# Meaning of extraction ratio $E=\frac{C_{Ai}-C_{Ao}}{C_{Ai}-C_{Bi}}$

My goal is to understand the meaning of the extraction ratio (artificial kidney, not other systems). I have the system in figure ($Q$ volumetric flow, $C$ concentration, the red line is a permeable membrane):

On my book (page 320 Cooney - Biomedical engineering principles) there is the definition of a parameter, the extraction ratio (note that all concentrations are evaluated at the same time):

$$E=\frac{C_{Ai}(t)-C_{Ao}(t)}{C_{Ai}(t)-C_{Bi}(t)}$$ where E is constant ($Q_A$ and $Q_B$ are constants).

In words, it say the extraction ratio "represents the amount of solute concentration change achieved relative to what would result from complete equilibrium with a very large supply" of liquid B "having a concentration $C_{Bi}$.

At page 549 of Saltzman - Biomedical Engineering, it say "The extraction ratio, E, is the solute concentration change" in the liquid A "compared to the theoretical solute concentration change that would occur if the" liquid A and B "came to equilibrium".

Mathematically what does mean equilibrium? Maybe $C_{Ao}=C_{Bo}$ or $C_{Ai}=C_{Bi}$ or $C_{A}=C_{B}$ for all value of $x$ or...?

Morover, how can I see mathematically the two terms of the comparison?

This ratio seem to express the actual concentration gained by A : $\Delta C_A = C_{A}(i) - C_{A}(0)$ divided by the maximum gain this system can have, which is given by : $\Delta^{MAX} C_A = C_{A}(i) - C_{B}(i)$.
To understand why the second is the maximum gain, consider the static case where you let the two fluid for an infinite time from $C_A(i)$ on the A side and $C_B(i)$ and the B side and assume that B is infinitely large, ie its concentration will not change as it is giving particule to the A side. In the end, the A side will be at equilibrium with the B side with a constration of $C_{A}(t=\infty) = C_B(i)$