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

Overall heat transfer coefficient 15 btuh ft2 of

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online