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

# Initial impurity concentration is uniform at wo

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Unformatted text preview: n density data in Fig. 17.37, which do not fit (17-38). 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/m-m3 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./m-m3 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./m-m3 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 (17-38), 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 (17-38). Assume that n in Fig. 17.38 has units of no./micron-L. Find: (a) Growth rate, G (b) Nucleation rate, Bo (c) Predominate crystal size, Lpd Analysis: (a) From (17-52), Bo L n= exp − G Gτ or in a straight-line 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/L-min. (c) From (17-49), Lpd = 3Gτ = 3(0.047)(5) = 0.705 micron Exercise 17.30 Subject: Melt crystallization of naphthalene from a mixture with benzene in a falling-film 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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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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