Power factor correction and I^2 R losses

[cbinla]

I'm having a disagreement with a fellow engineer regarding power factor correction and line losses. I feel I am right, but I'm getting tripped up over something he is stating that he says proves me wrong.

I believe that improving the power factor of a system can reduce the loss in the feeder to the load. For example, suppose there is a motor operating at some lagging power factor. I believe the magnitude of the current being delivered by the power source feeding the motor can be reduced by the addition of a capacitor near the load.

In other words, for a given power system, S = P + jQ, reducing Q (kvar) reduces S (VA). Because S, the overall power (complex power in VA) is reduced, the magnitude of current is reduced.

If the magnitude of the current is reduced, I believe the I^2 R losses are reduced.

Ok here is where I'm getting tripped up. Is the I^2 R losses in the system part of the real power P in the equation S = P + jQ? If the feeder is part of the impedance of the system, the reduction of the I^2 R losses means that P in the equation S = P + jQ is reduced, correct?

What he is saying is that I^2 R losses are not reduced. He says that since only kvar is being added to the system, the real power P remains the same. Since I'm saying that the power is reduced (I'm saying that I^2 R watts are reduced) he is saying this is impossible. Adding kvar to improve power factor can't reduce the watts.

Can someone help me with my understanding of the relationship between I^2 R losses and P in the equation S = P + jQ? This is very relevant to the PE exam I'm taking soon and though I feel I can solve the power factor questions, I'm getting lost in the meaning behind the equations.

Thanks. This board is great.

 

[Flyer_PE]

You are correct in thinking that adding a capacitor bank at the motor terminals will reduce the I2R losses in the cable.

The I2R losses in the cable are not determined by the real power dissipated in the load. They are determined by the apparent power.

Consider a single phase system where S = VI. A reduction in the reactive component will result in a reduction in current.

If you reduce current, you reduce I2R losses by definition.

 

[DK PE]

Flyer's answer is great but I'd suggest to settle this with your colleague you draw up a simple example using a long feed to say a 100HP motor with a lagging power factor of 0.8, then determine the line current. Then correct to 0.95 and determine the new line current. Then examine the difference in i2 R losses for your feeder conductors in each case. You probably won't see a lot of losses in each case but if you think of the total path all the way back to the source (generator, but remember some of these losses you pay for and some are the POCO's transmission wires being heated) and then add up all the i2 R losses for each of these uncorrected loads, you can see why POCOs hate (and financially penalize) those customers with poor power factors.

 

[Peele1]

Yes, improving the pf will reduce the current. As stated above, do the calculations yourself.

And, for additional documentation, NFPA 70, the National Electric Code, 460.9 states that the rating of the overload device shall be based on the improved power factor of the motor circuit. There is a question like this in the NCEES sample questions book.

 

[cbinla]

You are correct in thinking that adding a capacitor bank at the motor terminals will reduce the I2R losses in the cable.
The I2R losses in the cable are not determined by the real power dissipated in the load. They are determined by the apparent power.

Consider a single phase system where S = VI. A reduction in the reactive component will result in a reduction in current.

If you reduce current, you reduce I2R losses by definition.

Thanks for your responses. I understand that the losses in the line are reduced because the current is reduced.

But...

Flyer, with regards to your second paragraph, I'm still struggling with my understanding because I^2 R losses are in watts and P in the equation S = P + jQ is in watts, AND the feeder is part of the load (impedance) as seen by the source. I'm having trouble de-linking the line losses from P conceptually.

 

[Flyer_PE]

Ok. Let's try this:

For any impedance, be it cable or otherwise:

S = P + jQ

Impedance Z = R + jX

P = I2R

Q = I2X

 

[cbinla]

Ok. Let's try this:
For any impedance, be it cable or otherwise:

S = P + jQ

Impedance Z = R + jX

P = I2R

Q = I2X

Thank you all.

From considering your responses and pondering, I know how to clarify this for him.

P in the equation S = P + jQ is the real power and represents the capacity of the circuit to do real work in a particular amount of time.

The watts given off by the losses due to I^2 R are given off as heat and do not contribute to the work done by the circuit.

Like you all said, reducing the current due to the reactive load reduces the overall current. Current due to a reactive load, even for a purely reactive load, still flows through the conductors and contribute to the heat loss.

Thanks