Why does 3-phase panel power (2162W) differ from load power (1872W) with 2-pole 208V breaker?

Good afternoon,

I tried what I thought would be simple power calculations, but the calculations are not balancing. It seems that I'm doing something fundamentally incorrect. Here's a hypothetical situation that illustrates the problem.

There is a new 120/208 panel that has only one 2-pole breaker installed. The load is 208 and the neutral is not used by the load. The panel power meter shows line currents of 10, 10, 0. The power factor is .90. With the above readings, I believe the power for the panel (which represents only the one load) would be:

Average of line currents * line voltage * power factor * sqrt(3)

6.67 * 208 * .9 * 1.732 = 2,162 Watts

Then I go out to the field with my multi-meter and calculate power at the load:

Line current * line voltage * power factor (assumed to be same as panel)

10 * 208 * .9 = 1,872 Watts.

Why do I calculate ~15% lower power at the load?

Thanks,

Paul

Paul...Ok you have made 2 errors here.Firstly 120 / 208 power means you are getting two phases of a Y connected 3-phase transformer.Your sqrt 3 is that factor that is applied to get the voltage between two phases if you are given the single phase voltage.  So 120* sqrt 3 = 208 approx.So - if you are using 208 then you don't need the sqrt 3 factor, it has already been applied to get the 208V between phases.Secondly you have taken your average of all three line currents when in fact only 2 phases are used.  The 10A is flowing out of one phase into the other phase.  So your current is 10A, not 6.67. You'd only use the average if all 3 phases are in use, eg for a 3 phase motor, or a house wired to use all 3 phases on different circuits.  So your first equation becomes the same as the second   happy days!The reason you got a 15% difference was that you increased your actual  voltage by sqrt 3 (1.732) and decreased your actual current by 0.67.   1.732 * 0.667 = 1.155244.  QED.Here's a bit of a rough-and-ready phase diagram that might help explain it:
Does this make it clear?   Thanks // David

David,
Thank you for taking the time to answer.  Some of it I can make sense of, but it raises more questions.Exclude the 1.732 factor...  I believe this is because I have to use phase voltage (120) since I have phase current.
Do not average the currents... I can see why, but now I'm unclear as to the application of that formula.  It's "all over the Internet" and doesn't indicate phase loading is important in order to determine power.  I'm wondering if it is only intended to apply to a balanced 3-phase load and not to a panel which may also have single phase loads.For more confusion...The formula examples are usually using 480 (not 277) and still include the 1.732 factor....perhaps that's assuming a delta connection and that you have phase current and the factor is there to convert it to line current???Say I know the 3 line currents and the neutral current and magically the power factor (ok it will be a guess).  Is there a formula to determine the power, or can only a power meter do it (with the additional information it has)?Lot's of guessing here.  Any clarity appreciated.Thank you,Paul

I checked a couple of meters at work, and the my calculation only agrees with the meter if I use both the higher voltage, and the 1.732 term.  What could be going on here?Attaching metering image from a 208 volt branch breaker.

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Paul... OK the instance above does not correspond to your hypothetical case in your first post.  Here all three phases have currents and loads.  I can get a very similar result as follows:We have three individual voltages of 12o here, with 3 individual loads as shown by the currents.since the voltages are the same, we just add the currents5.8+2.7+7.4 = 15.9 Atotal power = 120 * 15.9 = 1908 WApply Power Factor 1908 * 0.88 = 1,679.04 WClose enough?Because your loads are directly on the phases here, you don't need to apply the sqrt 3 factor - you only do this to sum vectors for 2 phases.See below for my comments on your 2-phase problem.

Paul - ok, think of it this way.  in your first case, you have ONE voltage and ONE load, so forget about 3-phase.  The only way the phases come into it is that you know you're across 2 phases of a 120V 3-phase supply.  You DON'T have a neutral connection, remember?  So what you have is a 2-wire (ie effectively single phase) supply of 120 * 1.732 = 208V.Your formula examples - 480 does not make sense.... even using 230V mains the phase to phase voltage is usually 380-400.  so I think we are talking about something different here? Can you post a link or a scan of that example for me to look at?> Say I know the 3 line currents and the neutral current .....In a Y connection all you need are the 3 line currents - see my workings on your breaker display above.  If you want to calculate your neutral currents you could do so from the values given - do a vector addition and you'll get it.  Your breaker display is typical of what you'd get in a house  with heaters, lights and appliances on each of the 3 phases, which is not the same at all as having one load across 2 of the phases.When you're dealing with problems like this, get it down to its simplest terms.   Don't assume 3-phase unless you know you are dealing with 3-phase.  If you have one voltage across one load, treat it as a single phase problem.

