Determining water volume processed multiple times

The derivation of a formula for $V_k$ for $k$ larger than 1 is quite similar to the derivation of $V_1$. Already it has been observed that

$$V_k(0) = 0.$$ (12)

This equation is an initial condition that $V_k$ must satisfy.

A differential equation can be derived in a manner similar to the derivation of (6). Observe that the volume of water that has been processed exactly $k-1$ times is ($V_{k-1} - V_k$). That is, the volume of water that has been processed exactly $k-1$ times is equal to the volume of water that has been processed $k-1$ or more times minus that volume that has been processed $k$ or more times.

It follows then that $(V_{k-1}(t) - V_k(t))/V_0$ represents the fraction of tank water that, at any particular instant in time $t$, can possibly be converted to water that has been processed $k$ times. Therefore, the rate of change of $V_k$ is just the flow rate $F_0$ scaled by this fraction $(V_{k-1}(t) - V_k(t))/V_0$:

$$\frac{d V_k}{dt} = \frac{(V_{k-1} - V_k)}{V_0} F_0.$$

Letting $\alpha = F_0/V_0$, we find that

$$\frac{d V_k}{dt} + \alpha V_k = \alpha V_{k-1}.$$ (13)

Observe that (13) and (12) reduce to equations (6) and (7) when $k=1$. We have already shown that $V_1 = V_0(1-e^{-\alpha t})$. Now we find $V_k$ for $k \in \{2,~ 3,~ \ldots \}$.

Define the polynomial

$$P_{k}(t) = 1 + \alpha t + \frac{(\alpha t)^2}{2} + \ldots + \frac{(\alpha t)^k}{k!}.$$ (14)

Observe that

$$\frac{d}{dt}P_k(t) = \alpha P_{k-1}(t).$$ (15)

Next, define

$$\bar V_k = (1 - P_{k-1}(t) e^{-\alpha t} )V_0.$$ (16)

Observe that $\bar V_k(0) = 0$, so that $\bar V_k$ satisfies equation (12). It follows that

$$\frac{d \bar V_k}{dt} + \alpha \bar V_k = [-\alpha P_{k-2} e^{-\alpha t} + \alpha P_{k-1} e^{-\alpha t}] V_0$$

$$~ ~ +[\alpha - \alpha P_{k-1} e^{-\alpha t}] V_0$$

$$= \alpha V_{k-1}.$$

Since $\bar V_k$ satisfies the initial condition (12) and the differential equation (13), $\bar V_k = V_k$, and hence

$$V_k = (1 - P_{k-1}(t) e^{-\alpha t} )V_0.$$ (17)