Power Factor Calculation
Power Factor Calculation
(OP)
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.






RE: Power Factor Calculation
It's depend what is a power factor each one is represent.\
You have two options
DPF( displasment PF) and PF
DPF isn't include D=reactive power due to harmonics.
DPF is P/(sqr(P^2+Q^2)
PF is P/(sqr(P^2+Q^2+D^2).
Next point what is your CT/VT(PT) connections
For calculation you have few options:
1. Star connection of PT
2. Delta connection of PT
and ( I don't know IEEE names) Aron connection.
Please more information
Regards.
Slava
RE: Power Factor Calculation
RE: Power Factor Calculation
Slavag
My PT connection is a open delta PT connection using two VT's with 3 wires. The CT connection has all of the X2's of the CT's grounded and the X1's going into the relay inputs.
davidbeach
Why is power factor defined in terms of L-N and has no revelance in relation to L-L. How is the L-N derived if there is no physical nuetral to measure?
RE: Power Factor Calculation
It's classical 3-Wire Delta, 2 and half-Element, 2 VT(PT) & 3 CT connection. you need check, if your power meter have this connection option in the setting, same for protective relay . I don't know this relay, please check setting into relay for the P,Q,S, PF calculation.
Check your PT(VT) connection, as far, as I know in US/Canadien relays and meters you have 4 points for voltage connection V1,V2,V3,Vref in your case I assume you need connect V1 to A, V2&Vref to B, and V3 to C.
Regards.
Slava
RE: Power Factor Calculation
RE: Power Factor Calculation
to relay connected only phase to neutral current, and voltage is depend. for this in newer relay ( what i know, of course) you have option for P,Q,S,PF calculation according to voltage connection and of course calculation : assumed that load is symetrical. possible U1 and Il1 or U12 and Il3, etc..
Regards.
Slava
RE: Power Factor Calculation
ijl
How are the currents in the three phases caused by the phase-nuetral voltages? I thought they were based upon L-L voltages thus I=(pf)(VA)/(L-L V)(1.73). The L-L voltage term is used in this equation, so how does the L-N factor into everything.
Also how does using the L-N voltage cause the 30 degree phase shift that I mentioned?
RE: Power Factor Calculation
RE: Power Factor Calculation
RE: Power Factor Calculation
RE: Power Factor Calculation
A symmetrical three-phase load can be connected in a star or in a delta. When connected in a star, the load consists of three individual one-phase loads. The voltage in the neutral point is zero, so that the currents in the three individual one-phase loads are (clearly) caused by the phase-neutral voltage. If the load is connected in a delta, the currents in the phases are not literally caused by the phase-neutral voltage. But it is possible to solve the equivalent star-connected load that gives the same currents and power as the delta-connected load. If the load is considered as a black box, it is not possible to say from the currents and voltages, how the load is connected.
The power of an individual one-phase load is equal to the phase-neutral voltage times the current times the power factor. The total, three-phase power is three times this one phase power. The phase-neutral voltage can be expressed using the phase-phase voltage in this equation, Vpp = sqrt(3)Vpn, which gives the equation you are using.
And yes, the phase shift of 30 degrees is the result of the subtraction of the phase-neutral votages: Vab = Va-Vb, as vectors (or complex variables).
RE: Power Factor Calculation
RE: Power Factor Calculation
As I mentioned my system isa 4.16kv Wye configuration with a Low resistance ground on the nuetral. My loads on the system are both delta (substations and motors)with a few wye connected 5kV motors.
I'm having trouble understanding this whole L-L vs L-N thing with respect to currents. With wye connected loads it was mentioned above that current consisted of (3) single phase currents based off of the L-N voltage. I always thought however that even in a wye connected system the three phase currents were balanced by each other and returned on the other two legs. Is this the same concept involved with current and voltage differences btwn delta and wye connected transformers?
I'm assuming that since I dont have a neutral the relay/meter is either using Va-Vb to derive the L-N voltage or is using L-L*sqrt(3)? With that I am still confused as to where the 30deg comes into play. Can someone please draw a vector diagram, I may need a pic to see this.
Is there anywhere I can read up on all of this, I'm dying to get my head around this whole vector thing
RE: Power Factor Calculation
RE: Power Factor Calculation
RE: Power Factor Calculation
in additional to Stevenal's post.
Please see some "game" it's help you.
For both your Q's (thread Vectors)
Best Regards.
Slava
RE: Power Factor Calculation
RE: Power Factor Calculation
Gunnar's tool
thread237-214368: Short single, dual and three-phase demo.
RE: Power Factor Calculation
stevenal
Thanks for the diagram, a picture makes it so much easier to see. A few questions though
In my attachment in Figure 1 I'm assuming the 30deg is measured by placing the vectors like in the figure to the right and measuring the angle between them?
In Figure 2 I broke down the power triangle from your attachment and noticed that it is mad up of these two parts. I'm assuming the middle figure is the actual wye connection of the phasors and they are all 120deg apart. What does the outer triangle (figure to the right) represent and why are all of its angles 60deg apart? Is this 60deg signifigant?
Lastly in Figure 3, is the above diagram representation the same as solving the equation in this figure?
Do all relays/meters use L-N for power factor calculation or do some use L-L
Thanks for all of the references and tools guys!!!!
RE: Power Factor Calculation
RE: Power Factor Calculation
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.
RE: Power Factor Calculation
It helps if you add arrows. See the revised attachment.
Convention: Voltage from second subscript to first subscript, e.g. VBA is voltage from A to B.
RE: Power Factor Calculation
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?
RE: Power Factor Calculation
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.
RE: Power Factor Calculation
RE: Power Factor Calculation
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.
RE: Power Factor Calculation