Wind Turbine Efficiency Discussions

[Remundo]

the demo is here
renewable-energies/intuitons-ensemble-t17615-50.html

where I write "let's start from the beginning"

I would like to see the not obvious demonstration which demonstrates V0/3 as the maximum speed of training a body in the air at V0...

that's not what I'm demonstrating.

I repeat: all bodies will be taken to V0 asymptotically regardless of their Cx. But we know that and we don't care.

What interests us is the intermediate speed where a body receives the maximum power of the fluid which pushes it. Because it is another way of formulating the optimal speed condition of a wind turbine blade in drag.

This speed is worth V0/3 independently of the Cx.

To find this result, there is a fairly long preparation to express the power received by the body as a function of its speed V1 different from V0.

Then we look for dP/dV1 = 0 and we look for an extremum, which occurs if V1 = V0/3.

These calculations are based on the expression of the drag force proportional to V², this remains very valid for Reynolds which exceed 10.

[Christophe]

Yes ok I read too quickly...but does this fit with the simulation I did for you yesterday? Because it is also the optimal operating point...

It is enough to isolate an element of the blade, say at 2/3 of the blade (this is the aero standard) and see its relative speed V1 at 250 RPM compared to the incident V0 (12.5 m/s)... visually it will be bigger than V0/3...but I can be wrong...

Then, energy being a reversible concept, this would mean that the maximum efficiency of aeronautical propulsion is reached when the plane flies at V1/3? V1 = thruster output speed...sorry but it's wrong, maximum efficiency is reached when V0=V1...

I think we're mixing things up a bit on this subject...normally the question was twisted and unclear from the start!

I need to look at your equations in detail...

[Remundo]

It is enough to isolate an element of the blade, say at 2/3 of the blade (this is the aero standard) and see its relative speed V1 at 250 RPM compared to the incident V0 (12.5 m/s)... visually it will be bigger than V0/3...but I can be wrong...

there yes you are wrong. You are confusing the axial speed of the air with the orthoradial speed of the blades.

And you mix a result where the blades are supposed to propel with drag, with another case where the blades work with their lift.

According to a document like this:

The optimal three-blade speed corresponds to a tip speed ratio of 7: namely that the tips of the wind turbine blades must move 7 times faster than V0

So depending on the blade radius, this will correspond to R w = 7 x 12,5 m/s, I'll let you calculate w in rad/s then in rpm if necessary.

Then, energy being a reversible concept, this would mean that the maximum efficiency of aeronautical propulsion is reached when the plane flies at V1/3? V1 = thruster output speed...sorry but it's wrong, maximum efficiency is reached when V0=V1...

no no, you can't extrapolate so brutally.

The propulsive efficiency of the aircraft tends towards 100% if V1 is just above V0, but at the same time, the mass flow Dm = ro SV also tends towards zero, which leads to having very large propellers, i.e. is another subject.

I need to look at your equations in detail...

sure, bring an aspirin

[Christophe]

OK, I looked at your equations...

The energy E1 produced is therefore written:

E2 = 1/2.Ro.Cx.S.(V0-V1)^2.V1.T

Yeah let's go...(the sequel invalidates that)

POINT4: For V1=0, E2 is worth zero since the plate does not move back, this is normal. For V1=V0, E2 is also zero
and this is also normal since the plate moves back at the speed of the wind, it therefore has no force and cannot
provide no work. Between these two extremes (V1=0 and V1=V0) there must be a maximum since we
that the surface can provide non-zero work. The derivative of E2 with respect to V1 is written:

d(E2)/dV1 = 3/2.(V1-V0).(V1-V0/3).Ro.Cx.S.T

Ok mathematically (it was a long time, it's been 20 years since I played with derivatives...)

which cancels for V1=V0/3 and gives E2max = 4/54.Cx.Ro.V0^3.T

Not entirely ok, it also cancels out for V1 = V0. In this case E2max = 0 (while if the plate is accelerated to V1 it means that it has recovered 100% of the energy...)

This is starting to sound absurd, isn't it?

The curve has this shape with V1 = 12.5 m/s: https://fr.symbolab.com/solver/functions-line-calculator/f%5Cleft(x%5Cright)%3D%5Cleft(x-12.5%5Cright)%5Cleft(x-%5Cfrac%7B12.5%7D%7B3%7D%5Cright)?or=input

Screenshot 2024-02-05 at 14-03-35 f(x) (x-12.5)(x-(12.5)_3).png (148.7 KiB) Viewed 655 times

The energy curve E2 at 12.5 m/s looks like this: x(12.5-x)^2

https://fr.symbolab.com/solver/functions-line-calculator/f%5Cleft(x%5Cright)%3D%5Cleft(12.5-x%5Cright)%5E%7B2%7D%5Ccdot%20x?or=input

We obtain an energy which tends towards infinity at not even 20 m/s...Always making intermediate numerical applications allows us to see if we are correct or not...

In short, it becomes more and more ABSURD.

There is necessarily an error in reason: infinite energy at 20 or 25 m/s is not okay...

I think I found the error...

All this reasoning is based on point 2 which is false because it subtracts speeds in a calculation of energy or force.

POINT2: If S moves backwards at speed V1 (with V1
at the differential speed V0-V1, the force F' (with F'

F' = 1/2.Ro.Cx.S.(V0-V1)^2

This reminds me of the debate we had years ago on car safety: 2 identical cars which make a frontal impact at 100 km/h do not dissipate energy equivalent to a relative speed of 200 km/h but of 2 * 100 km/h... it is not at all the same thing in terms energy (and therefore strength)!

And as each car absorbs 1/2 of the energy, this frontal impact is equivalent to...a frontal impact at 100 km/h against a stationary obstacle!
I agree it’s not “logical” a priori!

