power factor in 3-phase current

jagec

Lifer
Apr 30, 2004
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I'm a little confused when it comes to power factor...

so basically, if you have a voltmeter, an ammeter, and a wattmeter hooked up to a motor, the voltmeter reading times the ammeter reading will be higher than the wattmeter reading by a factor of cos(phi), right? But the amount of energy the motor is actually consuming is what the wattmeter indicates?

I just stumbled across this effect, and it's pretty confusing...
 

pX

Golden Member
Feb 3, 2000
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Your'e right, the wattmeter only measures 'real' power. It doesn't measure apparent power (VAs) or reactive (VARs)...
When you measure current and voltage and multiply it's not the same as the wattmeter since they are not in phase, their peaks don't happen at the same time. The closer their peaks, the higher the real power. If they were completely in phase your real power would be equal to your apparent power.
The 'phi' the power factor angle varies based on the load.
Anyway, to summarize, here's some definitions:
Preal=Vrms*Irms*cos(phi) (watts) where phi = Vangle - Iangle
Preactive=Vrms*Irms*sin(phi) (VAR)
Papparent=Vrms*Irms= [(Preal)^2 + (Preactive)^2]^(1/2) (VA)

There is a way to measure reactive power with a normal wattmeter, but I can't remember the setup.
 

JSSheridan

Golden Member
Sep 20, 2002
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I concur with the above statement and would add a bit to it. Are you, the OP, familiar with the rms term? Basicly, it tells you what an AC voltage or current is equivilent to in volts/amps DC, so a 120 V(rms) AC source is equivilent to a 120 V DC source. This is just useful in analysis and should not be used as an actual substitute. When we say the wall socket is a 120 V source or the overhead transmission lines are 151kV, that is a rms measurement. When given a peak value for an AC voltage/current, you divide the value by sqrt(2) to get the rms value. So, be sure of what the numbers you are reading mean.

Now to deal with 3-phase power. The Preal, Preactive, and Papparent of the 3-P system is simply equal to the sum of the respective values for each phase. Peace.

Edit: pX, good luck with final exams.
 

Mday

Lifer
Oct 14, 1999
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since we're dealing with AC, is the voltmeter reading the pp or RMS value?
 

f95toli

Golden Member
Nov 21, 2002
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RMS always.

Of course, in principle there might be voltmeters with a "pp"-function for some strange reason, but I have never seen one (and I have seen a lot of voltmeters).
 

jagec

Lifer
Apr 30, 2004
24,442
6
81
Originally posted by: pX
Your'e right, the wattmeter only measures 'real' power. It doesn't measure apparent power (VAs) or reactive (VARs)...
When you measure current and voltage and multiply it's not the same as the wattmeter since they are not in phase, their peaks don't happen at the same time. The closer their peaks, the higher the real power. If they were completely in phase your real power would be equal to your apparent power.
The 'phi' the power factor angle varies based on the load.
Anyway, to summarize, here's some definitions:
Preal=Vrms*Irms*cos(phi) (watts) where phi = Vangle - Iangle
Preactive=Vrms*Irms*sin(phi) (VAR)
Papparent=Vrms*Irms= [(Preal)^2 + (Preactive)^2]^(1/2) (VA)

There is a way to measure reactive power with a normal wattmeter, but I can't remember the setup.

hmm, isn't there a sqrt(3) term for three phase power though?
 

bobsmith1492

Diamond Member
Feb 21, 2004
3,875
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If you hook a bridge rectifier and highish voltage capacitor up to the 120V wall AC, you can pull much higher than 120v - I've charged 'em up to 165 volts DC off a 120v AC connection.

fun stuff - discharge 5500uF worth of capacitors charged up to 165 volts :)
 

brentkiosk

Member
Oct 25, 2002
157
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The sqrt 2 comes in when relating the peak to the rms value for a sine function (either V or I). So there wouldn't need to be a sqrt 3 for three phase systems. And, it wouldn't be sqrt 2 for nonsinusoidal voltages and currents.
 

bobsmith1492

Diamond Member
Feb 21, 2004
3,875
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Also, voltmeters don't necessarily read RMS voltage - maybe newer ones are, but I remember seeing some advertised as reading 'true RMS', which implies there are some that don't show RMS voltage.
 

Mark R

Diamond Member
Oct 9, 1999
8,513
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On a 3 phase system the Phase-Phase voltage is equal to Sqrt(3) * the Phase-Neutral voltage. So 120 and 208 V in the US, and 240 and 415 V in Europe.

Ignore the following paragraphs if they cause too much confusion - power factor is quite a tricky concept, and unfortunately modern electronics can make it even more confusing.

Finally, caution is needed in defining power factor as Cos(phi) - this only applied when dealing with reactive loads (inductors e.g. motors) and not for non-linear loads (switching PSUs, computers, inverter-driven motors, electronic fluorescent lights). For non-linear loads phi is approx zero, but higher-frequency harmonics in the current waveform can seriously degrade the power factor.

Harmonic currents are a particular problem in 3 phase circuits, because they don't cancel out in the neutral conductor. It's quite possible for the neutral current to exceed the current in a single phase, in an otherwise fully balanced circuit, purely from harmonics generated by switching PSUs.
 

Geniere

Senior member
Sep 3, 2002
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To read true ?rms?, an instrument would have to have an integrating function to find the area under the curve. If the ac voltage is a perfect sine wave, is not distorted, a simple analog meter will provide the same result. In a residence, the simple NON-true-rms meter will usually suffice and may be preferred for troubleshooting. The digital meters usually have higher input impedance and may give a voltage reading when none exists. False readings occur often with an open neutral or grounding conductor due to capacitive or inductive coupling. Any device inserted in a circuit will affect the function of the circuit. It is always necessary to compensate for it when precision is required.
 

mAdD INDIAN

Diamond Member
Oct 11, 1999
7,804
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Originally posted by: bobsmith1492
fun stuff - discharge 5500uF worth of capacitors charged up to 165 volts :)

I'd rather not. :p

Related question to this whole concept of transmission of electricity. You know how we have trignometric functions in calculating power/voltage and all that, but is there really a physical case of angles coming into play? I understand their use in phase diagrams, to see if one source is in phase with other, but how can we know the exact angle between two signals?
 

jagec

Lifer
Apr 30, 2004
24,442
6
81
Originally posted by: mAdD INDIAN
Originally posted by: bobsmith1492
fun stuff - discharge 5500uF worth of capacitors charged up to 165 volts :)

I'd rather not. :p

Related question to this whole concept of transmission of electricity. You know how we have trignometric functions in calculating power/voltage and all that, but is there really a physical case of angles coming into play? I understand their use in phase diagrams, to see if one source is in phase with other, but how can we know the exact angle between two signals?

oscillascope.
 

JSSheridan

Golden Member
Sep 20, 2002
1,382
0
0
Originally posted by: mAdD INDIAN
Related question to this whole concept of transmission of electricity. You know how we have trignometric functions in calculating power/voltage and all that, but is there really a physical case of angles coming into play? I understand their use in phase diagrams, to see if one source is in phase with other, but how can we know the exact angle between two signals?
Phase angles are very important to know when designing a transmission system. We can measure the voltages at 2 buses and we know the properties of the line between those buses. Finding the current is easy after that. Also, if you were to measure the voltage as you traveled down a transmission line, its phase would change when compared to the voltage at the starting point. Peace.