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

# 0 05 10 15 20 aperture size mm 25 30 35 40 exercise

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Unformatted text preview: s = radius of particle in cm rc/rs = radius ratio corresponding to 50 to 100% of the particle volume = (1 – fraction recovered)1/3 De = effective diffusivity of the hydrogen ion of the H2SO4 = 0.6 x 10-6 cm2/s b = stoichiometric coefficient of the solid reactant, CuO in Eq. (1) of Example 16.8 = 0.5 MB = molecular weight of CuO = 79.6 cAb = concentration of H+ in bulk fluid = 0.001 mol/cm3 Substituting these values into (1), gives: r t = 94, 200 1 − 3 c rs 2 r +2 c rs 3 (2) Calculations for % recoveries of 50, 70, 90, and 100% give the following results: % Recovered 50 70 90 100 [1 – fraction recovered]1/3 0.7937 0.6694 0.4642 0.0000 Time, s 10,380 24,080 52,170 94,200 Exercise 16.15 Subject: Recovery of copper by the leaching of Cu2O from copper ore with sulfuric acid, by the method of Example 16.8, using the shrinking-core model. Given: Particles of 10 mm (1.0 cm) in diameter and an ore with 3 wt% Cu2O instead of 2 wt% CuO as in Example 16.8. Other data from Example 16.8. Assumptions: Applicability of the shrinking-core model. Cu2O is uniformly distributed in the particles. Pseudo-steady-state assumption is valid. Find: Time for 98% leaching of the copper. Analysis: Equation (16-34) is applied: ρB rs2 r 1− 3 c t= 6 DebM BcAb rs where, the leaching reaction is now: 2 r +2 c rs 3 (1) 1 1 + + Cu 2 O (s) + H (aq) = Cu (aq) + H 2 O (l) 2 2 Using cgs units and data from Example 16.8, the symbols are: t = time in s ρB = mass of Cu2O per volume of ore = 0.03(2.7) = 0.081 g/cm3 rs = radius of particle in 0.5 cm rc/rs = radius ratio corresponding to 98% of the particle volume = (1 – fraction recovered)1/3 = (1 – 0.98)1/3 = 0.271 De = effective diffusivity of the hydrogen ion of the H2SO4 = 0.6 x 10-6 cm2/s b = stoichiometric coefficient of the solid reactant, Cu2O in Eq. (1) of Example 16.8 = 0.5 MB = molecular weight of Cu2O = 143.14 cAb = concentration of H+ in bulk fluid = 0.001 mol/cm3 The only changes from Example 16.8 are ρB and MB. Therefore, instead of using Eq. (1), we can just ratio from the result in Example 16.8. Therefore, t = 21.4 hours 0.081 0.054 79.6 = 17.9 hours 143.14 Exercise 16.16 Subject: Leaching of a spherical particle where the rate is controlled by a first-order reaction at the interface of the shrinking core. Given: Spherical particle with a shrinking core as leaching proceeds. Find: Derive the expression: ρB rs2 r t= 1− c bM B kcAb rs (1) Analysis: The reaction occurs at the outer boundary of the shrinking core at r = rc , where the initial radius of the particle is rs. The rate of reaction is Rate = − kcAb ( 4πrc2 ) (2) where the term in parentheses is the surface area at the outer boundary of the shrinking core. The rate of reaction is also given by a mass balance on the shrinking core: Rate = ρB ( 4πrc2 ) dr bM B dt (3) Equating (2) and (3) and separating variables, followed by integration, gives: −ρB bM B kcAb Integrating (4) and rearranging the result gives (1). rc rs dr = t 0 dt (4) Exercise 17....
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