Free power ?

[WineGuy]

In my many hours of surfing this and other sites I have gleaned the following information for my design. Most of it is scattered, so I collected it here. I'm repeating it here for reference, so I can figure out where I went wrong.

Wind energy is kinetic energy, so

Kinetic energy is ½ mass times velocity squared

Ke = ½ M*V^2

Mass is the mass of the air going past the blades in a given time.

Sweep area of the blades is pi times the radius squared. Radius is the length of one blade.

A = pi * r ^2

Volume of air (per second) is the sweep area times the wind speed (velocity).

Volume = V * pi * r^2

Mass of the air is Volume * density of the air (Ad) (averages around 1.25 Kg per cubic meter at sea level, depending on temperature and barometric pressure)

M = V *  pi * r ^2 * Ad

Kenetic energy is ½ M* V^2

Ke = ½ (V * pi * r^2 * Ad) * V^2

Ke = ½  * pi * r^2 * Ad * V^3

This is the available power in the wind.

To find the power you can GET OUT of the wind you need to know the following:

Cp = rotor efficiency (usually about 0.4 from what I've read)

N = efficiency of the generator and system (typically about 0.7)

0.59 = the theoretical max of power you can extract from the wind (Benz limit)

So, power available to use = 0.59 * 0.4 * 0.7 * Ke.

Now to the real world.

I have an Ametec motor that measures 16.6 volts at 440 rpm, and 53 volts at 1400 rpm using my two speed cordless drill as input. I measured 1.3 ohms resistance. At 440 rpm this should put out 211 watts (V^2 / R).

I also found the formula RPM = V * TSR * 60/(6.28 * R)

       I assume that 6.28 * R is the circumference of the sweep area

       and V is velocity of the wind .

      Since Velocity is in meters per second, multiply by 60 seconds per minute.

I assumed a TSR of 6 for a 3 blade system.

I put the above formulae into a spreadsheet so I could play with the numbers. With a 1 meter blade at 16.7 mph wind (7.46 m/sec) I will get 219 watts at 400 rpm. If I go down to a 0.5 meter blade, I need a 26 mph wind to get the same power output, but with the rpm going up to 1332 rpm.

This kind of makes sense. You need air mass to generate energy. If you make the sweep area smaller, you need more wind speed to provide the same mass over a given time.

Question:

DC generator voltage increases with rpm. Looking at the above, using shorter blades requires three times the rpm to produce the same available power, but higher rpm should translate into higher voltages, which should mean more power. What am I missing?

[kurt]

i don't know what your fancy math says but real world slap a 4' prop on that amatek and you will get 10 amps into a 12v battery in a gale if you got a good one.

[vawtman]

Maybe just that the wind is only free part.The other formulas cost a bit.But fun.

[WineGuy]

I'm trying to figure out just what size of blades

and what configuration to use.

As you can see from the above numbers, I'm still not sure

how I can get the best design for the generator that I have.

Like the title says, I'm just a newbie

[jonas302]

remember the wind is always a different speed you might get a lot of rpm out of short blades but might have a hard time starting them I think everybody runs a 4 foot on the amtec a few hae cut them back with good results  there should be a few diarys on amtecs up and flying

[Flux]

I haven't had time to plough through all of this but I got the impression that your wind calculations are ok. I don't think the same applies to your electrical figures.

Your estimated figure for power into a 12v battery at 440 rpm is very wrong, I haven't checked anything else but that is almost certainly wrong if you used the same logic.

I assume you are intending to use a battery, if you are looking at heat then things may be different.

Flux

[ghurd]

The cool thing about Ameteks?

They are not the greatest thing since sliced bread,

but stick a 4' blade on the front, and it works.

G-

[fungus]

'I have an Ametec motor that measures 16.6 volts at 440 rpm, and 53 volts at 1400 rpm using my two speed cordless drill as input. I measured 1.3 ohms resistance. At 440 rpm this should put out 211 watts (V^2 / R).'

As I understand it, you only count the voltage above the battery voltage, so if the battery was at 13v then you'd have 3.6v more than the battery , about 10w.

[finnsawyer]

You wrote:

    "Kenetic energy is ½ M* V^2

    Ke = ½ (V * pi * r^2 * Ad) * V^2

    Ke = ½  * pi * r^2 * Ad * V^3"

You've equated kinetic energy to power.  They are not the same.  Power is the time rate of change of energy (W/t).  In this case the power available from the wind is the rate at which the kinetic energy of the air passes a given point.  You lost the time factor.  A better analysis is to start with a small mass of air dM = pdAdx, p is the density of the air and da is the frontal area of the small mass.  The kinetic energy of this small mass is dw = 1/2x(pdAdx)xV^2.  Now, the number of these little chunks of mass that go by a given point in a certain time t is given by Vxt/dx.  So, in a time t we find a total amount of energy gone by of W = 1/2xpxdAxV^2xVt.  The power available is simply P = W/t =1/2xpxdAxV^3.  If the air flow is uniform over some area A we may then write the power as:

    P = 0.5xpxAxV^3  

   

[WineGuy]

I was just looking at available power at the generator terminals, not loaded.

