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

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Unformatted text preview: 7 10.92 6.60 4.06 2.44 1.35 0.73 0.41 0.23 0.12 5.175 4.375 3.675 3.075 2.580 2.180 1.850 1.550 1.290 1.090 0.925 0.780 0.655 0.550 0.463 0.390 0.328 0.275 0.231 0.196 0.165 0.138 0.116 0.098 0.083 0.00 0.00 0.00 0.00 0.00 0.00 0.01 0.02 0.05 0.09 0.13 0.18 0.21 0.22 0.22 0.19 0.16 0.13 0.10 0.07 0.05 0.03 0.02 0.02 0.01 0.000 0.000 0.000 0.002 0.017 0.091 0.339 1.150 2.583 4.638 7.017 9.486 11.088 11.704 11.281 10.013 8.410 6.573 5.037 3.737 2.590 1.742 1.190 0.802 0.508 1.9142 100.0000 Exercise 17.26 (continued) (h) The following plots are obtained from the spreadsheet: 14.00 12.00 Differential Plot of Predicted Screen Analysis Differential Percent 10.00 8.00 6.00 4.00 2.00 0.00 0.0 1.0 2.0 3.0 4.0 5.0 6.0 Average Crystal Size, mm 100 90 Cumulative Mass, % 80 70 Cumulative Undersize of Predicted Screen Analysis 60 50 40 30 20 10 0 0 5 10 15 20 Crystal Size, mm 25 30 35 40 Exercise 17.27 Subject: Precipitation using the MSMPR model Given: Continuous precipitation of BaSO4 from the mixing of aqueous solutions of Na2SO4 and BaCl2 at stoichiometric ratio to give a NaCl concentration of 0.15 mol/L. Perfect mixing in a 1.8 L crystallizer. Data for mixing at 200 rpm fits the equation, ln n = 26.3 – 0.407 L, where n = number density of crystals in nuclei/micron-L, and L = crystal size in microns. Average residence time, τ, is 38 s. Assumptions: Equilibrium and applicability of the MSMPR model. Find: (a) no (b) G (c) Bo (d) mean crystal length (e) nc (f) whether the results are consistent with the trends in Example 17.12. (g) From the results here and in Example 17.12, predict G if no agitation. Analysis: (a) From (17-38), ln n = ln no – L/Gτ Comparing this to the given experimental correlation above, ln no = 26.3 Therefore, no = exp(26.3) = 2.64 x 1011 (b) Also from the same correlation, L/Gτ = 0.407 L Therefore, Gτ = 1/0.407 and G = 1/[0.407(38)] = 0.0647 micron/s (c) From (17-51), Bo = Gno = 0.0647(2.64 x 1011) = 1.71 x 1010 nuclei/L-s (d) From (17-42), Lmean = Gτ = 0.0647(38) = 2.46 microns (e) From (17-54), nc = noτG = (2.64 x 1011)(38)(0.0647) = 6.49 x 1011 crystals/L or 6.49 x 108 crystals/m3 (f) Comparison of results to those of Example 17.12: Rpm no, crystals/micron-L G, microns/s Bo, nuclei/L-s Lmean, microns nc, crystals/m3 950 1.11 x 1012 0.164 1.89 x 109 6.23 6.92 x 107 400 1.22 x 1011 0.0841 1.03 x 1010 3.20 3.91 x 108 200 2.64 x 1011 0.0647 1.71 x 1010 2.46 6.49 x 108 Consistent ? Yes Yes Yes Yes Yes (g) By extrapolation to an asymptotic value, as shown in the diagram on the next page, At 0 rpm, Lmean = 2 microns and G = Lmean/τ = (2)/(38) = 0.053 micron/s Exercise 17.27 (continued) 18 16 14 12 100 x Growth Rate, microns/s 10 8 6 Mean Size, microns 4 2 0 0 100 200 300 400 500 RPM 600 700 800 900 1000 Exercise 17.28 Subject: Precipitation of CaCO3 as calcite by mixing aqueous solutions of Na2CO3 and CaCl2 in an MSMPR crystallizer. Given: Precipitation conditions of 8.65 pH, an rpm of 800, and a residence time of 100 min. Crystal populatio...
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