Calculating Phase Currents for Unbalanced Loads

[Smart $]

So, are you going to back that statement up with the not roundabout way?

Yes. one can easily write,

Ia = Iab - Ica

Draw that expression as a phasor diagram to obtain 1/3 of your diagram. Repeat for Ib and Ic to get the other 2/3.

I did nothing different than what you suggest... only I did it graphically. How is that a roundabout way?

Besides, not everyone has comprehension of the matter to the degree an equation by itself suffices... and who was it that said a picture is worth a thousand words?

 

[Smart $]

Smart$, this was really well done, but I don't think that it works. In post 66 you mention that the accuracy depends on the "values adding to zero." While this should be true for the line currents Ia+Ib+Ic=0, using the triangle centroid also means that is necessarily true for the phase currents too Iab+Ibc+Ica=0. I don't believe that this should be the case.

For instance, imagine a 208V, 3ph system with phase voltages of Vab=208<-60, Vbc=208<180 and Vca=208<60. If you connect a 2080W (resistive) load between A-B and a 1560W (resistive) load between B-C, and connect no load between C-A, then you would have phase currents of Iab=10<-60 and Ibc=7.5<180 and Ica=0

We know that Ia=Iab-Ica (etc) so, Ia=10<0, Ib=15.2<145.3 and Ic=7.5<0. If you add Ia+Ib+Ic, you will find that the sum is zero. However, if you add Iab+Ibc+Ica, you will find that the sum is 9<-106.1

If you took the line vectors and graphed them, and then used the centroid lines for the phase currents, you'd have a value for Ica which was not zero. But in the example, there is no load between C-A, so Ica must be zero. I don't think the triangle centroid method will provide the correct phase currents in all cases.

Check your math. If there is no load across A-C, with Ia=Iab-Ica and Ica=0, then Ia=Iab. You show Iab=10<-60 but then Ia=10<0.

I do agree this seems to disprove the centroid, midpoint aspect... as even though the third load is zero, the premise should hold to infinitesimally small loads approaching zero. But I don't have time to evaluate at present... perhaps later, unless someone else cares to handle that challenge

 

[jghrist]

Smart$, this was really well done, but I don't think that it works. In post 66 you mention that the accuracy depends on the "values adding to zero." While this should be true for the line currents Ia+Ib+Ic=0, using the triangle centroid also means that is necessarily true for the phase currents too Iab+Ibc+Ica=0. I don't believe that this should be the case.

For instance, imagine a 208V, 3ph system with phase voltages of Vab=208<-60, Vbc=208<180 and Vca=208<60. If you connect a 2080W (resistive) load between A-B and a 1560W (resistive) load between B-C, and connect no load between C-A, then you would have phase currents of Iab=10<-60 and Ibc=7.5<180 and Ica=0

We know that Ia=Iab-Ica (etc) so, Ia=10<0, Ib=15.2<145.3 and Ic=7.5<0. If you add Ia+Ib+Ic, you will find that the sum is zero. However, if you add Iab+Ibc+Ica, you will find that the sum is 9<-106.1

If you took the line vectors and graphed them, and then used the centroid lines for the phase currents, you'd have a value for Ica which was not zero. But in the example, there is no load between C-A, so Ica must be zero. I don't think the triangle centroid method will provide the correct phase currents in all cases.

I agree, except that Ia=10<-60.

If you take Ia = 10, Ib = 15.21, and Ic = 7.5 and put it in Smart's equations, you get:

Iab = 8.21, Ibc = 7.26, Ica = 3

Clearly incorrect.

If you use symmetrical components to back-calculate the delta load currents, assuming no zero-sequence delta load current you get

IL0 = 0, IL1 = 10.1<-90?, IL2 = 5.2<16.1?

ID0 = 0, ID1 = 5.83<-60?, ID2 = 3<-13.9?

IAB = 8.21<-44.7?, IBC = 7.26<156.59?, ICA = 3<73.9?

