Simple cylinder fluid mechanics, draining from hole on bottom So my professor is having us work on an assignment where there is a cylinder full of water with a hole in the side near the bottom. Given the use of stopwatch, balance beam, beakers, etc, how can we determine the hole size in the side of the bottom of a cylinder? If I had the volume flow rate or mass flow rate this problem would be simple... However doesn't the ever changing flow rate due to height complicate this question??  Any input? 
 A: Since this is homework, I will just nudge you in the direction of an answer.
Assuming that the flow rate is reasonably slow, so you have time to measure the changing mass of the beaker over time, you should be able to plot the flow rate as a function of time. You expect the flow rate to depend on the (square root of the) height between the surface of the liquid and the (center of the) hole.
If you know the mass per unit time, you can figure out the flow rate. If you think about this, you will find that there is a straight line plot (what factors do you have to plot?) which has a slope equal to the hole size.
Can you figure it out from there?
A: Indeed it does complicate the situation but not so much that you will be unable to solve the problem. Considering you have used the bernoulli-equation tag, I am assuming you are familiar with it. I don't want to give a way too much so i will give hints. 
I would approach this the following way:


*

*Determine, from a mass balance, the rate of change of mass $M\left(t\right)$ of the system as a function of time $t$. As aptly pointed out by @Floris, this is the mass relative to when the flow stops. You should be able to find:
$$\frac{dM}{dt} = -\phi_{m,out} \propto -v_{out}$$
How is the mass flow out $\phi_{m,out}$ related to the outflow speed $v_{out}$? Hint: it involves the area of the hole $A_h$ and the density of the fluid $\rho$. At the moment we can't solve the equation without knowing $v_{out}$.

*In comes Bernoulli:
$$\Delta\left[\frac{p}{\rho}+\frac{1}{2}v^2+gz\right]=0$$
which gives us a relation between the speed $v_{out}\left(t\right)$ and height $h\left(t\right)$ if we choose the proper points on the streamline. Hint: Take one at the surface of the water and the other just outside the hole; what do you know about $p$, $v$ and $z$ at each position?. Since $M\left(t\right)=\rho A_c h\left(t\right)$, with $\rho$ the density of the fluid, $A_c$ the area of the cylinder and $h$ the height of the fluid in the cylinder, you should find:
$$v_{out}\left(t\right)\propto\sqrt{M\left(t\right)}$$
Substituting into the mass balance:
$$\frac{dM}{dt} \propto - \sqrt{M\left(t\right)}$$

*The solution to the equation above is of the form:
$$\sqrt{M\left(t\right)} \propto -t$$
Since you can measure the mass $M\left(t\right)$ as a function of time $t$, a plot of $\sqrt{M\left(t\right)}$ vs $t$ should give a linear graph with a negative slope. The slope and y-intercept of this graph are related to the hole size. Don't forget $M\left(t\right)$ is the mass relative to when the flow stops.


To make your life easier doing this experiment, I would suggest:


*

*Use a cylindrical beaker with a large area $A_c$

*Make the hole small and as symmetric as possible, remove any roughness and imperfections

