Power Factor Calculation

[rockman7892]

I am looking at a Multilin 750 relay and PQMII meter and trying to corrolate the power factor on each to the respecitve voltage and current phasors. With a power factor of .70 I expected to see the current phasor lag the voltage phasor by 45 degrees. However when I looked I saw Vab at 0 degrees and Ia at 80 degrees as an example.

When wondering why I saw 80 degrees instead of the 45 I expected I called Multilin and they explained to me that it was because the power factor was taken with respect to Van and therefore presented a 30 degree shift. By adding this 30 degree shift to my expected 45 degree angle I come up with 75 degrees which is close to the angle which I am seeing.

Why is the power factor calculated off of the L-N volage as opposed to the L-L voltage. Where does this 30 degree shift come from? Does is come from subtraction of Va-Vb? I have a wye LRG system so do not carry a nuetral into these relays. How is the neutral then calculated or derived.

I would appreciate any explanations or examples.

 

[rockman7892]

Woops sorry forgot the atachment!

 

[stevenal]

I don't like your equations. Va with no reference is meaningless, but is commonly used as shorthand for Van.

Van = Va
Vab = Van-Vbn.

Yes, you have 60 degrees between AB and CB. likewise, you have 60 degrees between AN and NC. Generally, though, we speak of phasors AB, BC, and CA which are displaced from each other by 120 degrees.

 

[jghrist]

rockman,

It helps if you add arrows. See the revised attachment.
Convention: Voltage from second subscript to first subscript, e.g. V_BA is voltage from A to B.

 

[rockman7892]

So when you see a particular rotation drawing just showing a,b,c is it assumned that these are ab,bc, and ac vecotrs or just a,b,and c vectors. Or are these just a general rotation drawing to show what direction rotation is?

When drawing the ab,bc,ca vectors does it matter if they are drawn 120deg apart, or drawn in a triangle at 60deg apart. I'm assuming that theoryetically they are 120deg apart and are drawn in a triangle at 60deg simply for problem solving purposes? Can an analysis be messed up if these two are confused?

So if my meter shows AB=0deg BC=240deg and CA=120deg, then my an, bn, and cn vecotrs are all 30deg behind these?

 

[stevenal]

1. As I said before, A is short for An.
2. ab, bc, and ca are 120 degrees apart, no matter if they drawn radially or as a triangle.
3. Yes your analysis will be wrong if you consider them to 60 degrees.
4. Yes, assuming you have abc rotation.

 

[jghrist]

I'm assuming that theoryetically they are 120deg apart and are drawn in a triangle at 60deg simply for problem solving purposes?

When drawn in a triangle, the angles between vectors are not obvious because the tails of the vectors are not drawn at the same point.

 

[sibeen]

When drawn in a triangle, the angles between vectors are not obvious because the tails of the vectors are not drawn at the same point.

Try this diagram. I've moved the line-line voltages to the same point as the line-neutral voltages. Hopefully it shows it better.

 

[pwrengrds]

The reason you saw the higher angle is because the relay uses delta PT's and shows the angle of the current from 0 degrees. The AB phase angle is shifted 30 degrees (draw a vector from A to B on a phasor diagram and note its 30 degrees positive. So the angle you were seeing referenced from 0 was 80-30=50 which is about right. It's just the way the Multilin shows angles when using a delta PT.