In a leaching operation, the rate at which solute goes into solution is given by: dM/dt = k(cs - c) kg/s where M kg is the amount of solute dissolving in t s, k (m³/s) is a constant and cs and c are the saturation and bulk concentrations of the solute respectively in kg/m³. In a pilot test on

Question

In a leaching operation, the rate at which solute goes into solution is given by:

[latex]d M / d t=k\left(c_s-c\right)[/latex] kg/s

where M kg is the amount of solute dissolving in t s, k (m³/s) is a constant and [latex]c_s[/latex] and c are the saturation and bulk concentrations of the solute respectively in kg/m³. In a pilot test on a vessel 1 m³ in volume, 75% saturation was attained in 10 s. If 300 kg of a solid containing 28% by mass of a water soluble solid is agitated with 100 m³ of water, how long will it take for all the solute to dissolve assuming conditions are the same as in the pilot unit? Water is saturated with the solute at a concentration of 2.5 kg/m³.

Accepted Answer

The mass of solute M, dissolving in t s is:

[latex]d M / d t=k\left(c_s-c ight) kg / s[/latex] (i)

For a batch of solution, V m³ in volume: dM = Vdc

and substituting for dM in (i): [latex]V d c / d t=k\left(c_s-c ight) / V[/latex]

Integrating: [latex]\ln \left(\left(c_s-c_0 ight) /\left(c_s-c ight) ight)=k t / V[/latex] (ii)

where [latex]c_0[/latex] is the concentration of the solute when t = 0.
For pure water, [latex]c_0=0 kg / m ^3[/latex] when t = 0 and hence equation (ii) becomes:

[latex]c=c_s\left(1-e^{-k t / V} ight) kg / m ^3[/latex] (iii)

For the pilot test, the batch volume, V = 1 m³, and [latex]c_s=2.5 kg / m ^3[/latex] at saturation. When t = 10 s, 75% saturation is achieved or:

[latex]c=(2.5 imes 75 / 100)=1.875 kg / m ^3[/latex]

Therefore, in equation (iii):

[latex]1.875=2.5\left(1-e^{-10 k / 1} ight)[/latex] and k = 0.138

For the full-scale unit, the batch volume, V = 100 m³. Mass of solute present [latex]=(300 imes 28 / 100)=84[/latex] kg and [latex]c=(84 / 100)=0.84 kg / m ^3[/latex].

Therefore, in equation (iii): [latex]0.84=2.5\left(1-e^{-0.138 t / 100} ight)[/latex] and [latex]t=\underline{\underline{297 s }}[/latex]

Question 7.P.4: In a leaching operation, the rate at which solute goes into ...

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