3 phase power calcs.

[gord99]

This question may seem trivial but could someone confirm my reasoning? In a 3 phase wye-connected system, with voltages at 120/208v, and feeding a 'balanced' 3 phase 'purely resistive'load, the power can be calculated by summing the power from each phase winding of the transformer.
ie. P_total = 3 ( E_phase)(I_phase)

since E_phase = E_line/(root 3) and
I_phase = I_line, in a wye system, then:

P_total = 3 [E_line/(root 3)][I_line]

simplified, this becomes:

P_total = 360 (I_line)

when E_line = 208 volts.

So far so good. What happens when a balanced purely resistive load is connected across only two of the phase legs? As I see it, the total power should now be 2/3 of the 3 phase power.

ie. P_2phase = 2/3(360)(I_line) = 240(I_line)

Is this reasoning correct?

An example of this would be a feeder to one suite in an apartment where the 208 volts is derived across 2 x 120 volt phase legs of a 3 phase 120v/208v wye secondary. The balanced portion of this load would be connected across the 208 volts. I'm not concerned with the unbalanced loads at this point, which connect from phase to neutral.

 

[Bung]

If each resistor has value R, are you connecting R+R in series between the two phase voltages (ie just lifting one leg of the three phase connection of the third phase), or are you putting just one of the single Rs between the phases? Using just two of the phase voltages, the votage available is the line voltage. The current drawn will be defined by the impedance of the load applied between these two phases.

Bung

 

[230842]

For pure resistive loads, power (real power) can always be calculated either as the product of voltage and current values on the load or the quotient between the applied voltage to the square and the load resistance values. And that for every load resistor.
Going to your particular case:

Three resistive loads connected between phase legs and neutral:
Total power: P = 3*120*120/R = 208*208/R or
P = 3*120*I where I = 120/R

One only resistive load connected to two phase legs:
Total power: P = 208*208/R
Line current (two concerned phases): I = 208/R.

As you can see, if, in both cases resistor values had the same value, total powers would be identical, while line currents would not. So your 2/3 rule is not valid.
Iven more, adding the products of phase voltages and currents, result is no correct:
2*120*208/R not equal to P = 208*208/R.

That is due to the fact that real power is depending on voltage and current and, besides, on actual, or apparent, power factor.
Julian

 

[Bung]

230842 is right, up to the point where he starts mixing the phase and line voltages in one calculation (2*120*208/R). Also, power factor has nothing to do with this case, if we are considering purely resistive loads. P=V^2/R applies regardless, provided you use the voltage applied to the resistor and the resistance and don't mix in other voltages from somewhere else.

Bung

 

[230842]

BUNG, there must be a misunderstanding. I apologize but my english is not too accurate. So I´ll try again:
In my country we define the power factor as the cosine of angle lag between voltages and currents. This angle can be "actual" when voltage and current, both, belong to the same load; but it can be an "apparent" angle when, as in this particular case, voltage is single phase voltage to neutral and current is line current corresponding to a resistor connected to two lines.
That is what I wanted to mean mentioning, with not too much fortune, the apparent PF, while for resistive loads the actual PF is the unity of course. Julian

 

[Bung]

Julian
I think we'll have to agree to disagree about "apparent power factor"! I don't really see what relevance the angle of a voltage measured across points X and Y has to the angle of a current flowing between points X and Z. Power factor of any type has always meant (to me at any rate) the ratio of real to apparent power in a device, so if you are measuring current X to Z, you must use voltage X to Z.

Doesn't apparent power factor refer to the apparent angle difference between V and I due to harmonic effects (current chopping in VSDs etc), and displacement power factor refer to the displacement of the V and I due to capacitive and inductive elements?

Bung

 

[230842]

BUND, we both are right, I think. My insistence on the "apparent power factor" is due to the following:
Most circuit analyzers equipments can display single powers (real and reactive) for every line (or phase), and total powers as well by simple addition of single powers. In the particular case of one resistor connected to two phase lines, circuits analyzers show a total real power corresponding to the actual resistor power and a nul reactive power; but for single phase powers they display, for the two involved phases, the half or real power each and, two reactive powers of same value but different sign. That can be explained by the "apparent power factor" and could excuse its "apparent inconsistency"
It's a pleasure to argue with you.
Julian

 

[230842]

To complete my former post. Aparent power factor, when related to distorded waves, is defined as the quotient between te real and the apparent powers, this latter defined, at its turn, as the product of RMS values of voltage and current. I think.
Regards.
Julian

 

[redtrumpet]

IEEE Std 100-1984 defines power factor as the ratio of total watts to the total root-mean-square (RMS) volt-amperes, in per-phase quantities.

There is no mention of "apparent power factor" in IEEE Std 100, at least not one that I could find.

