matching wind power and generator power

[apie]

I made a spread-sheet to illustrate the difficulty of matching the wind-power with the power from the alternator.  You can find it in my files.

The yellow cells need to be filled in:

• cut-in speed: the wind-speed (in m/s) at which the alternator starts to generate usefull power (reaches battery charging voltage)
  
• cut-in voltage: the voltage needed to charge a battery
  
• internal resistance: the resistance of the coils (in single phase this is the sum of the resistance of all coils)
  
• swept-area: the total area (in square meters) that is swept by the blades
  
• efficiency: the amount of power that is catched by the blades (max 59%, Betz limit)
  
  

The graph shows you:

• how much power the blades can catch from the wind
  
• how much power the alternator will be able to produce (using a load with the same resistance as the internal resistance)
  
• how much power the battery will use to load
  
  

You can clearly see that the last graph is linear!  Which clearly shows how difficult it is to catch power at higher wind speeds.

I also have a question about this: if the generator can deliver more power than the blades will provide.  What will happen?  In other words: do you need to make sure that the second graph is always beneath the first?

Any comments are welcome!

[georgeodjungle]

to me,, cut in is, the all most important.

i'd rather have bigger blades

and or an aggressive angle of attack.

it kills the top end but,

 when the batteries are low so is the wind.

to big of genny or to small blades works for high winds.

it's a give and take thing.

[jimovonz]

If the generator can deliver more power than is available and it is loaded such that it tries to, then it will just slow down. If your lucky the system will reach a point where both the available power and the load are matched before the blades stall - at which point there will be a considerable reduction in speed (and power).

[apie]

So, with a Savonius I think there is not a big problem as it gains torque when speed lowers.  But maybe it can kill a HAWT.  But if you want to catch the power on low speeds, you lose it on the high speeds.  Interesting!

I already learned not to design the thing for the average wind speed, but for about twice that value.  At that point, the wind has a lot more power and the frequency of those gusts is high enough to gain a lot.

[wind4Reg]

Hi apie, I was looking at your spreadsheet and it seems to me that there is something wrong with "power in the wind" calculation. For example for a swept area of 10m squared and at a windspeed of 10m/s and an efficiency of 25% I get the following:

Power Density in kW/m squared = ( 0.6125 * 1.91 * 10m/s * 10m/s * 10m/s ) / 1000

Power Density = 1.69875

power in the wind in kW = Power Density * Efficiency * Swept Area

power in the wind in kW = 1.69875 * .25 * 10

power in the wind in kW = 2.925 kW or 2925 watts

From your spreadsheet for the same values of 10m squared swept area, efficiency of 25% and at a windspeed of 10m/s, I get the following:

power in the wind in watts = 1525 watts

I noticed that in your equation you use 0.5 at the beginning as a constant for something, if that was removed the results would be very close. Seems to me like that 0.5 shouldn't be there, what is it for?

thx,

wind4Reg

[The Crazy Noob]

I put some extra things on there (Amps in battery, formula for swept area, thicker lines in graph...)

And I noticed that there's something wrong with the efficiency, it acts like it's efficinty^(-1) so a higher efficiency gives less power in generator, maybe it's got something to do with the code in cell D9: "=C9*C9/2/$B$3"

=> you get: "max power from generator" = "voltage of generator" * "voltage of generator" / 2 / "internal resistance"

that x/2/x thing may be the problem?

annyway, I don't feel like going through the formulas and i don't know the right ones either...

Look at my files for the version with the extra info (bug not fixed).

[apie]

Maybe I should explain the formula:

The power from the wind is calculated as:

1/2 * efficiency * air-density * swept-area * v^3 (this comes from E=1/2 * m * v^2)

air density is equal to 1.22 kg/m^3.

I don't know where your constants (0.6125 and 1.91) come from.  Has it to do with empirical units?

Or I could be wrong, of course!  Can anyone else check too?

[apie]

Thanks for editing the excel sheet.  I will take a closer look in a minute, but I was too enthusiastic to reply, so I did this first.

I think the formula's are correct (but I could be wrong).

If you increase the efficiency, the power from the wind increases.  If you look at the graph you will see that it seems that the power of the generator decreases, but look at the values on the Y-axis: they have changed!  In fact the power from the wind has increased, but the generator is still the same.  So the power from the generator is lower with respect to the power from the wind.  I hope this clarifies it!

For the value of 'power from generator':

The power of the generator is at it's maximum when the load is equal to the internal resistance.  That means: if I have a generator with an open voltage 'voltage' and an internal resistance 'resistance'.  And if I put a load with 'resistance' ohms, The current flowing through the load is equal to voltage / (2 * resistance).

Multiplying this with the voltage gives me the power.

[apie]

Thanks for the new version!

However I noticed a small error in cell E3.  The area is equal to Pi*radius*radius (not diameter).

But this illustrates the need for this in the spread-sheet.  Thanks!

[wind4Reg]

Hi apie, here is where my constants come from

Air Density Factor = 0.6125 (Based on an air density of 1.225 kg/m3)

Energy Pattern Factor = 1.91 (Rayleigh distribution)

When these two are multiplied together together it comes to 1.17 which is an attempt at a closer value than the standard 1.22. That is fine, I wouldn't split hairs over .05 difference in our air density numbers. The real question is where the 1/2 comes from at the beginning of your equation which is essentially cutting the power in the wind in half right from the start for some reason. My question was why do you cut the power in the wind in half by multiplying it by 1/2 at the beginning. You said your formula comes from E=1/2 * m * v^2 and have tried to find that formula using Google but can't get any results in English. What is that formula (E=1/2 * m * v^2) for? It seems like it must be for something else since the v is squared instead of cubed and the m looks like it represents mass instead of air density. The formulas I have from Paul Gipe's wind power book do not cut the power in half when calculating the power in the wind. All I really wanted to do was get to the bottom of why there are two different formulas for the same thing and which one is the right one? It is pretty bad if we don't have a standard formula for calculating the power in the wind for a given swept area, at a given windspeed, at a specific efficiency.

Don't get me wrong, I am not saying the formula I have is the right one, it just seems like there are two formulas out there for the same thing, and if your's is correct then that drastically lowers the power output that one can expect from a wind turbine.

wind4Reg

[The Crazy Noob]

oops, overlooked that, it's corrected now.

[apie]

wind4Reg, the formula I gave is that of kinetic energy.  My physic books tell me that kinetic energy of a moving mass is equal to 1/2 m v^2.

In this case, air is moving.  That means we have to know the mass of air that is moving through the wind-turbine in one second (as we want power = energy/second).  That is equal to the volume of the air times air density.  The volume of the air passing through the turbine in one second is equal to the swept-area times the speed of the wind (v).

If you combine all this, you get:

 power = 1/2 * mass * v^2 = 1/2 * density * volume * v^2 = 1/2 * density * area * v * v^2 = 1/2 * density * area * v^3

I hope this clarifies my formula.  I must say, I still don't get yours though.  What is 'Rayleigh distribution'?