Unbalanced Three Phase Load
Unbalanced Three Phase Load
(OP)
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.
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.






RE: Unbalanced Three Phase Load
Check the loads they may not be identical in magnitude and angle.
Check the currents they may not have the same angle
Beside that, the system has has a capacitance to ground that allows some current to circulate
RE: Unbalanced Three Phase Load
If the load is 3-wire and insulation is good shape, then the three measurements account for all current flow. There would be no zero-sequence component—only positive and negative.
{What are the respective ø-ø voltages? Is it a single load or a composite?}
RE: Unbalanced Three Phase Load
The current at each point of the delta must add to zero.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
Usualy for star/star high voltage Xfmr with tertiary winding, one corner of the delta winding is grounded to limit the surges transmitted from high voltage system.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
The unbalanced delta load, as stated by busbar, has no zero-sequence current. The positive- and negative-sequence currents come from the primary. If there were unbalanced wye-connected loads, the zero-sequence current would circulate in the delta-primary; the primary line currents would be positive- and negative-sequence.
RE: Unbalanced Three Phase Load
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 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
Transofrmer secondary Neutral is not grounded.
I have not checked the phase angles of the currents and voltages. I wish I knew what they were. Except for the transformer and the conductors from the transformer to the sheaves, I think it would be safe the say the system is primarily resistive and not reactive in nature. But I don't know for sure.
RE: Unbalanced Three Phase Load
The unbalance in the secondary is positive- and negative-sequence. This translates into the wye primary as positive- negative- and zero-sequence current.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
Is it possible to measure the other two phase conductors(not the grounded one) together inside one CT? If the amperage is equal to what you measure in the ground conductor, then you have another ground connection. If it isn't the same, you may have a faulty instrument or some other problem.
RE: Unbalanced Three Phase Load
Are you implying that only two of the three transformer secondary phases connected to the delta load are current carrying?
From load measurements of:
I leg1 = 1115A
I leg2 = 1000A
I leg3 = 1092A
You get calculated transformer secondary currents of:
Ix1 = 1931A
Ix2 = 1732A
Ix3 = 1891A
And calculated transformer primarmy currents of:
I L1 = 487A
I L2 = 477A
I L3 = 437A
x-former turns ratio = 480/70 = 6.857
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
I'm not sure the measurement suggested by DanDel is possible but I am going to look into it.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
By saying that one phase of the wye is grounded, which is how it appears, conjurs images of a phase-to-ground fualt condition. Is the grounded phase of the secondary wye (X2) physically the same as one of the primary delta corners (i.e. H2)?
RE: Unbalanced Three Phase Load
I'm getting increasingly confused about what you're talking about.....
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
Is the transformer winding ratio 480/70 or is this the voltage ratio? When the primary voltage is 480 volts phase-to-phase, what is the secondary voltage phase-to-phase? With a winding ratio of 480/70, the secondary phase to neutral voltage would be 70 volts and the phase-to-phase voltage would be 121 volts.
RE: Unbalanced Three Phase Load
Some utilities will furnish corner grounding of a 480V-wye secondary to get around the NEC 230-95 ground-fault-protection mandate. The problem is, with molded-case breakers used downstream, they must correctly be applied at their single-pole interrupting rating, that is often only 8.6-12KA even through their “nameplate” is stamped 42-65KA.
The 480V single-pole interrupting rating is typically buried in a manufacturers’ “frame book” and not likely any where else. This potential misapplication applies to non-slash-rated MCCBs as well as slash-rated devices.
RE: Unbalanced Three Phase Load
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. The voltage ratio of the transformer is 480:70 delta-wye, I do not know the turns ratio.
When the phase-to-phase primary (supply) voltage is 480V (wye connected) the secondary phase-to-phase voltage is 70V. The transformer secondary is wye connected, the load is delta connected. X2 of the transformer secondary is grounded.
RE: Unbalanced Three Phase Load
I suspect that the Ph. 2 current is split between the phase conductor and the ground path. This is how you are getting current in the ground connection. This is not zero-sequence current. With an isolated neutral, you would not have any zero-sequence current. This would only make sense if there were a connection to ground at both the load and the transformer. But if this were the case, I don't know where the 4.8v would be measured.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
gcaudill -- Apologies if I overlooked it, but can you describe the low-voltage 3ø load? With the wye secondary, it seems unusual that the secondary-winding neutral/wyepoint is floating. Is there a specific reason for that?
