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A new kind of wind turbine


RevI

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58 minutes ago, OceanBreeze said:

The video is garbage.

This man makes many videos by making something at his own cost. He has many viewers,subscribers and none has called his videos garbage yet.

59 minutes ago, OceanBreeze said:

Why are you so reluctant to provide that data?

I have already told that the torque generated by my machines at 6 m/s wind velocity is 1.52 Newton-Meter at 3600 RPM. Being a marine engineer, I think you can easily calculate the output from the data that I have provided.

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5 hours ago, RevI said:

This man makes many videos by making something at his own cost. He has many viewers,subscribers and none has called his videos garbage yet.

 

Maybe “garbage” is too strong a word to describe that video, but whenever I see a false claim such as any wind turbine exceeding Betz limit, “garbage” is really the first word that comes to mind.

I will say that the presentation is entertaining and the presenter is very personable and likeable, but that does not change the fact that the calculations that lead to an efficiency greater than Betz are laughably wrong.

Here is what he is doing: He is calculating the mass of the wind using a column of air moving at 2.4 m/s and hitting a rotor with a diameter of 150 mm (0.15 m). He calculates the area of the rotor to be ¼ π D2  =1.77 E-2 m2 so that the mass flow is 2.4 m/s x 1.77E-2 m2 x density of 1.225 kg/m3 = 0.052 kg/s mass flow rate. His calculated power of the wind is then ½ m v2 = 0.15 joule/s

After comparing the ratio of available wind power to the apparent power extracted in 10 seconds of 1.2 joules, the turbine/flywheel is collecting wind power at a rate of 0.12 joules/s, which is 80% efficiency and “beats" the Betz limit of 59.3%.

So, what is he doing wrong, and more importantly, is he aware of his “error”?

Notice the rotor is not a flat plate but it is a cone with a diameter of 150 mm and an apparent height of about 50 mm. The surface area of such a cone can be calculated here:

 

and it turns out to be 0.039 m2

Using 0.039 m2 as the true surface area that interacts with the wind column, yields a mass flow of

2.4 m/s x 0.039 m2 x 1.225 kg/m3 = mass flow of 0.114 kg/s. Now the true power of the wind hitting the rotor is then then ½ m v2 = 0.33 joule/s

The efficiency is 0.12 J/s / 0.33 j/s = ~ 36.4% which is consistent with most wind turbines and well below the Betz limit on turbine efficiency.

When some inventor or tinkerer claims their wind turbine has exceeded the Betz limit of efficiency, I immediately know the calculations were not done correctly and these claimants need to be corrected in order to stop the dissemination of “bad physics” regardless of how well-liked the presenter may be.

 

Quote

I have already told that the torque generated by my machines at 6 m/s wind velocity is 1.52 Newton-Meter at 3600 RPM. Being a marine engineer, I think you can easily calculate the output from the data that I have provided.

Now, you the one who is making the claim about your new WT. And you really have not provided enough information for me to calculate all the performance parameters.

Furthermore, why should I spend my valuable time doing your work to back up your claim?

Either you present your calculations or I will have to assume you have not done/cannot do them and you are making unsubstantiated claims. You will receive a warning and I will close this thread in 24 hours unless you post your calculations that I have requested many times.

 

 

Edited by OceanBreeze
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1 hour ago, OceanBreeze said:

Now, you the one who is making the claim about your new WT. And you really have not provided enough information for me to calculate all the performance parameters.

Furthermore, why should I spend my valuable time doing your work to back up your claim?

That's the easiest part of the whole process. Output = Torque X RPS (Round per second) X 6.28 - 1.52 X 60 X 6.28. Hope not a very complicated task to do so.

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On 1/1/2024 at 11:58 PM, RevI said:

That's the easiest part of the whole process. Output = Torque X RPS (Round per second) X 6.28 - 1.52 X 60 X 6.28. Hope not a very complicated task to do so.

Is there a reason why you didn’t fill in the blank?

Maybe you finally realized that it is a silly claim, for such a small turbine, of 0.35 m diameter, to produce 572 Watts of power from a 6 m/s wind, even equipped with the most efficient wind concentrator.

I will not waste much more of my time on this since you refuse to show me where your numbers of 1.52 Nm of torque @ 3600 rpm are coming from. It seems to me you are just pulling those numbers out of …the air.

