Determining processed water at any elapsed time

Now we make a modeling assumption. Suppose that after being processed, water that is put back into the aquarium rapidly becomes mixed up with the unprocessed water. That is, assume that processed and unprocessed water is uniformly mixed within the aquarium instantly. This simplifying assumption is true at least in an average way. At times the in-flow water may have a higher proportion of unprocessed water than does the water within the tank. These times are balanced by times when the reverse is true; that is, when the in-flow water has a lower percentage of unprocessed water. On average though, the percentage of flow into the device that is unprocessed matches the percentage of water in the tank overall that is unprocessed.

Given this modeling assumption, at any moment in time, the total gallons of unprocessed water in the tank is given by the amount of water that has been through the device zero or more times minus the amount of water that has been through the device one or more times: $V_0 - V_1$. The result of this subtraction is just the total gallons of water that has never been processed by the device. If we take this total, and divide by the total volume of the tank, we arrive at the fraction of water in the tank that has never been processed:

$$\frac{V_0-V_1}{V}.$$

Now $V = V_0$, so the fraction of water that has not been processed can be written as

$$\frac{V_0 -V_1}{V_0}.$$ (3)

The next question we ask: what is the rate at which unprocessed water is being converted to processed water? This is really just a question of how fast $V_1$ is getting larger. To put this a little more formally, so that we can apply techniques of calculus to solve this problem, let $\frac{d V_1}{dt}$ represent the rate at which water is being converted from unprocessed to processed. If the device that is processing the water has a flow rate of $F_0$, then unprocessed water flows from the tank into the device at a rate given by the expression

$$\frac{V_0 - V_1}{V_0} F_0.$$ (4)

The last expression is exactly the rate at which $V_1$ is changing. That is

$$\frac{d V_1}{dt} = \left (1 - \frac{V_1}{V_0} \right ) F_0 .$$ (5)

Let $\alpha = F_0/V_0$. Then equation 5 can be written as

$$\frac{d V_1}{dt} + \alpha V_1 = F_0.$$ (6)

We also know that initially, before the aquarium device is turned on, no water has been processed. Therefore,

$$V_1(0) = 0.$$ (7)

Equations (6-7) are referred to as an initial value problem. It is a mathematical fact that initial value problems such as (6 - 7) have one and only one solution. There are many techniques for finding the solution. One technique as good as any other is judicious guessing at the form of the solution. The reader should bear in mind that no matter how a solution is obtained, once it is demonstrated that a function satisfies equation (6) and equation (7) simultaneously, one concludes that this function must be the one and only solution to our initial value problem.

Consider the function

$$\bar V_1 = V_0 - V_0 e^{-\alpha t}.$$ (8)

Substitute $t=0$ into this expression. Observe that $\bar V_1(0) = V_0 - V_0 e^{\alpha \cdot 0} = V_0 - V_0 = 0$, which means the candidate solution function $\bar V_1$ satisfies equation (7).

If we show $\bar V_1$ also satisfies the differential equation (6), then $\bar V_1$ must actually be the one and only function that satisfies equations (6-7) and so $V_1(t)=\bar V_1(t)$.

Recall from calculus that $\frac{d}{dt} V_0 e^{-\alpha t} = -V_0 \alpha e^{-\alpha t}$, and that the derivative of a constant $V_0$ is zero. It follows that

$$\frac{d \bar V_1}{dt} = \frac{d}{dt} \left ( V_0 - V_0 e^{-\alpha t} \right ) = \alpha V_0 e^{-\alpha t},$$

and therefore that

$$\frac{d \bar V_1}{dt} + \alpha \bar V_1(t) = \alpha V_0 e^{-... ...( V_0 - V_0 e^{-\alpha t} \right ) = \frac{F_0}{V_0}V_0 = F_0.$$

This implies that $\bar V_1$ satisfies (6), and, because $\bar V_1$ also satisfies the initial condition (7), it must be that $V_1 = \bar V_1$; i.e.,

$$V_1(t) = V_0 - V_0 e^{-\alpha t}.$$ (9)

From the plot below, one can easily see that as $t$ increases (and so $\alpha t$ increases), $V_1$ approaches $V_0 = V$. Also note that $V_1(0) = 0$, as expected. In the plot, observe that when $\alpha t = 1$, $t = V_0/F_0$. This is tank volume divided by flow rate, and is at first glance a seemingly reasonable estimate of turn-over time. However, when $\alpha t = 1$, the plot below shows that there is about $40 \%$ of tank water left unprocessed! Time $t = V_0/F_0$ is clearly a poor estimate of turn-over time.