RE: Unbalanced Three Phase Load
It is a good thing that you don't really have a problem.
This thread is getting very interesting.
You did say that you have 70 amps measured on the load Delta ground condutor.Wouldn't it be your natural reason for the umbalance on such a system? I don't think you couldn have Balanced currents with this set up.
Thanks .
GusD
RE: Unbalanced Three Phase Load
Here is a description of the load again. This is a resistance annealer where the delta connected load is copper wire (3 segments between four sheaves (ph.1 & ph.3 "hot"), two sheaves grounded to ph.2).
Grounded Corner - one corner of the of the load delta is grounded at the transformer X2 tap for safety and equipment protection reasons.
Voltages and connections - 0-480VAC primary 0-70VAC secondary, 60Hz wye connected supply, delta-wye x-former, delta connected load. Ph.2 of the x-former secondary is grounded only.
I do not know why the neutral is left floating, always been done that way.
GusD,
You are correct that I do not have any problems. I am trying to understand the physics of the system. It is safe to assume that the load will always operate unbalanced due to 1.) differences in leg lengths 2.) differences in load resistance as the wire is heated along its length 3.) differences in cable supply lengths from the x-former to the sheaves. 10 250mcm cables feed each sheave assembly.
When the system is unbalanced, where does the unbalanced current circulate (or go); in the transformer, in the load, to ground?
RE: Unbalanced Three Phase Load
The unbalanced voltages could be caused by the phantom second connection to ground.
Unbalanced currents circulate in the transformer. The three phase currents in the wye tranformer legs do not have to be equal, but they must add vectorially to zero because there is no neutral connection.
RE: Unbalanced Three Phase Load
Why are you so certain there is a phantom connection to ground? The sheaves are all isolated so that the are not electrically connected to ground. The only connection point to ground is at X2 of the transformer secondary.
You had written previously:
"I suspect that the Ph. 2 current is split between the phase conductor and the ground path. This is how you are getting current in the ground connection. This is not zero-sequence current. With an isolated neutral, you would not have any zero-sequence current. This would only make sense if there were a connection to ground at both the load and the transformer."
One of my questions is this, would X2 of the x-former secondary stand at some voltage potential with no ground attached? If so, then by bolting it to ground it is essentially a continuous ground fault? If this is true, why don't I see a large short circuit in the ground conductor?
The rated impedance of the transformer is 5%. A bolted fault current of Is.c.=70V/.05=1400As.c. should exist on the ground conductor connected to X2?
RE: Unbalanced Three Phase Load
Without any ground connection on the transformer, the phase voltages to ground are determined by the relative capacitance of each phase to ground. There is some stray capacitive coupling to ground from cables and load equipment. Connecting one phase to ground would result in the stray capacitive current from the other phases flowing through the ground connection, but this will be very small at 48V.
If you were to ground the neutral and then ground the phase, you would have a bolted single phase-to-ground fault, but with the neutral floating, you would have no fault current.
RE: Unbalanced Three Phase Load
gc — Thank you for restating the application. {This thread is getting quite detailed.}
Small aside — Wouldn't secondary short-circuit current be based on the secondary-winding rated current and not secondary-winding voltage?
Given usual process resistive and reactive variables, the measured unbalance may be quite small and anticipated in the annealing-equipment design. There may not be any significant problem to address.
RE: Unbalanced Three Phase Load
I think what I need to learn from this thread is:
1.) Characteristics of the delta-wye connected transformers and effects of connecting/not connecting the neutral.
2.) Effects of capacitive coupling on the secondary side.
3.) Finding positive and negative zero sequence currents.
Anyone have any brief explanations (primers) on these three topics?
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.\\\
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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??
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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:
"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." 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.
RE: Unbalanced Three Phase Load
There are three heating "legs" in the annealer. When I say "legs", 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.?
RE: Unbalanced Three Phase Load
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 "grounded" sheaves connected to only terminal X2? This would require isolation of the "grounded" sheaves as well as the phase sheaves, How, then, are the annealed cables directed after passing through these sheaves?
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
The unbalance, i.e., 78 Amp, is coming from the quencher system, because it is not completely isolated from "ground"!
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 "open-delta load", 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!
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
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:
"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."
indicates that the corner of the delta of the load is grounded. O.K.
Your posting:
"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."
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.\\\
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
If the path for Ig is not created, all of the Leg 3 current would return to X2 through the cables and none throught the ground. Current from X2 to ground would be zero.
