Power Systems — Distribution

The power-factor penalty

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

A poor PF means the wires carry current that does no work — and the utility pays for those I²R losses, so they bill you for it. The current saved by correcting PF from 0.75 to 0.95:

P_{plant} := 200e3 W

= 200000 W plant load

I_{at_75} = \frac{P_{plant}}{\sqrt{3} \cdot V_{LL} \cdot 0.75}

= 320.75 A line current at PF 0.75

I_{at_95} = \frac{P_{plant}}{\sqrt{3} \cdot V_{LL} \cdot 0.95}

= 253.22 A same work at PF 0.95

I_{saved} = I_{at_75} - I_{at_95}

= 67.526 A ~68 A of pure waste, eliminated

That completes the Electrical track: DC fundamentals, AC behavior, electronics for signals, motors for motion, and the grid that feeds them all.