>Your formula examples - 480 does not make sense.... even using 230V mains the >phase to phase voltage is usually 380-400.  so I think we are talking about >something different here? Can you post a link or a scan of that example for me to >look at?Sorry, I introduced the 480 out of the blue.  The point there was to say on a different WYE connected panel which happens to be 277/480, that the factor 1.732 still has to be used *as well as* the 480 voltage to calculate power correctly (to agree with the meter power reading).  So it seems that the 1.732 is doing something other than converting the 277 (or 120 in the other example) into the higher voltage.  So the question would be: what is the 1.732 factor for?>In a Y connection all you need are the 3 line currents - see my workings on your >breaker display above.  If you want to calculate your neutral currents you could do >so from the values given - do a vector addition and you'll get it.  I tried my hand at calculating the neutral current.  I got 4.14 Amps, which seems like it could be correct.  Anyway it brings up another question...if there is no neutral connected to branch circuits, then where is this neutral current going or what affect is it having on the panel?I can appreciate the argument that there is only single phase power being used, so just use single-phase calculations.  My objection is that the meter doesn't know what combination of single and three-phase loads are connected to it, and it (presumably) calculates correctly.  So why can't I also calculate power correctly using one formula for all cases?Thank you

Hi Paul
Ok firstly let's see where the sqrt 3 = 1.732 comes from.  Below is a diagram of two phases, let's say with a voltage of 1.  They are 120 degrees apart.  We'll use Pythagoras here.  Phase B will have components in line with and at right angles to Phase A as shown - we know that angle is 60 degrees  (180 - 120) so the in line component is 0.5 and the right angle component is 0.866. So we now have a right angled triangle with sides of 1.5 and 0.866.  By Pythagoras's therorem Vab^2 = 1.5^2 + 0.866^2     =     2.25 + 0.75    =     3so Vab = sqrt 3 = 1.732So 1.732 is the factor between the single phase voltage and the phase to phase voltage.You'd ONLY use this when you have a load between two phases only as per your initial problem.(edit - but see below for further explanation)
Let's see if we can use your breaker currents for your first problem.You would have Vphase = 120, Ia=10, Ib=10, Ic=0.Because your load is between two phases, assuming a resistive load your power factor will be cos 30 degrees (the current would be 30 degrees away from both phases: (180 -120) / 2so use your three phase formula - 120 x sum of currents x power factor120 x 20 x 0.866 = 2078.4 WNow you stated that your actual load has a power factor of 0.9, the above calculations were for a resistive load, so let's now put that in2078.4 x .9 = 1870 - again, pretty close to what you get with your meter in your initial post.If you got the readings from your breaker on this, the power factor would show as 0.866 x 0.9 =0.7794So one place you have been going wrong is that you used the average of currents.  You should  use SUM of currents x single phase voltage x power factorI should have pointed out  more clearly that you should use the sum of currents rather than the average, sorry.
What it comes down to is that there are usually a couple of different ways to attack a problem.  Use whichever one makes more sense to you.BTW I calculated the neutral current on your breaker also using Pythagoras and got 3.62 A.  Close to yours, I did a quick and dirty calc and did not take the power factor into account which may explain the difference.On your 277 / 480 problem, if you can give me more specifics I might be able to make sense of it.  As above you should only use the 1.732 factor to get phase to phase voltage from single phase voltage, so it does not make sense to use it when you already have 3-phase voltage.  As above you may have been making two errors which did not quite cancel out...cheers // David