RE: Unbalanced Three Phase Load
jbartos,
If the path for Ig is not created, all of the Leg 3 current would return to X2 through the cables and none throught the ground.
///Please, would you clarify this in view of your posting:
"The ground current is a portion of the Leg 3 load current; part flows to X2 through the cable and part through the ground. "
and in view of the posting by "gcaudill (Electrical) Apr 17, 2003" indicating:
"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."\\\
Current from X2 to ground would be zero.
///This appears to contradict the test results stated in the "gcaudill (Electrical) Apr 17, 2003" posting, indicating 78A being measured between X2 and the ground bus in the switchgear.\\\
///Else, the calculation is impressive.\\\
RE: Unbalanced Three Phase Load
A clearer picture of the process would help!
1) Are the multi-cable feeders, the ones you call leg 1, leg 2, and leg 3, fixed?
Does the wire or cable being annealed go from some reel or spool thru sheave 1 to sheave 2?
Does another wire or cable being annealed from a second reel or spool go thru sheave 3 to sheave 2?
Then do the two annealed wires continue to the quench bath, thru sheaves 3 and 4, respectively?
Finally do they end up on a takeup reel or spool?
RE: Unbalanced Three Phase Load
Here's how I picture the situation:
The wire being annealed is fed into sheave 1 which is connected to X2. The wire then stretches from sheave 1 to sheave 2 (Leg 1). Sheave 2 is connected to X1. The wire then stretches from sheave 2 to sheave 3 (Leg 2). Sheave 3 is connected to X3. The wire then stretches from sheave 3 to sheave 4 (Leg 3). Sheave 4 is connected to X2 (with a separate cable than the X2-sheave 1 connection). The wire then goes from sheave 4 to a quencher.
Leg 1 is connected ø-ø from X2 to X1. Leg 2 is also connected ø-ø from X1 to X3. Leg 3 is connected ø-ø from X3 to X2. Sheaves 1 and 4 are meant to be at ground potential, but because of voltage drop in the two cables from X2, a voltage to ground exists.
Current through Leg 3 comes from X3 (through 10-250 MCM cables) and from Leg 2. In my example calculations, the current from X3 into Leg 3 is 2100A <80.3°. It adds to the load current in Leg 2, 1201A <-70.9°, resulting in a load current in Leg 3 of 1197A <51.4°.
The current in Leg 3 is supposed to all go back to X2 through the 10-250 MCM cables, but some (78A) goes through the wire to the quencher and then through some ground path back to the transformer ground and then to X2.
If the quencher (and the wire after it leaves the quencher) were isolated from ground, all of the current would be forced to get from Sheave 4 to X2 through the 10-250 MCM cables and there would be no current measured from X2 to the transformer ground.
RE: Unbalanced Three Phase Load
The current in Leg 3 is supposed to all go back to X2 through the 10-250 MCM cables, but some (78A) goes through the wire to the quencher and then through some ground path back to the transformer ground and then to X2.
///This sounds somewhat mysterious how that "some ground path” is in existence. Apparently, there must be some connection in the transformer from X2 to ground bus. This connection would lend itself for ground current measurements, yielding 78A. Then, everything is o.k. and the unbalanced currents path is established.
Then, contrary to the extensive calculations above, which is using zero sequence current equal to zero, there will actually be the zero sequence current flowing through the ground and it will be equal to 78A/3=26A.\\\
RE: Unbalanced Three Phase Load
There can be unbalanced phase current without zero-sequence current. If the three phase currents are unequal but add to zero, then there is no zero-sequence current. There can be no zero-sequence current if the transformer neutral is isolated. For instance, Ia = 1940.9 <-41.5°, Ib = 1970.3 <-156.6°, Ic = 2100 <80.3°. The currents are not equal (unbalanced), but Ia+Ib+Ic = 0.
There can be no zero-sequence current if the transformer neutral is isolated. It doesn't matter if there is or is not any ground current. If a delta transformer secondary has one corner grounded, a fault to an ungrounded phase would result in ground fault current, but no zero-sequence current.
RE: Unbalanced Three Phase Load
The "some ground path" is probably through building steel.
///This may be true, however, the 78A were measured between the transformer X2 terminal and switchgear ground bus as posted above. I was interested in this since this connection establishes path for the ground currents to flow. These are also referred to as common mode currents. They should be accounted for in the model by symmetrical components to have proper electrical energy balance and the model accurate. Else, the mathematical model is inaccurate, i.e. somewhat off.\\\
There can be unbalanced phase current without zero-sequence current.
