# How to Properly Rinse a Water Bottle

Posted by Jason Polak on 28. August 2015 · Write a comment · Categories: math

The other day we were camping and I had to rinse my Nalgene bottle (post not sponsored by Nalgene).

I only had a fixed amount of water, less than the capacity of the bottle, and I had to decide how to best use the water to minimise the concentration of contaminants. I could pour all the water in the bottle, give it a good shake, and then spill it out. Or, I could divide the rinsing water into several portions. I decided to choose amongst the methods, indexed by $n$, of dividing the rinsing water into $n$ equal portions, and then doing a sequence of $n$ separate rinses. The problem is, how should I choose $n$?

Of course, there are other schemes, such as use half the water for the rinsing, then use half the remaining water, and continue so that the amounds of water form the geometric progression $1/2, 1/4, 1/8, \dots$. However, practically, once the amount of rinsing water per rinse becomes small, there’s no point in continuing since the rinsing process itself becomes tedious.

Intuitively I felt I should make $n$ as large as possible without the process becoming impractical. So for instance, if I had a liter of water, I would probably divide it into $n=3$ or $n=4$ parts. So let’s see if we can come up with a reasonable model of this scenario to investigate whether dividing up the water is really worth it.

Let us suppose that $I_0$ is the initial concentration of contaminant, say in $g/L$, though the units don’t matter. It’s dissolved in some water, which is stil/el in the bottle. After turning the bottle upside-down to empty the contents, there is still a little water left after a fixed amount of time so let’s say we always turn the bottle upside-down for the same amount of time and that leaves a fixed volume $L$ so that the contaminant is dissolved in $L$ liters of water at the start (presumably, some water was left in the bottle and bacteria might have grown in it).

Let $V$ be the total fixed amount of rinsing water that we have on hand. Then, for each rinse we put it $V/n$ liters of water, and let’s assume that upon a good shake, the contaminated water and the rinsing water mix completely to give a completely homogeneous liquid where the concentration of contaminant is the same everywhere. If we let $\alpha = V/L$, then the new concentration after one rinse with $V/n$ liters of water is then
$$I_1 = I_0(1 + \tfrac{\alpha}{n})^{-1}.$$
Since we have $n$ portions of rinsing water, the final concentration $I_n$ after all the rinsing will be
$$I_n = I_0(1 + \tfrac{\alpha}{n})^{-n}.$$
Hence, if my theory of ‘increasing $n$ is good’ is correct, then we need to show that
$$(1 + \tfrac{\alpha}{n})^n$$
is an increasing function of $n$. Expanded this expression for the first few values of $n$ gives
$$n=1 : 1 + \alpha,\\ n=2 : 1 + \alpha + \alpha^2/4,\\ n=3 : 1 + \alpha + \alpha^2/3 + \alpha^3/27.$$
We can comparing the $k$th terms for $n$ and $n+1$: they are $\binom{n}{k}\alpha^k/n^k$ and $\binom{n+1}{k}\alpha^k/(n+1)^k$. If we can show that the $k$th terms themselves are increasing, this will certainly imply what we need; namely, that $(1 + \tfrac{\alpha}{n})^n$ is an increasing function of $n$. A little algebraic manipulation shows that it suffices to prove that
$$\frac{n!}{(n-k)!n^k} \leq \frac{ (n + 1)!}{(n+1-k)!(n+1)^k}.$$
This inequality is equivalent to
$$\frac{n}{n}\frac{n-1}{n}\cdots\frac{n+1-k}{n}\leq \frac{n+1}{n+1}\frac{n}{n+1}\cdots\frac{n+2 – k}{n+1}$$
which is true since $a/b$ for positive $a\leq b$ is less than $(a + 1)/(b + 1)$.

How does this actually work in practice? Suppose you have a cup of water (250mL) for rinsing and 5mL remains in the bottle after pouring the water out. Then $\alpha = 50$. With $n=1$ rinsing, you reduce the concentration by a factor of 51, but dividing the cup into $n=2$ rinsings already gives a reduction by a factor of 676! Three rinsinings ($n=3$) divides the concentration by about 5513.96 which is a little more than 108 times better than using the entire cup at once. Another way of looking at this difference is that to get the same rinsing effectiveness with a single rinse you would need a little more than 27.5 liters of water, assuming all of this could actually fit in the bottle at once!

The upper bound of reduction is the limit as $n\to \infty$, which is $e^\alpha$, which in this case is about $5.187\times 10^{21}$. Of course, this is not really achievable in practice since after a certain point, the effective rinsing and the good mixing properties of very small volumes of water diminish anyway.

## Other Practical Considerations

Of course, you may want to incorporate some practical things into your rinsing procedure, such as adding a small amount of soap into the bottle to promote a complete mixing of the contaminant into the rinse water, especially if the contaminant contains oily components, such as some organisms.

Disclaimer: Aleph Zero Categorical is not responsible for personal illness or injury caused by the mixing procedures advocated in this post.