Unformatted text preview: d the crystallizer.
Take a basis of 100 g of water in the entering feed.
From Table 17.5, the solubility of Na3PO4 at 40oC = 31 g/100 water.
At 20oC, the solubility is 11 g/100 g water.
Let x = g water exiting in the crystals
Then water left in solution = 100 – x and the amount of Na3PO4 in the exiting
solution = 11(100 – x)/100
MW of Na3PO4 = 163.97
MW of Na 3 PO 4 12H 2 O = 380.16
Therefore, the crystals are 163.97/380.16 = 0.431 mass fraction Na3PO4
A mass balance on Na3PO4 gives,
(100 − x ) + x 0.431
31 = 11
100
0.569 (1) Solving (1), x = 30.8 g water in the crystals with 30.8(0.431/0.569) = 23.3 g Na3PO4
Therefore, 100 g water in the feed gives 30.8 + 23.3 = 54.1 g crystals.
For a production rate of 4,000 lb/h of crystals, we need a feed containing 7,390 lb/h of
water and 2,290 lb/h of dissolved Na3PO4 or a total feed of 4.84 tons/h
(b) Now, compute the rate of heat to be removed. Assume a thermodynamic path of cooling
from 40oC to 20oC, followed by crystallization at 20oC.
From Table 17.5, heat of solution = +15,000 cal/mol of compound at room temperature
Therefore, the heat of crystallization is exothermic at 15,000(1.8) = 27,000 Btu/lbmol
Q to be removed = 9,680(0.80)[(40 – 20)(1.8)] + 27,000(4,000/380.16) =
279,000 + 284,000 = 563,000 Btu/h Exercise 17.23 (continued)
Now apply the heat transfer rate equation, Q = UA ∆TLM in American Engineering units
For countercurrent flow with linear temperature profiles,
∆TLM = ( 40 − 25 ) − ( 20 − 10 ) = 12.3o C = 22.1o F
( 40 − 25 )
ln
( 20 − 10 ) U = 20 Btu/hft2oF
Therefore, heat transfer area = A = 563,000/[20(22.1)] = 1,270 ft2
(c) The number of crystallizer units needed = 1,270/30 = 43 units. Exercise 17.24
Subject: MSMPR Crystallizer model.
Given: An aqueous feed of 10,000 kg/h of saturated BaCl2 at 100oC enters an MSMPR
crystallizer. Negligible evaporation. Magma leaves at 20oC with crystals of the dihydrate. The
volume of the vaporspacefree crystallizer = 2.0 m3. Experimental crystal growth = G =
4.0 x 107 m/s.
Assumptions: Equilibrium
Find: (a) The kg/h of crystals in the magma product.
(b) The predominant crystal size in mm.
(c) The mass fraction of crystals in the size range for U.S. Standard 20 mesh to 25 mesh.
Analysis:
(a) First carry out a mass balance around the crystallizer.
From Table 17.5, solubility of BaCl2 at 100oC = 58.3 g/100 g water, and
at 20oC, solubility = 35.7 g/100 g water
Therefore, the feed contains 10,000[58.3/(58.3 + 100)] = 3,680 kg/h of BaCl2, and
10,000 – 3,680 = 6,320 kg/h of water.
MW of BaCl2 = 208.27
MW of BaCl2 2H 2 O = 244.31
Therefore, the crystals are 208.27/244.31 = 0.852 mass fraction BaCl2
Mass balance for BaCl2 around the crystallizer. Let x = kg/h of water in the crystals.
35.7
0.852
(1)
3, 680 = ( 6,320 − x )
+x
100
1 − 0.852
Solving (1), x = 263 kg/h of water in the crystals.
BaCl2 in the crystals = 263(0.852/0.148) = 1,514 kg/h
Therefore, the product crystals in the magma = 263 + 1,514 = 1,777 kg/h
(b) Next calculate the volumetric flow rate of th...
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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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