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 106 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 shrinkingcore 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 shrinkingcore model. Cu2O is uniformly distributed in the
particles. Pseudosteadystate assumption is valid.
Find: Time for 98% leaching of the copper.
Analysis: Equation (1634) 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 106 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 firstorder 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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 Spring '11
 Levicky
 The Land

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