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

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Unformatted text preview: lving (3) for a range of crystal size, Dp, microns 0.01 0.10 1.00 10.00 Dp , m 0.00000001 0.0000001 0.000001 0.00001 c/cs 1.417 1.0355 1.0035 1.00035 c, g/100 g water 300 219.5 212.7 212.1 The results show that the effect on solubility is greatest for the smallest crystals. Exercise 17.17 Subject: Required supersaturation ratio. Given: Crystals of sucrose of 0.5 microns diameter. Properties of sucrose: MW = 342; crystal density = 1,590 kg/m3; and interfacial tension, σs,L = 0.01 J/m2 Assumptions: Spherical crystals. Temperature of 25oC. Find: Supersaturation ratio, S = c/cs required for the crystals to grow. Analysis: Use (17-16) with SI units: ln S = ln 4v σ c = s s,L cs υRTD p Substituting into (1) gives, 342 (0.01) 1,590 S = exp 0.5 1( 8315 )( 298 ) 106 4 = 1.007 (1) Exercise 17.18 Subject: Explanation for the experimental maximum in the crystal solubility curve with crystal size. Given: Surface energy of crystals may depend on surface electrical charge as well as on interfacial tension. Relationship for electrical charge effect. Assumptions: Spherical crystals. Find: A modification of (17-16) that predicts the maximum. Analysis: From (17-16), ln S = ln 4v σ c = s s,L cs υRTD p (1) To predict a maximum, modify (1) by incorporating the given electrical charge effect, as follows: 4vs σ s , L 2q 2 vs c (2) ln S = ln = − 4 cs υRTD p πKRTD p Equation (2) is of the form: ln v c A B =s −4 cs RT D p D p (3) Differentiate (3) with respect to Dp to find min/max: v d c A 4B ln = s − 2 + 5 =0 dD p cs RT Dp Dp Solving (4) for Dp, 4B Dp = A 1/ 3 2q 2 υ = πK σs,L 1/ 3 , which can be shown to be a maximum (4) Exercise 17.19 Subject: Effect of supersaturation ratio on rate of primary homogeneous nucleation. Given: Aqueous solutions of AgNO3, NaNO3, and KNO3 at 25oC, together with values for crystal density and interfacial tension. Assumptions: Spherical crystals. Find: Rates of primary homogeneous nucleation over the range of supersaturation ratio, S = c/cs , of 1.005 to 1.02. Analysis: Use (17-18) with SI units in the exponential term: B o , number of nuclei formed/cm 3 -s = 1030 exp The properties are: Component AgNO3 NaNO3 KNO3 υ 2 2 2 MW 169.6 85.0 101.1 ρ, ρ, kg/m3 4,350 2,260 2,110 −16πvs σ3 N a s,L 3υ2 ( RT ) ( ln S ) 3 (1) 2 σs,L, J/m2 0.0025 0.0015 0.0030 vs = MW/ρ 0.0390 0.0376 0.0479 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 AgNO3 NaNO3 KNO3 S = 1.02 4.37 x 1025 1.33 x 1029 4.20 x 1018 S = 1.01 5.43 x 1012 3.39 x 1026 8.72 x 10-16 S = 1.005 1.91 x 10-39 1.54 x 1016 4.50 x 10-150 Exercise 17.20 Subject: Primary homogeneous nucleation of BaSO4. Given: Crystal density of 4.50 g/cm3 and interfacial tension of 0.12 J/m2. Assumptions: Spherical crystals. Find: Effect of relative supersaturation on primary nucleation from an aqueous solution at 25oC. Analysis: Since from Table 17.6, BaSO4 is sparingly soluble in w...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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