///Certainly, if modeled so. See above comments.\\\
If the three phase currents are unequal but add to zero, then there is no zero-sequence current.
///Agreed. But if the "if" represents the reality.\\\
There can be no zero-sequence current if the transformer neutral is isolated. For instance, Ia = 1940.9 <-41.5°, Ib = 1970.3 <-156.6°, Ic = 2100 <80.3°. The currents are not equal (unbalanced), but Ia+Ib+Ic = 0.
///True, based on the "if".\\\
There can be no zero-sequence current if the transformer neutral is isolated.
///True, but what remains to be added, there is no any other ground connection, e.g. corner ground from transformer terminal X2 to ground as posted above.\\\
It doesn't matter if there is or is not any ground current.
///It just depends on the accuracy of modeling. How to account for the electrical energy balance accurately?\\\
If a delta transformer secondary has one corner grounded, a fault to an ungrounded phase would result in ground fault current, but no zero-sequence current.
///True, however, the load is also corner grounded as posted above, creating a ground path with 78A flowing through it. The transformer actually has an open wye connection since the transformer neutral is not grounded and the actual neutral is the ground path carrying three zero sequence currents 3xIo=78A. The symmetrical component modeling needs to be adjusted accordingly.\\\
RE: Unbalanced Three Phase Load
>This may be true, however, the 78A were measured between
>the transformer X2 terminal and switchgear ground bus as
>posted above. I was interested in this since this
>connection establishes path for the ground currents to
>flow.
The connection from X2 to the switchgear ground bus establishes only one end of the path for ground currents. There has to be another connection to ground or ground current will not flow. I surmise that Shortstub is correct that the other connection is at the quencher.
>They should be accounted for in the model by symmetrical
>components to have proper electrical energy balance and
>the model accurate. Else, the mathematical model is
>inaccurate, i.e. somewhat off.
Actually, I modelled the circuit using Kirchoff's voltage law and mesh equations, not using symmetrical components. I calculated the sequence currents from the phase currents after determining the phase currents by mesh equations.
>...the load is also corner grounded as posted above,
>creating a ground path with 78A flowing through it.
The load is corner grounded through the cable connecting it to X2. The only intentional connection to ground is at X2.
>The transformer actually has an open wye connection since
>the transformer neutral is not grounded and the actual
>neutral is the ground path carrying three zero sequence
>currents 3xIo=78A. The symmetrical component modeling
>needs to be adjusted accordingly.
If the transformer has an open wye connection, then there is no neutral path. If there is a ground path, it does not carry any zero-sequence current. Zero-sequence current is not equivalent to ground current. What I mean by ground current is any current that flows through the earth or things like pipes and building steel that are part of the grounding system.
RE: Unbalanced Three Phase Load
I think what I need to do is to take current and phase angle measurements on the conductor gangs that feed sheaves 1-4. This is easier said than done due to the number of conductors and their proximity to danger areas on the machine. I'm not sure if I can get phase angle measurements on the load legs (legs 1-3) as this is even more difficult.
I'm still not sure about the idea of current leakage to ground, via the quench process, causing the 78A measured in the grounding conductor connected to X2. The wire is in contact with the quench water for a very short length prior to contacting sheave #4. But the wire makes no contact with piping, metal, etc, prior to contacting sheave #4.
One last note on my theory of an unbalanced load causing the 78A measured in the grounding counductor connected to X2. If a different number of commutating brushes are used on sheaves #2 and #3, or in other combinations of different sheaves, the ground current more than doubles. For example, last week someone mistakenly installed twice the number of brushes on sheaves #1 and #3 as sheaves #2 and #4. This caused the current in the ground conductor to increase from 78A to 183A. This still leaves me to believe that an unbalanced load is causing the current in the ground conductor.
Questions:
Would a circulating current (due to load unbalance) in the load have to be a zero sequence current, or would it be a sum of the three currents at their respective phase angles?
Unbalanced currents would have one of three options?
- Find its way to the ground conductor at X2?
- Circulate in the load?
- Circulate in the transformer?
RE: Unbalanced Three Phase Load
Unbalance load currents can occur without zero-sequence currents or ground currents. Take this simplified example:
Zero impedance in connecting cables.
