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.

 

[Shortstub]

Jbartos,
The neutral terminal of the wye-winding is unused.

If you consider only the 3 phase terminals of the wye- winding then the supply to the annealer process is that of a corner-grounded delta. That corner is X2.

 

[RajT]

Stars galore for shots in the dark!

 

[Shortstub]

Gcaudill,

Your interpretation of my reasoning for the 78A is correct. Based on your measurement of 4.8 volts to ground, from phase 2 to ground, then, this voltage, or a part of it is also available at the sheave from which the "wire" is routed to the quench tank. It is that voltage drop that drives the 78A current from sheave to the quench tank, and then through the plant's ground-plane, as cicuitous as it may be, to the grounding conductor at X2!

 

[jbartos]

Suggestion to jghrist (Electrical) Apr 23, 2003 marked ///\\This is not a high resistance grounded system.
///Yes, agree, however, the ground current is small similarly to the high resistance grounding. Compare 10A to 78A and line currents over 1000A\\ The transformer has an ungrounded neutral.
///Never disputed and many times repeated\\ There will be no zero-sequence current.
///See Reference:
William D. Stevenson, Jr., Elements of Power System Analysis, 3rd edition, McGraw-Hill Book Co., 1975,
Equation: 12.18
Iao=(Ia + Ib + Ic)/3= constant different from zero
Let
Ia=Ix1
Ib=Ix2
Ic=Ix3
and from the original posting:
I leg1 = 1115A=Ia=Ix1
I leg2 = 1000A=Ib=Ix2
I leg3 = 1092A=Ic=Ix3
Now simply, Ia is different from Ib and Ib is different from Ic and Ic is different from Ia, therefore, Ia + Ib + Ic is not equal to zero. Ia,Ib,Ic are considered complex or phasors.
Transformer terminal X2 has a current Ix2 flowing to the load. It also has Ig=78A flowing into X2 as a ground current return. Then, winding X2 to transformer neutral carries Ix1+Ix3 currents. Therefore, X2 terminal becomes the transformer shifted neutral with currents Ix1, Ix2, Ix3, and Ig for symmetrical component modeling that includes the ground current 3 x Io.
Simply, for a star with 120degre angle shifts,
Io=0=(I/0deg + I/120deg + I/240deg)/3, which is not the case for:
(1115/0deg + 1000/120deg + 1092/240deg)/3=Io 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. The transformer wye connection neutral has Ia+Ib+Ic=0, however, it is an improper point to use for the modeling of the transformer with X2 grounded with Ig flowing into X2. I do not recommend the modeling that neglects Ig if it is different from zero. There may be significant ramifications to it in some applications.\\

 

[jbartos]

Suggestion to Shortstub (Electrical) Apr 23, 2003 marked ///\\Jbartos,
The neutral terminal of the wye-winding is unused.
///Agree, see my previous posting\\If you consider only the 3 phase terminals of the wye- winding then the supply to the annealer process is that of a corner-grounded delta. That corner is X2.
///Agree, see my previous comment regarding Ig and Io and the associated accurate application of Fortescu Symmetrical Components.\\

 

[Shortstub]

Gcaudill,
Can you provide approximate lengths for the three annealing-wire segments moving thru the four sheaves?

Jbartos,
My point is that the system is considered as solidly-grounded. The 78A current that is shown to be entering the grounded tramsformer terminal is flowing only because of an unintentional path to ground in the quench-tank water system.

My approach was not to use symmetrical components. Instead I converted the delta-load to a wye-load. If the 78A is ignored, then the result is a two-current loop model. When it is included, it becomes a three-loop model.

 

[jbartos]

Suggestion to the previous posting: Engineering has more than one way to do things. If your prerogative is to ignore some measured values, then the calculations and findings will be approximate only, which is often being done.
I never mentioned anywhere that the system is not solidly grounded. In fact, the 78 Ampere current is returning through the solid ground. If there is a short against the ground, there will be flowing a much higher current in the ground conductor.

 

[gcaudill]

jbartos and shortstub,

You both make convincing points.

