Wind energy

[FALCON_12]

Hello,


I tried to calculate the recoverable energy using a unit area plate S
placed facing a wind blowing at a speed V0 and retreating under the effect of the
force that it exerts on it at speed V1.

I used the formula F=aV^2 regarding the force the wind exerts on the plate.

The work of this force over a distance d is written E=a.(V0-V1)^2.d and since d=V1.T
this work, for a time T, is worth E=a.(V0-V1)^2.V1.T

If V1=0 the work is zero since the plate does not move back.
If V1=V0 it is also zero since the plate moves back at the speed of the wind
it therefore does not undergo any pressure and F=0.

Between these two extremes I calculated that the maximum is Emax = 4/54.a.V0^3.T for V1=1/3.V0

Does anyone see an error?

[Remundo]

your calculation seems correct, but it is also a little wrong because there is an error in reasoning when you say "the force of the air is equal to av^2/2".

It is necessary to make a dynamic assessment of the flowing air by choosing a "control volume" where the air mass is constant and apply the fundamental principle of dynamics to this air. By the principle of reciprocal actions, you will deduce the force of the air on the blades.

If you are looking for the optima, here they are
* impact turbine (Pelton type or deflector plane): the best deflection is 180° and the optimal bucket speed is half the fluid speed
* flow turbine: no deflection, the speed of the fluid must be divided by 3 and we touch then the Betz limit (16/27)

in your calculations you revolved a little around Betz's yield, but by mistake.

[FALCON_12]

your calculation seems correct, but it is also a little wrong because there is an error in reasoning when you say "the force of the air is equal to av^2/2".

It is necessary to make a dynamic assessment of the flowing air by choosing a "control volume" where the air mass is constant and apply the fundamental principle of dynamics to this air. By the principle of reciprocal actions, you will deduce the force of the air on the blades.

If you are looking for the optima, here they are
* impact turbine (Pelton type or deflector plane): the best deflection is 180° and the optimal bucket speed is half the fluid speed
* flow turbine: no deflection, the speed of the fluid must be divided by 3 and we touch then the Betz limit (16/27)

in your calculations you revolved a little around Betz's yield, but by mistake.

Why do you say that the formula F = aV^2 which gives the force F of a wind of speed V pushing on a perpendicular surface S
is not correct? Is this formula wrong for you? (1) or do you consider it good while contesting the correctness of its
employment in this problem? (2)

[Remundo]

I would have to give you a complete demo, but it's long.

dynamic assessments are very difficult to set up.

this page might give you some ideas

http://ambroise.brou1.free.fr/mdf_007.htm

compared to your notations, by calling alpha the angle of deviation (90° for you) and Dm the mass flow in kg/s on the blade (constant), we arrive at the following expressions
* for the force on the blade (which is not equal to "a V²" because Dm = ro S V0)
F = Dm (V0-V1) x (1-cos(alpha))

* for the power on the blade P = F x V1
P = Dm V1 x (V0-V1) x (1-cos(alpha))

the optimal power is that which maximizes V1 x (V0 - V1), this corresponds to V1 = V0/2 (find the max of x(1-x) with x = V1/V0

note that the best angle to maximize the power is alpha = Pi (the water must make a half-turn in the bucket's frame of reference).

This is why the Peltons have a bucket speed half that of the jet, and the shape of the buckets are spoons.

If you form the theoretical efficiency of a Pelton, here is eta=P/Pcin with Pcin = Dm V0^2/2 the incident kinetic power

He comes knowing that P = Dm/V0^2 (1-V1/V0) x V1/V0

eta = 2x (1-cos(alpha)) x V1/V0 x (1 - V1/V0)
which can reach 2 x 2 x 0,5 x 0,5 = 1 for the theoretical optimum.


Concerning the Peltons' "cousins" in the world of wind turbines, they are Savonius. Their performance, however, is far from 100%.

On the other hand, Peltons in hydroelectricity manage to exceed 90%. In fact the Peltons almost manage to "eat" all the incident kinetic energy of the jet (the water leaves the Pelton with almost zero residual speed).

[FALCON_12]

I would have to give you a complete demo, but it's long.

dynamic assessments are very difficult to set up.

this page might give you some ideas

http://ambroise.brou1.free.fr/mdf_007.htm

compared to your notations, by calling alpha the angle of deviation (90° for you) and Dm the mass flow in kg/s on the blade (constant), we arrive at the following expressions
* for the force on the blade (which is not equal to "a V²" because Dm = ro S V0)
F = Dm (V0-V1) x (1-cos(alpha))

Wait Remundo, I don't follow you anymore so let's try to start from a common base.

Concerning a blade exposed to the wind, my fluid mechanics course tells me this:

F = 0.5.Ro.Cx.V^2.S

T: Wind force [N]
Cx: Air penetration coefficient
P: Wind density or density [kg/m3]
v: Wind speed [m/s]
S: Surface exposed to wind [m²]

Do you think this formula is true or false?

[Remundo]

it is a formula which is true when the blade is fixed facing a wind of speed V.

It is also more inspired by aerodynamic drag.

But we cannot rely on this formula to estimate the force of the wind on the blade when everything is moving (both the wind and the blade).

It is necessary to carry out a dynamic assessment on the air contained in a “control volume” of constant mass.

[FALCON_12]

it is a formula which is true when the blade is fixed facing a wind of speed V.

It is also more inspired by aerodynamic drag.

But we cannot rely on this formula to estimate the force of the wind on the blade when everything is moving (both the wind and the blade).

It is necessary to carry out a dynamic assessment on the air contained in a “control volume” of constant mass.

Ok. Can you explain to me why we cannot consider that the blade experiences a wind of speed V0-V1 and use this formula?

Don't get me wrong, I'm not disputing it, I just need to understand what's going on. It is impossible for me to admit what I do not understand.

[phil59]

https://fr.wikipedia.org/wiki/Limite_de_Betz

Around 60% yield at best, if it's comparable.

[Remundo]

Ok. Can you explain to me why we cannot consider that the blade experiences a wind of speed V0-V1 and use this formula?

Don't get me wrong, I'm not disputing it, I just need to understand what's going on. It is impossible for me to admit what I do not understand.

Let's say that from the start, you and I do not think on the same approach.

My view is orderly flow with 90° deflection of the fluid stream.

Your point of view is an undeflected flow at 90° which "dissipates" as it goes around the blade in a turbulent manner, you are in the domain of aerodynamic drag (and not a momentum balance).

So under this assumption we can use your calculations using the relative wind to estimate the force. The Cx of a plate facing the wind is around 2 I believe.

[sicetaitsimple]

https://fr.wikipedia.org/wiki/Limite_de_Betz

Around 60% yield at best, if it's comparable.

60% is already huge. Only hydraulics do better, in pure electrical production.