Derivation of Fully Mixed Recovery Equation (Chapter 12, Equation 12.28)

Figure A.V.1 represents a continuously operating flotation cell with retention time τ. For convenience, in this derivation it is easier to refer to the concentration of particles rather than mass, so we have a concentration C₀ entering the cell with steady-state concentration C in the cell, which is also the concentration in the exit (nonfloats) as the cell is perfectly mixed.

The rate of recovery (by flotation) is then:

$\frac{\mathrm{d}R}{\mathrm{d}t}=\frac{C_0-C}{\tau }$

From the assumption of first order we can equate:

$\frac{C_0-C}{\tau }=kC$

From which we deduce:

$\tau =\frac{C_0-C}{kC}\to \frac{C_0}{C}=k\tau +1\to \frac{C}{C_0}=\frac{1}{1+k\tau }$

Since C/C₀ is the fraction left in the cell, then the recovery is:

$R=1-\frac{C}{C_0}=1-\frac{1}{1+k\tau }=\frac{k\tau }{1+k\tau }$

That is, the same as Eq. (12.28).