How does correcting power factor with capacitor banks affect I^2 R losses?

[biguelo]

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 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 having technical losses in the transformer and all the conductor. 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.

 

[FlexGunship]

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 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 having technical losses in the transformer and all the conductor. 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.

The important point here is that you've specified a "kWh" meter which does not account for out-of-phase power. In fact, a "Watt" assumes that power factor has already been taken into account. So, if you rectify your source with a capacitor bank and change the effective power factor, then your meter will read higher.

 

[Okefenokee]

The meter only measures current. It doesn't care if that current is carrying energy into or out of a load.

Nearly all major loads are inductive so it's well known that capacitors can improve your power factor and save money on your power bill. The capacitors must be between the meter and the load. It doesn't matter which side of the transformer that you put your capacitors on, as long as the capacitors have the appropriate power and voltage rating for their position. If the capacitors are on the secondary side of the transformer then all you have to do is convert the total secondary load (the actual load plus the capacitor bank) to the primary side to get the total impedance as seen from the meter.

$$Z_{in} = \frac{1}{N^2} (Z_l + Z_C)$$

 

[russ_watters]

This isn't really a EE issue it is a rate issue: the electric company measures and penalizes you for a low power factor.

 

[Bob S]

If you put the capacitor bank close to the motor (downstream of the transformer) then the reactive currents in the transformer coils will be reduced, and also the I-squared R power losses in the transformer coils. This will reduce the power meter readings.

Bob S

 

[mheslep]

I This will reduce the power meter readings.

Bob S

Not if the OP's "commercial power meter" is the traditional two coil spinning disk meter. Those measure instantaneous power, real power. The I^2R losses will lower some with the improved power factor as you suggest, but aside from those the real power displayed on the meter will not change.

 

[mheslep]

The meter only measures current. It doesn't care if that current is carrying energy into or out of a load.

This kind of meter measures both voltage and current, instantaneously.
http://en.wikipedia.org/wiki/File:Mechanical_electricity_meter_1965_(1).jpg

 

[biguelo]

Not if the OP's "commercial power meter" is the traditional two coil spinning disk meter. Those measure instantaneous power, real power. The I^2R losses will lower some with the improved power factor as you suggest, but aside from those the real power displayed on the meter will not change.

Hi, my concern is that since the current in the I^2R formula is made up of two components, the active current and the reactive. The reactive current is not in phase with the R component of the Z, and since the capacitor act only over the reactive component of the total current, then there won't be any reduction in the I^2R losses.

Regards.

 

[mheslep]

Hi, my concern is that since the current in the I^2R formula is made up of two components, the active current and the reactive. The reactive current is not in phase with the R component of the Z, and since the capacitor act only over the reactive component of the total current, then there won't be any reduction in the I^2R losses.

Regards.

So we are clear, where are your I^2 R losses now? If they are upstream from your load (the transformer?), then correcting the power factor of the total load (ie by placing the C downstream from the transformer in front of the load) will reduce the total current through the transformer and thus reduce the I^2 R losses there.