I beleive this would voltage ^2 / winding resistance.

Actual APPLIED power would depend on a whole bunch of different things like wire size. And Yes, applying it to a battery, I believe that only the voltage ABOVE 12 volts would be used to determine applied power.

I'm not saying I'm right, if fact I'm saying the opposite, and asking for someone to show me the errors of my ways.

What it "looks like" is that at 440 rpm I get 16.6 volts. At 1400 rpm I get 53 volts. So ........ using the same 1.3 ohms coil resistance would I get 211 watts at 440 rpm and 2160 watts at 1400 rpm? The math SOUNDS right, but looking at the unit, I find it hard to believe that it will EVER put out 2KW. So, where am I going wrong ?

[finnsawyer]

There is only voltage available at the terminals when open circuit.  The power depends on the load.  The numbers you have calculated would be the power dissipated in the internal resistance (that 1.3 ohms) when you short circuit the output wires.  That assumes you would have a mechanical source driving the generator that can provide the power.  But you are right about that 2k figure.  You would probably release the famous blue smoke well before that level of power is reached.  You need to get a book on basic circuit theory and study it.  

[commanda]

Using that 1.3 ohm figure works when dealing with an air-cored axial flux machine. The Ametek, like the F&P that I've actually done this test on, suffers from a whole heap of other internal losses related to the fact it has iron in the flux path.

You will find, in practice, that the apparent internal resistance will increase with rpm, and also with the current drawn. Using that figure for dc resistance and extrapolating that into any sort of figure for watts is just wishful thinking.

Amanda

[ghurd]

That is why they furl.  :-)

And there are about a billion other things, not included in the formula.

At 1400RPM, 53V, and 1.3 ohms...

Into a battery I get 53-14V= 39V.  39V/1.3 ohms= 30A. 30Ax14V= 420W.

Total Watts...

The 420W into the battery.

Plus (IxIxR) 1170W in the Ametek.

For a Grand Total of 1590W.

With a 4'Dia 3-blade prop, how fast is the wind going?

Remember Betz, CP, self-stall, TSR (related to the design speed:power), and almost a billion other things.

All together it means an Ametek with a 4' blade makes 80 to 100W.

Guessing that is about where you started?

Don't get overly lost in the math.  I did that too.

G-

[Flux]

Yes even with the air cored axial machines the effective resistance comes out to about 1.3 times the dc resistance. With iron cored machines the leakage reactance is significant. With dc machines such as the Amtek the ceramic magnets on the field will be influenced to some degree by armature reaction.

In real life I would expect the effective resistance to be nearer 3 ohms for predicting output, so you are never going to get any spectacular output.

Flux

[WineGuy]

Ding Ding Ding (light goes on)

Thanks everyone for all the usefull stuff.

Here's what I learned.

Yes the power is Voltage^2/coil resistance, but that is the energy that would be DISIPATED in the windings if I short circuited the unit.

Yes the unit would TRY to put out 2160 watts at 1400 rpm, but would probably FRY itself long before then due to I^2 * R losses.

Remember to use the voltage of the load (read battery) in calculating the USABLE power.

Albert Einstein never even MET me.

I'm getting smarter every day, cuter too!

Thanks again

[richhagen]

I wrote up my basic thoughts on power in the wind a long time ago here:  http://www.otherpower.com/images/scimages/742/02_power_in_the_wind.htm

I didn't plow through everything, but one thing I did notice is this:

'So, power available to use = 0.59 * 0.4 * 0.7 * Ke.'

The Betz limit isn't a factor that you would use in that equation.  It is the maximum theoretical value that you can obtain for your prop efficiency which you used the .4 for.  If you were to capture all of the kinetic energy from the wind, it would be stopped dead in its tracks relative to the turbine and no more wind could move past Betz figured the limit of the maximum percentage that could be removed.  Unfortunately, you are not going to remove that much.  In practice a good home built turbine can obtain an overall efficiency of converting 35% or so of the kinetic energy available in the wind passing through its swept area to electrical output, and that only for a narrow band of wind speeds for which it has been optimized, with lower efficiency outside of that.  For the best current commercial turbines the figure is likely around 40%.   Your Ametek mill will be lower.

Rich

[WineGuy]

I ran across your page a while back, but could never find it again.

Nice write up. Thanks for the additional info.