Generator Efficiency
From DIDEAS Wiki
Here several useful generator equations are derived.
Electrical Efficiency
Electrical efficiency as a fraction input mechanical power that is available as electricity the output terminals. Of course the voltage of the avaialbe power might not be optimal - but thats what switching power supplies are for.
E = Pout / (Pout + Ploss) Pout = I^2 * Rl and Ploss = I^2 * Rm thus E = I^2 * (Rl / (Rl+Rm)) let Rl = k * Rm then E = k/(1+k) for a given desired efficiency keep Rl >= Rm * E/(1-E) some results: k = 4 -> 80% k = 9 -> 90%
Power Efficiency
For a load that is a multiple of the generator resistance, a fraction of the maximum possible generator output power is obtained. If the load equals the internal generator resistance, then the maximum output power (Pmax) is obtained.
Let the power efficiency be: Pe = Pout / Pmax some math... Pe = 4k/(1+k)^2 64% -> k = 4 (80% electrical efficiency) 36% -> k = 9 (90% electrical efficiency)
Maximum Power
Its also possible to size a generator based on two fundamental properties of the generator. Here we derive the maximum generator power as a function of rotational velocity omega.
For a generator / motor: omega = Ks * Vemf (Vemf is back emf) or Vemf = omega/Ks Vm = Vemf - I * Rm (Vm is voltage at terminals) I = Vm / Rl let Rl = Rm for max output power then Vm = Vemf - (Vm/Rm)*Rm or Vm = omega/Ks - Vm so at max output power Vm = omega/2Ks output power is V^2/R so Pmax = Vm^2 / Rm or Pmax = omega^2 / 4 Rm Ks^2 lets define the power constant as Pk = 1/4RmKs^2 so Pmax = Pk * omega^2
For example: A generator might produce 70 volts at 2500 rpm. The internal resistance is 7.7 ohms. ks = 2500/70 (rpm per volt) Pk = 25e-6 units are v^2/(ohm rpm^2) at 1000 rpm maximum power is 25.5 watts at 2000 rpm 101.8 watts.
A 2nd example is this 490g outrunner motor:
Rs = 0.03, ks = 295 rpm per volt Pk = 1 / 4*0.03*295^2 = 96e-6 at 1000 rpm, max power is 95 watts