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Power Factor Calculation

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rockman7892

Electrical
Apr 7, 2008
1,171

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.
 
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Hi.
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
 
Power factor is defined in terms of L-N voltage and can not be meaningfully calculated using L-L voltage.
 

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?
 
Hi
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
 
As the name implies, the power factor is used to calculate the power of a load (or generator). Power = voltage over the load times the current through the load times the power factor. In order for this power to be meaningful, the voltage and current must be related just like this, voltage over and current through the load. In other words, the current must be "caused" by the voltage. In a three phase system, the power factor is mostly calculated using the phase-neutral voltage, because the currents in the three phases are "caused" by the phase-neutral voltage. The phase-phase voltage can also be used, but then also the phase-phase current must be used.
 
OK, theortecly yes, Ijl and David are right, but..
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
 

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?
 
On this rare occasion, I must disagree with Davidbeach. Power factor is defined as the ratio of real power to apparent power. To get there from voltages and currents, select like elements. For example use line currents with line to neutral voltages, or line to line currents with line to line voltages. Line to line current from inside a delta connection is not usually convenient to measure. 30 degrees represents the shift between line to line and line to neutral voltage on a balanced three phase system.
 
Since there is no neutral on the VT secondary, there is no line-to-neutral voltage. There may be a line-to-ground voltage by virtue of capacitive coupling or if one VT terminal is grounded, but that is irrelevant. You will have to use the line-to-line voltage and line-to-line current. How do you get line-to-line current? By subtraction of the line currents.
 
I agree that L-L currents could be used with L-L voltages to arrive at power factor. L-L current is not the easiest concept, so I omitted that combination and focused on L-N voltages and L currents. The relays/meters undoubtedly (if told the proper types of voltages and currents) make all of the necessary corrections and phase angle shifts and provide the correct values.
 
rockman

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).
 
A symmetrical three-phase load can be connected in a star or in a delta.
Nothing in the OP says the loads are symmetrical. With no neutral carried to the relays, I would assume that all loads are phase-to-phase, with no zero-sequence neutral current, but not necessarily balanced.

 

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
 

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!!!!
 
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