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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 solutefree (dissolved Na2CO3free) 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 solutefree 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 twophase 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.
 Spring '11
 Levicky
 The Land

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