Water change math

"Altum" wrote in message
.. .

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 an 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.

Thanks Altum, but my Differential Equations class was more years ago then I
care to remember. Sadly the dreaded Beer has erased all those brain cells.
I'm assuming 35g of initial Crap (Nitrates), .15 g/hr of additional
Nitrates, 140 gallons of water and 1 gallon per hour of additional
water/outflow. I'm also assuming perfect mixing, although hoping to do
better by adding the fresh water to the bottom of the sump (tall plastic
barrel) and taking the old water from the top of the "sump".

A math refresher would be appreciated if you have time.