Can the velocity in a funnel be higher than in freestream?

[John Mcrain]

Downstream of this object, the pressure is necessarily equal to the surrounding pressure. If the velocity is then higher than the free stream velocity, the total pressure has indeed increased.

Yes but topic-question was ,is speed in narrow part of funnel higher than freestream,not downstream of object..

 

[Arjan82]

If it is higher in the narrow part of the funnel, it is also higher downstream of it, due to inertia.

 

[John Mcrain]

If it is higher in the narrow part of the funnel, it is also higher downstream of it, due to inertia.

Yes I noticed that " inertia-paradox" in some of my posts..
But if outlet of funnel is in wake,than we get favorable pressure gradient to accelerate flow inside narrow section..

 

[Arjan82]

Ok, put a surface at the inlet of the funnel (the thick part). Measure the average velocity on that surface, also measure the massflow through that surface. Now do the same measurement at the outlet of the funnel. Necessarily, as per the continuum equation, the massflow at both surfaces need to be equal and the average velocity at the outlet surface of the funnel is higher than at the inlet surface, so there is indeed an acceleration of the flow inside the funnel.

However, the maximum velocity leaving the funnel is necessarily lower than free stream as per my earlier arguments. This means that the average flow velocity is definitely lower than free stream. This means that at the inlet the flow is blocked. So the flow is already lowering its velocity before it enters the funnel...

 

[Arjan82]

These surfaces I mentioned are virtual by the way, they do not affect the flow.

And just to be clear, the average velocity on the inlet surface is way lower than free stream (it seems you might be thinking it is equal to free stream). It is in fact so low, that even after acceleration through the funnel, the exit velocity of the narrow part is still lower than free stream.

 

[John Mcrain]

These surfaces I mentioned are virtual by the way, they do not affect the flow.

And just to be clear, the average velocity on the inlet surface is way lower than free stream (it seems you might be thinking it is equal to free stream). It is in fact so low, that even after acceleration through the funnel, the exit velocity of the narrow part is still lower than free stream.

If you work with CFD you can't post some pictures of this,if it is not complicated so much to do..
I have pitot tube and one big funnel so I can try this with car test..

 

[A.T.]

Increasing the flow speed to beyond the free stream velocity with an object of any shape without it adding energy to the flow, disregarding local flow accelerations (i.e. behind the object), is a perpetuum mobile...

But question is about local flow speed, not about a net increase overall.

 

[Arjan82]

If you want to increase the net flow overall, you at least need to increase the local flow speed right...?

 

[A.T.]

If you want to increase the net flow overall, ...

I don't. The question is only about the local flow through some kind of funnel, compared to freestream.

 

[Arjan82]

Oh, wait... your argument was the other way around

So, all streamlines entering the funnel come from an equal source, i.e. the total pressure (or the Bernoulli constant) on any streamline entering the funnel is equal. Since increasing total pressure means adding energy, nowhere, on any streamline does the total pressure increase. So when the static pressure has recovered to the free stream value just behind the funnel, and since the sum of static and dynamic pressure (i.e. kinetic energy) is the total pressure, the kinetic energy on any of the streamlines can be at most what it was before entering the funnel, i.e. it can at most be the free stream value.

The flow inside the narrow part of the funnel continues downstream of the funnel if there are no losses. This is because the geometry of the narrow part of the funnel nor the open space behind it will decelerate the flow (it doesn't constrict the flow in any way).

 

[Arjan82]

If you work with CFD you can't post some pictures of this,if it is not complicated so much to do..
I have pitot tube and one big funnel so I can try this with car test..

I might have a look this evening. We indeed have a bit of code that can do this quickly. It will not be an exact funnel shape, more actually like the second configuration you showed.

 

[A.T.]

So when the static pressure has recovered to the free stream value just behind the funnel, ...

The general question is not about behind the funnel, but anywhere within some kind of funnel. Consider a wind turbine duct shown below?

From: https://wes.copernicus.org/articles/3/919/2018/wes-3-919-2018.pdf

 

[John Mcrain]

The general question is not about behind the funnel, but anywhere within some kind of funnel. Consider a wind turbine duct shown below?