For comparison, I already demonstrated the math for a standard HAWT of 0.35 m diameter in a 6 m/s wind. The wind will generate a Force of ~ 2.12 N and the max theoretical power output is ~ 12.7 Watts. The actual max power output is limited by Betz to 7.5 Watts but I don’t even need to invoke Betz to show how ridiculous your claim is.

572 Watts is roughly 45 times the max theoretical output of the standard HAWT, (76 times the power at Betz limit). Nobody, anywhere, is making such a ridiculous claim!

How ridiculous is this claim? I will just offer some examples:

There is a 6.9 MW Vestas WT that has an impressive efficiency of 51% when operating in an average wind speed of 10 m/s. To generate 6.9 MW the blades are enormous, at 81 meters with a swept area of 20,000 square meters. Using your design with a 45 times improvement in power, the blade length can be reduced by a factor of 1/ (square root of 45), giving a blade length of ~ 12 meters. Do you have any idea how much money Vestas can make by building turbines with a blade length of 12 meters that produces 6.9 MW?

About the best performance I have found is for a very sophisticated (and very expensive) concentrator that can increase the wind speed at the turbine blades by 2.3 times the free wind velocity. This results in a power amplification of ~ 12 times. The output power of a turbine 0.35 m diameter at the point of concentration would be ~ 150 Watts, ignoring the Betz limit. At max Betz efficiency, the output is a very respectable 89 Watts. Actual test show that trying to exceed the factor of 2.3 results in too much turbulent resistance, forcing the free stream wind to follow the path of least resistance and flow around the concentrator instead of through it.

With such a concentrator I can possibly accept your Torque of 1.52 Nm but NOT at an rpm of 3600.

It is my contention that either your Torque is wrong or your RPM is wrong, or both, but you refuse to show how those figures are arrived at. I suspect you may be using the front-end torque on a gear box, and the output RPM of same gearbox. I hope I don’t need to tell you how wrong that is.

I should close this thread right now, but I will ask you one last time, to kindly show your full set of calculations or at the very least show a concentrator design that does better than 2.3 times wind speed at the turbine blades, validated by testing.

It would be great if you can prove me wrong with some calculations from first principles, or even better, some real-world data obtained by testing a prototype. If you can manage a 45 times improvement in wind turbine power, for some fixed diameter, you are going to be a very wealthy man; move over Musk and Bezos and make room for Revl

I hope that you will remember me, once the riches start pouring in.

 

 

 

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3 hours ago, OceanBreeze said:

It is my contention that either your Torque is wrong or your RPM is wrong, or both, but you refuse to show how those figures are arrived at. I suspect you may be using the front-end torque on a gear box, and the output RPM of same gearbox. I hope I don’t need to tell you how wrong that is.

It has been calculated by the ANSYS simulation, not me. So, if there is any mistake, then that has been done by the programing. If you have doubt, then I can share the CAD design against an NDA and you better run it on ANSYS in the similar conditions. In short, cross checking. If you are willing, kindly tell me.

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23 hours ago, RevI said:

It has been calculated by the ANSYS simulation, not me. So, if there is any mistake, then that has been done by the programing. If you have doubt, then I can share the CAD design against an NDA and you better run it on ANSYS in the similar conditions. In short, cross checking. If you are willing, kindly tell me.

I don’t need to see any proprietary design features, so there is no need for an NDA.

In order to form any opinion about the ANSYS simulation results, I just need to have a few key bits of data to plug into the well-established equations of wind turbine performance.

There are a limited number of ways the power output of a WT can be calculated:

 

1) 𝑃=1/2𝜌𝐴𝑉3𝐶𝑝𝜂

Where P = output power developed by the wind turbine in watts, V = free stream wind velocity  of 6 m/s, A = rotor swept of .0962 m2, 𝜌 = air density 1.225 kg/m3, CP = performance coefficient of turbine (max 0.59 Betz limit), and η = power multiplication factor added by the concentrator.

In this equation, I have all the data except for 𝜂. For our purposes, we can assume a value for CP that is very close to Betz, say 0.56

The equation reduces to: 𝑃= 7.13𝜂 Watts

For example, If 𝜂  = 22 (About the highest I have seen), Power = 157 Watts

What is needed: Please provide a value for 𝜂, if it is included in the ANSYS simulation data.

