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I took a bucket, drilled 2 different sized holes on the side near the bottom and filled it with water. The stream of water the proceeded from the larger hole traveled further than the stream from the smaller one. How does the size of the hole affect the distance that the water travels?

See also: Torricelli's Law.

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Were the holes at the same height? A hole lower down will shoot farther than a hole higher up. This is a consequence of the amount of liquid that is pressing down (and hence, out) at various heights. –  knucklebumpler Jul 21 '11 at 19:20
The centers of the holes were horizontal. –  JoeHobbit Jul 21 '11 at 22:31
Have you tried the experiment with two buckets? I predict that the smaller hole would then travel farther. I think that the current created by the water evacuating the large hole decreases the pressure against the smaller hole. Though I don't know how to calculate that. –  knucklebumpler Jul 22 '11 at 1:59
I am convinced the smaller hole merely has increased viscosity as user1631 pointed out in the accepted answer. –  JoeHobbit Feb 21 '13 at 5:38
Related: physics.stackexchange.com/questions/6341/… –  JoeHobbit May 7 '13 at 19:58
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4 Answers 4

up vote 6 down vote accepted

I believe the physically relevant parameters here are: 1. internal pressure P at the bottom of the bucket but not too close to the hole, units [$kg*m^{-1} s^{-2}$], the fluid density $\rho$ [$kg*m^{-3}$], hole area $A$ units $m^2$, and viscosity $\eta$ $kg m^{-1} s^{-1}$. First, let's ignore the viscosity, and find which combinations can give us a velocity, units [$m/s$]. The unique combination is $\sqrt(P/\rho)$, and so in the limit of vanishing viscosity the velocity will be independent of the hole size. Now if we add in viscosity, there is a dimensionless combination available, $\rho \eta^2 P^{-1} A^{-1}$. We could have any function of this dimensionless ratio, but common sense tells us that at least in the limit of small $\eta$ it should be a decreasing function, since viscosity should take kinetic energy away from the fluid. From this ratio we see that decreasing the size of the hole is equivalent to increasing the viscosity, and thus the smaller hole should have a smaller exit velocity than the bigger one. I guess this is a long-winded way of saying the big hole will have higher Reynolds number and be less effected by the viscosity.

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Ah, very nice! So now we have a good, clear candidate for an explanation of the direction of the observed effect. The thing that leaves me in doubt is that there are other dimensionless ratios we can form, such as the ratio of the diameter of the hole to the thickness of the wall of the bucket. The efflux coefficients are geometry-dependent, and because they give a huge effect on the result, even a tiny amount of geometry-dependence can produce a big change in the result. A simple test would be to change P. If it's viscosity, P should affect the result. If it's efflux coeff., it shouldn't. –  Ben Crowell Jul 21 '11 at 20:11
That's true, I was assuming negligible wall thickness. Of course in the opposite limit of very thick walls, we have the pipe flow problem, which is pretty well understood, and viscosity to pipe diameter ratio plays a big role there too. –  user1631 Jul 21 '11 at 21:07
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If the only forms of energy were the potential energy of the water in the bucket and the kinetic energy of the escaping stream, then by conservation of energy the velocity of the stream would have to equal $\sqrt{2gh}$, where $h$ is the difference in height between the hole and the surface. Since you observe that $v$ (or at least the component of $v$ parallel to the axis of the flow) depends on other factors, it's possible that in your experiment, there were other forms of energy involved. Two possible candidates are: (a) frictional heat, and (b) the kinetic energy of the water flowing toward the hole. We expect effect b to be minimal or zero, both because the velocity of this flow is small and because this KE represents a reservoir of energy that is either constant or only very slowly changing as the water level drops.

