Menelaos, sorry for the delay. Business is really busy these days.
Using the following assumptions for your genny:
Blade diameter: 6m
Rotor eff: 42%
TSR: 6
Poles: 16
RPM-to-Voltage ratio: 1.174 (unloaded AC)
Phase resistance: 0.95 Ohm (each Wye-leg of alternator)
The RPM-to-Voltage ratio is based on your statement of 1.66 V DC per RPM. My calculations work from the alternator's perspective, and I need unloaded RMS AC Voltage. I'm assuming that unloaded DC is very near to peak value of the AC waveform. Assuming a sine this means RMS AC value is DC / sqrt(2), or 1.174 in case of your alternator.
This results in the following values:
Wind (m/s) Watt Rotor RPM Hz Unloaded AC I phase Loaded AC P loss Pout Eff. % DC Voltage Watt Out
0 0 0 0.0 0 0 0 0 0 100% 0 0
1 7 19 2.5 22 0.2 22 0 7 99% 29 0
2 56 38 5.1 45 0.7 44 1 55 97% 58 12
2.5 109 48 6.4 56 1.1 54 4 105 97% 72 60
3 186 57 7.6 67 1.6 65 7 179 96% 85 130
3.5 293 67 8.9 78 2.2 75 13 280 95% 99 226
4 434 76 10.2 90 2.8 85 22 412 95% 112 351
4.5 615 86 11.5 101 3.5 95 35 580 94% 125 511
5 839 95 12.7 112 4.3 105 53 786 94% 138 706
6 1436 115 15.3 135 6.2 124 108 1328 92% 162 1222
7 2264 134 17.8 157 8.3 143 198 2066 91% 187 1923
8 3356 153 20.4 179 10.8 162 333 3024 90% 210 2833
9 4751 172 22.9 202 13.6 179 527 4224 89% 232 3973
10 6483 191 25.5 224 16.7 197 794 5689 88% 254 5364
11 8587 210 28.0 247 20.1 214 1152 7436 87% 275 7024
12 11100 229 30.6 269 23.8 230 1617 9483 85% 295 8969
First column is the wind speed. Second one is the power extracted by the rotor, based on efficiency and energy in the wind. This is the amount of power the load (inverter, plus losses) has to match for the rotor to neither speed up or slow down. Next is RPM based on the target TSR. Next is the frequency of the alternator, based on number of magnet poles and RPM. The unloaded AC column is volts from the rotor, RMS AC voltage, assuming it's a sine waveform. The phase current is next, based on the rotor power and unloaded voltage; P = V * I * sqrt(3). The loaded AC is derived from the unloaded AC and current, taking the voltage loss across the internal resistance for each phase leg into account; if you work out the equation the voltage loss is V = R * I * sqrt(3), and yes, this takes care of the resistance of both phase legs. Next are power losses in the alternator, based (only) on the internal resistance; it's simply P = 3 * I^2 * R (three times the resistive loss in each phase leg). Actual losses will be larger, since things like eddy current and mechanical losses have not been taken into account. This is just to have good idea of heat produced in the stator, to draw a line you don't want to cross (hopefully before things melt). The DC voltage derives from the loaded AC voltage, for 3-phase it's close to 1.33 * Vac. Finally the last column is what's coming out of the inverter, taking inverter losses into account.
There are a few refinements at work in the calculations, as well as some approximations. For power I have a factor that corrects for the rotor not quite following the cube law. Similarly, I assume alternators get a bit less efficient at the higher RPMs, and add that to the mix. A (large) approximation in all this is that I'm not taking phase angle of the current vs. AC voltage into account. A regular 3-phase bridge has a piss-poor power factor, and the actual AC current is actually quite a bit larger than what's indicated here. Also, wareforms for alternators are usually not quite a sine wave, which is assumed here. Still, it's a good place to start for a power curve, and most blade profiles are just not sensitive enough when run off-TSR to make much of a difference as long as the power curve is in the ballpark.
In case that wasn't obvious: The last two columns are what should be programmed into the inverter as the MPPT curve. For a 6kW Aurora wind inverter (PVI-6000-OUTD-US-W) I would cut it off at 264V DC and 6,200 Watt, at which point the wind speed should be around 10.5 m/s and you're running just shy of 1kW in losses in the alternator.
For the Aurora inverter I would set the Vstart voltage (when the inverter switches on) at 60V for this turbine, and put a first datapoint in the table with (60V,0W). The next one would be (72V,60W) etc. For Pramp (how fast the inverter responds) I would put in 10,000 Watt/sec; we don't want the inverter to lag events, and inadvertently set up oscillations, this value is as fast as the inverter will do and seems to work well in practise. A Tprot time (how long the inverter stays online after the voltage drops below Vstart) of 180 or 240 seconds is good.
You may wonder why this would work, and why this would do "MPPT" for a wind turbine: If the numbers are correct (ie. if the assumptions and calculations are correct), the load imposed by the inverter (plus losses along the way) at any given voltage will be exactly equal to the power extracted from the wind by the rotor. That means the rotor will neither speed up nor slow down. If it would speed up the voltage would go up, causing the inverter to increase its load, and something along the same lines happens if the rotor slows down. The net effect is that if we get it right it will keep the rotor RPM exactly in the sweet spot (the target TSR) for optimal power efficiency, regardless of the wind speed.
Hope this helps!
-RoB-