Power Consumption of 63 mVAR Bus Reactor In 400 kV Substation

Kindly turn off your Caps Lock feature. It is frequently read as yelling loudly.

You do realize that a reactor (a.k.a. an inductor) uses no power. There will be resistance in the wire, so Ohms law applies. Beyond that, you should turn to page 149 of your text book and start reading from there.

If you can afford to purchase, install and operate a 63 MVAR unit on 400 KV three phase feed system its power consumption is not worth worrying about.

Do you worry about the cost of power consumption for the operation of a USB charger when you own a million dollar mansion?

Consult the reactor nameplate for the impedance value.

The reactor is required to offset capacitance of the electrical system.

The very small amount of power consumed by heat losses will be determined by the amount of work the reactor has to do in order to counteract system capacitance.

The more radical the changes in system capacitance becomes, the more work the reactor will have to do therefore heat losses will increase.

Contact the reactor manufacturer for their power consumption table and other pertinent information.

Here is a link to very informative site.

http://top10electrical.blogspot.com/2014/08/fixed-shunt-reactor.html

Do the owners of your 250MW station realise just how incompetent some one of their staff are is?

Your question is one of a series. Questions that if I were in your situation I wouldn't be asking a group of guys on the internet, I would be talking to manufacturers and specialists in their particular field.

STOP SHOUTING!

I suspect that OP is a newbie technician in a generating station and is surrounded by personnel that are only concerned with how things work, not how they're engineered, and therefore has no access to any technical information.

That being said, more information is required before the answer is possible; i.e., is this is a shunt or series reactor, air/iron core, is it switched, how is it connected, and what is the nameplate information especially its impedance?

Once you know the above information you can use Ohm's Law, the Power Triangle, and the Per Unit system to easily calculate the losses in the reactor and the system feeding it. Hint: they're probably not trivial, but are considered "technical losses" which are just part of the cost of doing business as an electrical utility.

See [for instance]:

http://www.transformerscommittee.com/subcommittees/performance/F06-C57.21WG_Presentation-LossesOnShuntReactors.pdf