Hi again PaulOK, I did some fossicking around and think I finally have an answer as to what is confusing you I looked at various sites on the net and found your 1.732 = sqrt 3 used with the phase to phase voltage, now I know what is going on with that.if you have a 3-phase load the formula for the power is3 (phases) x 120 (line to neutral voltage) x Current (assuming same in all phases) x PF.Note that we are using the Line to Neutral voltage here (120)Now - say we want to do the same calculation but using the line to line voltage (208)To get to the line to neutral voltage we have to DIVIDE by 1.732 (sqrt 3) so the above equation becomes3  x  208/sqrt 3  x  I  x  PFNote we are multiplying by 3 phases and dividing by sqrt 3.  What is 3 divided by sqrt 3? Sqrt 3! so we can simplify this tosqrt 3 x 208 x I x PF  or generally   1.732 x VPH-PH x I x PF.This equation does work fine with the breaker display you give above, using the average of the currents.Right, so far so good.  WHY does it not work for your initial problem?  Although one phase is not used, the equation should still work.  It does, but we forgot one thing. I actually got the answer above but did not realise it.  Because in the instance of your initial problem, the load is connected across 2 phases only, it will change the apparent power factor, as above by a factor of 30 degrees or 0.866.If we apply this to the calculation it will come out right:1.732 x 208 x 6.67 x 0.9 (PF load) x 0.866 (PF due connection line-line) = 1872.83W
When you have line-line loads,  you can't just use the power factor of the load.If you had this situation on a 3-phase breaker it should show the power factor as 0.866 x 0.9 =0.7794 which is what it would be as seen from a phase.
It would be very interesting to know if a breaker with this load did display the power factor as 0.7794, but you said this was a hypothetical situation?  This has been driving me nuts, glad I've got to the bottom of it.....   

David, that's great!  I'll have to take time to assimilate all of this, but it looks complete.The 12o/208 situation was hypothetical, but we do have a 277/480 panel that can be manipulated.  This weekend I'm having a coworker read the meter before and after opens all breakers connected to L1.  I'll let you know how it turns out. Thank you,Paul

Paul, took a bit of getting there, but hope this does help.  It illustrates the fact that there are a lot of different ways of getting a result, and just as many ways of teaching them! Personally I would not use the line-to-line voltage when using only line-to-neutral loads, but there you go, it seems a lot of people do use line-to-line....Thanks for all the feedback and look forward to seeing your readings.  Happy to assist further if required.

Paul I was thinking about this some more - why would a delta load (ie connected line-to-line) have a different power factor from a Y load (line to neutral loads)?  Or, to put more bluntly what was worrying me, how can a resistive load have a power factor?The answer is that it won't.  If you have a balanced delta load (eg a motor) then yes the currents are 30 degrees out of phase with each line (even if they are resistive loads) BUT with current in all legs, the phase A-B current will be 30 degrees one way, and the phase C-A current will be 30 degrees the other way, so they will balance out.  See my diagram 4 posts up and imagine all 3 phases drawn
Your single load across two phases is a bit of a special case I think, and the 0.866 "power factor" is just due to the phase to phase current only.  But you still have to apply it to get the right answer when there is only 1 phase to phase load.I'm happy I have covered all bases now

David,
The test was run and below should appear a pic with the results.  It both clarified and confused.  It would have been better if there were 480 single-phase loads so it was more like the 120/208 question.I did the 480 volt panel calculation, the three individual 277 volt phase calculations and the neutral current.  They all agreed with the meter to within 1%.  Even accounting for rounding error, I couldn't calculate exactly the same as the panel readings, so I hope the 1% tolerance is not masking some other error of mine. The meter power factor never changed much and I don't understand why the .866 would be  applied as a power factor factor.  Could you perhaps speak more slowly, use smaller words, and draw a picture?

