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.

 

[gcaudill]

Stars for jghrist, busbar, and DanDel for sticking with me on this one.

 

[DanDel]

gcaudill, one thing that you must understand, current can't simply 'go to ground' and disappear, it only 'goes to ground' to complete a circuit and return to the source(in your case, the source is the wye secondary winding of the transformer).
To understand this, it really doesn't matter where your system is grounded, the neutral or a phase. The point that is grounded now sets up a voltage potential to ground for the other points in the winding. If the neutral were grounded, you would have a potential to ground at each phase connection which would be equal to each winding voltage. An example is a 208/120V system. There is 208V between phases, but 120V to ground because of the grounded neutral. If there were then an inadvertant connection to ground, ground fault curent would flow based on the voltage potential, the impedance of the connection, and the zero sequence impedance of the system.
If no point in the system were grounded, the only potential to ground would be due to capacitive coupling, and any resultant inadvertant ground connection would not cause ground fault(zero sequence) current to flow.
Since, in your system, one phase is grounded, you now have potentials set up to ground for the neutral and the other two phases. The potential to ground from your neutral would be equal to the winding voltage(phase-to-neutral), and the potential to ground for the other phases is equal to the phase-to-phase voltages.
If you measure any appreciable current flow(above capacitive current) in the ground conductor, this means that there must be another part of the system with a connection to ground to complete the circuit. This other connection can be anywhere in the load or wiring, but it is definitely present. It may represent circulating ground current caused by two connections to ground at different points in the grounded leg(one at the transformer, and one at the load).
It may instead be that there is an unknown high-resistance fault somewhere in the system.
It may be that your measurements are not accurate for any number of reasons.
Since your load currents are over 1000A, it is probably not practical to attempt my original suggestion to measure the other two phase conductors together inside one CT. Perhaps you could get a three-phase power monitor and hook up the CTs to see the vector currents. This will allow you to see the ground fault current.

 

[Shortstub]

You stated that the load was unbalanced. If you can isolate the load(s) from the transformer, measure the DC resistance for each:

o leg 1 to leg 2.
o leg 3 to leg 2.
o leg 1 to leg 3, if existing.

Then, if the resultant current by E/R calculation approximates what is measured, the load is resistive. At the very least you will determine the degree of unbalance.

 

[jghrist]

Just for grins, I made some assumptions of balanced primary voltages with unequal load resistances and calculated some currents and voltages. This will show what happens to unbalanced load currents where there is no zero sequence connection in the transformer.

Assumptions:
Primary voltage: 350 volts line-line
Load resistance: R1=0.042 ohm, R2=0.047 ohm, R3=0.045 ohm
Cable between transformer and load:
Resistance: 0.0012 ohm/phase
Reactance: 0.0005 ohm/phase

Calculated Values:
Secondary voltage: 29.5 volts line-neutral, 51.04 volts line-line
Load currents: I1=1121.7A, I2=1006.9A, I3=1048.4A
Secondary line current: IX1=1844.9A, IX2=1877.3A, IX3=1781.8A
Secondary sequence currents: Is0=0, Is1=1834.2A, Is2=55.9A
Primary line currents: IH1=275.2A, IH2=266.2A, IH3=261.3A
Primary sequence currents: Ip0=0, Ip1=267.5A, Ip2=8.15A
Voltage across loads: 47.11 volts (all identical)
Voltage between the Load Leg 1/Leg 3 connection and transformer X2: 2.44 volts


I have assumed only one connection to ground, so no ground current.

 

[jbartos]

Suggestion to gcaudill (Electrical) Apr 1, 2003 marked ///\\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?
///Please, would you clarify why you think that there is the unbalanced current.\\\

 

[gcaudill]

jbartos,

I think there is unbalanced current because I measure current in the ground that is connected to the transfromer secondary at X2. My thoughts were that current in the ground would indicated unbalanced currents in the load. After input from others in this thread regarding sequence currents, possible other inadvertant ground connections, and the like, I'm not sure where the ground current is originating.

 

[Shortstub]

gcaudill,
What is the current magnitude you measure in the "grounded" conductor between the transformer secondary X2, and the ground-plane or ground-bus, when the three "leg" currents are as shown, i.e., 1115, 1000, and 1092 Amps??

 

[gcaudill]

78A, 60Hz.

 

[Shortstub]

The key word was that this is an "annealing" application.

Essentially you have a 3-phase, 3-wire supply, feeding a two-phase load. Each load is resistive. Furthermore, they are independent of the other. One is fed thru "leg 1", the other thru "leg 3" and their combined current return is via the "grounded sheaves" which in turn are connected to "leg 2"! The loads are 120 degrees out of phase because their voltage sources are.

The ground return consists of two paths... one the groundED conductor connected to "leg 2", the other the groundING conductor (or ground plane). Because the groundING path is probably somewhat circuitous, and not in close proximity to the phase conductors, the bulk or major proportion of the current is in the groundED conductor, i.e., 1000 A. The groundING conductor will show a much smaller amount.

