Bill Stock wrote:
I'll be setting up an automatic water changer in a few weeks and I've been
having trouble with the math.
The reading I've done suggests that a 20% daily water change is still going
to leave the fish swimming in 82% crap (r = e ^ -.2) as opposed to 50% to
100% crap with large weekly water changes. BUT, the detailed formula I've
seen shows the crap remaining as r=(1-a/n)^n, where a is the percentage of
water changed and n is the number of changes. Now assuming I change 1 gph
and a drop is 1/3600 gallons, then I'm left with 99.99% crap because the
changes are so small. (One drop at a time) Am I missing something here or do
I need to remove some water first to make this work?

This kind of math problem is known as a mixing problem, and it's solved
with differential equations. The algebraic model you're trying to use
assumes that a is constant during the water change, but it's not.

If you assume perfect mixing, constant water volume during the change,
and no crap in the tap water, here's the basic system of differential
equations where Crap(t) is the amount of crap in the water at t hours
and gph is the rate you're changing water.

dCrap/dt = crap fish add/hour - gph * (Crap(t)/tank volume in gallons)
Crap(0) = initial amount of crap in the water

From there, you integrate, plug in the initial condition and solve for
Crap(t). However, I don't think you can assume perfect mixing. The
equations get messy with imperfect mixing and you need information about
the mixing rates to solve them. It's probably easiest to set the
changer up for 20% a day and measure nitrates at the start and end of a
week.

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