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?