Also, clearly incorrect. The intuitive assumption that there is no zero-sequence delta load current is incorrect. With the actual starting delta load currents, the sequence currents are:

ID0 = 3<-106.1?, ID1 = 5.83<-60?, ID2 = 3<-13.9?

I say intuitive assumption because I was thinking of the zero-sequence currents circulating in the delta and with one leg open, they can't circulate. My mistake. The zero-sequence currents can flow through the open leg because they are offset by the positive- and negative-sequence currents. The zero-sequence network does not have an open leg.

So where does that leave us in seeking a solution?

 

[david luchini]

Check your math. If there is no load across A-C, with Ia=Iab-Ica and Ica=0, then Ia=Iab. You show Iab=10<-60 but then Ia=10<0.

I do agree this seems to disprove the centroid, midpoint aspect... as even though the third load is zero, the premise should hold to infinitesimally small loads approaching zero. But I don't have time to evaluate at present... perhaps later, unless someone else cares to handle that challenge

Thanks for catching that, wasn't a math error, just a typo. Clearly 10<0 + 15.2<145.3 + 7.5<0 will not equal zero, but 10<-60 + 15.2<145.3 + 7.5<0 will equal zero. But as I thought about it more, it became clear to me that the answer to the question in the OP is - you can't calculate the phase currents knowing just the line currents. And here is why:

Imagine if you had two different systems side by side. On the first I connect loads in a delta configuration such that the phase currents are Iab=10<-30, Ibc=10<-150 and Ica=10<90.

On the second system I connect loads such that the phase currents are Iab=13.025<-36.9, Ibc=10.431<-131.44 and Ica= 7.392<76.26

I then measure the line currents on the first system and find Ia=17.32(<-60), Ib=17.32(<180) and Ic=17.32(<60), and ask you to calculate the phase currents from the measured line currents.

I also measure the line currents on the second system (Ia=17.32<-60, Ib=17.32<180, Ic=17.32<60) and ask your coworker to calculate the phase currents from the measured line currents.

I've given both of you the same line current values, but I am expecting each of you to provide a different value for the phase currents.:-? Clearly, you cannot calculate phase current from line currents alone.

 

[rattus]

I did nothing different than what you suggest... only I did it graphically. How is that a roundabout way?

It is roundabout because you threw in extra diagrams and polygons along the way.

 

[rattus]

Centroid of a triangle:

Centroid of a triangle:

The intersection of the medians of a triangle defines its centroid which does NOT in general coincide with the intersection of the phase current phasors in Smart's diagram. A few seconds at this website demonstrates clearly that the triangles formed by the medians and the line current phasors do NOT in general contain a 120 degree angle.

http://www.mathopenref.com/trianglecentroid.html

 

[Smart $]

It is roundabout because you threw in extra diagrams and polygons along the way.

I'll give you that aspect on your point... but I did so to carry out the "proofing". Just as in showing the calculation for the math proof you requested. After assembling the first equation (the one immediately after "thus"), I can jump mathematically to the last just by looking at the first. But in the proofing we're supposed to write out cancellation and transposition steps.

That said, it is much easier to critique than contribute. Take the example of a chef's dish vs. a gourmet critic. The chef can work painstakingly at preparing his favorite dish, while his critic can approve or disapprove with just a nibble.

 

[Smart $]

Thanks for catching that, wasn't a math error, just a typo. Clearly 10<0 + 15.2<145.3 + 7.5<0 will not equal zero, but 10<-60 + 15.2<145.3 + 7.5<0 will equal zero. But as I thought about it more, it became clear to me that the answer to the question in the OP is - you can't calculate the phase currents knowing just the line currents. And here is why:

Imagine if you had two different systems side by side. On the first I connect loads in a delta configuration such that the phase currents are Iab=10<-30, Ibc=10<-150 and Ica=10<90.

On the second system I connect loads such that the phase currents are Iab=13.025<-36.9, Ibc=10.431<-131.44 and Ica= 7.392<76.26

I then measure the line currents on the first system and find Ia=17.32(<-60), Ib=17.32(<180) and Ic=17.32(<60), and ask you to calculate the phase currents from the measured line currents.