 

[busbar]

Even further off topic… ))Read This To The End. If you follow the secret quadruple rot13-encrypted instructions, you could make millions.(( Measurement of AC power in instances where there are nonsinusoidal (harmonic-containing) waveforms is a bit tricky, and only recently has begun to be formally addressed in this blessing/curse of a digital planet. As has been mentioned in this thread, the IEEE dictionary definitions have holes in them. Since day one, apparent power has been [inconsistently] defined and calculated two ways: One is based on the Pythagorean relationship of apparent power being equal to the square-root-of-the-sum-of-the-squares of real and reactive power. Also, an accepted means of apparent power calculation is the product of rms volts and rms amperes. For zero-distortion (100.0% fundamental) cases, the two apparent-power definitions hold.

Where voltage and/or current is nonsinusoidal, then the calculations don’t match so well. For any given frequency (or ‘recipe’) you can measure apparent power with (per-cycle) rms volts and amperes, AND get real power from summing ‘time-slice’ multiplication of instantaneous, sub-cycle potential and current samples. These quantities are not frequency-dependent, so are considered inherently “wide band.” The problem shows up when you try to flush out reactive power. Var flow determined from doing a square-root-of-the-difference-of-the-squares with voltamperes and watts is almost always too high, especially for sizing capacitors well aside that they are harmonic-current sinks. The 90° potential phase shift needed is indeed a fixed time interval equal to ¼ cycle of the fundamental frequency. IEEE folks realized that there were some [long-standing] problems being reported by various researchers about limitations in AC-power measurement for complex waveforms. In 1990 they assembled a ‘tutorial’ collection of papers to draw attention to the practical and theoretical aspects of this task. Meanwhile the very real problems associated with detonating capacitor banks, enflamed transformers, fuse fatigue, overheating circuit breakers and smoking neutral busses wasn’t getting less evident.

There is an insane variety of terms that get assigned by various meter producers, like narrow versus wide-band reactive power, imputed current, scalar unary-minus subtraction, multifarious sliding-window Q-demand distortion power, fuzzy vars, mesh power and single-cycle k-factor computation using the first 126 harmonics. I have worked in various facilities that conducted an assortment of physics/chemistry/magnetics experiments, and believe me, those guys can find amazing ways to mutilate the 60-hertz stuff. It’s normal to think of sources of power as having comparatively low {conveniently zero} impedance. Things start to get especially grey when two “loads” are merrily using Alternating Current in their own weird [non-sinusoidal] ways, and interactions may not be fully understood or even recognized. At times, interactions are not discovered and analyzed until after a part of the system malfunctions or sustains damage.

Dr Alexander Emanuel at Worcester Polytechnic has authored a number of papers on AC-power definitions and measurements. Recenty, he’s also committeed a core ANSI document: {sorry} 1459-2000 IEEE Trial-Use Standard Definitions for the Measurement of Electric Power Quantities Under Sinusoidal, Nonsinusoidal, Balanced or Unbalanced Conditions.

http://www.ece.utexas.edu/~grady/POWERFAC.pdf

http://www.etec.polimi.it/harmonics/

http://ece-

http://www.colorado.edu/~pwrelect/book/slides/Ch15slide.pdf

http://kevf72.troja.mff.cuni.cz/xichty/lab7/downloads/reports/p&e_measurement.pdf

http://www.cinstrum.unam.mx:90/i&d/pdfv5n2/a survey.PDF

http://www.ipst.org/TechPapers/2001/IPST01Paper186.pdf

 

[jbartos]

Suggestion to gord99 posted March 21, 2002 marked ///\\So far so good. What happens when a balanced purely resistive load is connected across only two of the phase legs?
///Please, would you clarify if the balance resistive load is connected:
1. In delta that has voltage supply to its two corners and resistors Rab=Rbc=Rca
2. In star that has voltage supply to its two corners and resistors Ran=Rbn=Rcn
3. In open delta with Rab=Rac identical resistors
4. In open star with Ran=Rbn identical resistors\\ As I see it, the total power should now be 2/3 of the 3 phase power.
///Yes, for open delta and open star connections\\Suggestion to Busbar posting April 6, 2002:
1. The following link

http://www.cinstrum.unam.mx:90/i&d/pdfv5n2/a survey.PDF

is not possible to open. Please, could you email me the file?
jo1545@yahoo.com
2. Beside harmonics, there are also interharmonics that seem to be causing difficulties in power and energy calculations and metering. It appears that to meter the energy justly it would be better to convert the AC current and voltage to DC and to pay utilities for the DC metered energy consumption. This would be fair to everyone since the harmonic distortions may come from neighbors.

 

[busbar]

jbartos--

The link is no longer in my url history--sorry.

 

[jbartos]

To busbar. That is o.k. I have received the the *.pdf file. Thanks a lot.