Unbalanced 3-phase equation?
- Thread starterLighting Newb
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coldnorth57
Active Member
I think this is what you want. I'll give an example. You have 4 Mac Viper running on 208v. Using circuits 1 thru 4 on a breakout wired in the normal fashion. xy, yz, zx, xy, etc so your 4th one ends up on the same two legs as the first. The total load will be out of balance; nothing you can do about it.
Now, I'm trying some silly thing of hanging 4 lights and only running a breakout box powered by a single L21-20 circuit, because only power is an L21-20 in a theaters box boom position; It could happen. The specs say it draws 1190 watts, 5.8 amps. You want to know how much your total draw is per leg, Don't use amps, Just add up the wattage and divide by the legs. So each light pull 595 watts per leg.
With our 4 lights we are loading X 3 times, Y 3 times, and Z only 2 times. So
X=1,785/120= 14.875a
Y=1,785 /120= 14.875a
Z=1,190/ 9.92a
Notice once I have figured my wattage for each leg, only then do I calculate amperage per leg, so I can be sure my main power service is large enough for the rig, or in this example a triple 20A breaker. If its generator power, I stop calculating at total wattage, because that is how generators are sized.
Why use watts and not Amps? If you add amps per fixture at 208v you must then multiply bu root 3(1.73) because each of the 120v phases are 120 degrees apart, then divide that number by your three legs. Its an extra step, and furthermore, if you have fixtures running at 120v and 208v you have to add each one up separately. It's more work than it needs to be. With the exception of some variation in powerfactor, wattage does not change with voltage, so just add everything in watts and divide by your legs.
Now, I'm trying some silly thing of hanging 4 lights and only running a breakout box powered by a single L21-20 circuit, because only power is an L21-20 in a theaters box boom position; It could happen. The specs say it draws 1190 watts, 5.8 amps. You want to know how much your total draw is per leg, Don't use amps, Just add up the wattage and divide by the legs. So each light pull 595 watts per leg.
With our 4 lights we are loading X 3 times, Y 3 times, and Z only 2 times. So
X=1,785/120= 14.875a
Y=1,785 /120= 14.875a
Z=1,190/ 9.92a
Notice once I have figured my wattage for each leg, only then do I calculate amperage per leg, so I can be sure my main power service is large enough for the rig, or in this example a triple 20A breaker. If its generator power, I stop calculating at total wattage, because that is how generators are sized.
Why use watts and not Amps? If you add amps per fixture at 208v you must then multiply bu root 3(1.73) because each of the 120v phases are 120 degrees apart, then divide that number by your three legs. Its an extra step, and furthermore, if you have fixtures running at 120v and 208v you have to add each one up separately. It's more work than it needs to be. With the exception of some variation in powerfactor, wattage does not change with voltage, so just add everything in watts and divide by your legs.
JD
Well-Known Member
QUESTION CLARIFICATION:
It sounds to me like you are asking this question:
1) You have three conductors from a Wye power source (120/208)
2) It is a Delta load, meaning that the Neutral conductor is not in use.
3) you are measuring 50 amps flowing in the load that is tied to legs A and B
4) you are measuring 40 amps flowing in the load that is tied to legs B and C
5) you are measuring 60 amps flowing in the load that is tied to legs C and A
6) You are asking how many amps would be flowing from your power source through Leg A
7) You are asking how many amps would be flowing from your power source through Leg B
8) You are asking how many amps would be flowing from your power source through Leg C
9) You ask this question knowing that some current will be offset by traveling through the equipment to the opposing leg due to phase lag.
10) You are expecting there will be an equation that actually has three separate answers, or an equation that will need to be set up 3 different ways to provide the answers.
Lighting Newb
Member
Sorry, if there was some confusion but yes, that's a pretty accurate picture of what I'm looking for.
JD
Well-Known Member
Here's the long answer (prepare for headache): http://www.electrical-engineering-assignment.com/12-unbalanced-delta-connected-load
And here's the short answer: Since power is not created or consumed by configuration, each feed will be >40 amps and <60 amps, also, (AB+BC+CA) = (A+B+C) In other words , the sum total of current flowing on legs A, B, and C will equal the sum total of current flowing from A to B plus B to C plus C to A. If it is a gauging question, then you would simply base it on 60 amps as that would be the worst scenario if the other loads were removed.
Or you could set it up and slap an amprobe on it
And here's the short answer: Since power is not created or consumed by configuration, each feed will be >40 amps and <60 amps, also, (AB+BC+CA) = (A+B+C) In other words , the sum total of current flowing on legs A, B, and C will equal the sum total of current flowing from A to B plus B to C plus C to A. If it is a gauging question, then you would simply base it on 60 amps as that would be the worst scenario if the other loads were removed.
Or you could set it up and slap an amprobe on it
Dang, @JD couldn't you have found that in English, or at least a language all (some) of us could understand?
As I told @Lighting Newb privately, if it's that important to you (and it really needn't be) put all of the information into Paul Pelletier's Lighting Calculator, and trust the results. But there's probably only one or two CB members, and maybe 4-5 industry people I know, who can calculate problems like that. Certainly not something required of even an ETCP EE.
Me? I always got by simply distributing my 208V loads more evenly across the six circuits of my multi-cable s. Although the example given, 50-40-60, is fairly even, considering. If it were 500-400-600 I'd worry some. Actually, I'd worry a lot, as I wouldn't let a single load get above 350A or so, using 4/0 feeder cable.
