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Unformatted text preview: otal volumetric flow rate of exiting magma is:
1, 249
3,151
+
= 3.49 m3 /h
1.45(100) 1.20(100)
From Table 17.4, for 20 mesh, Lpd = 0.850 mm = 0.000850 m
From (1749), Lpd = 3Gτ
Therefore, τ = Lpd
3G = 0.00085
= 1.42 h
3 ( 0.0002 ) Therefore, volume of magma in the crystallizer = 3.49(1.42) = 4.96 m3 Exercise 17.26
Subject: Design of an MSMRP crystallizer
Given: Production of 2,000 lb/h of crystals of MgSO 4 7H 2 O with an Lpd of U.S. 35 mesh.
Magma is 15 vol% crystals. Crystallizer temperature = 50oC. Residence time, τ = 2 h. Crystal
density = 1.68 g/cm3. Mother liquor density = 1.32 g/cm3.
Assumptions: Equilibrium.
Find: (a) Volumetric flow rates of crystals, mother liquor, and magma.
(b) Crystallizer volume if vapor space = magma space.
(c) Crystallizer dimensions if height = twice the diameter.
(d) Required crystal growth rate, G, in ft/h.
(e) Necessary nucleation rate in nuclei/hft3 of mother liquor.
(f) Number of crystals produced /h.
(g) Predicted cumulative and differential screen analyses of product crystals.
(h) Plots of the predicted screen analyses
Analysis:
a) Volumetric flow rate of crystals = 2,000/[1.68)(62.4)] = 19.1 ft3/h.
Volumetric flow rate of mother liquor = 19.1(85/15) = 108.2 ft3/h
Volumetric flow rate of magma = 19.1 + 108.2 = 127.3 ft3/h
(b) Volume of magma in crystallizer = 127.3(2) = 254.6 ft3
Total crystallizer volume = 2(254.6) = 509 ft3 or 509/7.48 = 68.1 gal
(c) H = 2D , V = πD2H/ 4 = πD3 / 2 = 509 ft3
Solving, D = 6.87 ft and H = 2(6.87) = 13.74 ft.
(d) From Table 17.4, Lpd = 35 U.S. mesh = 0.0197 in = 0.00164 ft.
From (1749), Lpd = 3Gτ
Therefore, G = Lpd/3τ = 0.00164/[3(2)] = 0.000273 ft/h
(e) From (17.60), assuming a shape factor, fv = 0.5
Bo = 9C
9(2, 000)
=
= 1.80 × 108 nuclei/hft 3
3
3
2 f v ρ pVML Lpd 2(0.5)[1.68(62.4)][108.2(2)] ( 0.00164 ) (f) Crystals produced per hour = 1.80 × 108 (108.2)(2) = 3.9 x 1010 crystals/h Exercise 17.26 (continued)
(g) For predicting the screen analysis, use Table 17.9 for the third moment, where,
z2 z3
xm = 1 − 1 + z + +
exp(− z ) for Cumulative fraction smaller
26
dxm z 3
= exp(− z ) which must be normalized for 100% total
dz
6 (2) where z is a dimensionless particle size.
z = L/Gτ = L/[0.000273(2)] = 1,832 L in ft or 6.01 L in mm
Using a spreadsheet, the following results are obtained using (1) and (2): U.S.
Mesh Aperture, z,
Dimensionless
Length, Number Dp, mm mm 3.5
4
5
6
7
8
10
12
14
16
18
20
25
30
35
40
45
50
60
70
80
100
120
140
170
200 5.600
4.750
4.000
3.350
2.800
2.360
2.000
1.700
1.400
1.180
1.000
0.850
0.710
0.600
0.500
0.425
0.355
0.300
0.250
0.212
0.180
0.150
0.125
0.106
0.090
0.075 33.656
28.548
24.040
20.134
16.828
14.184
12.020
10.217
8.414
7.092
6.010
5.109
4.267
3.606
3.005
2.554
2.134
1.803
1.503
1.274
1.082
0.902
0.751
0.637
0.541
0.451
Total Cumulative
Analysis
xm,
Mass
Percent (1) Differential Analysis smaller Average
Particle
Size,
mm Mass
Percent
dxm/dz Differential 100.00
100.00
100.00
100.00
100.00
99.96
99.77
99.12
96.80
92.29
84.97
74.99
61.69
48.61
35.39
25.41
16.7...
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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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