The pressure in a water spout and Bernoulli's equation

Question

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 $$P+\frac{1}{2}\rho v^2=\rho g h_2\longrightarrow h_2=\frac{P}{g\rho}+\frac{v^2}{2g}$$

Answer 1

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.

Answer 2

The pressure in the stream must be approximately atmospheric, which causes the two solutions to be identical.

Fluid will naturally want to accelerate down a pressure gradient; there's more force on one side than the other, so there's a net force and thus the fluid will accelerate.

Viscosity can prevent this acceleration by providing a counteracting force but it can only do this in the direction of flow, and only if there is a wall to transfer the momentum to.

So if you look the the pressure in the center of the free stream, and then track the pressure as you get closer to the air the pressure must be constant, as there is no acceleration, nor viscous pressure losses.

Then at the water/air interface the pressure difference would be determined by surface tension, but that difference would be so small as it could probably be ignored. Any other pressure difference would not be balanced by an opposing force, so it would cause acceleration, but we don't see transverse acceleration of the stream, so we know there must be a negligible pressure difference.

Answer 3

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:

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