Power Factor & kWH?

[Richard J Kinch]

No.

[Richard J Kinch]

No, there is still an economic penalty to the current capability of the generation plant being used up for no delivered power. Capital is tied up without producing power for the customer.

This may be more costly than the line losses today.

It is not much different than the peak demand charges that even small commerical customers now endure, and which ought to be paid by everyone. The true cost of electric power is not just fuel and transimission, but the capital investment for peak capacity.

[Bud--]

In a motor. the current creates a magnetic field which varies in magnitude and direction during the cycle. Some energy is stored in the magnetic field for part of the cycle and sent back toward the utility on another part of the cycle. The 300 watts of reactive power is from energy being stored and returned 120 times per second and is not power that disappears. Because of the properties an inductor the current is shifted and the current flow sine wave is later than the voltage sinewave and the reactive power is 90 degrees out of phase with the real power.

A lightly loaded motor will have a lower PF than fully loaded motor because the real power of the fully loaded motor will be higher and will be a higher percentage of the total power. My memory of dim teachings is that the reactive power doesn't change much with loading.

The higher current caused by the magnetic field interchange does cause real I squared R losses in wire as detailed in other posts.

In a capacitor energy is stored in an electric field and is also stored and released 120 times per cycle. But in a capacitor the current leads the voltage. Power factor correction capacitors cancel the effects of inductance.

Over correction - too much capacitance to balance the inductance can cause resonance (if I remember right), which can cause the voltage to rise as in the post by Beachcomber. This can be made worse by harmonics which can be generated by VFDs, power supplies, dimmers.

bud--