 

2) 𝑃=1/2𝜌𝐴(𝑉)3𝐶𝑝

Where P = output power developed by the wind turbine in watts, V = free stream wind velocity  of 6 m/s, A = rotor swept of .0962 m2, 𝜌 = air density 1.225 kg/m3, CP = performance coefficient of turbine (max 0.59 Betz limit), and = wind speed multiplication factor added by the concentrator.

In this equation, I have all the data except for . For our purposes, we can assume a value for CP that is very close to Betz, say 0.56.

The equation reduces to: 𝑃= 0.033(6∫)3 Watts

For example, If   = 2.8 (About the highest I have seen), Power = 157 Watts

 

What is needed: Please provide a value for , if  it is included in the ANSYS simulation data.

 

3) 𝑃= Tω

Where P = output power developed by the wind turbine in watts, Τ is the torque developed by the turbine blades, and ω is the angular velocity of the turbine blades in radians/sec.

For this equation, you have provided a value for torque of 1.52 Nm and given an rpm of 3600. If that rpm is at the turbine blades it equates to (3600 / 60) = 60 rev / s x 6.28 rad / rev = 376.8 rad / s, multiplied by the torque of 1.52 Nm results in an impossibly high value of 573 Watts!

If the torque of 1.52 Nm is correct, it must be taken at a point where ω is about 103 rad / s. That corresponds to a blade rotation of 987 rpm for a power output of 157 Watts, to agree with the other two methods.

What is needed: Please determine where in the drivetrain the value of 1.52 Nm exists, and what is the rpm at that same place in the drivetrain. It simply isn’t possible for the torque to be 1.52 Nm at a point where the rotation speed is 3600 rpm.

I feel sure ANSYS will be happy to provide the missing data and clarify the performance specifications for your turbine if you ask.

 

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19 minutes ago, OceanBreeze said:

For example, If 𝜂  = 22 (About the highest I have seen), Power = 157 Watts

Before answering, kindly tell me whether you have considered the diameter of the turbine to be 35 cm or not. I can only answer your questions after knowing that.

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2 hours ago, RevI said:

Before answering, kindly tell me whether you have considered the diameter of the turbine to be 35 cm or not. I can only answer your questions after knowing that.

Yes. You can see that the swept blade area in all equations is 0.096 square meters.

Since swept area =  ¼ π D2 , clearly I am using a diameter of 0.35 m, = 35 cm.

 

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Well, those wind velocity contours do shed some light on the ANSYS simulation results.

I should say right at the start that the simulation is idealized and the actual real-world performance of this turbine will not even come close to the simulation results.

Although it is not shown in the illustrations, the wind profile appears to show a conical concentrator at the front end which produces a wind speed of ~ 21 m/s at the tip of the cone. All I have to go by for free stream velocity is the 6 m/s you told me. Therefore, the wind velocity increases in the concentrator by a factor of 3.5. This is a very high factor that cannot be achieved anywhere except in a wind tunnel.

The problem is, wind cannot be concentrated the way many people think. At the low velocities being considered here, air acts like an incompressible fluid. Due to conservation of mass, the air that passes through the rotor blades cannot slow down at that point or it would block the air coming behind it. That is why the velocity of the slowed wind is taken far downstream of the rotor. In reality, energy is extracted at the rotor due to the pressure drop at that point. There is a high pressure at the front of the rotor and a low pressure at the back. (You can see this clearly in the wind profile you sent me) The point I am trying to make is simply that this high pressure in front of the wind turbine rotor deflects some of the upstream air to flow around the turbine. This is a real issue and it is why wind concentrator designs rarely, if ever, work as well in the real world as they do in a wind tunnel.

Normally, these conical concentrators achieve a nominal doubling of wind velocity.

Continuing: For the purpose of our deliberations, a mass flow rate @ 21 m/s confined to a blade swept area of 0.096 m^2 ~ 2.5 kg/s. Then Power = ½ mv^2 = 550 Watts close enough to the 572 Watts arrived at earlier.