View attachment 275794

View attachment 275795
From: https://wes.copernicus.org/articles/3/919/2018/wes-3-919-2018.pdf

from text " It was observed that the velocity at the duct inlet face was about 10–15 % above the upstream reference value across the entire speed range. "

 

[Arjan82]

Yes, and "a handheld anemometer was positioned at various locations in the flow field to gain some understanding of the velocity field near the rotor" So these are local flow velocities.

I explicitly excluded local flow accelerations, which can indeed be higher than free stream (in which case the pressure is lower than static free stream pressure, to keep total pressure in check). But also in the paper the average flow velocity, averaged over the entire inlet, cannot be higher than free stream.

Personally, I think this paper is bullocks... They claim significantly higher than Betz efficiency (2.65 times more!) which is highly controversial. With a duct it might be possible to increase the effective diameter of the rotor a bit, such that in theory you could exceed Betz if you base it on the rotor diameter (instead of the total diameter of rotor + duct). In practice you would likely still be below Betz because most turbine don't exceed Cp = 0.5. But in no way can this lead to a factor of 2.65. This thing is also a far cry from the funnel, without a any turbine in it, of the OP.

So, can you have a local acceleration of the flow at certain parts inside a duct exceeding free stream? Sure, that kind of diverging shape this might be possible.

But the OP actually showed a funnel. A funnel has a narrow part which has either constant cross section or is contracting even further. In that case there will also not be local flow inside the funnel with a velocity higher than free stream.

I also implicitly assumed the simple case of a narrow part with constant cross sectional area. If there are no losses, then the flow inside this narrow part, once established, continues forever, also downstream of the funnel. This means that if the local flow velocity inside the narrow part exceeds free stream, it will also do so downstream of the funnel.

Lastly, you are now talking about very local effects. Nowhere in the OP or in any subsequent posts of him did I understand that he would like to know about these kind of effects (which depend on many things, for which indeed no general statement can be made). I have interpret them as at the very least averaged over the entire cross section.

 

[A.T.]

I have interpret them as at the very least averaged over the entire cross section.

Lets say you vary the diameter of the opening in the duct shown below. Can the flow speed, averaged over an entire cross section of the opening, ever be greater than freestream speed?

 

[Arjan82]

Ok, let me first say that we go off in a bit of a tangent here, this configuration is a far cry form the OP's funnel.

I also have to come back on the paper, it is less bullocks than I thought when reading it a bit better. They don't claim exceeding Betz by 2.65, they even explicitly state that others do make this claim which doesn't help their cause. They also note that the configuration with duct does not exceed Betz if you take the outlet diameter of the duct as reference (which, as I mentioned in my previous post, is indeed what I think you should do), so that's good. Their efficiency is still higher compared to an open turbine, which sounds plausible to me. So sorry for the confusion there.

I'll be honest, I do not understand how the averaged inflow velocity to a ducted turbine as shown in this paper can exceed free stream. But that is exactly what the paper claims by measurement and computation. So I'm a bit lost, I would certainly want to understand this better.

 

[John Mcrain]

I'll be honest, I do not understand how the averaged inflow velocity to a ducted turbine as shown in this paper can exceed free stream. But that is exactly what the paper claims by measurement and computation. So I'm a bit lost, I would certainly want to understand this better.

If they exceed freestream speed ,that doesn't mean they exceed total pressure of freestream.
If speeed goes up then static pressure goes down and total pressure is same again.

Why do you generally think that incereasing speed is perpetuum mobile?

 

[John Mcrain]

Lets say you vary the diameter of the opening in the duct shown below. Can the flow speed, averaged over an entire cross section of the opening, ever be greater than freestream speed?View attachment 275796

I think for sure can be.
This is how F1 diffusers work,airflow under car bottom is faster then freestream,static pressure drop and you get dowforce meassured in tons!
But F1 diffuser stay in opposite direction as my funnel in this topic..

