5 replies to this topic

[kana]

Dear All,

 

There is an existing installation which has already been equipped with automatic power factor correction capacitors to correct the overall system power factor to 0.95 lagging and the capacitor bank is doing its job well.  

 

I understand that very little kW saving is achieved due to the reduced cable losses between the distribution transformer and the main switchboard where the capacitors are being installed. Theoretically the reduced cable losses shall be reflected in the electricity bill however due to the saving being extremely little, the payable amount for kWH still remains as per the usage.  

 

In order to see substantial saving in several kW by reducing the cable losses, the plant engineer suggested to install static power factor correction to the existing loads in addition to the existing capacitor bank in the main switchboard.

 

Just for the information, the existing cables (supplying loads located approx. 30 to 100m from the main switchboard) were critically sized and it appears that static correction in each feeder of large loads can result savings in the range of 0.5kW to 1kW. The engineer is of the opinion that kVA reduction in each feeder this will provide a cumulative saving due to the reduced cable losses.

 

I'm of the opinion that no savings can be achieved because the existing capacitor bank is decently supplying the loads with the required kVAR. No doubt the loads are drawing KVA however the additional current due to low power factor in which accounts for the I2R losses is now compensated by the capacitor bank. By installing static correction,the I2R losses on the cable which has no effect at all on the system will be saved. I appreciate some advice on this.  

 

Thanks,

 

Regards,

Kana  

[marke]

Hello Kana

 

You will only save KW in the cable between the point where the KW usage is metered and where the power factor correction is connected.

If the correction is connected at the main switchboard, and the power consumption is measured at the main switchboard, you will make no KW savings at all.

 

If you can move the correction closer to the load, then you will increase the KW saved, but the cable losses should be low, so the savings should be even less.

If the cable losses are significant, then you probably need to look at the cable rating and possibly increase its size.

 

The current between the load and the connection of the power factor bank, is not reduced. The current is only reduced between the power factor correction connection point and the supply.

 

Best regards,

Mark.

[kana]

Hello Mark,

 

I agree with your statement that the current between the load and the connection point of the power factor bank, is not reduced. 

 

Is any saving in terms cable losses (that can be measured by the energy Meter at the incoming) achieved by installing static capacitor to reduce the current between the load and the existing capacitor bank?   

 

Regards,

Kana

[marke]

Hello Kana

 

Yes, if you apply static correction at the load, you will reduce the current flow in the cable to the load and therefore reduce the copper losses in those cables.

The addition of static correction at the load will reduce the reactive current (KVARs) flowing at the switchboard, and therefore the number of banks switched on the bank corrector will reduce.

 

The magnitude of the energy saved is dependent on the KVAR reduction and the KW losses in the cables.

 

Best regards,

Mark

[kana]

Hello Mark,

 

Let's say that the power factor correction is done at the main switchboard instead of the load end. The load will be drawing current I, in which the I cos (theta) will be supplied by the transformer and the I sin (theta) will be supplied by the capacitor. The cable losses (kW) measured by the energy meter will be due to the current I cos (theta) and not I because the reactive current is supplied by the capacitor. Please advice if my understanding is correct.  

 

Thanks,

 

Regards,

Kanagaraj    

[marke]

Hello Kana

 

At any point on the electrical supply to the load, the KW loss in the cable is the square of the resultant current passing through the cable, times the resistance of the cable.

If you have some reactive current, typically inductive current, then that will contribute to the resultant current.

 

If we break it down to Ir (resistive current) and Ix (reactive current) then the current through the cable is the square root of (Ir x Ir + Ix x Ix)

 

Adding power factor correction at the supply point, will not alter Ix flowing down the cable to the load.

Adding power factor correction at the load will reduce the current flowing down the cable to the load by reducing Ix.

 

If we have a power factor of say 0.7, then the Ix is equal to Ir and so the total current is 1.414 x Ir. The power loss in the cable is 2 x Ir x Ir x R.

If we apply correction to 1.0, then the total current is equal to Ir. (Ix = 0) so the power loss is Ir x Ir x R. We have halved the power loss in the cable, but the power loss in the cable should not be significant relative to the load power, so the losses should be small and the energy saved even smaller.

 

Best regards,

Mark.