NapkinCalc

Power Systems — Distribution

The power-factor penalty

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

ELI5: a poor power factor means the wires carry current that does no work — yet the utility still pays for the I²R heat it causes, so they bill big customers for it. Correcting PF from 0.75 to 0.95 sheds a startling amount of wasted current.

Pplant:=200e3WP_{plant} := 200e3 W = 200000 W plant load
Iat75=Pplant3⋅VLL⋅0.75I_{at_75} = \frac{P_{plant}}{\sqrt{3} \cdot V_{LL} \cdot 0.75} = 320.75 A line current at PF 0.75
Iat95=Pplant3⋅VLL⋅0.95I_{at_95} = \frac{P_{plant}}{\sqrt{3} \cdot V_{LL} \cdot 0.95} = 253.22 A same work at PF 0.95
Isaved=Iat75−Iat95I_{saved} = I_{at_75} - I_{at_95} = 67.526 A ~68 A of pure waste, eliminated

That completes the Electrical track — DC fundamentals, AC behaviour, electronics for signals, motors for motion, and the continent-sized grid that feeds them all.