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 to jghrist (Electrical) May 2, 2003 marked ///\\\
I finally got around to reviewing your Stevenson reference drawings. All show ground connected to the neutral. Do you know of any references that show zero-sequence equivalents of corner grounded systems? Either transformer windings with an isolated X0
///Reference:
1. IEEE TRANSACTIONS ON INDUSTRY APPLICATIONS, VOL. 36, NO. 6, NOVEMBER/DECEMBER 2000 1741
The Use of Low-Voltage Current-Limiting Fuses to
Reduce Arc-Flash Energy
Richard L. Doughty, Fellow, IEEE, Thomas E. Neal, Terry L. Macalady, Vincent Saporita, Member, IEEE, and
Kenneth Borgwald
Includes:
"Another grounding scheme that was quite popular, and is still being used, is the corner-grounded delta system. One corner of the delta is intentionally grounded, and has no overcurrent protection in that phase. (A fuse can be used in the grounded phase if it is providing motor running (overload) protection, or a three-pole circuit breaker can be used if all three poles open simultaneously.) The advantage of such a system is that 1/3 of potential ground faults are eliminated because one phase is already intentionally grounded. Faults from either of the other two energized phases are at full line-line voltage and require overcurrent devices capable of handling the phase–phase voltage across only one pole. While the corner-grounded delta system offers help by eliminating 1/3 of the phase-to-ground faults, and no help for energized phase-to-phase or energized phase-to-ground faults, the addition of current-limiting fuses for phase-to-phase
and three-phase arcing faults also minimizes arc-flash hazards associated with these systems."
Evidently, if a line-to-line to ground fault happens such a way that the conductors to the load will be ungrounded, then there will be current flowing through the ground to the corner grounded delta. This ground current will be equal to the zero sequence current Io multiplied by 2 since there will be no current flowing from the grounded delta corner to the load.\\

 

[jbartos]

I beg your pardon, correction: The last sentence should read: This ground current will be equal to the zero sequence current Io multiplied by 3. However, there will be no current flowing from the grounded delta corner to the load, and zero sequence current will be:
Io=(Ia + Ib + 0)/3=Ig/3
since the ground current Ig:
Ig=Ia + Ib + 0

 

[Shortstub]

Jbartos,
Sounds like special math 101. 78-Amperes was measured flowing between system ground (earth) and the intentionally grounded transformer tap, X2. If your definition of Ia, Ib, and Ic, are the phase currents between the transformer and the annealing sheaves, then, how does Ig relate to Ia & Ib? What happened to Ic?

Thus, there must be a connection somewhere in the load circuit and ground, that is, system ground (earth), and not the two sets of groundED phase conductors between X2 and sheave 1, and X2 and sheave 4!

Gcaudill,
Have you given up, capitulated, or lost interest in further discussion or measurements?

 

[gcaudill]

Shortstub,

I have not given up on this, just been otherwise distracted. The system is running with no apparent operational problems. I was attemting to better understand the application, and have learned plenty from the posts.

There are no intentional ground connections between the load circuit and ground. But as was pointed out in previous posts, it is possible there are unintentional leakage paths via the quench water and other electrical isolation points in the system.

Ic is alive and well, not forgtten by any means. In my post on April 9 I listed some values for which you correctly characterized as the phase currents between the transformer and the sheaves. They were as follows:
X-former secondary amps
Conductor Gang 1 (Ph. X2 to sheave 1) = 866A
Conductor Gang 2 (Ph. X1 to sheave 2) = 1300A
Conductor Gang 3 (Ph. X3 to sheave 3) = 1153A
Conductor Gang 4 (Ph. X2 to sheave 4) = 606A

Ic is made up of Gangs 1 & 4.

Interestingly enough, we switched the Ph.A and Ph.C leads of the voltage controller on the primary side of the transformer. When we did this the secondary and currents X1 and X3 values (and voltages measured to X2) swapped accordingly. Seems to me that Ph.C of the controller is a littly "lazy". I am still thinking one bad SCR could possibly be to blame, against some opinions.

We have more than one of these applications and the currents are not nearly so unbalanced on others. This one has been checked forwards and backwards for isolation integrity. They all have the same quench water setup. There must be some explanation.

 

[jghrist]

If you swapped øA and øC on the primary and X1 and X3 swapped accordingly, this indicates that the unbalanced load is not because of differences in the load impedance but because of differences in the primary voltage. You have never given measurements of the primary voltage. This would be quite useful. Also a measure of the current in the wire between sheave 4 and the quencher. If this current is the same as the ground current it just about clinches the source of the ground leakage. If it is zero, then we are back to ground zero, so to speak.

 

[gcaudill]

Shortstub,

Here are some new measurements. These numbers can be compared with the April 29 post. This is after switching Ph.A & Ph.C on the voltage controller output. These measurements are for a heavier product (thus higher voltages and currents) but the unbalances are farily proportional. There is no way to measure any current in the wire in the quench process, the wire is not accessible.

