NapkinCalc

AC Motors

Power-factor correction

continues from lesson 4 — values defined earlier in the course stay live here

ELI5: that 0.86 power factor means the line carries extra current that does no useful work (it just sloshes back and forth magnetizing the motor). A capacitor bank supplies that reactive current locally, unloading the wiring upstream.

PFtarget=0.95PF_{target} = 0.95
Qc=Pin(tan(acos(PF))tan(acos(PFtarget)))Q_{c} = P_{in} \cdot \left(\mathrm{tan}\left(\mathrm{acos}\left(PF\right)\right) - \mathrm{tan}\left(\mathrm{acos}\left(PF_{target}\right)\right)\right) = 2181.4 W capacitor bank size (VAR)

Real-world hook: factories add capacitor banks because utilities bill a penalty for poor power factor — correcting it pays for itself. (Try p_poles := 2 to re-derive a 3600 rpm machine.) Where next: Power Systems scales all this to kilovolts.