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

986 was ln n 4008 exp000124 l 006075 l05 the following

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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 (17-49), 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/h-ft3 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 (17-49), 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/h-ft 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.

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