I also measure the line currents on the second system (Ia=17.32<-60, Ib=17.32<180, Ic=17.32<60) and ask your coworker to calculate the phase currents from the measured line currents.

I've given both of you the same line current values, but I am expecting each of you to provide a different value for the phase currents.:-? Clearly, you cannot calculate phase current from line currents alone.

While I must agree the centroid solution is not the ultimate solution when considering extremely unbalanced phase currents, this latest example steps outside the boundaries of the criteria, which require the line currents total zero, as must the phase currents. Your second set of phase currents do not sum to zero.

I now believe the centroid solution fails as the unbalance approachs the extreme. Perhaps when any one of the relative angles between line currents exceeds 90?, since my formula was based on the law of cosines. 90? is where the cosine function transitions for positive to negative.

But getting back to the OP, if the concept is to determine when a balanced 3? motor load is approaching unbalance beyond nominal tolerance, I don't see where using the centroid calculation is going to hurt in approximating moderately unbalanced phase currents. How accurate must the phase currents values be? According to jghrist's info, the centroid calculation seems to be equivalent to symmetrical components less the zero-sequence.

 

[mivey]

That said, it is much easier to critique than contribute. Take the example of a chef's dish vs. a gourmet critic. The chef can work painstakingly at preparing his favorite dish, while his critic can approve or disapprove with just a nibble.

Sting! :grin:

 

[rattus]

That said, it is much easier to critique than contribute. Take the example of a chef's dish vs. a gourmet critic. The chef can work painstakingly at preparing his favorite dish, while his critic can approve or disapprove with just a nibble.

My compliments to the chef; the meal was very good even if the entree was a bit over done.

 

[david luchini]

While I must agree the centroid solution is not the ultimate solution when considering extremely unbalanced phase currents, this latest example steps outside the boundaries of the criteria, which require the line currents total zero, as must the phase currents. Your second set of phase currents do not sum to zero.

I must have missed the criteria that requires the phase current to total zero. My point was that the centroid method would necessitate phase currents summing zero, but that the OP or other situations wouldn't require the phase current to total zero, so the centroid method wouldn't work in a general sense.

But getting back to the OP, if the concept is to determine when a balanced 3? motor load is approaching unbalance beyond nominal tolerance, I don't see where using the centroid calculation is going to hurt in approximating moderately unbalanced phase currents.

I don't think the OP's question involved a balanced motor load approaching unbalance (although he did mention "eg. and induction motor.") It involved a much broader point of calculating phase currents for a delta connected load where the load "may or may not be balanced and the phase impedances are unknown." What is known is the "line currents and the line to neutral voltages from the supply."

As for the chef, I say well done! I was trying to consider a simple way to solve multiple equations with multiple unknowns in order to solve for the phase currents (although I didn't put too much effort in.) But it was checking into your triangle centroid method that led me to understand that it is impossible to determine the phase currents for a delta connected load (which may be balanced or unbalanced) knowing only the line currents and the line to neutral voltages from the supply. So thanks for saving me from a huge waste of time trying to find a mathematical solution.

 

[rattus]

But it was checking into your triangle centroid method that led me to understand that it is impossible to determine the phase currents for a delta connected load (which may be balanced or unbalanced) knowing only the line currents and the line to neutral voltages from the supply. So thanks for saving me from a huge waste of time trying to find a mathematical solution.

FWIW, one can easily solve for the line current phase angles with the "half-angle" formulas.

Now, if the load is balanced, the line currents will simply be 1.732x the load currents.

But, if the load is unbalanced with the same PF, it seems there should be a olution, but I don't know what it is.

 

[Smart $]

FWIW, one can easily solve for the line current phase angles with the "half-angle" formulas.

Now, if the load is balanced, the line currents will simply be 1.732x the load currents.

But, if the load is unbalanced with the same PF, it seems there should be a olution, but I don't know what it is.

Wouldn't unbalanced with the same power factor put us back to the First Fernat Point solution?

With the same power factor, the phase current angles would be 120? out-of-phase, regardless of balance, right?

