 Subject: Crystallization from a mixture of benzene and naphthalene. Given: 6,000 lb/h of 40 wt% benzene and 60 wt% naphthalene in liquid state at 50oC t...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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