Another issue is that kinetic energy is independent of the direction of motion, and the flow of the stream as it exits may not be entirely parallel, so the velocity of flow in the direction parallel to the axis of flow will in general be less than $\sqrt{2gh}$. The stream contracts significantly on its way out, so quite a bit of its KE is in the form of motion that's transverse to the axis. This leads to something called an efflux coefficient. There is a discussion of both the frictional effect and the non-parallel effect in the Feynman Lectures, section 40-3. The non-parallel effect can be quite large. Feynman gives some examples for different types of holes and discharge tubes, and the efflux coefficients are typically factor-of-two effects.

Another way to approach this is through conservation of mass. When you put your thumb over the end of a garden hose, the cross-sectional area $A_1$ of the aperture under your thumb is less than the cross-sectional area $A_2$ of the hose. Since water is being neither created nor destroyed, we must have $v_1/v_2=A_2/A_1$. This is the opposite of what you observed, and the two situations are not really analogous. They would be analogous if there was a piston pressing down on the surface of the water at the top of the bucket so that the level was constrained to go down at a fixed rate.

[Edited the above answer to improve it based on comments by Omega Centauri.]

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"(b) the kinetic energy of the water flowing toward the hole" If my understanding of hydraulics is correct, this is wrong. In the local vicinity of the hole, higher velocity corresponds with a lower pressure, which is possible due to the liquid acceleration. At the point you reach the hole, the liquid potential in the form of pressure (which comes from gravitational potential) has been converted to velocity, represented by your $\sqrt{2gh}$ term. In other words, the stagnation pressure in the absence of friction is constant until it leaves the container. –  AlanSE Jul 21 '11 at 19:28
@zass More simply, once the flow in the vicinity of the hole reaches a quasi-steady state, the only thing the kinetic energy of flow (within the bucket) represents, is stored kinetic energy. The time rate of change of energy is still the mass rate of flow out the hole, time velocity squared over 2. –  Omega Centauri Jul 21 '11 at 19:36
@Zassounotsukushi: Maybe I'm misunderstanding, but what you wrote doesn't seem to connect to what I wrote. It is certainly true that there are pressure gradients in the water, but that has nothing to do with gravitational potential energy, which is simply $\int \rho g y dV=(1/2)\rho g A h^2$ for a cylindrical column of height h and cross-sectional area A. It sounds like you have in mind a description in terms of Newton's laws, which is fine, but doesn't relate directly to a description in terms of energy. We do expect the energy (b) to be small, since its v is small and $K\propto v^2$. –  Ben Crowell Jul 21 '11 at 19:37
@Omega Centauri: Good point. This is a much better reason than my $K\propto v^2$ for ignoring this effect. However, note that the level of water in the bucket is falling, so the reservoir of KE inside the bucket is not quite constant. By the time the water level gets down to the hole, the reservoir of KE vanishes. Your argument would prove that the effect exactly vanishes if we were continuously keeping the bucket topped off. –  Ben Crowell Jul 21 '11 at 19:40
There is some good discussion here, but I would advise limiting the discussion to a constant water level in the bucket (and small hole compared to the bucket). There are plenty of complicating factors even with this assumption, and I think that the distance the streams travel will still be different. It's nice b/c we don't even have to assume the pressure is from gravity. It could equally be a closed container with a pump maintaining constant pressure. –  AlanSE Jul 21 '11 at 19:46
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user 1631 is correct in his/her thinking. It is based on Poiseuille's law.

Flow Q ~ (dP/dL)*r^4

where dP is the pressure differential across the small length L of the hole (imagine the hole had a small pipe of radius r. Now we know that dP/dL is same because it is the pressure difference in the (fluid - P_atm).

Therefore, Q is proportional to r^4 while velocity is Q/Area which is proportional to r^2, Therefore, higher velocity from a larger hole... Of course, I have left out the viscosity term in the proportionality equation. It should be always be there because it is viscosity that causes this effect as pointed out earlier by others. But it is a constant hence I just left it out as it goes away in ratioing.

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I like the notation from Ben in his answer. So we will compare two situations with equal water column height but different areas of the hole, $A_1$ and $A_2$. We find that the distance the water squirts is different due to friction. I propose that the friction is from three different sources.

