Unbalanced 3 phase load and line current

[Ragin Cajun]

Now that I have read the posts and gathered what wits I have left, I can look at it a slightly different way.

If there were three single phase loads, the connected KVA would be:
480V*60A =28.8 KVA * 3 = 86.4 KVA TOTAL 3 PHASE LOAD.

Thus, line current is 86.4KVA/[sqrt 3 * 0.480KV] = 103.9A per phase.

The current at "??" would also be 103.9A. The other two legs would be 60A.

Backhanded way to avoid vexing vectors - at least in this case!

Unbalanced 3 phase loads are so much fun. ;-(

Beaucoup thanks for all your help!

RC

 

[LarryFine]

I think if that phase is not loaded the current in ?? goes to 104A.

Better yet, say that it stays at 104a. Losing one load doesn'tchange the current in the opposite line.

 

[Smart $]

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?

That is correct.

However, note the order in which you state the current: A to B, B to C. One is measured coming into B while the other is measuured leaving B. We could use 120? apart phasing, but we would have to vectorially subtract A to B from B to C... because the direction of measurements are inversed with respect to the common point B. To vectorially add, we have to flop the direction of measurement on one or the other, and that results in a 60? phase shift between the two wave forms, given identical power factors.

 

[rattus]

Simply put:

Simply put:

The L-L currents may be written as:

60A @ -60 and 60A @ -120, then

I?? is simply the sum of these currents,

60A @ -90

Note that the 60 degree separation results from the definition of the known currents and does NOT conflict with the 120 degree separation usually employed to describe 3-phase currents.

 

[steve66]

The L-L currents may be written as:

60A @ -60 and 60A @ -120, then

I?? is simply the sum of these currents,

60A @ -90

Note that the 60 degree separation results from the definition of the known currents and does NOT conflict with the 120 degree separation usually employed to describe 3-phase currents.

Rattus:

Is that a typo: We were starting to get a consensus that the answer was I?? = 104 amps.

Steve

 

[mivey]

Rattus:

Is that a typo: We were starting to get a consensus that the answer was I?? = 104 amps.

Steve

Then I guess we are fortunate that we do not solve math problems by consensus. :grin:

 

[rattus]

Rattus:

Is that a typo: We were starting to get a consensus that the answer was I?? = 104 amps.

Steve

Not really a typo, just my brain shutting down last midnight. Let me restate the solution:

the L-L currents may be written as,

60A @ -60 = 60A(0.5 - j0.866)
and
60A @ -120 = 60A( -0.5 -j0.866)

I?? = -j2*0.866*60A = 104A @ -90

The 60 degree separation between the L-L currents results from the choice of reference and does NOT conflict with the 120 degree separation usually employed.

 

[gmtt]

First of all, I don't understand the current measurement technique you have shown in the picture. Are you trying to measure current across phases? At least that?s how you have shown in the picture. Current is measured thru a line, not across a line. In that case ?? is the true line current measurement , and what are the measures of 60, 60 and 0, I don?t get it. Can you reword your question?

 

[mull982]

Attached I have shown the phasors I used to arrive at an answer of 104A for ?? current.

The first sketch is the phasor diagram that I used with the A-N current of 0deg being the reference. The phasor diagram then shows all angles for L-N and L-L currents.

So since we are solving for the current on "C" we use all the L-L current angles pointing towards "C" From the phasor diagram we see that these two angles are the L-L BC phasor at -270deg and the L-L CA phasor at -210deg.

Adding these two phasors together vectorally as I did in the bottom sketch you will arrive at 104A for ??.

 

[rattus]

Even simpler:

Even simpler:

This is nothing more than breaking the phase/line current from one leg of a wye into its L-L components in a balanced wye. The fact that there are only two L-L currents is immaterial as is the fact that the presence or absence of a neutral is also immaterial.

By inspection we can write,

Iphase = Iline-linexsqrt(3)

 

[gmtt]

With no information on L-L or L-N impedance, I don't know how can you think of calculating current. You can measure RMS value of the line current but to calculate, you must know phasor value of impedance. Since we know the voltage values and their phase angles(0,120 and ?120 deg), we can devide the two phasors (V/Z) to get the I phasor.

 

[steve66]

Then I guess we are fortunate that we do not solve math problems by consensus. :grin:

True, but it's probably nice for the original poster if a few of us agree once in a while

 

[gar]

091211-1056 EST

Consider this approach:

Two sine waves of the same frequency, and amplitude, but separated by 120 deg. Let the peak current of each be Ip. Note: the RMS current is also proportional to the peak for a sine wave.

Remember from another post of mine that the sum of two sine waves of the same frequency is another sine wave of the same frequency and some resultant phase. If you graphically draw two displaced sine waves and by hand add them you will see the result is a sine wave.

Back to the basic problem. The sum of two equal magnitude sine waves 120 deg apart produces a sine wave with a peak midway between the two original waves. This is by observation and symmetry.

This peak occurs 60 deg ahead of one source and 60 behind the other. This is really a 30 degree point on a sine wave. What is the sine of 30 deg? 0.5 . Thus, we have 0.5*Ip + 0.5*Ip at this phase angle which equals Ip.

Thus, the answer to the original question is 60 A at the ?? leg. Note: the assumption is resistive loads, or the same power factor including sign for both loads. Otherwise there are all sorts of answers.

.

 

[gar]

091211-1400 EST

Revision to my last post. For the calculation we need to subtract one current from the other. Or, for addition we need to consider peaks 60 deg apart (180-120= 60) and thus the sum is 0.866*Ip + 0.866*Ip = 1.732 * Ip. The result is the 103.9 A group.

Eliminate the third secondary and reverse the phase of one secondary and the result is the 60 A.

.

 

[mivey]

True, but it's probably nice for the original poster if a few of us agree once in a while

Well, you pretty much cleared up the 60 amp confusion in post #16 and after that the consensus still could have went either way, but the math don't lie:

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.

 

[rattus]

BTW, this is wrong!

BTW, this is wrong!

The L-L currents may be written as:

60A @ -60 and 60A @ -120, then

I?? is simply the sum of these currents,

60A @ -90

Note that the 60 degree separation results from the definition of the known currents and does NOT conflict with the 120 degree separation usually employed to describe 3-phase currents.

Made a mistake once. Thought I was wrong. This is number two.

 

[rattus]

With no information on L-L or L-N impedance, I don't know how can you think of calculating current. You can measure RMS value of the line current but to calculate, you must know phasor value of impedance. Since we know the voltage values and their phase angles(0,120 and ?120 deg), we can devide the two phasors (V/Z) to get the I phasor.

All three L-L currents are given. Assuming the same PF for the 60A load currents, we can calculate (as has been shown) I??.

 

[skeshesh]

All three L-L currents are given. Assuming the same PF for the 60A load currents, we can calculate (as has been shown) I??.

Obviously engineering is not absolute physics - an important aspect is to be able to make estimations and eliminate elements/information that are difficult to obtain to be able to come to a reasonable answer quickly and effectively. I think that's what we've done here (I agree with rattus).

 

[rattus]

Where's Charlie?

Where's Charlie?

Charlie, do you now agree that the unknown current is 104A?

If not, let us discuss this further.

 

[Ragin Cajun]

Thanks for all the postings.

I looked at it the next day and . . . why didn't I SEE that??!! Simple.

Of course the leg in question is 104A and the other two 60A!

CRS strikes again! More like "advanced CRS.

Again, thanks.

RC