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

[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 ,because 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 ,because 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.