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.//

If I understand you correctly, you indicated the neutral is shifted to X2 and zero sequence current does exist based on your math above? The zero sequence current (78A) is due to the unbalanced currents? Does the zero sequence current flow from X2 to ground?

shortstub,
I agree with your point that the 4.8V measured from the isolated sheaves to ground can cause current to flow at the quench tank. The medium would only be the quench water which is deionized. But, are you indicating that the 78A flows from ground into X2 via the ground conductor, or the other way around?

 

[jghrist]

jbartos wrote:
>(1115/0deg + 1000/120deg + 1092/240deg)/3=Io different
>from zero. Therefore, the Ig is different from zero and
>equal to Ig=3Io

Where did you get the phase angles for the currents? Unequal phase currents can add to zero if the phase angles are different from 0, 120, 240. See the simplified example in my Apr 21 posting.

If (current from X0 to X1) plus (current from X0 to X2) plus (current from X0 to X3) do not equal zero, where does the current come from? With no connection to X0, the sum has to equal zero. There is no zero-sequence current if there is no connection to X0. B-(

You are confusing ground current with zero-sequence current. In this case the ground current is merely current flowing through a parallel path from the quencher to X2. Part of the Sheave 4 to X2 current flows in the 10-250 MCM cables and part flows through a ground path.

The neutral is not really shifted to X2 when X2 is grounded. If this were the case, then the voltage of the X2-X0 winding would equal zero. The voltage of the X2-X0 winding is equal to the turns ratio of the transformer times the ø-ø voltage of the primary.

 

[Shortstub]

Gcauhill,

My reason for requesting length data was so that I could make per-unit calcs.

Reur query the 78A. It is caused by the excitation voltage of 4.8V. It flows into X2 via the grounding conductor, or "bonding" cable if you prefer to call it that.

Reur your query about "neutral" It doesn't exist, neither in point or conductor form . The closest you could come to defining one would be the Displacement Neutral Voltage (DNV), as follows:

If the annealer delta-conected load were balanced, then the load neutral point would be the centroid of the equilateral trangle formed by the load terminals i.e., the point with equal voltages from each of the terminals. Normally the voltage, with respect to system ground, would zero. But, because the source delta is grounded at X2, there is a voltage between the load neutral point and system ground!

Now consider the case of an unbalanced delta-connected load. The neutral referred to in the balanced case shifts from the neutral point noted above, by an amount equal to the Displacemnt Neutral Voltage. This point, would also show a voltage with respect to system ground.

 

[jbartos]

Answers to jghrist (Electrical) Apr 24, 2003 marked ///\\jbartos wrote:
>(1115/0deg + 1000/120deg + 1092/240deg)/3=Io different
>from zero. Therefore, the Ig is different from zero and
>equal to Ig=3Io

Where did you get the phase angles for the currents?
///The wye connection with grounded (not floating neutral) has normally fixed angles in three phase symmetrical system to 0 deg, 120 deg and 240 deg. even if the load is unbalance. This is an advantage of the grounded star transformer on the secondary. It preserves this angle symmetry.\\ Unequal phase currents can add to zero if the phase angles are different from 0, 120, 240.
///Yes, if the neutral is floating, e.g. delta connection fictitious phase currents.\\ See the simplified example in my Apr 21 posting.
///I am addressing a different aspects than you are trying to justify with Ig = 78A being neglected. So-called approximate calculation.\\
If (current from X0 to X1) plus (current from X0 to X2) plus (current from X0 to X3) do not equal zero, where does the current come from?
///It comes from the ground path return, i.e. 78A. Incidentally, this is a fundamental question in the original posting, i.e. "Can someone tell me where the unbalanced current will go?" which was downplayed by approximate calculations neglecting Ig=78A\\\
With no connection to X0, the sum has to equal zero.
///True, this was address already in the above posting. Never disputed.\\ There is no zero-sequence current if there is no connection to X0.
///True, never disputed, addressed in above postings. No need to repeat it.\\\

