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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?

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    $\begingroup$ The time required for a container to empty itself through a hole in the bottom is a standard problem and solutions can be found on the Net. Very similar problems have also been treated here on SE. I suggest using the search box. $\endgroup$ – Gert Oct 21 '15 at 15:14
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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?

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  • $\begingroup$ still slightly confused and having a hard time getting started. This is an ungraded exercise so don't feel like your giving it away. I've been following this example on this website which seems similar, I think im just overthinking this concept. montessorimuddle.org/2012/05/01/… $\endgroup$ – Delano1090 Oct 21 '15 at 17:18
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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:

  1. 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 $v_{out}$.
  2. 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: $$\frac{dM}{dt} \propto - \sqrt{M\left(t\right)}$$

  3. 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:

  1. Use a cylindrical beaker with a large area $A_c$
  2. Make the hole small and as symmetric as possible, remove any roughness and imperfections
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    $\begingroup$ Under your point 2, there will be an offset in the velocity/mass curve since the hole is not quite at the bottom. If you define M relative to "mass when flow stops" it would be valid. Worth pointing out? $\endgroup$ – Floris Oct 21 '15 at 16:48
  • $\begingroup$ still slightly confused and having a hard time getting started. This is an ungraded exercise so don't feel like your giving it away. I've been following this example on this website which seems similar, I think im just overthinking this concept. montessorimuddle.org/2012/05/01/… $\endgroup$ – Delano1090 Oct 21 '15 at 17:19
  • $\begingroup$ You may well be overthinking this. Did you try plotting the relationship that nluigi recommended? $\endgroup$ – Floris Oct 21 '15 at 17:43
  • $\begingroup$ @Floris - good point, I have edited my answer to reflect this. $\endgroup$ – nluigi Oct 21 '15 at 22:58
  • $\begingroup$ @Delano1090 - what specifically is confusing you? Even if it is an ungraded exercise, understanding the approach to solving the problem will be more valuable to you in the long run than knowing the exact answer. $\endgroup$ – nluigi Oct 21 '15 at 23:00

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