# The water ejected from the two hoses collides in the air. Will the pressure of the collided water be higher than atmospheric pressure?

The water ejected from the two hoses collides in the air. Will the pressure of the collided water be higher than atmospheric pressure? I think it will be higher than atmospheric pressure, and this pressure is also related to the angle and speed of the collision. But as long as it collides, it will be higher than atmospheric pressure. Am I right?

• I edited that more. Just saw all the comments a minute ago. Only changes are the “edit” section added in the beginning, and note 1 Commented Aug 19, 2021 at 22:52

Edit: Rather than argue in comments, I will note here that a dozen or so comments are arguing that the answer is something that doesn’t even provide enough force to stop the jets.

Clearly we cannot have the following facts:

A. The max pressure is $$P_{max}=\rho V^2$$.

B. It decreases radially from there.

C. There is sufficient force to stop the stream axially.

These cannot all be true because $$F =\int P dA$$ wouldnt be enough to stop the axial momentum. There is no axial impulse transfer at any radii above the jets’ original radii, at $$r>R$$.

Here is prof Skakya discussing a jet hitting a wall: https://youtu.be/kNL5w73owzU

Finally, it is true that pressure of a fluid at the boundary with the atmosphere is normally $$P_{atm}$$, but it is not true when the fluid is being accelerated. The entire cross section, including at $$r=R$$, is accelerating axially. This motion requires a pressure difference to cause it. Because the acceleration is uniform across the area, the pressure driving it is. (As is standard even in pretty precise engineering situations, this analysis does indeed omit pressure-driven radial acceleration, which is very low compared to axial, as the outflow circle has high diameter and does not need high velocity. And more importantly the flow goes to higher areas adding the $$1/r$$ term which is high at low $$r$$. But that only increases pressure a little at various points; it cannot eliminate or reduce the pressure that’s needed at every point of the cross section to stop the fluid in the axial direction.)

I think what’s being missed is that the entire cross section of the stream hits the entire cross section of the other stream, and the axial impulse (meaning assumed-instantaneous momentum transfer) creates the pressure. This can be solved by Bernoulli also, which the other answer correctly did, only to dismiss the result. I also added the new, second half to note 1. Remainder unchanged.

End of edit

Yes. If we assume two identical jets with cross-sectional area $$A$$ are traveling in opposite directions, and they hit and stop due to the impact (which I think is reasonable* since it is water), then the pressure is providing the force to transfer momentum.

The force to stop a column of water (from $$v=V$$ to $$v=0$$) of length L is $$F= (m) (\frac{dv}{dt})=(\rho L A)(\frac{\Delta v}{\Delta t})$$ $$= (\rho L A)(\frac{V}{\tfrac{L}{V}}) = \rho A V^2$$ $$P=\frac{F}{A}=\rho V^2$$

Of course $$L$$ was arbitrary and drops out. (I used $$L=V \Delta t$$ to get the denominator.) Note that the area doesn’t matter either. Two low-radius jets and two high-radius jets will have the same pressure.

Notes:

1. *Incidentally that would not be reasonable at all if a gas. In fact gas jets hardly interact at all. If you point a fan at your face and point one perpendicular to it, where the second one isn’t hitting you at all, then you will notice very little difference with and without the second fan - unless you put the crossing fan right against the main stream, but that will be a change due to room pressure gradient dynamics not streams colliding. (I softened this statement which previously almost implied there will never be any difference. Secondly, it will not be true if you use a hand fan as the crossing fan because that creates large, discrete pressure waves.)

2. Fluids technically do not, in any normal sense, exert impact or weight forces directly and/or aside from pressure force. Fluids only exert force with pressure, usually pressure difference.

• This result is not correct. The area is not A at the collision interface, so the pressure you calculate is not even the average pressure over the contact plane (which has fluid spread vertically over its entire surface). The most you can say is that the impact force is $F=\rho V^2 A$. The pressure is maximum at the centerline of contact, but drops off with radial position away from the centerline. Moreover, at the free surface between the fluid and the atmosphere, the pressure is, of course, atmospheric. Commented Aug 18, 2021 at 11:28
• Furthermore, along the center streamline, the fluid velocity at the impact plane is zero, so, from the Bernoulli equation, the pressure at the centerline at the impact plane is only $\frac{1}{2}\rho V^2$. This confirms that the "effective" area of contact is > A. Commented Aug 18, 2021 at 12:06
• I'm also very skeptical of your first footnote. In fact, I just tested it; I don't have two desk fans, but pointing an electric desk fan at my face and waving a hand-held paper fan near (but not in) the airstream definitely diverted it away from my face. Anyway, your claim that "gas jets hardly interact at all" should be easy to conclusively test with two smoke machines and a video camera. And I'd be surprised if someone hadn't already made such a video, although my web search skill haven't turned one up yet. Commented Aug 18, 2021 at 12:31
• @enbin: No, because the flow is not being stopped at the outlet of the hose; it continues on, and in fact would continue into infinity at constant speed were there no subsequent viscous friction with the surrounding air, a whole different phenomenon. Not to mention gravity bending it down, but we're neglecting that here. Commented Aug 18, 2021 at 14:37
• I strongly support what @RalfB is saying, and contend that you are the one who is wrong. Commented Aug 18, 2021 at 18:54

## Jet hitting a frictionless wall

The idealized situation here is fully symmetrical; we can replace it with a single jet hitting a (frictionless) wall, a boundary condition of zero axial velocity. Each jet may be considered to have a constant uniform axial velocity transverse to the wall (actually, the jet will interact with the surrounding air, entraining it---but we can neglect that). At the wall, i.e. the meeting plane of the jets, there will be no axial velocity at all. At the center of this plane, there will be a stagnation point---no velocity at all, a zero vector.

Comparing the energy of the flow in the free jet with the condition at this stagnation point, we see that all the dynamic pressure (a measure of kinetic energy, loosely speaking) of the flow has been converted to static pressure. The classic equation describing it is p0 = p + rho * u^2 / 2, where p0 is stagnation point pressure. So , the pressure at the center of the meeting plane will increase by exactly rho*u^2 / 2 with respect to the center of the free jet (where ambient atmospheric pressure may be assumed). Note that rho is the density of the liquid, 1000kg/m^3 for water, and u is the axial speed of flow in one jet (not the sum of their velocities).

• You say that A. the max pressure is rho V^2. B. It decreases radially from there. C. There is sufficient force to stop the stream axially. |||| These cannot all be true because $F =\int P dA$ wouldnt be enough to stop the axial momentum. Here is prof Skakya discussing a jet hitting a wall youtu.be/kNL5w73owzU Commented Aug 19, 2021 at 22:06