NapkinCalc

AC Motors

Power-factor correction

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

That 0.86 power factor means the line carries current that does no work. A capacitor bank supplies the reactive part locally, unloading the wiring upstream. The size needed to reach a 0.95 target:

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)

Where next: the Σ library has three-phase power, motor current and RLC impedance formulas — and try p_poles := 2 above to see the whole worksheet re-derive a 3600 rpm machine.