Capacitor Bank Energy Consumption

No. The capacitors will have a very tiny amount of parasitic resistance, but the current (I assume) you are seeing is reactive current, it is not being pulled from the line source as additional load. Do a study of "reactive current flow" to learn more.

thanks Jraef. Does it means that the MCCB protecting the capacitor bank is for the reactive current flow?

Yes.

The capacitor does not consume energy. The meter will see it in combination with your other load. There is a small amount of I²R watts is consumed in the wiring to the caps but is is usually very small.

thanks a lot

let us make it simple.

all capacitor manufacturers state that the capacitors must be mounted and connected as specified. they add that some of the mounting and spacing constraints are intended to avoid overheating. now very simply - for a lay person - if any device, other than a heater, undergoes a rise in temperature then there is a energy loss occurring. it may not always be possible to eliminate it completely - but one must try to reduce it to the best possible extent in order to avoid spending money on wasted (lost) energy.

further - the cables (conductors) used to connect the capacitor also add to the resistive watt loss of the system in proportion to the current in the capacitor.

another consideration worthy of attention is the location of the capacitor in the system. if the capacitor is located nearest to the inductive load then it will reduce the watt loss in the cables feeding the load substantially. if on the other hand it is located at the point of supply then it will correct the power factor at the point of metering but not bring any real energy saving to the consumer because he will still pay for cable losses in feeding the load. this approach of locating the capacitor nearer the load also makes it possible to do correction in smaller steps as required.

Your answer assumes a single load. Most capacitor banks cover a multiple load situation. Adding a capacitor specifically sized to each load is more efficient because the capacitor is only switch in when that load is live. But the space requirements and capital cost in most factories make this approach unrealistic.

In most cases the power loss in feeder cable is several orders of magnitude lower than the load itself and can be ignored.

Capacitor banks generally don't run very hot. The power factor savings will far outweigh the small heating load they impose, but nothing runs fro free.

Quote Jhhassoociates " In most cases the power loss in feeder cable is several orders of magnitude lower than the load itself and can be ignored."

Not always. This point has been discussed many times on this forum. Systems with large motors running 24/7 can benifit greatly with addition of capacitors located at the motor. The saving in I²R losses more that make up for the installed cost of the caps.

Power has two components Active & Reactive. The Capacitor is taking reactive power which diminishes the reactive power of the inductive load, thereby improving power factor. You should see the total power ie KVA & Total energy KVAhr. With improvement in power factor your total KVA will come down & hence KVAhr. will also come down i.e. you save on both demand & energy.

Debasis Nandy

Dear Gigabolt,

As rightly said by many , the power factor you are seeing in the panel is for the load and not for the capactor.

In the name plate of the capacitor there will be rating mensioned as tan-delta. The value will be in the range of 0.004. This is the power facto value of the capacitor unit.

suppose the voltage power rating of a cacotor is 100kvar, 3 ph, 50Hz, 400v,144A,tan d= 0.004 , then the loss will be 100 x 0.004 =0.4 KW =400 w (i.e 4 w / kvar ).

For elaboration refer any standard book on Capacitors.

Dear gigabolt,

It should not be calculated like that. The capacitor will take , theoritically LEADING CURRENT, by a phase angle of 90 Degrees, but in practise, it will be 82 to 85 Degrees. The small resistance of the Capacitor, bank, will consume a very little power and that is to be calculated by the Impedence which is Sq.root of R^2 + (1/Xc)^2 where Xc is 1/ 2.Pi.f.C, where f is the frequency, and C is capacitance in farads.

DHAYANANDHAN.S