HT Power Factor Compensation

I am not sure what you are asking. Could you re-phrase the question?

7 transformers(1600KVA each) are connected to a 33 kv line.During no load we would like to improve the PF at the HT side (becos of trafo inductance).

The power factor contribytion of the transformers are supposed to be very low and considering you still have the core losses even on no load, I don't think any significant savinge will be achieved by putting the HT compensation.

However check the load pf and you will know whether you need the LT side compensation or not.

In fact since it is still under project stage, find out from your transformer supplier/ data supplied about the no load data and you will know. Typically the noload currents should be of the order of only 0.2% of FLA and at pf of say about 0.5 or so. Since the total current itself is so low, the effect on the total system average pf will not be significant even if the loads are off 50% time.

However i will put more significance to load compensations (ie LT side). It may typically come out much cheaper and the saving in power on HT may not prove more economical.

A simple question: since power factor represents the timing of the amps compared to the voltage, and the only current at "no load" is through the resistance of the transformer coils only, what power factor are you talking about?

There is no lagging PF if there is only resistive loads.

In point of fact the only reason you can connect 33 kV across a thick length of copper wire without it melting is the fact that it is coiled, and so generates equal and opposite inductive voltage to buck the 33 kV, except for the minuscule current through the resistance alone which produces a small amount of heat and a faint hum.

Think about what removing that inductance would look like.

regards, CJM

so if the no load losses are 5 KW for each trafo ,i would get some value of Pf capacitor which should compensate for the lagging pf.

Please understand there is a huge difefrence between no load loss and pf.

Let us say at no load

I_nl = I_nl(0^o) + I_nl(90^o)

The

I_nl(0^o) Corresponds to the losses in the transformers (cu losses, Eddy losses etc)

I_nl(90^o) Corresponds to the inductance of the transformers ( the current required to generate the magnetic flux)

Based on these two, you can gave a Capacitor which compensates by supplying

I_nl(-90^o) Thereby making

I_nl = I_nl(0^o) + I_nl(90^o)+ I_nl(-90^o)= I_nl(0^o)

However when there is a secondary load

I_Ls = I_ls(0^o) + I_ls(90^o)

I_Lp = The total current in primary (corresponding to I_Ls )

= n (I_ls(0^o) + I_ls(90^o)) , n = transformer ratio

= I_lp(0^o) + I_lp(90^o)

Now the total primary current to be compensated is

I_p = I_nl(0^o) + I_nl(90^o) + I_lp(0^o) + I_lp(90^o) = (I_nl(0^o) + I_lp(0^o)) + (I_nl(90^o) + I_lp(90^o))

And the capacitor has to compensate by providing a current

I_nl(-90^o) + I_lp(-90^o)

This current is going to be much more dependant upon load simply because

I_nl(90^o) << I_lp(90^o)

Hence we go for compansating for load specifically and not much bother about comenasting the transformer per se.

The added advantage of compesating load (on secondary) is there when you have loads distributed - the line losses can be reduced for individual compensation at a poin nearer to load (this is already being discussed elsewhere)