Power Factor

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  • #1
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Main Question or Discussion Point

Why is it that the utility will penalize a customer for a low power factor.

I cant imagine why they would penalize you if a purely reactive load puts all its power back into the mains.
 

Answers and Replies

  • #2
russ_watters
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Because it doesn't put all its power back into the grid. Reactive power has no capacity to do work because the voltage and current are out of phase with each other. So essentially, when you take reactive power, you are taking generator capacity (apparent power) that isn't showing up on your meter (real power).
In an electric power system, a load with low power factor draws more current than a load with a high power factor for the same amount of useful power transferred. The higher currents increase the energy lost in the distribution system, and require larger wires and other equipment. Because of the costs of larger equipment and wasted energy, electrical utilities will usually charge a higher cost to industrial or commercial customers where there is a low power factor.
http://en.wikipedia.org/wiki/Power_factor
 
  • #3
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A purely reactive load still draws current and gives rise to undesirable joule heating losses and voltage drops in the power distribution network conductors and components (such as transformer windings). That loss is not recoverable. Allowing very low power factor conditions to prevail could require the system to have much bigger conductor sizes to reduce the losses and voltage drop - thereby increasing the capital cost.
 
  • #4
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Why is it that the utility will penalize a customer for a low power factor. I cant imagine why they would penalize you if a purely reactive load puts all its power back into the mains.
You are correct. A low power factor puts more current in the lines. The utility charges you only for real power, not the reactive part. The extra current in their lines just loads them down with extra I2R losses, which you do not pay for. On the other hand, they may pay YOU for some capacitive power factor. You could put a synchronous capacitor on line (a salient pole synchronous motor w/o load).
 
  • #5
Averagesupernova
Science Advisor
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Because power is wasted in the wires carrying current to the load. Do a search on this. It has been covered here on PF numerous times.

Edit: WOW. Three posts within minutes of each other.
 
  • #6
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had a feeling that, that was the reason, but wasn't sure....thanx
 

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