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Disclaimer: I’m not trying to violate the conservation of energy here; I know that’s impossible for us mortals. I think I’m either wrong or not considering something, I just want someone to point it out in as nice a way as possible so I can learn from my mistake.

So, my question originally started when I learned that as wind velocity doubles, power is cubed as given by:

$$ P = \frac1 2 \cdot v^3 \cdot r^2 $$

Whilst considering this I found the notion a little difficult to grasp, my conclusion to begin with was that if velocity doubles so does the energy. I know this is connected and explained by newtons second law of motion but I don’t necessarily have a comprehensive understanding of that although I accept it as fact.

My trouble really started when I began to consider funnels, So, in a funnel, if the inlet is twice the area of the outlet and the air velocity is 5 knots at the inlet, then the outlets air velocity should be 10 knots. I checked that… and I’m 99% certain that’s correct.

Then I figured if I generated power without a funnel with a turbine, I would get a portion of that power, according to Betz limit no more then 59.3% of the kinetic energy. However, if I funnelled that air, I wondered could I get more power without violating Betz limit. By my (very possibly incorrect) calculations you can and a lot more.

I got the formula for calculating wind power from turbines from here: https://www.raeng.org.uk/publications/other/23-wind-turbine

$$ P=\frac 1 2\cdot Density\cdot (radius^2*\pi)\cdot velocity^3 \cdot \text{Power Coefficient} $$

So…

With a swept area of 10m and a velocity of 10m/s with a power coefficient of 0.4 = 2459 Watts.

With a swept area of 5m and a velocity of 20m/s with a power coefficient of 0.4 = 9838 Watts.

With a swept area of 1m and a velocity of 100m/s with a power coefficient of 0.4 =245,975 Watts

If this math is correct (I have my doubts), funnelling a given air flow at 10m/s, the inlet 10 metres squared and the outlet 1 metre squared you can increase your power yield by 100 times. This sounds very wrong. I’m almost sure this is wrong.

To conclude my confusion, If the air speed is 10 m/s and you have three turbines in an enclosed space with one turbine active at a time.

Turbine placed at A yields = 2459 Watts

Turbine placed at B yields = 9838 Watts.

Turbine placed at C yields = 245,975 Watts

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So, I know on this website you like for users to show how they’ve tried to work through homework like questions before asking questions some might consider to be silly. I did my best. I googled the advantages and disadvantages of ducted wind turbines and found some interesting articles about systems like the Invelox turbine.

enter image description here

The inventors claimed there were significant advantages to using ducted wind turbines especially in areas with low windspeed but the overall consensus was that the cost of per unit power was too high to give these systems a significant hold over the conventional wind turbine.

This made me wonder if my math was actually correct, these systems were better they just weren’t better from an economic stand point. Given that my math told me these systems could be 100 times more powerful. I scrapped that idea pretty fast.

My next point of call was Bernoulli’s equation, which makes the observation that as fluid velocity increases in a funnel, pressure decreases, however I know these pressure decreases can be very insignificant when dealing with a fluid like air and applies much more appropriately to water which has 1000 times the density of air. So, I don’t think its that either.

My next thought was perhaps the skin friction in funnel was so significant that it robbed the wind of its power at an ever-increasing rate as it travelled further into the funnel. I know from other things I’ve done that the friction losses in funnels aren’t that bad when it comes to air, I’m referring to Ansys fluent simulations with funnels.

Perhaps my understanding of the relationship between power and energy is incorrect. In my mind this understanding would violate the conservation of energy because the same air flow is yielding 100 times the joules per second which sounds impossible.

I’m really at a loss at this point, my best guess is that my calculations are simply incorrect or taken out of context or perhaps air struggles to enter the proposed funnel for some reason I don’t understand. Hopefully someone can explain this to me because I’m determined to exit this rabbit hole somewhat enlightened.

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    $\begingroup$ You finally got there in the end. Yes. air struggles to enter the proposed funnel. If you read about the Betz limit, then you should be familiar with the actuator disk model and the pressure and flow assumptions at the disk itself. The funnel can be regarded as an actuator disk - it impedes flow like any semipermeable disk would. So you can use a ducted fan. Theoretically, if the duct and fan combined have the ideal Betz permeability, they would also have the same Betz limit. This means the fan in the duct, assuming accelerated air, must have a lower pressure drop than it would in open flow. $\endgroup$ – Phil Sweet Sep 28 '20 at 20:25
  • $\begingroup$ It's not necessarily a bad idea, though, it just doesn't get you any free power. $\endgroup$ – Phil Sweet Sep 28 '20 at 20:28
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    $\begingroup$ The flaw in your logic is that there is nothing to increase the outlet velocity of the air, except for energy coming from the air itself. What actually happens is that the outlet velocity doesn't increase, but the inlet velocity reduces and the "excess" air flows round the outside of the funnel instead of going through it. $\endgroup$ – alephzero Sep 28 '20 at 21:43
  • $\begingroup$ So not only is the funnel going to not increase the power but it will significantly reduce it. Or will it remain the same? Thanks for the answers by the way. $\endgroup$ – David Nolan Sep 28 '20 at 21:51
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    $\begingroup$ The other thing you need to consider is every time the air changes flow diameter ( restrictions or expansions) or changes direction (the 90 degree bend) shock losses occur & the air loses some kinetic energy. Also, the longer the duct the more energy the air loses because of skin friction (roughness). $\endgroup$ – Fred Sep 28 '20 at 22:09

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