You are confusing ground current with zero-sequence current.
///No, see the reference which I posted above and many other text books.\\ In this case the ground current is merely current flowing through a parallel path from the quencher to X2. Part of the Sheave 4 to X2 current flows in the 10-250 MCM cables and part flows through a ground path.
///Essentially correct except direction of the current flow. Ig flows into X2 and Ix2=Ib flows from X2 to the load.\\
The neutral is not really shifted to X2 when X2 is grounded.
///X2 creates a point where Ia=Ix1, Ib=Ix2, Ic=Ix3 and Ig meet by Kirchhoff Current Law.\\ If this were the case, then the voltage of the X2-X0 winding would equal zero.
///Voltage is a different aspect and analysis. The voltage between X2 and X0 must be different from zero since the path has impedance and the 2xI0 common-mode current must flow from X2 to X0 and 1xIo from X0 to X1 and 1xIo to X3.\\ The voltage of the X2-X0 winding is equal to the turns ratio of the transformer times the ø-ø voltage of the primary.
///Not quite. There is a factor of sqrt3 involved since the transformer connection is delta-wye.\\///As an example, try to solve a grounded corner delta secondary supplying the load, that may be either in wye or delta connection with a single fault to ground from one of those ungrounded lines.\\\

 

[jbartos]

Suggestion to the original posting: Please, would you provide measured angles with posted currents:
I leg1 = 1115A
I leg2 = 1000A
I leg3 = 1092A
to enable the accurate calculation:
My approach assumes 0deg, 120deg and 240deg angles among currents. This may be inaccurate if the load is asymmetrical. It is needed for the following equation:
Iao=(Ia + Ib + Ic)/3= Ig/3 = 78/3 = 26A constant different from zero

 

[gcaudill]

I am looking into getting phase angle measurements for the load. But as I said before, this is difficult, and somewhat dangerous.

 

[jghrist]

Answer to jbartos:

///The wye connection with grounded (not floating neutral) has normally fixed angles in three phase symmetrical system to 0 deg, 120 deg and 240 deg. even if the load is unbalance. This is an advantage of the grounded star transformer on the secondary. It preserves this angle symmetry.\\ Unequal phase currents can add to zero if the phase angles are different from 0, 120, 240.
///Yes, if the neutral is floating, e.g. delta connection
fictitious phase currents.\\\

The angles are fixed for voltage (if you neglect source impedance), not current. In my Apr 21 example with an unbalanced delta load, it makes no difference if X0 is floating or grounded. The currents will be as calculated and they will add to zero. If X0 were grounded, with a delta connected load, there is no other connection to ground so still nowhere for Io to go so it has to be zero.

With no connection to X0, the sum has to equal zero.
///True, this was address already in the above posting. Never disputed.\\ There is no zero-sequence current if there is no connection to X0.
///True, never disputed, addressed in above postings. No need to repeat it.\\\

You say "never disputed" but you also say in your next post
Iao=(Ia + Ib + Ic)/3= Ig/3 = 78/3 = 26A constant different from zero

You also say
///Essentially correct except direction of the current flow. Ig flows into X2 and Ix2=Ib flows from X2 to the load.\\\

I consider Ix2=Ib to be the current flowing in the X0-X2 winding, that is, the ground current plus the current flowing in the 20 cables connected to X2.

You are confusing ground current with zero-sequence current.
///No, see the reference which I posted above and many other text books.\\\

Can you supply the particular section from Stevenson? I suspect my Second Edition, 1962 probably has the same section as the Third Edition that you youngsters used.

The voltage of the X2-X0 winding is equal to the turns ratio of the transformer times the ø-ø voltage of the primary.
///Not quite. There is a factor of sqrt3 involved since the transformer connection is delta-wye.\\\

Note I said turns ratio, not voltage ratio.

