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enter image description here

The image is taken from this video.

We notice that the hose in this video is always curved and never straightened. It gives us the feeling that there is always a force perpendicular to the outlet section and opposite to the flow direction applied to the fluid at the outlet section.

enter image description here

How did this force come from? As shown in the figure, the fluid flows in the hose and the hose generates pressure. Affected by the pressure, the downstream diameter of the hose is small and the upstream diameter is large. Therefore, "vena Contracta" is generated at the outlet of the hose. In this way, the pressure at the outlet section of the hose will be higher than atmospheric pressure. This pressure results in a force perpendicular to the outlet section and opposite to the flow direction. This force, together with the force generated by the bending of the hose, causes the hose to swing back and forth.

Is my explanation right?

Reference: https://en.wikipedia.org/wiki/Vena_contracta

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    $\begingroup$ @J.Murray Sorry, I want everyone to analyze only my explanation. $\endgroup$
    – enbin
    Commented Aug 22, 2021 at 3:18
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    $\begingroup$ @AlBrown Do you think the hose will produce vena? Will vena cause the force in the outlet section? $\endgroup$
    – enbin
    Commented Aug 30, 2021 at 22:10
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    $\begingroup$ @enbin Yes. Force from a tapered section happens. The force will be in the axial direction. The force will manifest at the location of the tapering, not especially right at the outlet. (For example we could imagine a tapered section in the middle.) The force would be wherever the tapering is. I calculated the magnitude in my answer $\endgroup$
    – Al Brown
    Commented Aug 30, 2021 at 22:19
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    $\begingroup$ I updated the answer to say all that in an organized way. $\endgroup$
    – Al Brown
    Commented Aug 30, 2021 at 22:33
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    $\begingroup$ @AlBrown When there is vena, does the outlet section also have a backward force? The outlet section is not atmospheric pressure, is it? If there is no vena, the outlet section will be atmospheric pressure, right? $\endgroup$
    – enbin
    Commented Aug 30, 2021 at 22:53

2 Answers 2

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In the video, by the looks of it, the tube is overall hanging down. Even when wriggling the general orientation is still down.

I don't think the vena contracta aspect makes any difference here.

How I would approach this:
I imagine the water flow initially very slow, and the rate of flow is gradually increased, to explore the entire range of rates of flow.

The lower the flow rate the lower the exit velocity of the water.

When water exits a nozzle (more generally, when any fluid/gas exits a nozzle) there is a recoil force. If the nozzle would be secured to some cart then the recoil force would propel the cart.

At low flow rate the force of gravity is larger than the recoil force at the nozzle. The larger the flow rate, the larger the recoil force. At some flow rate the recoil force will be enough to lift the nozzle against gravity.

The tube is flexible, so the tube bends. That makes the nozzle point sideways. With the nozzle pointing sideways the tube is moved sideways, and gravity has opportunity once again to pull the nozzle down.

The result is that the nozzle is violently swinging from side to side, flexing the tube all the time.

In addition there is the inertia of the water flow in the tube itself. When the water is forced to move along a bend in the tube the inertia of the water will tend to straighten the bend, contributing to the overall chaos of the motion.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Buzz
    Commented Aug 23, 2021 at 3:01
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    $\begingroup$ @Dale Your so-called straight nozzle does not exist in reality, so it can be said that any straight nozzle has vena Contracta. Therefore, it cannot be said that the vena Contracta will not cause the reaction force of the straight pipe. It can be said that the same is true of diverting nozzles. $\endgroup$
    – enbin
    Commented Aug 23, 2021 at 3:06
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Intro/Summary

The exiting per se does not provide force. But this only refers to crossing the boundary from just inside the hose to just outside. The diameter tapering does provide force inside the hose. Could “tapering” outside the hose cause pressure and force?

Bends, and changes in diameter.. both cause forces on the hose for the reasons discussed below.

Finally, I ultimately agree that there is a region outside the hose that’s above atmospheric. Understanding that might be easier after understanding tapering forces. So consider those first. They are also important because they affect the hose at the end as the tapering is located there.

Forces from Internal Diameter Changes

Bends push the hose out, out along the direction of the radius of the bend. Reductions in diameter push the hose backward, in the axial direction opposite to the flow direction.

Both forces come from changing the velocity of the fluid. Changing velocity is acceleration, $\vec{a} = \frac{d \vec{v}}{dt}$. To accelerate the fluid requires force, $F=ma$. The reaction force to accelerate the fluid is on/from the hose. This force tends to move the hose of course.