Paul - OK this still does not correspond to the case in your first post.  Why?  because the currents in your 3 legs are so vastly different.  If you had one load between B and C legs you'd get the same current in them, and you have a large neutral current in both cases.  So I suspect your load here is all or mostly single phase loads and, from the power factor, mostly resistive.
OK here's my vector diagram from above, with some extra info.  Let's say Va is positive - current Iab will flow out of A and into B.  You will appreciate from the workings with the original diagram that Vab will be 1.732 x the system voltage (here I've just shown it as 1).Now there is 120 degrees between the phases so the other corners of that triangle will be 30 degrees each.  If your load between A and B phases is purely resistive (no power factor) the current Iab will be out of phase with both Va and Vb by that 30 degrees.So it will LOOK LIKE the PF is cos 30 = 0.866.  It's NOT a power factor (as the load is resistive) but it will appear as one to any measuring equipment.  So my guess is that if your breaker gave you values for this case it would look like this if the load was 1 ohm:Ia = 1.732   Ib = 1.732   Ic = 0PF A = 0.866  PF B = 0.866   PF C = 1Your total power would be 1.732 x Iab and that's what you should see in the breaker display at the top.  What you'd see on individual power displays like your table above I'm not sure, but we can try and plug some values into our formula.  Lets say the current is 10A. We know the load is resistive and the voltage across it is 1.732, so power has to be 1.732 x 10 = 17.32 W.Treating the phases individually, each would have 1 (V) x 10 (A) x 0.866  (apparent PF)= 8.66W   on each phase so total power = 2 x 8.66 = 17.32.   Nice!Now taking the 3-phase case and taking 1.732 as the voltage (phase-to-phase)P = 1.732 (from equation)   x 1.732 (VPH-PH) x 6.67 (AAV) x 0.866 (apparent PF)   = 17.32.  Nice again! But whether your breaker would try and average the PFs I don't know, that would result in errors.I hope the 1.732V is not confusing, I should maybe have chosen a different voltage, but I hope you can see how it works?  If your load has it's own power factor, that will have to be put in as well as the 0.866 apparent PF.I repeat that with a 3-phase load, the apparent power factors on each side of the phase will cancel out so the power factor will appear to be 1 for a resistive 3-phase load.How's that?Cheers / David

David,I've mulled this over for a few days and it seems that I'm just not familiar enough with three-phase power and phasors to really understand this.  Firstly I couldn't see how there was a 30 degree shift until I re-worked the phasor.  Would this be a correct alternative rendering?  I subtracted A from B so the result remained on the origin.
And the statement below loses me because if the current is out 30 from the A and B phases, wouldn't the single-phase voltage also  be out by 30?  So no shift.  I see that the three and single phase power are out of phase, but not understanding why it is relevant.
> If your load between A and B phases is purely resistive (no power factor) the current Iab will be >out of phase with both Va and Vb by that 30 degrees.So it will LOOK LIKE the PF is cos 30 = >0.866.   

Hi Paul - yes your diagram is absolutely correct. it's just drawn slightly differently.  The angles are the same.  I prefer my diagram because the load is not actually connected to the origin (ie neutral) but it's pretty academic really.So you had your resultant voltage coming from the origin - fair enough, but look at that resultant voltage.  It's bigger than VA and VB  (by our good old factor of sqrt 3 or 1.732).It's also at an angle of 30 degrees to both of them (one will be lagging and the other leading I think).  SO if you have a resistive load on that VAB, the current is going to be in phase with VAB which means that it too (the current) will be 30 degrees out of phase with both VA and VB. So if you put a power factor meter on the A or B phase, it's going to see a power factor of 0.866 on each.  Or so I assert!  Your table a couple of posts up shows the power factor for each phase, so I think this is what you'd see there.  On your breaker display neat the top, which just lists one power factor, I'm not sure what you'd see.  IF you can arrange it, connect a couple of 1KW or 2KW household radiant heaters in series across 2 phases of that breaker and see what you get. I repeat again that this is not a true power factor, but it's what it would look like viewed from the VA or VB phases, and my calculations 2 posts up bear this out, you calculate the power on the phases, put in a power factor of 0.866, multiply by 2 (phases) and you get a correct resultI did all this stuff a few years ago - much more complicated stuff in fact - but I don't use it day to day so I was a bit rusty.  But I did ask a couple of guys at work who should be more au fait with it and they weren't much better than I was.  So if you're starting out on this, you have my sympathy.....

David,It finally clicked in my head what you have been explaining. The current is out of phase with the voltages on the three-phase side, but in-phase at the single-phase resistive load..so a non-unity power factor results in a lower kW calculation result on the three-phase side.  I had never heard of apparent power factor before. If your whole facility was was somehow wired like that, I wonder if the utility company would charge for a low power factor....which I guess could not be corrected by capacitors since it is not really reactive. I'll see if I can set up a test to see how the power meter interprets it.Thank you for your patience. 

Paul - great, it's not an easy concept.  Again, with a balanced 3-phase delta load, the currents to each of the other phases  are  30 degrees one way and 30 degrees the other way, and balance out, so with a 3-phase load you would't get this effect.  But calculating it as 2 single phase loads with low power factors you get the right KW, so I suppose that's what the utility company would do.  I work for one so I'll ask the metering guys about this.> Thank you for your patience. And thank you for yours, this is not my strongest suit so I hope I did not "lead you down the garden path" too much .  And it's always great to have someone who interacts...it's been a pleasure.