 

[Arjan82]

If they exceed freestream speed ,that doesn't mean they exceed total pressure of freestream.
If speeed goes up then static pressure goes down and total pressure is same again.

Why do you generally think that incereasing speed is perpetuum mobile?

Indeed. Speed goes up, static pressure goes down. No perpetuum mobile there. I did not claim a perpetuum mobile with generally increasing speed. I claimed perpetuum mobile when, after pressure recovery (i.e. pressure equal to surrounding pressure) aft of whatever you are looking at which is not generating energy, the velocity is still higher than free stream. Also for the turbine of the paper this is not the case.

 

[Arjan82]

I think for sure can be.
This is how F1 diffusers work,airflow under car bottom is faster then freestream,static pressure drop and you get dowforce meassured in tons!
But F1 diffuser stay in opposite direction as my funnel in this topic..

Conclusion, if you invert your funnel (and make the expansion gradual enough) then indeed the flow inside the narrow section (now upstream) can be higher than free stream .

 

[Lnewqban]

... bonh3ad wrote " There will almost always be some inviscid core flow through such a funnel in which Bernoulli is perfectly applicable" as respons to Lnewquban post. So he implies that using Bernulli I will get answer...

Lnewquban wrote that speed will be higher then freestream(opposite from yours),bonh3ad didnt complain nothing about it,he just add it that I can use bernulli for core flow..

You could imagine both extremes:
1) Outlet having same area as inlet.
2) No outlet.

Naturally, molecules of air will always try moving from volume of high concentration to lower concentration.
There is energy moving the funnel, which movement disturbes the homogenous concentration of the air molecules in its path.

Some molecules will get trapped inside the funnel and will be forced to occupy a decreasing volume (higher concentration).
Outside the funnel, downstream the inlet, the trapped molecules are missing; therefore, less concentration.

Trying to re-establish balance of concentrations, some molecules will rush trhough the outlet, if they can.
Some will go back the inlet, trying to find an easier path out.
The shape of the funnel and its speed trough relatively calmed air has a lot to do with how many molecules take each path.

All of them will overflow if the outlet is closed, none, if the outlet and inlet have same cross-section area: there will be stagnation inside the funnel.

 

[boneh3ad]

Increasing the flow speed to beyond the free stream velocity with an object of any shape without it adding energy to the flow, disregarding local flow accelerations (i.e. behind the object), is a perpetuum mobile...

But you can't do that. Your assumption here is that the external pressure in the free stream is equal to the external pressure at the exit, and the shape of the funnel would seem to imply that this is a poor assumption in general. If, as is probably the case, the pressure in the funnel's wake is lower than the free stream, then the exiting flow can certainly be faster than the free-stream flow without breaking conservation of energy rules. At that point the question will be just how much total pressure is lost inside due to viscous dissipation.

The bottom line here is that it's an ill-posed question. There is a lot more going on here than what meets the eye (as is often the case in fluid dynamics).

 

[Arjan82]

But you can't do that. Your assumption here is that the external pressure in the free stream is equal to the external pressure at the exit,

I do not assume that for this perpetuum mobile statement since this assumption is not necessary (I did however assumed it in other posts), but it is however usually quite accurate. It is not necessary because the pressure will recover to its free stream value at some point. The statement I am making is only valid if the pressure has indeed recovered to free stream pressure.

If you think that this takes a lot of time/space to recover the pressure, look at this post of mine:
https://www.physicsforums.com/threads/sorry-another-bernoulli-equation-paradox.995389/post-6412522

The case is not the same, you have a high pressure area connected with a low pressure area via a tube. The pressure at the centerline is plotted in a graph. There you see that the pressure is equal to the surrounding pressure pretty much at the exit of the tube. This is also an assumption that is often applied in Engineering problems of piping ending in large containers or other open spaces when determining flow rates. I think this must also happen for the funnel in open flow.

and the shape of the funnel would seem to imply that this is a poor assumption in general.

I don't see why this is the case. The funnel has a piece of straight tube at the end (at least that's how I interpreted the shape) which makes that the flow can straighten out. I think this shape is actually very suitable for the assumption of pressure equal to surrounding pressure at the outlet. A very large diameter compared to the length of the funnel would make it more problematic.