 

[Shortstub]

gcaudill,
The circuit is definitely not "resistive". If it were then the current in leg 2 would be Sqrt(3) times that of either leg 1 or leg 3, regardless of transformer connection. Therefore, the fact that the three legs currents are approximately the same indicates that mutual reactance of the 10-conductor phase and return circuits must play an important role!

Can you provide a better description of the physical configuration. How are the (10)x 250MCM conductors connected between the transformer terminals and the sheaves. That is, are there 10 conductors between X1 terminal and ph-1 sheave? 10 conductors between X3 terminal and ph-3 sheave? And 10 conductors between X2 terminal and the grounded sheaves associated with each of the phase sheaves? Or, is there another distribution?

Also, please include some typical lengths.

 

[jghrist]

Shortstub,
If you are right about the two-phase connection (I'm not familiar with the annealing process), and I understand it correctly, then the current in the third leg would be the vector sum of the two other currents. The vector sum of two equal currents 120 deg out of phase is of the same magnitude. 1000A <0 deg plus 1000A <120 deg = 1000A <60 deg.

What confuses me is gcaudill's description:
&quot;On the secondary X1 (ph.1) is connected between leg 1 and leg 2, X2 between leg 1 and leg 3, X3 between leg 2 and leg 3.&quot; This makes it seem like there are three load legs, connected phase-to-phase, although describing the phases connected between the load legs instead of the load legs connected between phases is confusing.

From your description, I picture two loads from phase-to-ground, and then the ground connected to the third phase.

 

[gcaudill]

Shortstub, jghrist,

There are three heating &quot;legs&quot; in the annealer. When I say &quot;legs&quot;, I am refering to the legs of the delta connected load. The x-former secondary taps (X1, X2, X3) are connected as I indicated earlier and as jghrist has stated.

There are 10 250mcm cables feeding each sheave from the x-former secondary. There are 20 cables attached to the x-former X2, and 10 cables attached to X1 and X3. The first and last sheave are connected to X2.

Distances:
x-former X2 to sheave #1 - 10'
x-former X1 to sheave #2 - 35'
x-former X3 to sheave #3 - 15'
x-former X2 to sheave #4 - 35'

I am certain that due to reactances and resistances the system is unbalanced. I can even see that when looking at transformer primary currents. What I am still seeking is what is the effect of this unbalance? Does the unbalanced currents exit via the ground, cirucilate in the load, in the transformer, etc.?

 

[Shortstub]

In response to the person that asked why the wye (excuse the pun) would have one terminal grounded. One scenario follows:

If the transformer was used in a conventional way, with phase-to-ground connections for the process, the output voltage would have been only 70/Sqrt3 or 40V. This is obviously too low for the present requirement. Although unusual in the conventional sense, there is nothing wrong with the present arrangement.

gcaudill,
Are the &quot;grounded&quot; sheaves connected to only terminal X2? This would require isolation of the &quot;grounded&quot; sheaves as well as the phase sheaves, How, then, are the annealed cables directed after passing through these sheaves?

 

[gcaudill]

Shortstub,

The grounded sheaves are only connected to x-former terminal X2. All four sheave assemblies are isolated. The wire exits the last (#4) sheave, is quenched and goes onto to packaging. The theory of having the first (#1) and last (#4) sheaves grounded (connected to X2) is to keep the wire at ground potential outside of the annealer.

 

[Shortstub]

gcaudill,
The unbalance, i.e., 78 Amp, is coming from the quencher system, because it is not completely isolated from &quot;ground&quot;!

jghrist,
You are correct. In fact, using leg 1 current (1,115A @ 0deg) as reference, and adding leg 2 (1,092A @120 deg) vectorially, yields a total current of 1,103A @59 deg. This is close to the measured 1,000 Amp in leg 2, plus the ground return of 78 Amp, for a total measured current of 1,078 Amp. Although the calculation was based on a 1-2-3 sequence, it wouldn't have mattered for the reverse sequence because of circuit symmetry.

For reference, I viewed the circuit as a balanced transhormer supply, connected to an &quot;open-delta load&quot;, i.e., L1-Load-L2, and L3-Load-L2, and no load between L2 and L3! The connection from the sheaves to the quencher accounts for the additional 78A unbalance! Presumably thu the water piping system!

 

[gcaudill]

Maybe so, but the quench process is after the last leg (leg 3) and at less than 50V to ground is the water is a very good conductor? And lastly, why would 78V flow from my ground into the transformer X2 and to the quench box/water?

I hope I made it clear before, the 78A in the ground was measured between X2 and the ground bus at the switchgear, not between X2 and the #1 and #4 annealer sheaves.

 

[Shortstub]

gcaudill,
You will have to answer the question about the water. But presuming it is distilled, what about the piping? Are all connections isolated from every structure that serves as a ground-plane. Furthermore, what happens to the conductor after it passes thru the quench stage. At that point is it isolated from the electric supply?

