Unbalanced Three Phase Load 6

[gcaudill]

I have an electrical system which consists of a delta-wye connected transformer supplying a delta connected load. One corner of the delta is grounded. I have the following load current measurements:
I leg1 = 1115A
I leg2 = 1000A
I leg3 = 1092A

Can someone tell me where the unbalanced current will go? Circulate in the load or exit to ground at the grounded corner? The ground is well connected, low impedance, and terminated to the main grounding bar at the switchgear.

 

[jbartos]

Suggestion: Visit

http://engr.astate.edu/jdg/Circuits/Lab/PhaseMeasurement.html

http://www.tech.purdue.edu/Eet/Courses/eet207/Labs/Phase_Measurements.pdf

http://www.stclairc.on.ca/people/pages/dholmes/index.htm

(for oscilloscope quick phase measurement)

http://www.ee.utulsa.edu/labs/ee2001/exp8.html

http://www.hameg.de/en/_pdfs/507e.pdf

http://www.metec.co.nz/Products/HEG/HEG.html

(for PLA 10A/100A, good for 3-phase measurements)
etc. for more info

 

[Shortstub]

Jghrist,
Single-phase power measurements are possible. The kW meter's potential leads are connected from X1-Xo, and the phase-current is I1. Then determine the kVAR for that phase. Repeat for the other two phases, X2, and X3. The phase angle for each phase can be calculated. However, while fruitful for acedemic reasons, its results will be impractical, for the following reasons:

o It would be wrong to average phase-pf results to obtain load-pf.
o Two of the anneal-wire segments, are in series with the feeders connected to their sheaves, namely sheave 1 and sheave 4.
o Feeder losses must be subtracted from the phase measurements.

An option to determine each phase-angle would be a graphical representation using six current measurements, i.e., the three line currents and the three delta-connected segment currents. This approach requires construction of a quadrangle, but, once again this will be unwieldy because of the series-feeder impedances.

Another option would be to use the four voltage measurements, sheave 1-2, 2-3, 3-4, 4-1. A similar graphical presentation is used, but it too results in construction of a quadrangle.

Jghrist,
Can you modify your schematic to show the effect of SCR control in the transformer's primary?

Gcaudill,
As a means of improving "control" have you considered joining sheave 1 to sheave 4?

BTW, how are transformer's secondary line-currents and the delta-connected segmental-currents determined now? By fixed panel metering? Clamp-on metering?

 

[Shortstub]

Jghrist,
1st paragraph, kVAR should have been kVA!

 

[jghrist]

Shortstub,
Assuming X0 is available, your procedure should work for the currents labelled Ia, Ib, Ic in my diagram without worrying about the feeder cable losses. It should also work for separately measured current in the feeders to sheave 1 and sheave 4 and to ground. These add to be Ib.

I don't know anything about the SCR control. I have assumed that it will provide essentially sinusoidal voltages 120 degrees apart. If the output isn't sinusoidal, the problem becomes a lot more complicated. You'd have to solve for each separate harmonic and consider the lack of zero sequence (triplen harmonics) path through the transformer.

Sheaves 1 and 4 are essentially connected together now, through the feeder cables to X2. If you made a direct connection, the length (and voltage drop) may still be nearly as much as in the feeder cables.

 

[gcaudill]

The SCR controller has a typical 6 SCR, anti-parallel connected, output. I have measured harmonics on one particular system (not this one) and found that the fith harmonic was 85A. This is the same current I measured in the ground conductor connected to X2. Coinsidence?

 

[jghrist]

I think the relationship between 5th harmonic output of one system and the ground current is a coincidence. I'm not familiar with SCR controllers, but according to IEEE std 519, control of power to loads by phase control of thyristors (SCRs) will create harmonic currents. IEEE-519 discusses the thyristor-controlled reactor (TCR), indicating that triplens will not be present in the line currents. A TCR is a special case of an SCR controller with a purely reactive load, so I suspect there will be odd harmonics but no triplens. In any case, any triplen currents would be blocked by the Dy transformer.

Harmonics will affect the calculations because the impedances will be different for each harmonic, but the analysis for each harmonic would be done in the same manner.

 

[jbartos]

Suggestion: 6-pulse 3-phase converter or rectifier will generate 5th, 7th, 11th, 13th, 17th, 19th, etc. harmonics
12-pulse 3-phase converter or rectifier will generate 11th, 13th, 23rd, 25th, etc. harmonics. Visit

http://www.paptac.ca/eleceng/handout.pdf

for: Basics of Harmonics

 

[busbar]

A basic instrument like a Fluke 39-series could be used for phase-angle measurements, with currents referenced to voltage X1-X0 or X1-X3. Only one 250kcmil cable could be tested at a time, but that shouldn’t be a limitation. Or, there may be in-place CTs on cable groups or buses that would permit measurement in the secondary circuit. {With the higher models, harmonic content could also be determined.}

 

[Shortstub]

SCR control simply delays the turn-on time of both the positive and negative portions of an AC wave. Although harmonics are generated they are not as well identified as those expected with a multi-step converter.

