Calculating Torque Over a Surface Area

If I remember my college aerodynamics classes correctly, any given airfoil (your turbine blade should qualify) has a center of pressure. Just like you may assume dynamic forces act on the mass center of gravity for an object, aerodynamic forces may be assumed to act on the center of pressure.

Knowing the pressure, total effective surface area, and distance of the center of pressure from the turbine hub, an effective torque may be derived.

Alternatively, if your CP location is not known, you can integrate the force x area along the length of the blade. If you do not have a mathematical expression for the blade area as a function of distance from the hub, you can perform the integration numerically.

Thanks , I will look into it. it's been along time myself.

Found this....

= 41.65 Nm

https://www.raeng.org.uk/publications/other/23-wind-turbine

From the sited paper...

"... Hence, the power coefficient needs to

be factored in equation (4) and the extractable

power from the wind is given by: ..."

.

What power coefficient did you use to arrive at the 41.65 Nm figure......or is that answer correct because it is really 42 in unrounded disguise?

Haha, no calculation just the inch lbs to Nm he asked for.....I think....I didn't check his calculations...

It looks to me, that the conversion being requested is from lbs in² to Nm... which would be a neat trick if convincing.

I'm not sure of the units intended for the second of the two provided figures multiplied in the OP..

"...

1440 sq in x .00256(10^2) = 1440 x .256 = 368.64 lbs inch^2

..."

, but I guess 0.256 should be lbs. but I don't know what multiplying those does for you (maybe it helps calculate acceleration through a fluid of a massless/inertialess object with an incompressible nonzero cross section area?).

If you divided the force by the area over which the force was applied, you get a pressure...

Unless you are designing a vertical axis Savonius type wind turbine, you are barking up the wrong orthogonality if you want to "... determine the size of generator ...".

For any wind turbine that predominantly utilizes lift (instead of drag) to spin the turbine, the calculations needed for accurately determining the appropriate generator size would be a bit more complex than your initial stab.

Lift and drag of the foil at some given wind speed and tip speed will need to be determined. That is no trivial calculation. If you doubt this, consider something as basic as determining just the angle of attack. You can't begin to do that without knowing tip speed for the wind speed you are calculating, as well as specifics of the foil geometry. *Cue Chubby Checker.

This is going to be far easier to flesh out empirically than attempting to get very close number crunching from theory...though it can be done.

A funny thing is that the surface area and wind speed isn't ever enough to allow you to know the loads the structure the turbine is mounted on will need to handle, because generally the higher the tip speed, the higher lateral force on the hub, since as lift and drag scale with the square of wind speed (apparent wind speed in the case of the spinning blades) the faster the turbine turns, the more thrust (the sum of the lift and drag vectors) will be produced.

.Frankly I would just buy a kit....

In any case the pulley drive ratio, the drag of the generator under different loads, the targeted average wind speed of the location and many other factors will come into play here....so the generator load will have to be well below the torque capability of the turbine at minimal wind speeds, and probably have the load capacity control to control the rotor speed in higher wind speeds...So it seems the control system is the key to success here, everything else can be ballparked...

" The average wind efficiency of turbines is between 35-45%." So if you know the number of blades and the length of the blades, then you should be able to approximate the generating capacity..5 blades...52" rotor size = ~ 1200 watts...

free online simulator

https://www.quora.com/What-methods-are-used-to-turn-wind-into-usable-energy

If a wind turbine is between 35% and 45% efficient, when do you enter the number of blades into the equation.

Wind speed would definitely be something needed for such an equation. As an example, for a 52" diameter rotor wind turbine with efficiency in the 35% to 45% range (and one blade if you must know...) the wind speed needs to be up around 30 mph to put 1200 watts into spinning the generator. If you want to get 1200 watts out of the generator the wind needs to be blowing even harder.

5 blades, 20 mph wind, 35% of time active generation average...

The rating of the generator determines the size, pitch, number and length of the turbine blades.

Maximum power generated (a function of the generator), divided by overall efficiency, divided by angular velocity, determines the torque that the driveshaft will experience and the stress that the mechanical components will need to endure.

Not the other way round.

It sounds like he already owns a turbine and knows the wind velocities, now he just needs to figure out how big of a generator it can support.

One can carry out an energy balance calculation to do that.

Torque is a notion that supports the mechanical design of the rotor.

Thank you all for the knowledge and support.

We are currently working on new generator technology

to adopt to our existing Wind technology at www.be-wind.com.

thanks for the support.

The first rated power claim I saw on the be-wind.com sight looks very optimistic.

The EOW -100 given the measurments provided has a total frontal (wind facing) area of 80" by 70", or roughly 3.6 m²

The website claims that in 11 m/s wind the rated output is 1.3 kW.

11 m/s wind at sea level has a little over 800 watts per square meter available to make power.

If we go on the very generous assumption that the entire frontal area of the EOW-100 is the swept area, this suggests the total wind power at 11 m/s the 100-EOW can convert from is a little over 2900 watts.

Getting 1.3 kW of electricity out of a little over 2900 watts in the wind is a commentable feat. That is about 44% efficiency inclusive of the losses in things like gears, wires, generator, rectifier, etc.

Doing this with a small Savonius type turbine should make anyone who knows just a little about wind power to at least raise an eyebrow...pretty high.

.

Even if this is absolutely correct because this is a miracle breakthrough turbine, it would probably be a good idea to scale back those rated power output claims lest the firm suffer from a perception that begins with ill concealed laughter of those in the know.

If the performance claims are true, the word will get around rapidly enough.

George

You are absolutely correct. the chart needs to be modified to the current new generators we are working with. that data is approx 5 years old and not correct, we need to revise it. I will look at updating the power specification chart to correctly represent the units and power production. Since 90% of our sales are for Battery Storage systems. The lower voltage generators and power curves would have to drop slightly.

Again , thanks for the observation and feed back.

Mike B