I have a pump that will (according to the instructions) pump $450$ litres of water per minute.
Water is pumped into a box that can hold $85$ litres. This box has $3 \times 1.5$ inch holes in a row on one side, near the bottom of the box. One on the left. One in the center, and one on the right. :)
I'm trying to solve a problem of the box overflowing after so many minutes/seconds of the pump being switched on. Due to this, the pump then needs to be switched off again for a few minutes to allow the box to drain slightly, before resuming.
Question: What size box do I need, in order to handle this amount of water? Preventing the need of switching the pump off after so long?
In order for it to be unnecessary to stop the pump, the flow out of the box must be greater than or equal to the flow into the box. And the flow out of the box depends to a large degree on the height of the water.
To a first approximation we can use Torricelli's Law where we find that the speed of water flowing out of a hole in a container is given by $$v = \sqrt{2gh}$$
where $v$ is speed, $g$ is the gravitational acceleration near the Earth's surface and $h$ is the height of the water above the hole. The flow out of the container $Q_{out}$ is then $$Q_{out}=Av$$
where $A$ is the area of the hole, or holes in this case. The flow into the container $Q_{in}$ is given. Setting $Q_{out}=Q_{in}$ and solving for $h$ we find $$Q_{out}=Q_{in}$$
$$=> \; A\sqrt{2gh}=Q_{in}$$ $$=> \; \sqrt{h}=\frac{Q_{in}}{A\sqrt{2g}}$$ $$=> \; h=\frac{Q_{in}^2}{2gA^2}$$
Plugging in the numbers, with $Q_{in}= 0.0075 \; \frac{m^3}{s}$, $g=9.81 \;\frac{m}{s^2}$ and $A=0.00342 \; m^2$, we find $$h=0.245 \; m$$
To a first approximation, your box should have at least this height above the holes. In reality, I would suggest multiplying this by a safety factor of at least $2$ as the flow through a hole depends a lot on the shape and smoothness of the hole.