Well, I'm not an engineer, I was simply giving an example of a real world application; rough calculation and rules of thumb are often all that is needed for everyday application. Chris, seeing as you are an engineer, I would appreciate the more scientifically/mathematically accurate explanation, an I'm sure others would as well.
I never bother to calculate how much power my 120v conventional lamps will draw at 208v, rather I just calculate how much overnight shipping will be for replacement lamps.
JD
Well-Known Member
Need to correct my post (#9 above.) I based it on a diagnostic I did in the 1980's for an electric feed to a 3 phase duct heater. I believed it to be Delta at the time. There was question as to if one of the elements was open, but when I measured it, the amperage was right on the money based on V*A for the rating of the unit. I now question if it was in fact a Wye wired heater as once the numbers came up right, I was sure all three elements were working (end of diagnostics.) Never checked for a neutral. In PM's outside of this thread, I was reminded that in Delta, you have to multiply the amperage by 1.73.
A quick "thought experiment" shows why: Since each phase leg lags the prior leg, the current flow will not align with the peak voltage of a given leg as it would in a single phase loop. The result is the actual RMS current will exceed the expected draw in a way reminiscent of power factor equations. However, unlike PF, it is normal and does not need correction.
As for the unbalanced equation, it appears the consensus is that a root canal would be more pleasant to endure.
A quick "thought experiment" shows why: Since each phase leg lags the prior leg, the current flow will not align with the peak voltage of a given leg as it would in a single phase loop. The result is the actual RMS current will exceed the expected draw in a way reminiscent of power factor equations. However, unlike PF, it is normal and does not need correction.
As for the unbalanced equation, it appears the consensus is that a root canal would be more pleasant to endure.
YesItWillWork
Member
The others here are right, if you want to work this out in the generic case from basic principles then you are in for a bad time if you aren't familiar with complex numbers and three phase analysis.
If we take JD's clarification of what is being asked and add an additional caveat that the load is purely resistive (meaning the current phase angles are exactly 120° apart) then the math can be simplified down to the following reasonably simple equations:
Remember this is valid only under the circumstances specified and should not be relied upon in any other case.
If we take JD's clarification of what is being asked and add an additional caveat that the load is purely resistive (meaning the current phase angles are exactly 120° apart) then the math can be simplified down to the following reasonably simple equations:
Remember this is valid only under the circumstances specified and should not be relied upon in any other case.
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Applying @YesItWillWork 's formulae (isn't Excel the best thing since sliced bread ever?),
A=95.4
B=78.1
C=87.2
So now we know:
1. How to do it, and
2. That we will likely never need to.
Ain't book learnin' grand?
I'm writing late at night, but I'm not convinced we need to constrain the maths to purely resistive loads.
If we broaden the constraint to all loads needing to share a common power factor, regardless of whether that is 1 (a resistive load), or something less than 1, then because we have a consistent voltage / current phase relationship on all loads, the 2 pi /3 shift between phases holds, and thus the formulas presented above hold true...
YesItWillWork
Member
That's true, as long as there is 120° between phase currents then the maths holds. I figured that since we are considering the unbalanced case then it could be a bold assumption to make to assume that is true - it's all very dependent on what the physical loads are.
If we are to be pedantic about terminology then it is the phase angle which is of importance as to the applicability of these equations not the the power factor. True power factor consists of two components, the displacement PF, and the distortion PF. Displacement power factor is what most people think of when referring to power factor - the phase angle between voltage and current as a result on non-resistive loads, while a distortion power factor of less than one results from the presence of harmonics.
To the average person does this really matter? Almost certainly not. To me who spends my days when I'm not on a gig working on a Phd in electrical engineering - it's an important distinction to make.
JD
Well-Known Member
It is true that the three legs would have to have the same type of power factor distortion for the formula to hold. A lower than 100 power factor can have several causes, for example, a magnetic ballast would have a classic phase lag in ampacity, whereas a standard phase-chopped dimmer would score its low power factor in a different way and vary all over the place depending on the dimmer setting, and lastly, a computer power supply would score its power factor due to the poor distribution caused by the fact that a load only exists when the diodes are in forward conduction (Line voltage greater than supply cap charge.)
In our line of work, if the loads were ALL magnetic ballasts, or ALL electronic ballasts, the formula (in my opinion) would hold. If you were to mix and match, the results would vary.
In our line of work, if the loads were ALL magnetic ballasts, or ALL electronic ballasts, the formula (in my opinion) would hold. If you were to mix and match, the results would vary.
Dionysus
Well-Known Member
And THIS is why we focus on MAXIMUM values than actual ones, rounding up really is a thing in electrical in the "real world". There really is no sense trying to re-invent the wheel in order to attempt to squeeze more onto a line or service or such. There are always things you can't adequately account for as well that won't rear their head until you put everything to practice (especially in our industry). Err on the side of using larger conductors. There is a reason why there are large sections of the code books dedicated to this.
Wow. this thread has really gone FAR to the technical side. Let me give you a real world example of how to apply the above in an actually entertainment job setting. Also I have included some shorthand for doing all the complex math.
"Hey, Is a 65kw geni big enough for that street fair?"
"hmm. probably, but I think it will be close"
"Okay, I'll just order a 100kw then."
"Hey, Is a 65kw geni big enough for that street fair?"
"hmm. probably, but I think it will be close"
"Okay, I'll just order a 100kw then."
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