Given:
- a cuboid container, of dimensions h into w into l
- a rate of flow of liquid per unit time through an inlet, in
- a rate of flow of liquid per unit time from an outlet, out, when the container is full
How would we find the amount of liquid in the container after time t?
This sort of problem is often found in middle school math textbooks, where the approach is "if the container empties in three hours, then a third of the container empties in one hour" but this is obviously wrong as the container empties fast at first but slows down as it does so.
You can show by applying the Bernoulli's Theorem that the volume output rate at anytime is
$out_{x} = -A \sqrt{x}$
Here, $x$ is the vertical distance from the water surface to the outlet. $A$ is a constant depending on the physical parameters of the system. The constant can be determined in terms of the given full container flow rate $out_{max}$.
You can show the volume output to be:
$out_{x}=out_{max}\sqrt{\frac{x}{h}}$
The volume input is always $in$
We can write a differential equation: $\frac{dV}{dt}=\frac{d(wlx)}{dt}= in- out_{max}\sqrt{\frac{x}{h}}$
If you integrate and find an expression for the height of water $x$ as a function of $t$, you are done