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Unformatted text preview: n density data in Fig. 17.37, which do not fit (1738).
Assumptions: Equilibrium
Find: (a) An empirical equation that does fit the crystal population density data.
(b) Whether and how nucleation rate and growth rate can be determined from the data.
Analysis:
(a) From Fig. 17.37, values of n in nuclei/mm3 for crystal sizes L are read and listed in
the table below. Using the nonlinear regression feature of Polymath, a number of different
expressions were tried for fitting the data. The best fit, which was not entirely satisfactory,
giving an R2 of 0.986, was:
ln n = 40.08 exp(0.00124 L – 0.06075 L0.5)
The following spreadsheet compares the data points to the predictions of the correlation. Data:
Crystal
size,
L, microns Correlation:
n,
no./mm3 ln n ln n n 3
4
6
9
12
18.5
26
37
53
74 1.30E+16
1.90E+15
8.80E+14
4.50E+14
8.80E+13
4.80E+13
3.00E+13
6.00E+12
6.20E+11
2.00E+11 37.10
35.18
34.41
33.74
32.11
31.50
31.03
29.42
27.15
26.02 36.21
35.67
34.80
33.78
32.96
31.58
30.37
29.00
27.50
26.05 5.32E+15
3.10E+15
1.29E+15
4.67E+14
2.06E+14
5.18E+13
1.54E+13
3.92E+12
8.80E+11
2.06E+11 A comparison of the data to the prediction by the correlation is given on the next page. Exercise 17.28 (continued)
1.E+17 Crystal Population Density, no./mm3 1.E+16
____________ Data
        Correlation 1.E+15 1.E+14 1.E+13 1.E+12 1.E+11
0 10 20 30 40 50 60 70 80 Crystal size, microns (b) Because the correlation is not of the form of (1738), the theory in Section 17 .6
cannot be used to predict nucleation rate and growth rate. Exercise 17.29
Subject: Precipitation of Mg(OH)2 from mixing of aqueous solutions of MgCl2 and Ca(OH)2 in
an MSMPR crystallizer.
Given: Crystallizer volume = 1 L. Operation at 450 rpm and 25oC. Measured crystal size given
in Fig. 17.38 for 5 minutes residence time.
Assumptions: Crystal size is proportional to exp(L/Gτ) as in (1738). Assume that n in Fig.
17.38 has units of no./micronL.
Find: (a) Growth rate, G
(b) Nucleation rate, Bo
(c) Predominate crystal size, Lpd Analysis:
(a) From (1752), Bo
L
n=
exp −
G
Gτ or in a straightline form, Bo
L
log n = log
−
G
2.303Gτ (1) Take end points of the line in Fig. 17.38 as: n = 90 at L = 0.55 and n = 3 at L = 1.35
Substituting these two data points into (1) with τ = 5 minutes, and solving, gives
G = 0.047 microns/min
(b) With this value of G, substitution into (1) for either data point gives,
Bo = 44 nuclei/Lmin.
(c) From (1749),
Lpd = 3Gτ = 3(0.047)(5) = 0.705 micron Exercise 17.30
Subject: Melt crystallization of naphthalene from a mixture with benzene in a fallingfilm
crystallizer.
Given: Feed of 60 wt% naphthalene and 40 wt% benzene at saturation conditions. Coolant
enters at top at 10oC. Physical property data from Example 17.13.
Assumptions: Equilibrium. Crystallization on a planar wall.
Find: Crystal layer thickness for up to 2 cm as a function of time.
Analysis: Assume the major resistance to heat transfer is the crystal layer building up on the
wall. Let ri  rs = thickness of the crystal layer
F...
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 Spring '11
 Levicky
 The Land

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