If, as is probably the case, the pressure in the funnel's wake is lower than the free stream,

For a diverging duct I can see why the pressure could be lower at the exit, since for this case indeed the pressure inside the duct must be lower than its surrounding. However, for a converging duct, or a funnel, I don't see how any location within the funnel can have a pressure lower than the surrounding. The flow entering the funnel must be accelerated and pushed radially inward, it takes pressure to do this.

Along the outside however the flow must converge inward, and there you will have some low pressure areas. And maybe at the lip/edge at the inlet of the funnel, very locally. But definitely not in the narrow section.

 

[John Mcrain]

Conculusion is: it seems we cannot agree on the final answer..

 

[boneh3ad]

I do not assume that for this perpetuum mobile statement since this assumption is not necessary (I did however assumed it in other posts), but it is however usually quite accurate. It is not necessary because the pressure will recover to its free stream value at some point. The statement I am making is only valid if the pressure has indeed recovered to free stream pressure.

I am not claiming that the pressure coming out of the funnel is lower than the ambient pressure it encounters. That only makes sense in a compressible flow. Listen to what the question is asking. "Can the exit velocity be higher than the free stream?" The answer is yes. If the external pressure at the outlet is less than the pressure in the free stream (due to interaction with the body), then it breaks no rules for the exit pressure to be below that in the free stream and the velocity can be higher than that in the free stream provided there is not too much dissipation (which is geometry specific).

Of course the pressure will eventually (and fairly rapidly) return to equilibrium, but it's not immediate.

I don't see why this is the case. The funnel has a piece of straight tube at the end (at least that's how I interpreted the shape) which makes that the flow can straighten out. I think this shape is actually very suitable for the assumption of pressure equal to surrounding pressure at the outlet. A very large diameter compared to the length of the funnel would make it more problematic.

For a diverging duct I can see why the pressure could be lower at the exit, since for this case indeed the pressure inside the duct must be lower than its surrounding. However, for a converging duct, or a funnel, I don't see how any location within the funnel can have a pressure lower than the surrounding. The flow entering the funnel must be accelerated and pushed radially inward, it takes pressure to do this.

Along the outside however the flow must converge inward, and there you will have some low pressure areas. And maybe at the lip/edge at the inlet of the funnel, very locally. But definitely not in the narrow section.

That's not what I said. There is a distinction here between free-stream pressure and the local external pressure at the outlet. The pressure coming out of the funnel will be equal to the external pressure at the outlet, but that is not necessarily equal to the free-stream pressure.

 

[Arjan82]

I am not claiming that the pressure coming out of the funnel is lower than the ambient pressure it encounters. That only makes sense in a compressible flow. Listen to what the question is asking. "Can the exit velocity be higher than the free stream?" The answer is yes. If the external pressure at the outlet is less than the pressure in the free stream (due to interaction with the body), then it breaks no rules for the exit pressure to be below that in the free stream and the velocity can be higher than that in the free stream provided there is not too much dissipation (which is geometry specific).

This I agree with. As long as everyone realizes that this geometry would look quite different than a funnel. Also, this higher than free stream flow seizes to exist the minute the pressure is recovered to free stream pressure, which, as you also mention, is rather quick.

That's not what I said. There is a distinction here between free-stream pressure and the local external pressure at the outlet. The pressure coming out of the funnel will be equal to the external pressure at the outlet, but that is not necessarily equal to the free-stream pressure.

Ok, this I also agree with. But this is a local effect. Again, the pressure recovery would be rather quickly.

Now I can also place your earlier remark a bit better, that external flow is important. Because if the pressure at the outlet of the funnel is not (close to) ambient, it is mainly due to the flow around the funnel.

So, I was answering the following question: if you put a funnel out of the window of a car, would you expect the flow coming out of it to be higher than free stream. What I then see as a potential error in thinking about this configuration is that people think the flow is forced through the funnel and because of the contraction, there will come a jet out of the funnel blowing downstream (higher than free stream velocity), analog to a nozzle at the end of a water hose.