ø-ø secondary voltages: Vba (X2 to X1) = 50 V <0°, Vcb (X3 to X2) = 50 V <-120°, Vac (X1 to X3) = 50 V <120°
Leg resistances: Leg 1 (X2 to X1) = R1 = 50 milliohm, Leg 2 (X1 to X3) = R2 = 50 milliohm, Leg 3 (X3 to X2) = R3 = 100 milliohm
Current in Leg 1 = I1 = Vba/R1 = 1000A <0°
Current in Leg 2 = I2 = Vac/R2 = 1000A <120°
Current in Leg 3 = I3 = Vcb/R3 = 500A <-120°
Phase a (X1 to Sheave 2) current = Ia = I2-I1 = 1732A <150°
Phase b (sum of X2 to Sheave 1 and X2 to Sheave 4) current = Ib = I1-I3 = 1323A <19.1°
Phase c (X3 to Sheave 3) current = Ic = I3-I2 = 1323A <-79.1°
Zero-seq current = (Ia + Ib + Ic)/3 = 0A
Positive-seq current = (Ia + Ib*1<120° + Ic*1<240°)/3 = 1443A <150°
Negative-seq current = (Ia + Ib*1<240° + Ic*1<120°)/3 = 288.7A<150°
Current from X2 to ground = 0
RE: Unbalanced Three Phase Load
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.
///In view of your last posting, where it is indicated that the X2 at the transformer wye is grounded, would you clarify the above "one corner of the delta is grounded." It can easily be understood that one corner of the load delta is grounded at the load, not at the transformer wye connection X2 to switchgear bus. This would then somewhat correct the situation and interpretation of the problem. However, the ground currents in form of the common-mode current, perhaps leakages would still exists, as generally admitted in above postings, and form the 3xIo current. The symmetrical component calculation should be adjusted accordingly.\\\
I have the following load current measurements:
I leg1 = 1115A
I leg2 = 1000A
I leg3 = 1092A
RE: Unbalanced Three Phase Load
The fact that the 78A current changes (although corresponding changes in legs 1,2,3 were not noted) indicates that brush contact is an influence. Is there a way you can measure the current in the cable routed to the quench tank, by means of a large "window" clamp-on ammeter?
RE: Unbalanced Three Phase Load
I believe, that for safety reasons, the "wire" being annealed moves as follows:
1) From wire moves from the source reel to sheave 2.
2) From sheave 2, it moves to sheave 1.
3) From sheave 1, it moves to sheave 3.
4) From sheave 3, it moves to sheave 4 (conductively connected to sheave 2).
5) From sheave 4, it moves to the quench tank.
6) from the quench tank it moves to a takeup reel.
The path described insures that the beginning and end of annealed "wire" loop occurs at sheaves, 2 and 4, respectively. And, because they have the lowest exposure volts to ground, 4.8 Volts, personnel risk is minimized.
Now, then, there are two possible "sneak" paths to ground... one being the quench tank, the other the supply reel. If the feeding reel is well insulated from "ground", then there will be only one sneak path. The test I suggested earlier should confirm it!
RE: Unbalanced Three Phase Load
Close in your routing, but not exact.
1.) Wire moves from source to sheave 1.
2.) From sheave 1 to sheave 2 (x-former X1).
3.) From sheave 2 to sheave 3 (x-former X3).
4.) From sheave 3 to sheave 4.
5.) Quench process after sheave 4.
Sheaves 1 and 4 are electrically connnedted to x-former secondary X2. Tap X2 is connected to ground.
RE: Unbalanced Three Phase Load
>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'
that sheaves #1 and #4 are connected to the grounded X2 and that for safety reasons, this is where the wire enters and exits the annealer.
RE: Unbalanced Three Phase Load
Now then, can we further agree that there is only one intentional ground/earth connection. And, it is at X2, not at the sheave 2/4 corner of the delta formed by the annealed wire path through the four sheaves! If so, then the only path for the mysterious 78A is the quench tank path.
RE: Unbalanced Three Phase Load
Seems like we are on the same page. There is only one intentional ground, at X2 of the transformer. All four sheaves are to be isolated.
Your synopsis is that the only cause for current in the ground is stray current through the quench tank? Not due to any load or current unbalance?
RE: Unbalanced Three Phase Load
Measure the current in the wire before sheave 1, between sheave 4 and the quencher, and after the quencher. If there is 78A in the X2-ground connection and 78A in the wire between sheave 4 and the quencher, and nothing in the wire after the quencher, then we have pretty much proven that the quencher is the source of the ground current.
