The turn-over formula

Suppose an aquarium device processes $F_0$ gallons of tank water per minute. If the processed water is returned to the tank, how long will it take for (nearly)¹ all the water to be processed?

To facilitate a discussion, suppose $V_0$ is the volume of water that has passed through the device zero or more times. That is, $V_0$ is simply equal to the original tank volume $V$. Let $V_1$ be the volume of water that has passed through the device $1$ or more times, $V_2$ the volume of water that has passed through $2$ or more times, etc., so that $V_n$ indicates the volume of water that has passed through the device $n$ or more times.

As time $t$ increases and for each $k \in \{1, 2, \ldots n\}$

$$V_k(t) \rightarrow V.$$ (2)

Of course, $V_0 = V$ for all $t$ because all the water has been processed zero or more times. Equation (2) will provide a useful ``sanity check'' on the expressions that we derive below for these functions of time $t$. Whatever answer we get, we should see that at $t=0$, $V_k(0) = 0$ for each $k$ bigger than 1. Then, as $t$ gets large, $V_k$ should get close to $V_0 = V$.