Since this would definitely not happen, I was very keen to point out that the flow exiting the funnel cannot be higher than free stream (and I thus mean after pressure recovery, which is not very far downstream of the exit of the funnel) and is much more likely to be lower than free stream. This is somewhat counter intuitive since you see a contraction which would instinctively result in acceleration of the flow and a jet.

 

[A.T.]

Conculusion is: it seems we cannot agree on the final answer..

The funnel configuration shown in the OP is not optimal for maximizing the flow in the funnel. Turning it around and/or replacing the thin walls with airfoil shapes might achieve an average flow speed in a funnel cross section greater than freestream. But there is likely no extended jet of faster than freestream air coming out of the back.

 

[John Mcrain]

But there is likely no extended jet of faster than freestream air coming out of the back.

bonhad shows that probably is ,becuase of low pressure in wake.

 

[A.T.]

But there is likely no extended jet of faster than freestream air coming out of the back.

bonhad shows that probably is ,becuase of low pressure in wake.

He says that it would not violate conservation laws. Not that it probably happens with your setup.

 

[256bits]

I'll be honest, I do not understand how the averaged inflow velocity to a ducted turbine as shown in this paper can exceed free stream. But that is exactly what the paper claims by measurement and computation. So I'm a bit lost, I would certainly want to understand this better.

Because an area of incoming air larger than the inlet area of the duct is entering the duct.
The air has to speed up to maintain the same mass flow, and its pressure has to drop in tandem.
You can consider the interface between the air that does not enter and that which does enter the duct as being an 'imaginary' extended surface, or part of your control volume. Air crosses the control volume at a surface perpendicular to the velocity of the incoming air, with that surface somewhat parallel to the duct inlet area.

It should be the same principal for a venturi.
Upstream has velocity P1, V1 and since theoretically there are no losses, the downstream alos has P1, V1.
The venturi throat has P2, V2 where P2<P1 and where V2 > V1.
The throat of the venturi area has a velocity greater than freestream.

Consider two funnels being placed to resemble a venturi, and one gets the idea.

 

[Arjan82]

Because an area of incoming air larger than the inlet area of the duct is entering the duct.
The air has to speed up to maintain the same mass flow, and its pressure has to drop in tandem.
You can consider the interface between the air that does not enter and that which does enter the duct as being an 'imaginary' extended surface, or part of your control volume. Air crosses the control volume at a surface perpendicular to the velocity of the incoming air, with that surface somewhat parallel to the duct inlet area.

It should be the same principal for a venturi.
Upstream has velocity P1, V1 and since theoretically there are no losses, the downstream alos has P1, V1.
The venturi throat has P2, V2 where P2<P1 and where V2 > V1.
The throat of the venturi area has a velocity greater than freestream.

Consider two funnels being placed to resemble a venturi, and one gets the idea.

Alright, this was the kind of reasoning I was trying to caution the reader about the whole time .

First, what I didn't understand were diverging ducts, you are talking about a contracting duct / funnel. (I thought about it a bit better now, and they start to make sense to me now)

What you've proven in your argument is that V2 > V1, that I totally agree with. But what you have not proven is that V2 > V_freestream... My point is that V1 is much lower than freestream, the flow is blocked. It is so much lower than freestream that even V2, being bigger than V1, is still lower than, or at best equal to freestream.

 

[256bits]

Alright, this was the kind of reasoning I was trying to caution the reader about the whole time .

First, what I didn't understand were diverging ducts, you are talking about a contracting duct / funnel. (I thought about it a bit better now, and they start to make sense to me now)

What you've proven in your argument is that V2 > V1, that I totally agree with. But what you have not proven is that V2 > V_freestream... My point is that V1 is much lower than freestream, the flow is blocked. It is so much lower than freestream that even V2, being bigger than V1, is still lower than, or at best equal to freestream.

Thanks for your remarks.
I can't really comment on your reply as it leads away from the body of my discussion.

Hopefully the OP has had enough replies to his query.