 

[rattus]

Wouldn't unbalanced with the same power factor put us back to the First Fernat Point solution?

Maybe. If I ever knew about First Fermat, I have forgotten it.

With the same power factor, the phase current angles would be 120? out-of-phase, regardless of balance, right?

Right!

 

[Smart $]

I must have missed the criteria that requires the phase current to total zero. My point was that the centroid method would necessitate phase currents summing zero, but that the OP or other situations wouldn't require the phase current to total zero, so the centroid method wouldn't work in a general sense...

Well my thinking cap must not be on, or I must be missing something. If the line currents sum to zero, how could the phase currents not. In my mind, only with circulating or ground fault current could they not sum to zero. The phase current is the same current as the line current and therefore must sum to the same zero value. It's easy to think up vectors which depict a non-zero sum... but can you make it a reality?

 

[Smart $]

Maybe. If I ever knew about First Fermat, I have forgotten it.

...

This thread. Starts at about Post #9. You jumped in at #13 :roll:

 

[david luchini]

FWIW, one can easily solve for the line current phase angles with the "half-angle" formulas.

Now, if the load is balanced, the line currents will simply be 1.732x the load currents.

But, if the load is unbalanced with the same PF, it seems there should be a olution, but I don't know what it is.

And if you don't know if the load is balanced or unbalanced? And if you don't know what the power factors of the loads are? Then how could you solve for the phase currents (knowing only the line currents)?

As I demonstrated in an earlier post, you could have different phase currents which will give you identical line currents. So how would two people given only the same line currents be able to accurately determine what the phase currents are in their systems when their load impedances are completely different?

 

[david luchini]

Well my thinking cap must not be on, or I must be missing something. If the line currents sum to zero, how could the phase currents not. In my mind, only with circulating or ground fault current could they not sum to zero. The phase current is the same current as the line current and therefore must sum to the same zero value. It's easy to think up vectors which depict a non-zero sum... but can you make it a reality?

Well, double check me on this...but if I had a system with Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect an inductive impedance of 15.969<6.9, between B-C, I connect a capacitive impedance of 19.941<-18.56, and between C-A, I connect an inductive impedance of 28.139<13.74. My phase currents should be:

Iab=13.025<-36.9
Ibc=10.431<-131.44
Ica=7.392<76.26

In rectangular form:

Iab=10.416-j7.820
Ibc=-6.904-j7.820
Ica= 1.756+j7.180

Iab+Ibc+Ica = 5.268-j8.460 = 9.97<-58.09

Since Ia=Iab-Ica, Ib=Ibc-Iab & Ic=Ica-Ibc, we get

Ia=8.660-j15.000
Ib=-17.320+j0.00
Ic=8.660+j15.000

Ia+Ib+Ic= 0+j0 = 0

This is an unbalanced (and unequal power factor) delta connected loads, where the sum of the line currents equals zero, but the sum of the phase currents does not.

 

[rattus]

And if you don't know if the load is balanced or unbalanced? And if you don't know what the power factors of the loads are? Then how could you solve for the phase currents (knowing only the line currents)?

As I demonstrated in an earlier post, you could have different phase currents which will give you identical line currents. So how would two people given only the same line currents be able to accurately determine what the phase currents are in their systems when their load impedances are completely different?

If the phase currents are not separated by 120 degrees, there is no unique solution if we have only the line currents.

 

[jghrist]

Well, double check me on this...but if I had a system with Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect an inductive impedance of 15.969<6.9, between B-C, I connect a capacitive impedance of 19.941<-18.56, and between C-A, I connect an inductive impedance of 28.139<13.74.

Or, a simpler example:

Vab=208<-30, Vbc=208<-150 and Vca=208<90 and between A-B, I connect a 10 ohm resistor. B-C and C-A are open. Iab = 20.8<-30. Ibc=Ica=0
Iab+Ibc+Ica=Iab=20.8<-30 not =0

Ia=20.8<-30, Ib=20.8<150, Ic=0, Ia+Ib+Ic=0