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

[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.