 

[jbartos]

Comment on answers from jghrist (Electrical) Apr 25, 2003
marked ////\\\
The angles are fixed for voltage (if you neglect source impedance), not current. In my Apr 21 example with an unbalanced delta load, it makes no difference if X0 is floating or grounded.
////True, never disputed. In fact, I provided postings above that would state this with respect to the grounded load, which would create the ground path.\\\ The currents will be as calculated and they will add to zero. If X0 were grounded, with a delta-connected load, there is no other connection to ground so still nowhere for Io to go so it has to be zero.
////True, never disputed.\\\
With no connection to X0, the sum has to equal zero.
////True, this was address already in the above posting. Never disputed.\\\ There is no zero-sequence current if there is no connection to X0.
////True, never disputed, addressed in above postings. No need to repeat it.\\\
You say "never disputed" but you also say in your next post
Iao=(Ia + Ib + Ic)/3= Ig/3 = 78/3 = 26A constant different from zero
////This equation is valid if Io path, e.g. ground path, or neutral path is in existence, as always claimed.\\\You also say
///Essentially correct except direction of the current flow. Ig flows into X2 and Ix2=Ib flows from X2 to the load.\\
I consider Ix2=Ib to be the current flowing in the X0-X2 winding, that is, the ground current plus the current flowing in the 20 cables connected to X2.
////True.\\\
You are confusing ground current with zero-sequence current.
///No, see the reference which I posted above and many other text books.\\
Can you supply the particular section from Stevenson? I suspect my Second Edition, 1962 probably has the same section as the Third Edition that you youngsters used.
////I first learned symmetrical components in mid sixties. I use the Third Edition as a convenient reference. I will provide this information.\\ The voltage of the X2-X0 winding is equal to the turns ratio of the transformer times the ø-ø voltage of the primary.
///Not quite. There is a factor of sqrt3 involved since the transformer connection is delta-wye.\\
Note I said turns ratio, not voltage ratio.
////This is why I address it since the turn ratio is number of turns on one winding over number of turns of other winding, unless defined differently.\\\

 

[Shortstub]

Seem like the like a classic "PEing" contest! Or, much adieu about "zero!"

 

[jbartos]

Suggestion: Reference:
1. William D. Stevenson, Jr., Elements of Power System Analysis, 3rd edition, McGraw-Hill Book Co., 1975,
Figure 12.15 (e) Zero-sequence current paths
Figure 12.17 (c) Zero-sequence network for Y-connected loads
Figure 13.1 Circuit diagram for a single line-to-ground fault on phase a at the terminals of an unloaded generator whose neutral is grounded through a reactance
Figure 13.6 Circuit for a double-line-to-ground fault on phase b and c the terminals of an unloaded generator whose neutral is grounded through reactance
Figure etc.
It is clear from the above Figures that the ground path, when in existence by a ground connection at the source, e.g. wye, is allowing the ground current to flow back to the source. All these currents are possible to measure, and the measurements tend to be in agreement with calculation results. Also, software is requesting the above discussed data to be inputted to produce the results.

 

[jbartos]

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.//

If I understand you correctly, you indicated the neutral is shifted to X2 and zero sequence current does exist based on your math above?
////Yes. The neutral point is at X2.\\\ The zero sequence current (78A) is due to the unbalanced currents?
////Yes. However, there may be other unbalances that are analyzed over the positive and negative sequences if the ground path has high ground impedance and more than 78A would be needed to account for the unbalanced current return. This is why I mentioned that posted currents in the original posting should better have angles measured and posted. This would ease the calculation substantially and provide more accurate calculation results.\\\ Does the zero sequence current flow from X2 to ground?
////Zero sequence current flows into X2. In this application, it leaks in the power distribution and load downstream from the transformer.\\\

 

[jbartos]

A little reading on a side:

http://eivind.imm.dtu.dk/staff/kjn/icassp98_kjn.pdf

which addresses the ground current and zero sequence current

http://www.bowest.com.au/library/electric.html#09

for summary of symmetrical components. It incidentally includes the equation I posted above

http://www.bowest.com.au/library.html

http://www.selinc.com/techpprs/6066.pdf

 

[jghrist]

Shortstub,

Trouble with old f**ts like jbartos and I, when we start a p***ing contest, sometimes we can't stop.

Suggestion to gcaudill,

Try the more direct approach I mentioned a while back. Measure the current in the wire before it enters sheave 1, between sheave 4 and the quencher, and after the quencher. I'll bet that this will show where your ground current comes from.