Does the Bernoulli Effect Apply to the Jet Reaction Force in a Leaking Bucket?

[Danger]

A 10 m deep bucket is getting pretty far afield, but regardless... the water is not pressurized relative to the top of the bucket (because there isn't one). The pressure is unevenly (but laterally symmetrically) distributed to the walls and bottom. If all of those forces are balanced, as they must be, there is no thrust. That would occur if there were an imbalance between the top and bottom, and again there is no top. Would a rocket work if you cut the top off?

 

[Danger]

they apparently define "bucket" as bucket without water.

That's pretty obvious from his enhanced translation, if not the first couple of questions.

 

[A.T.]

The pressure is unevenly (but laterally symmetrically) to the walls and bottom. If all of those forces are balanced, as they must be, there is no thrust.

The forces on the water aren't balanced if there is a hole in the bottom, so there is thrust.

 

[Danger]

so there is thrust.

Against what...?

 

[A.T.]

Against what...?

Explained many times. See for example post #23.

 

[Danger]

Explained many times. See for example post #23.

And as you said your very own self in that very same post, the thrust is upon the water, not the bucket, which is exactly what I've been saying all along. You just made my argument for me.

 

[A.T.]

...which is exactly what I've been saying all along.

Unfortunately along with some completely wrong stuff.

 

[Dale]

It doesn't matter if the thrust of the jet is directly against the water or the bucket. Either way, there is a different net force on the bucket with the jet than without.

 

[sophiecentaur]

Against what...?

I introduced Bernouli on the last page but no one took it up., What does Bernoulli say about the pressure change when a fluid flows into a narrow aperture? The pressure in the narrow bit is less than the pressure in the wide bit. So there is a higher pressure just above the hole than if there were no flow. So the rate of flow will be less than if the diameter of the bucket were the same as the diameter of the hole. (Naturally the water level would soon drop with such a narrow tube but just consider the initial situation) Is that not what's responsible for the 'mysterious' reaction force?

 

[Dale]

A 10 m deep bucket is getting pretty far afield, but regardless...

Nothing far afield whatsoever. It is just a big bucket to make the point clear. There is without a doubt a pressure greater than 1 atmosphere at the bottom of a 10 m bucket of water.

the water is not pressurized relative to the top of the bucket (because there isn't one).

Danger, this is simply factually wrong. If the top of the bucket is at 1 atm then the bottom of the bucket will be at 1 atm + the hydrostatic pressure of the water column. If you have spent any time diving or swimming then this should be clear to you.

Whether it is a hole or a nozzle, there is atmospheric pressure on the outlet and above-atmospheric pressure inside the bucket. This pressure difference causes the water to accelerate. That is what a jet is, regardless of whether the nozzle is a complicated tube or a simple hole.

Newton's 3rd law does the rest. The force that accelerates the water has a 3rd law partner which is equal and opposite. This is the thrust.

 

[Danger]

Sophie, perhaps I'm misunderstanding Bernouli; I admit that I've never designed a carburator, although I've rebuilt a few. I can't quite see how a hole in a piece of sheet metal is long enough to be considered a venturi, but I'll take your word for it. Still, my understanding is that a venturi doesn't increase the pressure ahead of it; it just converts some of it to speed. In either case, though, the water instantly reverts to the same atmospheric pressure as is experienced at the surface of the reservoir. How can thrust be imparted to the bucket in that instance? If Dale cuts the top off of his rocket bottle before launch, will it still fly?

 

[Dale]

If Dale cuts the top off of his rocket bottle before launch, will it still fly?

I never claimed that the thrust from the bucket's jet is greater than the weight, just that it is non zero.

The point of the 10 m bucket and the water rocket is to emphasize physical features that you are ignoring, but that are already present in a standard bucket with a hole.

 

[sophiecentaur]

I never claimed that the thrust from the bucket's jet is greater than the weight, just that it is non zero.

The point of the 10 m bucket and the water rocket is to emphasize physical features that you are ignoring, but that are already present in a standard bucket with a hole.

Yes. That's the whole point. The existence of a force doesn't imply it is a great force.
There is an intuitive feel about this that is misleading people. It can 'feel right' that a hole in the side of the bucket would push it sideways with a considerable force but there is no reason to think it actually would be so. So we are, in fact, only looking for a small force. (Bernoulli based) How would this force reveal itself when in the vertical direction? It would be that the measured weight (bucket hung on a sensitive balance) would be a bit less than expected (say against a volume scale on the side). The difference would be greater, the higher the water level in the bucket and would approach zero as the bucket empties. A control experiment, run beside the leaking bucket, could consist of small masses (ball bearings?) released by a subtle system of magnets of shutters from the bottom of the bucket. Without the fluid flow, the 'weight' of the control bucket would always be a bit heavier if the rate of mass release were kept the same as the jet of water.

 

[A.T.]

A control experiment, run beside the leaking bucket, could consist of small masses (ball bearings?) released by a subtle system of magnets of shutters from the bottom of the bucket. Without the fluid flow, the 'weight' of the control bucket would always be a bit heavier if the rate of mass release were kept the same as the jet of water.

That's complicated. Just fill the bucket to different marked levels, and note down the scale readings. Then open the hole and
compare the scale readings at the same levels.

 

[sophiecentaur]

That's complicated. Just fill the bucket to different marked levels, and note down the scale readings. Then open the hole and
compare the scale readings at the same levels.

