Power Factor correction

[Gerry de Pedro]

Our system's primary is 11KV coming from several sets generator connected to HV switchgear to transformer of 1500 KVA, 50 hz.,11KV pri - 415/230V Secondary. Don't we really need to install our Capacitor Bank of 400 KVAR for PF at the secondary because they said that since our source are Generator sets, we don't have to connect it because it already has a PF of 0.8? The only thing I know that corrects PF are Capacitor Banks and Synchronous Motor.Is the generator enough to correct the power Factor?

 

[charlie b]

Re: Power Factor correction

Originally posted by Gerry de Pedro: Is the generator enough to correct the power Factor?

The generator will not correct for power factor, it will supply whatever power factor the system demands of it. Most of the loads that we encounter are inductive in nature. Too much of that inductive load causes the power factor to drop too low. That is when it starts making sense to look into power factor correction.

But it is always a question of how low you can tolerate, and how much you are willing to pay to get the problem corrected. I am guessing that a 0.8 power factor on the primary side of your transformer would be too low. So I would look into placing some capacitors on the primary side.

As to the secondary side, if you use a capacitor bank it should be as close to the load as you can get. It makes no sense to place a capacitor bank right at the secondary side of a large transformer. It will gain you nothing.

The only thing I know that corrects PF are Capacitor Banks and Synchronous Motor.

If you and overexcite a synchronous generator, it becomes a synchronous condenser, and that will counteract the effects of inductive loads. But generators are not generally operated in an overexcited mode.

 

[Gerry de Pedro]

Re: Power Factor correction

Charlie,
Thanks for your reply.

How is the Capacitor Bank at the secondary side be of no sense to use? Am I right to say that KVAR would just go back and forth from the inductive loads to Capacitor Bank and would increase the capacity of the Generator Sets as well as the Transformers thereby correcting the PF to an ideal and acceptable value? I am concern with the transformer as well the generator whose capacity can be used in other future loads if PF is corrected at the secondary.Or maybe your are thinking that because the genset is ours and that we are not using a source from a utility company.

 

[bob]

Re: Power Factor correction

It makes no sense to place a capacitor bank right at the secondary side of a large transformer. It will gain you nothing.

Gerry
I think that Charlie was suggesting that you install the capacitors at the motor loads. However
you can install them at the transformer if you want. They will reflect back to the primary and improve the PF. Suggest you remember harmonic loads and Caps do not go together well.

 

[charlie b]

Re: Power Factor correction

Let's talk "inductive" for the moment. The basic concept of "reactive power" is that it represents an energy exchange between the magnetic field of the generator (or in your case, the transformer) and the magnetic field of the motor. This energy exchange is above and beyond the energy exchange that permits the motor to do its work: to drive the pump so that it can push water down a pipe.

This exchange of energy causes some additional current to flow in the conductors between the transformer and the motor. This causes two problems:
</font>

• <font size="2" face="Verdana, Helvetica, sans-serif">First, the additional current will result in additional I*2R losses, and represents a waste of energy.</font>

<font size="2" face="Verdana, Helvetica, sans-serif"> </font>

• <font size="2" face="Verdana, Helvetica, sans-serif">Secondly, the additional current will result in a greater voltage drop within the transformer (i.e., due to the resistance of its windings). Therefore, the voltage available at the transformer terminals will be lower, and that will be felt throughout the system.</font>

<font size="2" face="Verdana, Helvetica, sans-serif">Now let us add a capacitor bank. What will happen is that you will now have an energy exchange between the magnetic field of the motor and the electric field of the capacitors. If you pick the right sizes, then the transformer will not longer play a role in this energy exchange. It takes place downstream of the transformer. There will be the same type of "additional current flow" that I described above, but it will take place between the motor and the capacitors.

Starting at the point at which the capacitor bank in installed, and looking downstream to the motor, there will be some additional I*2R losses and some additional voltage drop (again, "additional" meaning above and beyond what is needed to make the motor do its work). Thus, the advantage of installing the capacitor bank close to the motor is that you minimize the I*2R losses and the voltage drop.

But if you put the capacitor bank close to the transformer, then there will be more I*2R losses and a greater voltage drop along the entire length of conductors. That is what I meant, when I said that placing a capacitor bank close to the transformer is of no real value.

But Bob has a good point. Let us start at the point at which the capacitor bank in installed, and looking upstream to the transformer. The energy exchange going on because of inductive and capacitive stuff further downstream will not be seen by the transformer. The additional current flowing because of the inductive and capacitive stuff will not flow from (or within) the transformer. Therefore, there will be no additional voltage drop within the transformer, and its output voltage will be higher. So yes, the power factor seen at the transformer secondary will be higher. Also, that will reflect to the primary side, so there will be a power factor improvement on the primary side.

So let me revise my statement about placing the capacitor bank at the transformer. Let me replace the "It will gain you nothing" with "It will gain you something, but not as much as you could gain."

 

[rattus]

Re: Power Factor correction

Wherever you place them, the caps should reduce or eliminate the power factor penalty imposed by the poco.

[ September 07, 2005, 10:23 AM: Message edited by: rattus ]

 

[laidman]

Re: Power Factor correction

If there is one piece of equipment that is causing the power factor problem. Placing the capacitors near the source could result in a swing of power factor in the opposite direction when the equipment is not running. I would keep the caps as close as possible to where they are needed.

 

[petersonra]

Re: Power Factor correction

Originally posted by laidman:
If there is one piece of equipment that is causing the power factor problem. Placing the capacitors near the source could result in a swing of power factor in the opposite direction when the equipment is not running. I would keep the caps as close as possible to where they are needed.

it is not a matter of distance. it does not change much if you wire the PFC caps so they are always energized, but install them close to an inductive load. They need to be switched with the load.

 

[laidman]

Re: Power Factor correction

you are correct that was what I meant. but my hands get ahead of my brain at times

Thanx
Ted

 

[Basra123]

Re: Power Factor correction

Charlie,

you mentioned that additional current will ensue from the exchange of energy between the capacitive and inductive loads or between the magnetic fields of the transformer and the motor. My question is what is the source of this additional current? how it is generated?

 

[rattus]

Re: Power Factor correction

Basra,

Look at it this way:

Current in an inductive load lags the voltage and comprises a real component and a reactive component. The reactive part contributes nothing to the work being done, but it does generate I^2*R losses, and I*R voltage drops which are undesirable.

Current in a capacitive load leads the voltage and also consists of a real component and reactive component.

These reactive components are 180 degrees apart and tend to cancel each other when added. If we add the right amount of capacitance in parallel with the inductive load, we can eliminate the reactive current all the way back to the generator.

You need to brush up on your trig first and then read about power factor before you can get a handle on this. This subject has caused much grief and frustration for engineering students over the years, and it still does.

 

[Basra123]

Re: Power Factor correction

Rattus,
thanks for yoru explination! I understand this relationship well but what really throw me off is the term "additional Current"...thanks again

 

[charlie b]

Re: Power Factor correction

That term was intended to be a simple way of saying this: A load with both inductance and resistance will draw more current than the resistance would draw by itself. In other words, the "additional current" is in "addition" to the current that the resistance would draw by itself.