How much prime mover horsepower is required to generate kVARs? 2

[gentech1975]

At work we have several different opinions ref the horsepower required to generate kVARs. I know there is a direct relationship to Kw and prime mover horsepower. I have read many articles ref kVAR and that there is no relationship to almost no required horsepower to deliver kVAR.
I am looking for a formula to convert kVAR to Hp. Much the same idea as the formula to convert Kw to Hp.
I've been researching this for the past 3 weeks and have came up with a multitude of explanations that there is no Hp required to generate kVAR. So I am convinced and so are most of the others at work. But I will need a mathimatical formula to convince the last 2 remaining hold outs. Thanks.

 

[alehman]

In a mathematical sense apparent power (kVA) is represented by a real component, kW and an imaginary or reactive component, kVAR. kVAR can be inductive or capacitive.

P (real power in watts) = V * I * cos p
Q (reactive power in VAR) = V * I * sin p
S (apparent power in kVA) = sqrt(P^2 + Q^2)

where p is the phase angle by which current lags voltage

hp = P / 746

You'll note there is no Q in the above equation because no horsepower is required to generate reactive power (KVAR) component, because it is not REAL power. Therefore you will not find a formula to convert kVAR to hp.

Now, the reality is that the reactive power requires current to flow which causes a small amount of REAL power to be dissipated as heat. This loss is one reason it is good to maintain operating power factor ( cos p ) close to 1.

In a theoretical sense, you are correct. In reality, a small amount of real power results from flow of reactive power.

This is all explained in any power systems text book.

 

[waross]

This is a two part answer, and your question has a challenging wording.
How much power is required to produce VARs. None.
How much power is required to deliver VARs. It depends.
VARs are Volt-Amps-Reactive. The current, or amps must be carried by the generator windings, transformer windings and conductors. The basic formula of I^2 R operates. The reactive current flowing in the conductors and windings produces heat and the equivalent energy must be supplied by the prime mover.
If there is 100 amps reactive current and no active current then losses or the energy to deliver the VARs will be based on 100 amps.
If there is 100 amps real current and no reactive current then the losses will still be based on 100 amps.
Here's where it gets a little more interesting.
If there is 100 amps reactive current and 100 amps real current then the resultant current will be 141 amps.
The losses will be based on 141 amps. Half of this will be caused by the reactive current and half will be caused by the real current.
However if we now reduce the reactive current by half of 141, the reactive current will be 29.5 amps and the resultant line current will still be above 100 amps. The line current will be the vector sum of 29.5 amps and 100 amps acting at an angle of 90 degrees to each other.
And by the way, horse power is real power, KW is real power and KVARs are not real power. There is no formula to convert KVARs to HP. The horsepower to DELIVER KVARs (as opposed to GENERATING KVARs) depends on the current and the resistance of the circuit.
I hope your group can all agree on this.
respectfully

 

[ScottyUK]

There are a number of reasons for real power consumption while supplying a purely reactive load. Usually they can be neglected, but on a big machine it might well equate to a few MW. Even a machine in service as a synchronous condenser has to draw field current from somewhere. Then there are windage, stator iron, and stator copper losses, again all consuming real power.

Perhaps this is getting too deep into real world behaviour.


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