Hi Jason,

If you look at Penny's response to a similar problem you will see an expression for the Volume of the tank. I am going to change her notation slightly and let $L$ be the length of the tank and $h$ the height of the water in the tank. With this change the expression is

\[V = L \left(r^2 cos^{-1}\frac{r - h}{r} - (r - h) \sqrt{2 h r - h^2} \right). \]

In this expression $L$ and $r$ are constants and $V$ and $h$ vary with time. The rates at which $V$ and $h$ vary with time are measured by their derivatives with respect to time $t$.Differentiating both sides with respect to $t$ and some simplification gives me

\[\frac{dV}{dt} = L \frac{dh}{dt} \left( \frac{r^2}{\sqrt{2 h r - h^2}} + \sqrt{2 h r - h^2} - \frac{(r - h)^2}{\sqrt{2 h r - r^2}} \right). \]

I think this is the expression you need, $\large \frac{dV}{dt}$ is the rate of change in the volume (the flow rate, which is negative) and $\large \frac{dh}{dt}$ is the rate of change of the height of the liquid. You need to take care with the units however when you use it.Let me suppose that you are measuring $L, r \mbox{ and } h$ in inches then you need to measure the flow ($\frac{dV}{dt},$ rate of change in the volume) in cubic inches per minute. It looks like you are measuring the flow in gallons per minute and there are 231 cubic inches in a US gallon so you need to multiply the rate in gallons per minute by 231.

I hope this helps,
Harley