Power Factor Correction Benefits

[bgl2010]

Hi,

I have read that one of the benefits to increase the power factor through capacitor banks is the reduction of technical losses.

Here is my doubt, if I have a comercial kWh-kW meter in medium voltage and it's metering the active energy of a facility with a medium voltaje transformer (500 kva) and low pf (0.7), hence the technical losses in the transformer and all the conductor are been metered. If I install a capacitor bank, in the low voltage side of the transformer, that rise the power factor to (0.95) will this effect the amount of energy active energy in the meter.

My concern is that since the capacitor will only reduce the reactive component of the total current, and the reactive component is not in phase with the transformer and conductors resistents, the active energy should not be effected.

Regards.

 

[rbulsara]

See thread238-277959.

Yes, active "power" is not affected. What is technical losses? Only losses affected are copper losses that is I^2*R.

Rafiq Bulsara

http://www.srengineersct.com

 

[waross]

Simple example;
Reactive current 3 Amps.
Active current 4 Amps.
Total current 5 Amps.
If you correct the power factor to unity the total current will be reduced from 5 Amps to 4 Amps.
Your I²R losses in the transformer will be reduced by a factor of 16R/25R or 36% in this example.
The power factor in this example is 80%. The I²R losses in the transformer will depend on the power factor.
Before you get too excited, the I²R losses are only part of the losses in the transformer.
The transformer may be 95% efficient. By correcting the power factor you may be saving 36% of part of 5%.
BTW I don't know what technical current is. Do you have another word for "technical" or can one of our multilingual members comment? This may be a "Lost in translation" issue.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[bgl2010]

"Reactive current 3 Amps.
Active current 4 Amps.
Total current 5 Amps.
If you correct the power factor to unity the total current will be reduced from 5 Amps to 4 Amps."

Hi waross,

But why in first place you have to use the 5 Amps. to calculate the losses (I^2R). I thought that only the active current was in phase with R, so before and after the power factor improvement the current to calculate the loss should be the active current (4 Amps.)

Regards.

 

[desrod]

Hi waross
Technical losses are losses associated with the process of generating, transmitting and distributing electricity over non perfect devices.

It is in opposition with non technical losses that are related with human factors like using electricity by tapping on the wrong side of the meter, bypassing the meter, not paying the bills etc.

In some countries non technical losses are up to 30%....and technical losses are kind of irrelevant.
I heard of distribution transformers overloading from such unauthorized electricity usage.

 

[stevenal]

Resistance is not sinusoidal quantity that has any kind of phase relationship with a sinusoidal current flowing through it. The full magnitude of the current causes the I^2*R loss.

For a pure resistance, the voltage drop across the resistance is in phase with the current through it no matter what the rest of the circuit looks like.

 

[rbulsara]

bgl2010:

Just two points:
1. PF correction does not change real power consumed by the load.
2. The current I in the R of your transformer will see total "apparent" amps. Taking Bill's example, it is 5 amps before the correction and 4 amps after the correction. The reduction in losses is only the difference between those two I^2*R values.

If this is far too confusing, I am afraid this forum cannot be a substiture for a formal electrical engineering course. Perhaps you can look up some text books or even do a careful web search. Not everything posted on internet is accurate though, so review with a good electrical friend or a sr. engineer where you may be working.

May be drawing a circuit diagram and phasor diagrams would help.


Rafiq Bulsara

http://www.srengineersct.com

 

[electricpete]

waross:"Reactive current 3 Amps.
Active current 4 Amps.
Total current 5 Amps.
If you correct the power factor to unity the total current will be reduced from 5 Amps to 4 Amps."

desrod: But why in first place you have to use the 5 Amps. to calculate the losses (I^2R). I thought that only the active current was in phase with R, so before and after the power factor improvement the current to calculate the loss should be the active current (4 Amps.)

I agree with comments from the others. And yes it is basic but I will take a stab to explain.

Let's look at the system:[TT]

+ ============Rseries===============
Vs Zload
-===================================
[/TT]
Where
Vs = 100 vac (angle zero)
Z = 16 + j*12
Rseries = 0.01 represents transmission or distribution loss.

To determine current, we can practically neglect Rseries:
I = 100 / (16-+j*12) = 4 - j3

How to determine transmission losses in Rseries:
|I|^2*Rseries = 5^2 * 0.01 = 0.25 watts.

Notice the angle of I makes no difference. When you talked about angle of I you were comparing it to the source voltage.... that has no relevance. The angle of current with respect to that voltage across Rseries (call it V_Rseries) is the relevant voltage drop. Because Rseries is purely resistive, it's voltage drop is always in phase with its current.

Change the angle of Zload (while keeping magnitude the same) and you will see it does not change losses in Rseries. The reason it works this way is because Rseries << Zload. So for purposes of calculating losses in Rseries we can effectively replace the rest of the circuit with a current source I = V/|Zload|. Angle of Zload has no significance to this calculation because V_Rseries will always be in phase with I_Rseries

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[LionelHutz]

Around here, the only benefit of power factor correction with a possible pay back is the reduction of the demand or power factor penalty charges on the bill.

 

[bgl2010]

Thank you all for your replies.

Regards.