How about showing the gen connected to a battery?
For simplicity we will use 12v for the battery, no rectifiers losses, OK?
Battery impedance is << gen impedance. So will ignore it also.
I = (24 - 12) / 0.054 = 222.222 amps
Power into battery is 12 * 222 = 2,667watts.
How much power is dissipated in the generator?
2,667watts.
With high winds the generators output voltage can go over 48v.
I = (48 - 12) / 0.054 = 667amps now were smoking!
Power into battery is 12 * 667 = 8kw
Power in generator is (48 - 12) * 667 = 24kw.
To keep the generator from burning up it must furl before this point.
This shows why targeting a low cut in speed will limit the high end because of excessive voltage, forcing early furling to protect the generator. Or use limiting generator impedance, which will limit generator output at all points.
The use of star / delta can help, by reducing the voltage by 1.73 in higher wind speeds. But is it enough?
Have fun,
Scott.