Generator Efficiency
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Electrical Efficiency
Efficiency as a fraction input mechanical power that is available at the output terminals:
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 output power is obtained. If the load equals the generator resistance, then the maximum output power 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
Maximum generator power as a function of 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