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This is a conceptual question about the application of Bernoulli's equation to a water spout.

There is a classical problem found in many physics texts which goes something like "you have a garden hose with a nozzle which flares inward so the radius is smaller at the end. How high does the water shoot into the air?"

So there are obviously details for the exact problem (like what is the angle of the hose, pressure or velocity, etc) but I am specifically interested in the applicability of Bernoulli's equation to the water which has left the hose. It would seem to be that after the water has left the hose, it is no longer satisfying the conditions for Bernoulli's equation. I can't quite put my finger on why; I can't exactly see the the flow is not laminar but since the pressure outside the flow (the air) is certainly not at the same pressure as the water, it's certainly doesn't seem to be steady.

In my understanding, you would treat the water in the nozzle with Bernoulli's equation (or continuity, depending on exactly conditions) and then simply treat the water as droplets acted on by gravity. If that's true, can someone clarify exactly what conditions of Bernoulli's equation are being violated?

Alternatively, if I am wrong, can you convince me that the water can still be considered a "fluid" for the purposes of applying Bernoulli's equation?

EDIT: A specific example in response to a comment. This shows that assuming Bernoulli still applies is equivalent to assuming the pressure of the stream is the same as the atmospheric pressure. Vertical oil pipe of height $h_1$, oil (gauge) pressure at its base of $P$ and fluid velocity $v$. How high into the air down the oil shoot?

Solution 1) Bernoulli fails when the fluid exits the pipe. The pressure when the fluid leaves is equal to 0 (gauge) so we have Bernoulli's equation at the top of the pipe

$$P+\frac{1}{2}\rho v^2=\frac{1}{2}\rho v'^2+\rho g h_1\longrightarrow v'^2=v^2+\frac{2P}{\rho}-2gh_1$$ Then using energy conservation we have $$mgh_1+\frac{1}{2}mv'^2=mgh_2\longrightarrow h_2=h_1+\frac{v'^2}{2g}$$ $$h_2=h_1+\frac{1}{2g}\left( v^2+\frac{2P}{\rho}-2gh_1 \right)=\frac{v^2}{2g}+\frac{P}{g\rho}$$.

Solution 2) Bernoulli holds throughout the motion. The (gauge) pressure at the very top (and throughout the entire spout) is zero, so Bernoulli gives us $$P+\frac{1}{2}\rho v^2=\rho g h_2\longrightarrow h_2=\frac{P}{g\rho}+\frac{v^2}{2g}$$

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2 Answers 2

If you'd do the full theoretical derivation of the Bernouilli-laws you will see that there are two situations in which the law applies (aside from being steady, incompressible and free of viscosity-effects):

  1. If you look at a particle along a streamline.

  2. If the flow is irrotational

Within a nozzle I believe they usually assume that the second condition is met. As soon as your liquid leaves the nozzle condition 2 ceases to exist and it's no longer possible to apply Bernouilli.

In this case you should resort to the classical laws of kinetics that you know to discribe the motion of the fluid, to find the height you simply apply:

$mgh = \frac{1}{2}m v^2$.

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well see this is how I was originally thinking of it, but particles still move along streamlines. The streamlines may "spread out" because the pressure on the outside is no longer the same but each little chunk of mass $m$ from the kinematic equations has a streamline which seems isomorphic to the parabolic motion, right? –  levitopher Jun 25 '13 at 1:53
    
Yes, Bernouilli simply describes the classical motion, only in the case of Bernouilli we go to a continuüm-limit. But ordinary Newton etc should still hold. The mass in the equation above even cancels out, so it gives an easy, effort-free equation for the heigth :). –  Nick Jun 25 '13 at 12:08
    
See my edit above; picking a random case I see that the assumptions behind Bernoulli's equation apparently hold throughout the motion. Can you find a counterexample? –  levitopher Jun 25 '13 at 14:04
    
I understand that Newton's equations will always hold; Bernoulli is just a macroscopic description of them - I have added a specific example to the question which shows this. Can you give me a counterexample, where you get a different answer with Bernoulli? –  levitopher Jun 25 '13 at 14:05
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I think they usually assume that the streamlines leaving the nozzle are parallel, which means the different layers wouldn't feel normal stress (i.e. pressure) but only shear stress (i.e. viscosity) which we are not considering. –  Nick Jun 27 '13 at 10:15

It is an empirically observable fact that subsonic jets (of which your water spout is an example) do in fact exit into a quiescent medium at the pressure of that very medium. The water will leave the nozzle at precisely the ambient pressure if the exit Mach number is less than 1.

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So you are saying that the gauge pressure of the water is 0 as soon as it exits the hose, and my solution #1 is correct? It sounds like you have a specific experiment in mind - any online data or video that we can check out? –  levitopher May 27 at 5:19
    
I do not have a video in front of me at the moment, but this is one of the very first things you learn in an introductory course on jet engine propulsion. There is a pressure-times-area component of the thrust that is always ignored for subsonic relative exit Mach numbers. Conceptually, you can think of it like this: If the exit pressure were not equal to the ambient pressure, compression or expansion waves could propagate upstream, which would adjust the upstream conditions until the pressure was matched. If the flow is supersonic this won't be possible, due to zero upstream influence. –  Bryson S. May 27 at 5:26

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