In my view there are two parallel grounding circuits between the sheaves and X2. Call one the neutral-circuit, and the other the ground-bus. The first is the multi-cable feeder (leg 2) between the grounded sheaves and X2. The second is an equivalent circuit between the quench tank (water/piping) and the grounding connection at X2. The former is ovbiously very low impedance. But the latter, because it is so circuitous, can be considered a relatively high impedance! Therefore, the measured current split, i.e., 1,000A, for the former and 78A for the latter is not only possible, but indeed, probable. The difference between calulated amps, 1103A, and measured 1078A (1000+78) can be attributed to meter error!

BTW, did any of the old-timers concur with my scenario regarding why one leg of the transformer's wye-side is grounded?

 

[jbartos]

Suggestion to gcaudill (Electrical) Apr 17, 2003 marked ///\\Maybe so, but the quench process is after the last leg (leg 3) and at less than 50V to ground is the water is a very good conductor? And lastly, why would 78V flow
///78V cannot flow, but 78A can\\ from my ground into the transformer X2 and to the quench box/water?

I hope I made it clear before, the 78A in the ground was measured between X2 and the ground bus at the switchgear,
///How was this 78A measured between the X2 and ground bus? Did you connect X2 and ground bus by any conductor?

Your posting:
&quot;gcaudill (Electrical) Apr 1, 2003
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.&quot;

indicates that the corner of the delta of the load is grounded. O.K.

Your posting:
&quot;gcaudill (Electrical) Apr 2, 2003
Application - this is a resistance annealer where the delta connected load is copper wire (3 segments between four sheaves (ph.1 & ph.3), two sheaves grounded).

Grounded Corner - one corner of the load delta is grounded for safety and equipment protection reasons.

Voltages and connections - 0-480VAC primary 0-70VAC secondary, 60Hz supply, wye-conneced primary, delta-wye transformer, delta connected load.

At the given currents the voltages were as follows:
Ph.1 to gnd - 48VAC
Ph.2 to gnd - 4.8V (this is the grounded corner)
Ph.3 to gnd - 46VAC
Currents:
I leg1 = 1115A
I leg2 = 1000A
I leg3 = 1092A

Transformer secondary Neutral is not grounded.&quot;

O.K.

Now, you have a source of ungrounded wye supplying delta load with one corner grounded.
Therefore, there cannot be any ground current since the ungrounded wye power source can position voltages such a way that the unbalance current does not exist (there is no path for it). If you connect the conductor at transformer X2 to ground, then you create a ground circuit that has voltage 4.8V mentioned above and depending on the ground impedance, the ground current over your added conductor to X2 can be 78A.\\

 

[jghrist]

OK, I think I finally got this thing figured out. You have 4 sheaves with wire between them. This wire is being heated by current from contact with the sheaves which are energized by the transformer. X1 is connected to sheave 2, X2 to both sheaves 1 and 4 (on the ends) with separate cables, X3 is connected to sheave 3.

I think Shotstub is right about the source of the ground current.

I made some calculations. While they do not exactly match your measurements, I think they can demonstrate where the current goes. They show that you get unbalanced currents even with balanced load resistance. My values for cable reactance are for balanced three-phase load; there may be considerable error in these values.

Assumptions:

Resistance of 10-250 MCM cables in parallel: 0.0054 ohm/1000'
Reactance of 10-250 MCM cables in parallel: 0.11 ohm/1000'
Primary voltage = 350 volts ø-ø
(hey, thanks busbar - I didn't realize you got different symbols. alt248 gives °, alt0248 gives ø)
Reference angle 0° is the primary ø-neutral voltage.
Resistance of wire between sheaves = 43 milliohm
Resistance of ground path from quencher to X2 = 60 milliohm

Calculations:

Current in Leg 1 = I1 = 1071A <171.9°
Current in Leg 2 = I2 = 1201A <-70.9°
Current in Leg 3 = I3 = 1197A <51.4°
Current in ground (away from X2) = Ig = 76.5A <-45.1°

Current in secondary phase a (X1) = Ia = 1941A <-41.5°
Current in secondary phase b (X2) = Ib = 1970A <-156.6°
Current in secondary phase c (X3) = Ic = 2100A <80.3°

Ia = I2 - I1, Ib = I1 - I3, Ic = I3 - I2

Volts to ground (X0):
Sheave 1 V1g = 1.18 V <-100.9°
Sheave 2 V2g = 46.0 V <-9.6°
Sheave 3 V3g = 50.0 V <55.3°
Sheave 4 V4g = 4.6 V <138.5°

Sequence currents:

zero-seq Ia0 = 0, pos-seq Ia1 = 2003A <-39.3°, neg-seq Ia2 = 98.8A <-168.9°

Primary line currents:

IA = 277.8A <-8.8°, IB = 301.4A <-127.2°, IC = 297.3A <108.1°

There is no zero-sequence current, which is the sum of the phase currents. There cannot be because the sum of the phase currents has to flow through the neutral X0 and there is no connection to X0. The ground current is a portion of the Leg 3 load current; part flows to X2 through the cable and part through the ground.

 

[jbartos]

Suggestion to the previous posting: There were some postings above where the current from X2 to ground is 78A. However, this appears to be the case, when that ground path for Ig is created. What if that path for Ig is not created?