You mentioned that changing the number of commutator brushes changed the amount of ground current. Did it also change the amount of load current?
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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!
RE: Unbalanced Three Phase Load
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.\\\
RE: Unbalanced Three Phase Load
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.\\
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
>(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.
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.
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.\\\
RE: Unbalanced Three Phase Load
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
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
///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.
RE: Unbalanced Three Phase Load
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.\\\\
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.\\\\
RE: Unbalanced Three Phase Load
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
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
Suggestion to previous posting: That is what the eng-tips are all about.
2.
Suggestion to gcaudill (Electrical) Apr 25, 2003 ///\\\
I am looking into getting phase angle measurements for the load. But as I said before, this is difficult, and somewhat dangerous.
///It just needs to sense the currents over conductors from X1, X2, and X3 transformer terminals to loads, respectively, e.g. by suitable probes, e.g. CTs and monitor the voltage on shunts by a three channel oscilloscope, and notice the angle shifts between currents Ix1, Ix2, and Ix3. The rest will be accurate and easy then.\\\
RE: Unbalanced Three Phase Load
How about a breather? Let's see if we can a least agree on the load configuration:
Gcaudill (please confirm) has stated that the delta-connected load has three legs which have been identified as leg 1, leg 2, and leg 3. For clarification I will call them segments A, B, and C, and the resultant delta arrangement follows:
o A segment has as terminal points sheave 2 and 3.
o B segment has as terminal points sheave 1 and 2.
o C segment has as terminal points sheave 3 and 4.
Now then, that means segments B and C are in series with their respective feeder cables, while segment A has no additional series impedance:
o B has a 10'x5/C feeder cable from its sheave 1 end to transformer terminal X2.
o C has a 35'x5/C feeder from its sheave 4 end to the same transformer terminal X2.
o A and B, have one of their common corners, connected to transformer terminal X1, via a 35'x5/C feeder.
o A and B, have their other common corner connected to transformer terminal X3, via a 15'x5/C feeder.
Gcaudill, please also confirm what the "leg" currents are. Are they measuring the delta-load segments? Or, are they measurements of the feeder conductors between the transformer and the sheaves to which they are connected?
Jghrist & Jbartos,
The 250MCM cable X/R ratio seems to be too high, unless you are somehow compensating for mutual reactance.
RE: Unbalanced Three Phase Load
o The common corner of segments A and B, are connected to transformer terminal X1, via a 35'x5/C feeder.
o The common corner of segments A and C, are connected to transformer terminal X3, via a 15'x5/C feeder.
RE: Unbalanced Three Phase Load
You are right about the X/R ratio. I tried to guesstimate the reactance from NEC Table 9, but used ohm/km instead of ohm/1000' but did use ohm/1000' for resistance. The reactance is problematic because each phase conductor is a bundle of 10 (not 5 according to gcaudill's Apr 17 posting) cables. I calculate a reactance at one foot spacing of 0.0576 ohm/1000' for the bundle if it comprises three layers, 3/c bottom, 4/c middle, and 3/c top of 250 MCM RHH copper wire. If the three phase bundles were laying side by side, the reactance per phase would be 0.0315 ohm/1000'. If the equivalent spacing were two feet, the reactance would be 0.0735 ohm/1000'.
The resistance would be 1/10 of 0.052 ohm/1000' if not in metallic conduit.
We don't know the spacing, but we do know that the mutual reactances between phases are not equal, so things are a bit messy. One of the phase connections (X2) is actually two separate bundles going to different sheaves. With my corrected smaller reactance, I feel better about ignoring the unbalanced cable configuration. The reactance of a 35 foot length would be around one milliohm if the cables were lying side by side.
I'm not sure what you mean by "segments B and C are in series with their respective feeder cables, while segment A has no additional series impedance". The segments are connected between phases, not in series with the feeder cables. Your segment A is between X1 and X3, segment B is between X1 and X2, and segment C is between X2 and X3.
RE: Unbalanced Three Phase Load
Call the three annealed-wire element impedances, Za, Zb, and Zc:
o Za, located between sheaves 2 and 3.
o Zb, located between sheaves 1 and 2.
o Zc, located between sheaves 2 and 4.
Identify the four feeder elements taken from the transformer terminals as Z1, Z2, Z3, and Z4:
o Z1, located between sheave 1 and X2.
o Z2, located between sheave 2 and X1.
o Z3, located between sheave 3 and X3.
o Z4, located between sheave 4 and X2.
