It can be calculated, but it takes a lot of measurements through the whole the system to compute an answer (atmospheric conditions, rotor blades, electric generator configuration, power conversion system).
It's often easier to just measure it.
Mechanical power = (Torque)X(RPM)/(conversion factor)
Electrical power = (Volts)X(Amps)
The aerodynamic conversion is 50% efficient at best, but often less. Mechanical conversion to electricity can vary between 70% and 20% depending on the generator. The one I'm using varies between about 50% at low speeds to 40% at higher speeds. I pay a price for robustness. I remember when Kitestrings published info about his alternator build (about 5 years ago) I was impressed by its efficiency. Though I don't remember the exact value it was quite high.
When converting mechanical to electrical energy, the majority of energy lost goes to resistance heat. If the windings have 10 ohms resistance, then passing 10 amps through them releases 1000 Watts as heat. If the same generator had reached 100 Volts as a result of that work, then it generates 1000W of useful electric energy. The sum of the two outputs (useful and non-useful energy) requires 2000 Watts of mechanical power at the driveshaft to make it. And my example generator would be 50% efficient.
I simplified my example so that I wouldn't have to consider 3-phases and rectifiers. They have a role to play in practical generators, but I hope the example is complete enough to get the idea across.
