Unbalanced 3 phase load and line current

[Ragin Cajun]

My brain(?) will not function today. In the diagram below, what is the current indicated by "??" below? It's two line-to-line loads of 60A on a three phase system.

Thanks

 

[charlie b]

I won't attest to the functionality of my brain this morning, but my answer is 60 amps. Consider the point at the bottom of the diagram, and apply Kirchhoff's Current Law. Without loss of generality, you can assign one of them an angle of 0 degrees. Thus, you have 60 amps at an angle of 0 plus 60 amps at an angle of 120. For the sum of three currents at that point to be zero, the third current has to be 60 amps at an angle of 240.

 

[LarryFine]

Another way to see it: Looking at the three conductors that form a Y in the diagram, the two upper lines would each carry 60a, the individual load currents.

Like any previously-balanced Y load, any current you deduct from one line is manifest as an equal increase in the neutral current. So, as Charlie said, 60a.

 

[jghrist]

60?sqrt(3) = 104

 

[mull982]

I believe that you would need to add the two current vectorally on the ?? phase to arrive at this current. If we call from top right going around counter clockwise A, B, C then we are looking for the current on the C phase. This current on the C phase will be a combination of the currnet contributions from the B-C connected load of 60A and the C-A connected load of 60A.

So adding these two currents vectorally I believe you would have:

60A @ -270deg + 60A @ -210deg = ????

Not sure if I did all of the math right but I come up with 38.4A

 

[steve66]

60?sqrt(3) = 104

I'll second that answer with the assumption that all the load currents are 120 deg. out of phase (a standard 3 phase system with the same PF on each leg.)

In a balanced 3 phase system, the line currents are sqrt (3) times the load currents. In this particular case, it doesn't matter that we have one phase current that is zero. It doesnt' affect the phase we are looking at.

Steve

 

[Smart $]

60?sqrt(3) = 104

I'm going to jump on this wagon.

Assuming the "Y" represents a wye-configured xfmr secondary, the amount of current through the winding between the"??" is the vector sum of two 60A currents 60? out of phase.

 

[steve66]

Without loss of generality, you can assign one of them an angle of 0 degrees.

Agree.

Thus, you have 60 amps at an angle of 0 plus 60 amps at an angle of 120.

Almost, but I think you forgot one thing.

If we draw the reference arrows using the same standards that say the phase currents are 180 deg. out of phase, one will point into our node, and one points away from the node. So you have to invert on of you vectors to use Kirchoffs in the form of Ia+Ib+Ic = 0.

Try 60 at angle 0 plus 60 at angle 300. I think that turns out to be 60* sqrt(3).

Steve

 

[charlie b]

If we draw the reference arrows using the same standards that say the phase currents are 180 deg. out of phase. . . .

Any two phase currents will be 120 degrees apart from one another. Any two line currents will be 120 degrees apart from another. Why are you describing them as being 180 degrees apart? :-?

 

[charlie b]

This current on the C phase will be a combination of the current contributions from the B-C connected load of 60A and the C-A connected load of 60A.

That is essentially the same thing I said, when I applied Kirchhoff's Current Law. The third current, when added to the vector sum of the first two, gives a result of zero.

So adding these two currents vectorally I believe you would have:

60A @ -270deg + 60A @ -210deg = ????

Why are you describing the two currents as being 60 degrees apart from one another?

 

[charlie b]

60?sqrt(3) = 104

That would work for a balanced system with equal loads on all three legs. This is not the same situation.

 

[charlie b]

Assuming the "Y" represents a wye-configured xfmr secondary, the amount of current through the winding between the"??" is the vector sum of two 60A currents 60? out of phase.

You are also calling the two currents 60 degees apart. Where is this coming from? The current from A to B and the current from B to C will be 120 degrees apart, will they not?

 

[skeshesh]

That would work for a balanced system with equal loads on all three legs. This is not the same situation.

I think you're statement is correct, but the result you're getting from that statement is not. Take a look at the diagram. In this case since the two legs are equal (60) and we're assuming the three phases are 120deg apart, the line to neutral current indicated by ?? on figure from the OP will end up being 104A as stated by jghirst.

 

[mull982]

[
Why are you describing the two currents as being 60 degrees apart from one another?

That is my mistake I was again putting the vectors tail to tail instead of head to tail. I usually plot it in CAD when drawing these vectors.

Can someone refresh my memory on how to arithmitically add two vectors head to tail?

 

[jghrist]

You are also calling the two currents 60 degees apart. Where is this coming from? The current from A to B and the current from B to C will be 120 degrees apart, will they not?

But the current from B to A is 180? from the current from A to B, making it 180?-120?=60? apart from the current from B to C. You are adding the current from B to A, not the current from A to B.

This assumes equal power factor for the two currents.

 

[steve66]

Any two phase currents will be 120 degrees apart from one another. Any two line currents will be 120 degrees apart from another. Why are you describing them as being 180 degrees apart? :-?

I can't tell you how many times I have been puzzled by this same problem because I didn't draw the reference arrows in before writing the equations.

Looking at the node between the (2) 60 amp currents, one 60 amp current flows into the node, and one flows out. (That's if we use the 120 deg. phase seperation we all are used to using). Assume one is Iab and it flows into the node. The other is Ibc and flows out. Also assume the phase current flows into the node and is IB.

That means our Kirchoffs equation is IB+Iab = Ibc.

If we want to use Charlies equation: IB+Iab+Ibc = 0, we have to reverse the reference arrow on the Ibc current. In order to do that, we also reverse the sign of the current Ibc, or add another 180 degrees to its phase. So in Charlies equation, the two currents are actually 120 + 180 = 300 degrees apart, which is the same as 60 degrees apart.

Steve

 

[steve66]

That is my mistake I was again putting the vectors tail to tail instead of head to tail. I usually plot it in CAD when drawing these vectors.

Can someone refresh my memory on how to arithmitically add two vectors head to tail?

The easy way is to take one and divide it into its horizontal and vertical components using the sine and cosine functions. I have to draw pictures when I do this to keep all the components straight. Then you just add or subtract the ones that are going the same direction. And use sqrt(a^2+b^2) to find the result.

 

[Ragin Cajun]

There is no neutral involved. These are single phase, line-to-line 480V loads. The "??" is the line current back to the panel.

Interesting.

RC

 

[LarryFine]

There is no neutral involved. These are single phase, line-to-line 480V loads. The "??" is the line current back to the panel.

Don't holler at us, Mr. Ragin! We didn't draw the neutral in the diagram, the OP did. (J/K )

If there were three loads, the line current would be 104a, so the real question is whether the shared line's current changes with the removal of the third load.

Unless I'm mistaken, the two non-shared lines' currents would equal that of those two loads. I envision this as a wye now, or is it an open Delta? Hmmm...

Added: As we know, the source type (wye or Delta) doesn't matter with a Delta-wired load. What happens to line currents when a phase opens with a Delta source?

 

[skeshesh]

Don't holler at us, Mr. Ragin! We didn't draw the neutral in the diagram, the OP did. (J/K )

If there were three loads, the line current would be 104a, so the real question is whether the shared line's current changes with the removal of the third load.

Unless I'm mistaken, the two non-shared lines' currents would equal that of those two loads. I envision this as a wye now, or is it an open Delta? Hmmm...

Added: As we know, the source type (wye or Delta) doesn't matter with a Delta-wired load. What happens to line currents when a phase opens with a Delta source?

Well, I don't know if that's true, the system characteristic doesnt turn into open delta just because one leg is not loaded, it has to be open delta connected at the transformer. I'll stick to my take from post #13, I think if that phase is not loaded the current in ?? goes to 104A.