How to differentiate positive Vars from negative Vars?

[il102]

I've read through a lot of old threads on this, and I admit I still do not fully understand. I just want to double check this.

S = P + jQ. The sign of Q is completely dependent on whether the load is leading or lagging, AND which convention you take the sign of P to be in.

For example. For the sake of simplicity lets assume single phase.

Let's say we have a 10MVA, .8 lagging load.
The real power consumption is 10MVA * .8 = 8MW.
The reactive power is sqrt(10^2 - 8^2) = 6MVAR.

The question is, does the load consume 8 + 6J MVA or 8 - 6J MVA. My gut tells me that a lagging load consumes watts and consumes vars, making this an overall load that consumes 8+6J MVA. (alternatively, -8-6J MVA if you arbitrarily reverse the convention of the positive power flow).

For argument's sake, let us prove this. Let's pretend our system is running at 100kV. The current into this load is (10MVA/100kV) = 100 amps. We said the load is lagging at .8 so that means the angle between the current and the voltage is -36.9º. I = (100∟-36.9º) assuming V = (100∟0º).

That means S = V x (I*). S = 10MVA at an angle of positive 36.9º since we took the conjugate. That means S = 8 + 6 MVA. Meaning for a lagging load, the sign is the same for watts and vars. Everything here checks out just fine. Our load is consuming 8 MW and 6 MVAR.

Now lets look at one generator supplying these vars and watts. My gut, which I now think is wrong, would lead me to believe that you need a LEADING generator to supply a LAGGING load. I think I'm about to prove myself wrong but let's see.

Clearly our generator is supplying 10MVA. It also needs to supply MVAR. If we said the power flow into our load is +10MVA that means the power flow out of our generator is -10MVA, right? So our power flow out of the generator is -8-6j MVA I would think. That means S = 10MVA∟36.9º meaning I = 100∟-36.9º. So our generator is lagging as well.

I think I just cleared this up for myself, but I'd appreciate a sanity check.

Bottom line: On an isolated system - a lagging Generator produces VARS that a lagging load consumes. For a lagging generator/load - the sign is always the same for the watts and the vars. For a leading generator/load, the sign is always different for the watts and the vars.

Capacitors could be treated as a LEADING load that produces vars, while a LAGGING induction generator consumes those vars. However, a generator would need to be LAGGING to produce vars that an induction generator consumes.

The subtle difference I believe that was tripping me up was the difference between generator/load.

 

[stevenal]

"+10MVA that means the power flow out of our generator is -10MVA, right?"

No. Since the reference direction is consistently toward the load, both values are positive. I agree your generator is supplying VArs, but these are all positive as well.

Seems like you might be trying to apply conservation of power here. To do so, pick a boundary crossing for your reference direction. Summing the power flow into the conductor will yield zero, meaning if power flow into the generator end of the conductor is positive (out of the generator) then power flow into load end of the conductor (out of the load) is negative.

 

[il102]

Gotcha.

So if someone told you there was a 10MVA load consuming 8+6J MVA, and asked you what S produced was for the generator at the other end of the line, you would say 8+6J?

Easy enough.

 

[stevenal]

Agreed.

 

[electricpete]

For a lagging generator/load - the sign is always the same for the watts and the vars. For a leading generator/load, the sign is always different for the watts and the vars.

Yes, I agree, assuming we use the same fixed reference direction of power flow for real and reactive power (which is perfectly rational). When analysing transmission line problems where the reference direction is not necessarily obvious, some people say that for lagging conditions, the vars flow in the same direction as the watts, while for leading conditions the vars flow in opposite direction of the watts. Inherent in this latter teminology is an assumption that “vars” and “watts” flowing are always positive... they change direction rather than sign (which is sort of the opposite of your assumption that we pick a fixed direction and considered either positive or negative quantities.....both of course equivalent views of the world.)


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(2B)+(2B)' ?

 

[rasevskii]

When talking about VARS and generators, a simple concept is "overexcited=VARS OUT" and "underexcited=VARS IN" as seen by the generator itself. If one refers to the capability diagram of the generator, as you swing to the right of the Zero VARS line (vertical)(1.0 PF)you are putting out VARS to the system, and the DC excitation current is increasing. If you go left, VARS are coming from the system, and thr DC excitation is decreasing.

rasevskii

 

[DTR2011]

I like Mr. Rasevskii's definition. I have used something similar to this when explaining to a generation owner (SCADA person) meter readings on the generator vs. metering at the Inter tie connection.

 

[cranky108]

Vars out of a generator do not always equal vars into the transmission system. Frequently the vars consumed in the Step-up transformer are not consitered.

So it is possible the generator could be producing vars, and the system could be supplying vars. The middle factor is the step-up could be consuming from both sides.

 

[tlrols]

Ha...an easy question. The one that always made my head hurt was measuring the pesky things...