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

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

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

This is the end of the preview. Sign up to access the rest of the document.

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...
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