This will also make sense for the power to be calculated using torque of 1.52 Nm x ω, where ω is (60 rev/s x 6.28 rad/rev) =377 rad/s. Power = 573 Watts

However, the Betz limit still applies regardless of turbine design and the actual limit of 59% is never reached.

For our purposes, let’s assume a factor of 59% so that the max theoretical power extracted is ~ 325 Watts.

A far more likely real-world value, would result from a wind concentrator value of 2.3 times 6 m/s at the tip of the cone. Mass flow is now 1.6 kg/s and power is 90 Watts Betz-legal.

My conclusion: The design looks very difficult to implement and not very practical. One leaf blowing in the wind can jam the rotor. Besides that, the simulation is overly optimistic. I would not place a lot of faith in it. To get 90 Watts out of a 6 m/s wind, a simple tried-and-true HAWT with a swept blade area of approximately 1 m^2 will do nicely.

As the inventor, you are naturally biased, perhaps to the point of suspending critical judgment regarding the simulation. I am just giving you the viewpoint of an unbiased technical observer. If you decide to move ahead with a production model, I can only wish you luck.

Now I am done with this thread. I should close it but I will give you the courtesy to post one more reply before I do.

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  • OceanBreeze locked this topic
  • 7 months later...

I have made a prototype (photo and video available). I used a common market available fan of 255 mm diameter blade and which can generate 6.2 m/s wind velocity. I placed my machine before the fan and fitted the fan to the inlet of my machine in such a way that all the flow generated are forced to enter the turbine. Initially the RPM was lower (around 190), but as time goes on (probably due to adjustment of the bearings), RPM started to increase and at present I have got 350 RPM (video available). My machine has a 35 cm inlet diameter. The kinetic energy contained by the flow is 14.294 W. And, as per Betz law, the maximum extractable energy from the flow for a conventional turbine is 8.57 W. But, now I can confidently say from the RPM, it can be said that the possible energy generation can be far greater than that. I will soon test the maximum possible output with an alternator and will update the results with others here.

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By now, Revi, you are probably wondering why you can only get about 2 Watts of power out of your machine; am I right?

If you don’t mind, I will try to explain why that is happening:

1)     Your calculation of the kinetic energy of the wind, coming from the fan, is off by a factor of two.

2)     You did not show your calculations, but I will show mine.

The wind velocity at your fan is 6.2 m/s

Diameter of the fan blades is 0.255 m

Area is π/4*Diameter^2 = π/4*0.065025m^2 = 0.05107m^2

Mass flow from the fan = 6.2 m/s x 0.05107 m^2 x 1.225 kg/m^3 = 0.388 kg/s

 

                                                                                                                                                                                   

Kinetic energy of the wind is just ½ m v^2:

½ * 0.388 kg/s * 38.44 m^2/s^2 = 7.458 Watts

This is about half of what you calculated, and the Betz limit has not yet been applied.

The biggest problem is the way you have set up the fan to drive wind into your machine.

Since the fan has a lesser diameter than the turbine (0.255 m vs 0.35 m) Instead of a concentrator, you now have a diffuser!

Since air mass is a conserved quantity, the same mass of air that passes through the fan, also passes through the turbine.

At the turbine, we can write: 0.388 kg/s =1.255 kg/m^3 * 0.0962 m^2 * Vel m/s

Solving for velocity at the turbine: Vel = 0.388 kg/s / 0.117845 K/m = 3.29 m/2

From Power = ½ pAV^3, you should see that:

Output power varies with the square of diameter and the cube of the velocity.

By using the smaller fan to drive the turbine, you have defeated the concentrator nad created a diffuser.

This is reducing the available power by a factor of (0.35/0.255)^3 = 0.3867

7.458 Watts x 0.3867 = 2.88 Watts! After applying Betz, you will only achieve an output power of less than 2 Watts.

To fix this, I suggest you use a fan which has a diameter at least as large as the diameter at the input to the concentrator.

Just a suggestion.

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14 hours ago, OceanBreeze said:

By now, Revi, you are probably wondering why you can only get about 2 Watts of power out of your machine; am I right?

 

No. I haven't measured the output yet with an alternator attached to the shaft. What I have done so far is to measure the RPM of the machine at this velocity. The RPM is 350 and I am expecting much more than 2 Watts from the machine.

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