 

[jbartos]

Suggestion: Since the phase angles are needed only, they could be obtained by winding a few turns over the conductors leaving X1, X2 and X3 transformer terminals. If three conductors, e.g. #16 are cut of the same length and then the same number of turns wound over each line conductor, and the each conductor ends connected to the same load, e.g. bulb, then the three phase oscilloscope would measure the current angles, i.e. Ix1 vs Ix2, Ix2 vs Ix3 and Ix3 vs Ix1 via signals tapped at the bulbs. The measurement results would be reasonably accurate since the angles and relativity of angles are sought. Some experimentation may be needed with the amount of turns and conductor length needed. This could lead to some savings on doughnut CTs.

 

[Shortstub]

I hate to see this tread end with a zero (excuse the pun) for effort. Here is a simpler mathematical approach using trigonometry instead of additional measurement, and without symmetrical compomnents:

1) Construct an uneven-sided triangle using the magnitudes of the original 'given' currents as sides. Let L1, L2, and L3 represent the annealer segment currents, i.e., 1,115 Amp, 1,078 Amp (including quench ground-return), and 1,092 Amp, respectively.

2) Label the interior angles, A, B, C, as follows:
A = the angle between sides L2 abd L3.
B = the angle between sides L1 and L2.
C = the angle between sides L3 and L1.

3) Then, using the Law of Cosines results in the following values:
A = 59.70 deg.
B = 61.82 deg.
C = 58.40 deg.

4) Construct a phase-sequence diagram using L2 as reference. The resultant phasors are:
L1 = 1115A @ 118.2 deg (leads L2).
L2 = 1078A @ 0.0 deg.
L3 = 1092A @ 239.7 deg (lags L2).

5) Calculate the corresponding transformer secondary line currents:
Ix1 = L2 - L1 = 1,882 @ -31.5 deg.
Ix2 = L1 - L3 = 1,926 @ +89.3 deg.
Ix3 = L3 - L2 = 1,882 @ -149.9 deg.

6) Calculate the associated primary-side line currents using the turns ratio, i.e., n = sqrt(3)xN, the voltage ratio, as follows:
Ia = (1/n) x (Ix1 - Ix3) = 272A @ -0.7 deg.
Ib = (1/n) x (Ix3 - Ix2) = 277A @ -120 deg.
Ic = (1/n) x (Ix2 - Ix1) = 278A @ +119 deg.

A current balance check results in an almost equal primary and secondary loading of about 229 kVA.

Another check would be to determine the SCR firing angles for each phase and then calculate the equivalent resultant rms voltage.

Have a happy holiday!

 

[jghrist]

Shortstub,

Why must the annealer segment current vectors form a triangle? This implies that they add to zero. The secondary transformer line currents have to add to zero but not the segment (delta connected load) currents.

Consider a simplified case where transformer impedances and connecting cable impedances are negligible, balanced voltages, and unequal load resistances. The segment current vectors would be 120° apart (in phase with the ø-ø voltages), but unequal so they couldn't form a triangle (a triangle with all 120° exterior angles is equilateral).

Also, why include the ground current in L2?

 

[Shortstub]

Jghrist,
This approach does not imply that the vector sum equal zero. It does, however, require that the magnitudes be such that a triangle is possible. For example, for the case where the magnitudes are 1,2,3, then this approach can't be used because construction of a triangle is not possible. In fact the Law of Cosines will yield zero for the interior angles.

In this particular case the vectoral sum is virtually zero, only because the triangle is close to being an equilateral triangle. The unequal segment currents are probably due to their lengths, and not the feeder impedances. This coincidence probably led to your conclusion. Also, KCL requires the sum of branch currents into a node must equal zero, while KVL requires the sum of loop voltages must equal zero.

Regarding your question about the quench-system ground return current. It too is flowing into terminal X2, and therefore added to current Ix2. The saving grace, here, is probably due to the fact that the load is virtually resistive (a 180 deg turn fom my original postulate way back when)! In fact, there seems to be a negligible effect from the feeder cable reactances!

If you're still interested, I can supply the results for the case with the quench contribution omitted. The results don't vary very much.

 

[jghrist]

Shortstub,
Maybe I'm not understanding what you are doing.

First of all, I don't get the same angles as you do with the law of cosines. With L1=1115, L2=1078, L3=1092, I get A (between L2&L3) = 61.83°, B (between L1&L2) = 59.7°, C (between L1&L3) = 58.47°.

