Question 7.P.4: In a leaching operation, the rate at which solute goes into ...
In a leaching operation, the rate at which solute goes into solution is given by:
d M / d t=k\left(c_s-c\right) kg/s
where M kg is the amount of solute dissolving in t s, k (m³/s) is a constant and c_s 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.
The mass of solute M, dissolving in t s is:
d M / d t=k\left(c_s-c\right) kg / s (i)
For a batch of solution, V m³ in volume: dM = Vdc
and substituting for dM in (i): V d c / d t=k\left(c_s-c\right) / V
Integrating: \ln \left(\left(c_s-c_0\right) /\left(c_s-c\right)\right)=k t / V (ii)
where c_0 is the concentration of the solute when t = 0.
For pure water, c_0=0 kg / m ^3 when t = 0 and hence equation (ii) becomes:
c=c_s\left(1-e^{-k t / V}\right) kg / m ^3 (iii)
For the pilot test, the batch volume, V = 1 m³, and c_s=2.5 kg / m ^3 at saturation. When t = 10 s, 75% saturation is achieved or:
c=(2.5 \times 75 / 100)=1.875 kg / m ^3
Therefore, in equation (iii):
1.875=2.5\left(1-e^{-10 k / 1}\right) and k = 0.138
For the full-scale unit, the batch volume, V = 100 m³. Mass of solute present =(300 \times 28 / 100)=84 kg and c=(84 / 100)=0.84 kg / m ^3.
Therefore, in equation (iii): 0.84=2.5\left(1-e^{-0.138 t / 100}\right) and t=\underline{\underline{297 s }}