Force from a tapered section will occur at the location of the tapering, not especially right at the outlet. (For example we could imagine a tapered section in the middle. In that case the force would appear at the middle.) The force will be wherever the tapering is. I calculate the magnitude below:

Imagine if you go fairly quickly from diameter of 1 to diameter of 0.5. The fluid velocity increases fourfold. This change in momentum is similar to accelerating a fluid around an elbow-bend, or a curve - except axial.

$$F= (\frac{dm}{dt}) ~ (\Delta V) = (V_i \rho A_i) ~ (V_f-V_i)$$

$$ = (\rho Q) ~ (3V_i)= \frac{3 \rho Q^2}{A_i}$$

$Q$ being volumetric flow rate, like gpm. That was for $A_f=\tfrac{A_i}{4}$. More generally for incompressible:

$$ F =( \tfrac{1}{A_f} -\tfrac{1}{A_i} ) \rho Q^2$$

And that axial force is distributed along the length of the taper where the force added per unit length $\tfrac{dF}{dz}$ is of course higher with more rapid tapering (with higher $\tfrac{dA}{dz}$), but is also higher at lower areas, and higher flows. It’s actually proportional to $\tfrac{dA}{A^2}$, which can be seen by rewriting the previous as:

$$ F =( \frac{A_i-A_f}{A_f~A_i} ) \rho Q^2 \approx ( \frac{A_i-A_f}{A^2} ) \rho Q^2$$

Pressure Region beyond Hose

To consider your explanation, people should note we are focusing on the section of fluid outside the hose. Not the hose tapering leading up to that. Although a common and usually safe assumption, it is not a law of physics that dynamic, especially colliding, unconfined fluids will always be at atmospheric pressure. (See this answer, especially the prose including note 2. At a minimum it is a solid counterexample to pressure having to stay atmospheric.)

The pressure of a fluid at the boundary with the atmosphere is normally 𝑃𝑎𝑡𝑚, but it is not true when the fluid is being accelerated. Here the water is accelerating radially. This motion requires a pressure difference to cause it. We know it’s accelerating radially because when leaving the nozzle, it has a velocity component in the direction of lower radii, but this is less volume ($\frac{ \partial (\pi r^2 L)}{\partial r} >0$), which is why $v$ increases at a taper generally. So in the external vena, one or both of the following must happen, quickly:

  1. The radial component of velocity must be quickly changed, decreased.

  2. The jet must get faster axially.

This is to balance mass flows upstream and downstream of the point where the jet’s diameter stops decreasing. That requires acceleration which requires force which requires pressure gradient. Of course water jets always change shape, but their velocity vectors don’t normally change anywhere nearly this quickly.

Yes absolutely it is above atmospheric. This is similar to the two jets colliding. Even knowledgeable people will argue that it must be atmospheric outside of the hose, but we’ve already seen that’s untrue for colliding. So I do think there is a force transmitted along the fluid outside the hose.

In summary, I think your explanation is right and interesting. But the axial force inside the hose matters too.

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  • $\begingroup$ @enbin For incompressible. Finally, this is for force, using momentum conservation. Bernoulli uses energy conservation and more about fluid properties. I dont think pressure at outlet is the way to think of it. Is right now. My original answer was opposite of the case. Nuther good q. Force is in contracting section; outlet doesnt make force. Changing the velocity of a fluid (in a bend or contraction or anything) is acceleration ($ \vec{a}=\frac{d \vec{v}}{dt}$) and requires some force on the fluid, and reaction force by the hose. $\endgroup$
    – Al Brown
    Commented Aug 30, 2021 at 19:10
  • $\begingroup$ @AIBrown Pure straight pipe has no reaction. Straight pipe with decreasing diameter has reaction, right? $\endgroup$
    – enbin
    Commented Aug 31, 2021 at 22:15
  • $\begingroup$ Do you think the axial force inside the pipe is exerted on the pipe or on the fluid? $\endgroup$
    – enbin
    Commented Sep 1, 2021 at 5:32
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    $\begingroup$ @enbin it is on both. Just like the bending hose. The hose has to provide the reaction force that accelerates the water. $\endgroup$
    – Al Brown
    Commented Sep 1, 2021 at 5:33
  • $\begingroup$ Can we create a pipe without reaction? $\endgroup$
    – enbin
    Commented Sep 1, 2021 at 20:37

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