Incoming line voltages
L1-L2 - 480
L2-L3 - 477
L3-L1 - 475

These voltages are after the voltage controller on the primary side of the transformer.
T1-T2 - 348V T1 connected to Ph.C of the controller
T2-T3 - 372V T2 connected to Ph.B of the controller
T3-T1 - 356V T3 connected to Ph.A of the controller

X-former primary amps
I ph.1 = 401A
I ph.2 = 407A
I ph.3 = 380A

X-former secondary amps
Conductor Gang 1 (Ph. X2 to sheave 1) = 1300A
Conductor Gang 2 (Ph. X1 to sheave 2) = 2250A
Conductor Gang 3 (Ph. X3 to sheave 3) = 2092A
Conductor Gang 4 (Ph. X2 to sheave 4) = 1154A

X-former secondary volts
X1 to gnd - 65VAC
X2 to gnd - 0V
X3 to gnd - 60VAC

Load amps (delta)
Currents:
I leg1 = 1350A
I leg2 = 1202A
I leg3 = 1067A

X2 grounding conductor is carying 78A

For the same process in a different machine, the load currents are:
I leg1 = 1115A
I leg2 = 1000A
I leg3 = 1092A
Much more balanced than above. These are typical values.

 

[Shortstub]

I have produced a schematic of my interpretation of the system. How can I "post" it?

 

[gcaudill]

I don't know, or else I would have posted mine alreadly.

 

[Shortstub]

Someone must know how to post a schematic!

Alternatively, can you provide an e-mail address? Or contact me at EpsiconInc@aol.com!

 

[jghrist]

I finally figured out how to use my Webspace. See my calculations (with schematic) at (I hope this works)

http://home.pipeline.com/~maximilian/Calculations/DLoad.htm

I don't get exactly the measured values, but with some assumptions, including equal load resistances, I get:

Primary voltages (input) VAC=356V, VBA=372V, VCB=348V
Primary amps IA=417A, IB=420A, IC=400A
Secondary amps Ia (X1 to sheave 2) = 2311A
Ic (X3 to sheave 3) = 2327A
X2 to sheave 1 = 1363A
X2 to sheave 4 = 1445A
Xfmr sec volts VX1grd = 63.5V, VX3grd = 62.7V
Load amps I1=1363A, I2=1335A, I3=1380A
Ground amps Ig=78.8A

Xfmr secondary sequence amps I0=0A, I1=2353A, I2=70.7A

 

[gcaudill]

jghrist,

Thanks. I will compare with my calculations. I have mathcad but have never setup my equations in it. I have put together the equations on paper. I will compare notes. Anyway to attach your mathcad file on your website?

 

[jghrist]

gcaudill,

Anyway to attach your mathcad file on your website?

Not that I know of. Give me your email address and I'll send it to you. What version of Mathcad?

 

[gcaudill]

gregory_caudill@superioressex.com

I have v7.0 and v6.0.

 

[Shortstub]

Jghrist,
Let me be one of the first... excellent work!

I always said that EE (power-side) is the only true discipline. All others are "black magic"... especially ChE!

 

[jbartos]

Suggestion: The angles measured among:
Conductor Gang 1 (Ph. X2 to sheave 1) = 1300A
Conductor Gang 2 (Ph. X1 to sheave 2) = 2250A
Conductor Gang 3 (Ph. X3 to sheave 3) = 2092A
Conductor Gang 4 (Ph. X2 to sheave 4) = 1154A
would enable one to apply the Fortescu symmetrical components, rather than to settle for:
"I don't get exactly the measured values, but with some assumptions, including equal load resistances, I get:"

 

[jghrist]

Shortstub,
Thanks. I already had the basics in previous Mathcad calcs. It was interesting figuring out how to get it onto a web page.

jbartos,
I really would be interested in seeing how this could be set up for symmetrical component analysis. I haven't found any references that show equivalent circuits for corner grounded transformers. If you treat the grounded corner as the neutral point, then you only have two phases to neutral. Symmetrical components is a method for analyzing unbalanced three phase systems. Any suggestions on how gcaudill could measure the phase angles? Maybe with an oscilloscope using one of the voltages as a reference. If he doesn't have a scope, could a power meter be used to find real power and calculate the angle to a voltage from that? What voltage?

 

[gcaudill]

I have a Fluke power meter but it does not measure phase angles. I am thinking my oscilloscope is the only way to do it as Shortstub mentioned. That should work well.

 

[Shortstub]

Gcaudill,
Fluke kW-meter may allow you to measure kVAR in addition to kW. Then, pf, hence, phase-angle(s) can be derived. Perfect opportunity to apply the 2-wattmeter configuration since there is no "neutral" available at the load.

 

[gcaudill]

Good point.

 

[jghrist]

A straight-forward 3ø power measurement would not give the phase angles of the individual phase currents. You might be able to devise a way to measure 1ø power for each phase and then divide by volt-amps to get pf, but what voltage to use?