I guess that could do it; small hole and bigger hole and bigger hole etc. I think the bucket with the whole bottom taken away would be subject to 'experimental problems'. The ball bearing method is a bit lumpy, I have to admit but I just wanted to eliminate the effect of the fluid dynamics.

 

[Danger]

That's complicated. Just fill the bucket to different marked levels, and note down the scale readings. Then open the hole and
compare the scale readings at the same levels.

It's nice to see that you've been paying so much attention that you attributed to me a quote that was written by Sophie. How can you even do that with a "Reply" function button?

Anyhow, now that you're fighting amongst yourselves, I'm going to get the hell out of this thread and leave you to it. I still say, however, that if this thing was in microgravity it wouldn't go anywhere. (And there is nothing in the question, if you'll notice, that says gravity exists in the experimental location...)

 

[Dale]

The whole OP wouldn't work in microgravity.

 

[256bits]

I guess that could do it; small hole and bigger hole and bigger hole etc. I think the bucket with the whole bottom taken away would be subject to 'experimental problems'. The ball bearing method is a bit lumpy, I have to admit but I just wanted to eliminate the effect of the fluid dynamics.

Your ball bearing analogy is valid, as a liquid can be considered of being composed of tiny molecular balls of mass dm. Pressure is just the energy gained by a dropping a mass from a height h, so we get mgh, and it doesn't matter if we initially drop the mass dm from the height h, or squirt the mass dm out by the pressure - the velocity at exit is the same.

The vectoring is a side-shoot. Any change in momentum, or a change in direction of velocity, will of course have a associated forces. While a stream of water coming out of a bucket and directed to the side will produce a sideways force from the velocity ( from mgh ) and direction ( just to be clear even though velocity includes direction ) of the stream being altered, both on the bucket and velocity stream, it is not so clear on intutive grounds whether an L-shaped tube with a downward exit, would would supply a or thrust on the system as a whole, any different than that from a hole exiting directly out th botom. If that were so, though, we should be able to configure a system with many twists, turns and lengths of pipe, where internal forces can defy physical laws to produce an anti-gravity machine, which no one would consider to be likely. Certainly a torque is accomplished, but that just shifts the centre of moment. An added complication is whether the pipe will droop or rise upward - ie the weight of the fluid in the pipe verus the exit momentum change.

 

[256bits]

I would guess you guys do know you are talking about the same thing, no, from different perspectives.
One is the weight of bucket as a whole which changes due to dm/dt.
The other is the force from an ejected dm/dt at the jet.

 

[A.T.]

And there is nothing in the question, if you'll notice, that says gravity exists in the experimental location...

Wrong again. The question clearly suggests that gravity exists in the experimental location:

If in the bottom of the bucket with water we made a hole, then water flows out of it.

 

[sophiecentaur]

I would guess you guys do know you are talking about the same thing, no, from different perspectives.
One is the weight of bucket as a whole which changes due to dm/dt.
The other is the force from an ejected dm/dt at the jet.

But there is a significant difference between just letting a ball drop down off the bottom (just a dm/dt) and ejecting water through a hole (with a dx/dt, as well as a dm/dt). The ball leaves the bottom at zero velocity (which was my reason for doing the ball thing) but the jet of water has velocity. If you want to get a sideways force from a dropping ball, you would need to run it down a curved track, inside, to give it a velocity - but that's not my experiment.

 

[256bits]

The ball is not just dropped off the bottom, but released from a height equal to the surface of the liquid.
Molecules are just really tiny little balls after all.

The dx/dt comes from the pressure which is a function of the height of liquid.
The exit velocity of the stream of water is √2gh, where h is the depth of water.

Similarily, for a ball dropped from a height h, the velocity of the ball will be √2gh.

 

[Wucko]

Have you tried using Bernoulli equation? Here is mine solution of the problem. The difference is I haven't used any derivations in my equation. Hope it will help and excuse me if my english is bad. And yes, there is no jet force. Hope you will get the picture

 

[sophiecentaur]

Have you tried using Bernoulli equation? Here is mine solution of the problem. The difference is I haven't used any derivations in my equation. Hope it will help and excuse me if my english is bad. And yes, there is no jet force. Hope you will get the picture

That's fine and it involves Energy conservation (when there is laminar flow and no energy loss in the fluid). If the exit velocity is (ideally) √2gh, that must involve a step change in velocity as the molecules speed up on exit. So there must be a force on each one, to accelerate it. This involves no energy transfer to the bucket as there is no vertical movement - so no Force times Distance. There is no violation of any conservation law, even though a reaction force appears. People are looking for a paradox when there is none.

If the water were to leak out of the bottom, at the same rate, via a sponge bung in a larger hole, there would be no upward force as the water would be starting from zero speed on leaving the sponge.

 

[sgphysics]

Would one not need to consider also the change ov momentum of the water insise the bucket? The change d(mv)=mdv+vdm. For a straight bucket the first term is zero, but the second is not. It would counteract the momentum at the outlet, so if the bucket was hanging in a rope, the weight woluld not be lessened by the jet, only by the loss of water.

 

[sophiecentaur]

This has all been discussed earlier but the thread is long enough for you to have missed it. There is a change of momentum of the water as it passes through the constriction. Bernouli described the effect, which accounts for an increase in velocity. This change in momentum means that an impulse (downwards) is imparted to every drop of water that is ejected. There must be an equal and opposite reaction against the water in the bucket - making it ( ever so slightly) lighter.

Is there anyone who can argue that the Bernoulli effect does not happen in this circumstance?