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

431 31 11 100 0569 1 solving 1 x 308 g water in

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Unformatted text preview: ater, a high supersaturation ratio will be needed. Try values of supersaturation ratio, S, of 50, 40, and 30; noting that the relative supersaturation, s, from (17-17) is s = S – 1, so that in this case the two supersaturations are approximately the same. Use (17-18) with SI units in the exponential term: B , number of nuclei formed/cm -s = 10 exp o The properties are: Component BaSO4 3 υ 2 MW 233.4 30 ρ, ρ, kg/m3 4,500 −16πvs σ3 N a s,L 3υ2 ( RT ) ( ln S ) 3 (1) 2 σs,L, J/m2 0.120 vs = MW/ρ 0.0519 Na = Avagadro’s number = 6.022 x 1026 molecules/kmol R = 8315 J/kmol-K T = 298 K Use of a spreadsheet gives the following results from (1): Rate of primary nucleation, Bo , nuclei/s-cm3 Component BaSO4 S = 50 1.38 x 108 S = 40 2.59 x 105 S = 30 12.0 Exercise 17.21 Subject: Crystal growth and the controlling resistance. Given: Data in Example 17.9, except that solution velocity past crystal face = 1 cm/s. Assumptions: Crystal growth controlled by mass transfer to the face, surface reaction, or a combination of the two. No primary or secondary nucleation. Find: Crystal growth rate and its controlling mechanism. Analysis: A review of the results of Example 17.9 shows that with a solution velocity past the crystal face of 5 cm/s, crystal growth is controlled mainly by surface reaction because the measured growth rate was given as 0.005 mm/min, while a rate calculated on the basis of control by mass transfer is higher, namely: 0.029 mm/min for the largest crystal size of 400 microns. Since, the solution velocity in this exercise is lower by a factor of 5, maybe mass transfer will be more important. First, calculate new mass transfer coefficients from (15-62), using ratios from the results in Example 17.9. NSh = kc D p D = 2 + 0.6 D p vρ 1/ 2 µ µ ρD 1/ 3 The only change from Example 17.9 is the velocity, v, in the Reynolds number. For each crystal diameter, the Reynold’s numbers here are 1/5 of those in Example 17.9. The resulting mass transfer coefficients are as follows: Crystal size, microns 50 400 kc, cm/s in Ex. 17.9 0.019 0.006 kc, cm/s here 0.011 0.003 Take the smallest value, kc = 0.003 cm/s. As in Example 17.9, the growth rate is, dr = 0.008kc = 0.008 ( 0.003) = 0.000024 cm/s = 0.0144 mm/min dt This growth rate is still larger than the measured value of 0.005 mm/min. Thus, the growth rate will be more at the lower solution velocity. To determine more quantitatively the effect of mass transfer on the growth rate, use the results of Example 17.9 to compute an approximate value of ki in (17-23). From (17-23), rewrite the growth rate in terms of an overall coefficient, dr = 0.008 K c = 0.005 mm/min dt Exercise 17.21 (continued) Therefore, Kc = 0.005/0.008 = 0.625 mm/min = 0.00104 cm/s From Example 17.9, the average kc = (0.019 + 0.006)/2 = 0.0125 cm/s From (17-23), 1 1 K c = 0.00104 = = 11 1 1 + + kc ki 0.0125 ki Solving (1), ki = 0.00113 cm/s. Now assume that this value of ki holds for this exercise. For this exercise, the average kc = 0.007 cm/s. From (17-13), Kc = 1 11 + k c ki =...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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