If the tail of the L1 vector is drawn at the arrow end of the L2 vector, and the tail of the L3 vector is drawn at the arrow end of the L1 vector, and the arrow end of the L3 vector is at the tail of the L2 vector, this defines a triangle. Vectorially, it also means L1 + L2 + L3 = 0. This applies whether or not the triangle is equilateral. Lets say L1=1000, L2=2500, L3=2000. Law of cosines gives A = 22.33°, B=49.46°, C=108.21° as you defined A,B,C.

Using L2 as a reference, L1 = 1000 < 130.54°, L2 = 2500 < 0°, L3 = 2000 < 202.33°. L1+L2+L3=0

KCL tells us that Ix1+Ix2+Ix3=0 because they are all of the currents into node X0. This is also obvious because with:
Ix1 = L2 - L1
Ix2 = L1 - L3
Ix3 = L3 - L2
Adding Ix1, Ix2, and Ix3 together cancels L1, L2, and L3 regardless of their value.

On the quench system ground return current (if this is where it comes from), this is as you say added to current Ix2, but you added it to L2 which I thought was the current in segment 2 between X1 and X3.

 

[jbartos]

Suggestion to Shortstub (Electrical) May 25, 2003 marked ///\\\
I hate to see this tread end with a zero (excuse the pun) for effort. Here is a simpler mathematical approach using trigonometry instead of additional measurement, and without symmetrical compomnents:
///Yes, indeed; however, it appears to be an approximate approach. The approximate approaches are o.k. if they satisfy the Client and hardware.\\1) Construct an uneven-sided triangle using the magnitudes of the original 'given' currents as sides. Let L1, L2, and L3 represent the annealer segment currents, i.e., 1,115 Amp, 1,078 Amp (including quench ground-return), and 1,092 Amp, respectively.
///Good; however, where is the 78A ground current?\\2) Label the interior angles, A, B, C, as follows:
A = the angle between sides L2 abd L3.
B = the angle between sides L1 and L2.
C = the angle between sides L3 and L1.
///O.k.\\3) Then, using the Law of Cosines results in the following values:
A = 59.70 deg.
B = 61.82 deg.
C = 58.40 deg.
///O.k.\4) Construct a phase-sequence diagram using L2 as reference. The resultant phasors are:
L1 = 1115A @ 118.2 deg (leads L2).
L2 = 1078A @ 0.0 deg.
L3 = 1092A @ 239.7 deg (lags L2).
///O.k.\\5) Calculate the corresponding transformer secondary line currents:
Ix1 = L2 - L1 = 1,882 @ -31.5 deg.
Ix2 = L1 - L3 = 1,926 @ +89.3 deg.
Ix3 = L3 - L2 = 1,882 @ -149.9 deg.
///O.k.\\6) Calculate the associated primary-side line currents using the turns ratio, i.e., n = sqrt(3)xN, the voltage ratio, as follows:
Ia = (1/n) x (Ix1 - Ix3) = 272A @ -0.7 deg.
Ib = (1/n) x (Ix3 - Ix2) = 277A @ -120 deg.
Ic = (1/n) x (Ix2 - Ix1) = 278A @ +119 deg.
///O.k. as far as the approximations are concerned. I also looked into the approximate shifts in currents on the secondary side, i.e. 0deg, 120deg and 240deg in:
&quot;&quot;&quot;jbartos (Electrical) Apr 25, 2003
Answers to gcaudill (Electrical) Apr 24, 2003 marked ////\\\jbartos wrote:
//(1115/0deg + 1000/120deg + 1092/240deg)/3=Io (is) different from zero. Therefore, the Ig is different from zero and equal to Ig=3Io, and Ix1+Ix2+Ix3=Ia+Ib+Ic=Ig different from zero at X2.//&quot;&quot;&quot;&quot;
However, I claim that the angles should be measured instead of approximated by 0deg, 120deg, 240deg, to have the problem well posed. Textbook examples are often given with the angles given. This is of course for students exercise only.\\A current balance check results in an almost equal primary and secondary loading of about 229 kVA.
///Good, up to &quot;almost&quot;.\\Another check would be to determine the SCR firing angles for each phase and then calculate the equivalent resultant rms voltage.
///What about the angle shifts between Vab, Vbc and Vca?\\\

 

[Shortstub]

Jghrist,
Man, you're good!
1) I erred in matching a side to its associated opposite angle.
2) I also erred in adding 78-Amps to Leg2. It should have been added to Leg3.

Jbartos,
1) The calculated primary and secondary kVA values are 229.2 and 228.9, kVA, respectively, hence my &quot;almost&quot; comment.

2) The use of measured vs calculated phase-angles values, is unwarranted. The &quot;anticipated&quot; improvement, is at best, marginal. Thus far four solution methods have been presented:

o MatLab.
o Symmetrical Components.
o Delta-Wye transformation.
o Graphic solution.

I believe the original question about the grounding current flow has been adequately explained.

 

[gcaudill]

This thread has been a lot of fun. Thanks for everyone's input. I think 157 responses must be a record for one thread.