  1. Fluid internal friction as it approaches the hole
  2. Boundary layer friction around the perimeter of the hole
  3. Air friction after it leaves the hole

I would actually imagine that the 3rd one would be the most significant. Some meaningful relations for this could be deduced exclusively from the hole size. If we assume the first two sources of friction are small, then the initial velocity leaving the hole is the same for both cases, thus we have a certain flow rate and velocity for each. If we assume the holes to be circular and extend this assumption to the stream itself, we can find the perimeter interacting with air and the linear density of the stream of water (close to the hole).

Then the problem is similar, but not exactly the same, as that of a projectile with air resistance. The metric that matters is the ratio of the drag coefficient to the mass, but the linear mass density in the case here. I'll try to come back and put some equations up here later so we can maybe even get a general figure of merit for where deviation from a parabolic path of the water jet will be observed.

Local forms loss of the hole

I realized that the parts of 1&2 is really a pretty standard engineering problem.

  • K = Local forms loss coefficient
  • H = fluid head, used in place of pressure

$$\Delta H = \frac{v^2}{2 g} + K \frac{v^2}{2 g}$$ $$ v = \sqrt{ \frac{2 g h}{1+K} }$$

Now, we won't be able to avoid the inevitability of finding Reynolds number, so here we are.

  • $\rho = 1000 kg / m^3$
  • $\mu = 0.001 Pa s $

$$Re = \frac{G D}{\mu} = 2 \frac{v \rho}{\mu} \sqrt{ \frac{A }{\pi} }$$

If we're talking about in the neighborhood of a hole $1 cm^2$ and a velocity $1 m/s$ then we find Reynolds number to be around 1,000. Now, I found no choice to use an actual academic paper to get the loss coefficient. See Accurate Evaluation of the Loss Coefficient and the Entrance Length of the Inlet Region of a Channel.

$$K = 4.183 \times 10^{-5} Re + 0.152$$

So K will be in the neighborhood of 0.2. This is the major difference for the two cases regarding 1&2 types of losses. I tried some numbers, like 1 m for the head, and found without the forms loss you get about 4.4 m/s exit velocity, with 1 cm^2 hole, you get 2.8 m/s. Increasing the size of the hole 10x I get 2.1 m/s and decreasing it 10x I get 3.4 m/s. This isn't what I expected. I thought that a small hole would make the fluid velocity smaller, but maybe I've messed something up with the signs although I can't find it right now. I know how I would approach #3 as well, but I don't have the time. I certainly don't think anyone has sufficient evidence to say one type of loss is greater until some more work is done.

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I don't believe that air resistance is going to be a significant effect. The stream is presumably steady, so it's not analogous to a projectile. Air is displaced from in front of a projectile and has to flow around it. This is completely different from a steadily flowing stream of water, which doesn't displace any air. There will certainly be a small amount of air friction, but I doubt that it will be anywhere near enough to matter. Since we know that the efflux coefficients give factor-of-two effects, and they depend on geometry, I would suspect them as the main reason for the effect. –  Ben Crowell Jul 21 '11 at 19:58
In short, yes, of course I agree that drag coefficient is different, but I am unsure if the form of the drag force will still be $C v^2$. I am not convinced, however, that friction forces 1&2 combined will be greater than 3. I don't have strong arguments either way, but at least 3 is calculable easily. Consider the fact that both 1&2 require introducing viscosity. –  AlanSE Jul 21 '11 at 21:05
Number 1 and 2 are only relevant in case of very small pores, or as one states in hydrodynamics: at very low Reynolds numbers in "potential flow" . For rge scope of this question your answer is wrong. –  Georg Jul 22 '11 at 9:12
@Georg I don't see anything wrong with what you said, but how it shows what I wrote in my answer is wrong... is completely beyond me. What is rge? –  AlanSE Jul 22 '11 at 17:12
Rge is a stump, the rest.....si tacuisses –  Georg Jul 22 '11 at 17:29
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