Even if it is difficult to imagine. This is physical reality. I think we are in the same type of reasoning bias here!

So who's the boss?

ps: I have a bit of a headache anyway...

[Christophe]

It is enough to isolate an element of the blade, say at 2/3 of the blade (this is the aero standard) and see its relative speed V1 at 250 RPM compared to the incident V0 (12.5 m/s)... visually it will be bigger than V0/3...but I can be wrong...

there yes you are wrong. You are confusing the axial speed of the air with the orthoradial speed of the blades.

And you mix a result where the blades are supposed to propel with drag, with another case where the blades work with their lift.

No no I'm not wrong...when I say isolate it means isolating a section of the square with its fluid vein around it by digital simulation.

The blade tip speed of the 10 m wind turbine from the simulation at 250 RPM at the blade tip is 2*Pi*250/60 *5 = 131 m/s (this is in the screenshot in more: 471 km/h...)...so a ratio of 11...

So on your curves we wouldn't be far from the intersection of 3 blades and 1 blade...I don't really know why.
The difference with 7 undoubtedly comes from the calculation of the ideal profiles for a given operating point.

There are surely other effects that the simulation took into account.

When is your doc? Do you know the profiles?

sure, bring an aspirin

It's done, I'm a little confused, but I seem to have identified the error...

POINT 2 is biased...see the example of cars in frontal impact...in physics, a calculation of force or energy is NOT done on a relative speed...but a difference in force calculated on absolute speeds

[Christophe]

This is where we are. It remains to be understood what is happening!

It's done: your biased point 2...

[Remundo]

Christophe,

you have drawn the derivative, it does not tend towards infinity if V=20 m/s.

Furthermore, beyond 12,5 m/s, the layout no longer has any connection with physics, because the speed limit is 12,5 m/s. The dragging plate will not move faster than the incident wind.

The validity interval of the curves is V1 = 0 up to V1 = 12,5 m/s, it must be limited to that.

Concerning aerodynamic forces, it is always the relative speed of the wind in relation to the moving element that counts.

[Christophe]

I was talking about the function of energy: https://fr.symbolab.com/solver/functions-line-calculator/f%5Cleft(x%5Cright)%3D%5Cleft(12.5-x%5Cright)%5E%7B2%7D%5Ccdot%20x?or=input

But you are right beyond 12.5 m/s it no longer means anything (mea culpa) except that at 12.5 m/s it goes to 0...while the energy should be maximum. This is obviously false.

Furthermore, the fact that the Cx does not intervene in the shape of this curve is necessarily false. The strength of the “sensor” will of course affect the optimum operating point!

The mathematical reasoning is right, the physical reasoning is wrong because point 2 is biased.

Read carefully and understand the example of cars. It's almost the same here.

It is necessary to calculate the difference in forces, not a difference in speed in the calculation of the force.

The error in reasoning is to consider that there is no force against the movement of the plate. Physically it is false.

ps: yes for the speed limits...I haven't found how to fix that...

[FALCON_12]

OK, I looked at your equations...

The energy E1 produced is therefore written:

E2 = 1/2.Ro.Cx.S.(V0-V1)^2.V1.T

Yeah let's go...(the sequel invalidates that)

POINT4: For V1=0, E2 is worth zero since the plate does not move back, this is normal. For V1=V0, E2 is also zero
and this is also normal since the plate moves back at the speed of the wind, it therefore has no force and cannot
provide no work. Between these two extremes (V1=0 and V1=V0) there must be a maximum since we
that the surface can provide non-zero work. The derivative of E2 with respect to V1 is written:

d(E2)/dV1 = 3/2.(V1-V0).(V1-V0/3).Ro.Cx.S.T

Ok mathematically (it was a long time, it's been 20 years since I played with derivatives...)

which cancels for V1=V0/3 and gives E2max = 4/54.Cx.Ro.V0^3.T

Not entirely ok, it also cancels out for V1 = V0. In this case E2max = 0 (while if the plate is accelerated to V1 it means that it has recovered 100% of the energy...)

This is starting to sound absurd, isn't it?

I don't have the impression that this subject interests you Christophe. There is acrimony
in your words, so what’s the point? I am here to learn and exchange, not
to fight with someone who is angry, even with speckled foils.

I proposed this subject because I believed that this question would fascinate and amaze.

It seems that only Remundo received things like this.

Fortunately this gentleman exists.

[Christophe]

Your answer doesn't really surprise me...is it so difficult to notice your mistake (plus it was you who asked...)? So we attack the messenger?

I spent 2 hours on your equations and writing answers on your subject today (and I don't just have that to do...) and a good hour yesterday doing the wind turbine simulation...so if the subject interests me (interested...because we've covered it) and I'm not upset, I'll explain...

I found your error, point 2 is physically wrong. And Remundo didn't see the mistake either. No problem, to err is human. The error is to persist in the error...

If you don't understand with the example of cars it will be difficult to make things clearer.

If you think it's logical that a recoverable energy curve decreases with speed (when it should only increase)...
If you think that it is logical the shape of the object in the wind does not affect the shape of the recoverable energy curve...

So I can't do much more for you...sorry...

I still suggest you apply for the Nobel Prize in Physics. You never know, maybe you'll be able to convince them...

Here I'll give you the link: https://fr.symbolab.com/solver/functions-line-calculator/f%5Cleft(x%5Cright)%3D%5Cleft(12.5-x%5Cright)%5E%7B2%7D%5Ccdot%20x?or=input

Now do what you want, believe what you want...except that physics is not about belief but about facts and correct reasoning.

ps: keep us informed about the Nobel...