An estimate of turn-over time

The plot in the previous section shows that actually all of the water cannot be processed in finite time. This because $1-e^{-\alpha t}$ is never equal to 1 for finite t, and hence $V_1 = V_0(1-e^{-\alpha t})$ can never, for finite time $t$, equal $V_0$. Theoretically, we would have to wait forever to process each and every molecule of water. It is possible, however, to determine a time $t$ which makes the amount of water that goes unprocessed an arbitrarily small proportion of total volume. Therefore, to calculate turn-over time, we ask the question: how long will it take for 99.9 % of all tank water to be processed? Formula (9) enables us to quickly answer this question.

So how much time will it take to process $99.9$ % of the water moves through the device? We can answer this question by determining the value of $t$ when $V_1/V_0 = .999$. Observe that for all $t$,

$$\frac{V_1}{V_0} = 1 - e^{-\alpha t}.$$ (10)

We want to know the value of $t$, say $T$, that makes $1-e^{-\alpha T} =.999$. A little algebra shows that

$$T = \frac{V_0}{F_0} \ln \left ( \frac{1}{.001} \right ) = 6.90776 \frac{V_0}{F_0} = 6.90776 \cdot \alpha.$$ (11)

Therefore, when $T$ is chosen so that $T = 6.90776 \cdot \alpha$. $99.9 \%$ of tank water will have been processed after the device has run a time of $T$ hours or more. This time $T$ is defined as the turn-over time for the device. It indicates the amount of time it takes to circulate most all tank water through the device.

Example.

Suppose a 100 gallon aquarium has a cannister filter with a flow rate of 50 gallons per hour. How long before $99.9 \%$ of the tank water has been filtered?

Answer:

Apply formula (11) with $F_0 = 50$ gallons per hour and $V_0 = 100$ gallons, to find that after an elapsed time of

$$\frac{100}{50} 6.90776 \mbox{~hours},$$

$99.9 \%$ of the water will be filtered. Thus, after about 14 hours of running the filter, only $1/10^{th}$ of a gallon, or less than a pint of water, will remain unfiltered.