Remember, Gcaudill said one corner of the delta load was groundED. He meant the corner was connected via X2, the intentionally grounded terminal of the transformer. Not the"corner" of the delta elements!
Thus, the three delta-load segment impedances have to be "adjusted" to include feeder impedances. They are, ZA, ZB, and ZC, where:
o ZA = Za.
o ZB = Zb + Z1.
o ZC = Zc + Z4.
The final impedance to consider is the effect of the quench system. I called this ZQ between sheave 4 and ground.
RE: Unbalanced Three Phase Load
I think I see your picture. I modelled the circuit with each of the cables and wire elements separately instead of combining them, and then wrote four loop equations. One of the loops just accounted for the ground path and really could have been handled as a parallel path to Z4. My loops, in terms of your nomenclature are:
X0-X3-Z3-sheave 3-Za-sheave 2-Z2-X1-X0
X2-Z1-sheave 1-Zb-Sheave 2-Za-sheave 3-Zc-sheave 4-Z4-X2
X0-X2-Z1-sheave 1-Zb-sheave 2-Z2-X1-X0
X2-Z4-sheave 4-ZQ-X2
Any theories on why changing the number of commutator brushes on sheaves 1&3 doubles the ground current? See gcaudill's Apr 21 post. Makes me think that there is a significant contact resistance between the sheaves and the wire which is reduced by increasing the number of brushes.
RE: Unbalanced Three Phase Load
I beleive you have noted that the controls for this annealer have SCR's. No one to date has acknowledged harmonics. If the load had as much as 2.5% current harmonic at 1000A in each of the three phases of the load, the harmonic components would total approx 75A. This current would not add to zero in the normal sense.
RE: Unbalanced Three Phase Load
Are you sure that sheaves 1 and 4, are not somehow connected together? If they are truly connected only at X2, then the the two feeder impedances are in series with their respective annealed-wire segments!
Weggi,
Could you identify the response showing the the annealer is SCR controlled!
Jghrist,
You can reduce the configuration to just two loops by delta-wye conversion of the annealer load, and maintaining a delta configuration of the transformers wye terminal X1, X2, and X3.
RE: Unbalanced Three Phase Load
The voltage controller is a SCR type. I am looking into measuring the current in the grounded conductor connected X2 to see if the current (78A) is at 60Hz or at some harmonic.
RE: Unbalanced Three Phase Load
X-former primary amps
I ph.1 = 216A
I ph.2 = 224A
I ph.3 = 246A
Seems okay.
X-former secondary amps
Conductor Gang 1 (Ph. X2 to shwave 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
X-former secondary volts
X1 to gnd - 56VAC
X2 to gnd - 0V
X3 to gnd - 48VAC
Seems like a problem?
Load amps (delta)
Currents:
I leg1 = 836A
I leg2 = 673A
I leg3 = 529A
Seems very unbalanced?
X2 grounding conductor is carying 46A
Comments?
RE: Unbalanced Three Phase Load
Does the primary-current imbalance affect other processes or equipment operation? What portion of the total plant load is represented in the annealer 480V, ±250A 3ø current?
For the given dynamitic load, I don’t think the primary-current unbalance is that unacceptable. Whether the SCR firing is based on a constant current, constant voltage or constant power, it will still vary with the moving product. Even if you redesigned the controls to use an individual 3ø full-wave bridge rectifier {and maybe separate transformers if SCR gating is done on the primary side} with DC out for each segment of the annealing process, the currents may be less unbalanced, but there will still be fluttering/variations observed in primary current. The only fix at that point may be huge LC components, probably at a prohibitive price.
Maybe I have misunderstood, but there a no zero-sequence component in the circuit serving the annealer, and no staggering negative-sequence current either.
RE: Unbalanced Three Phase Load
I am somewhat concerned by the differences in the secondary voltages measured to ground. These voltages are actually measured X1-X2 and X3-X2. X2-ground measured 0V.
RE: Unbalanced Three Phase Load
Shortstub, I decided not to do the star-delta transformation so that I could calculate directly the current in each transformer winding and show that it adds to zero. Either way is equivalent.
wggei, With the delta-connected load and a delta-ungrounded wye transformer, there is no zero-sequence path so third-order harmonics would be suppressed. As gcaudill noted, the SCR is on the primary, so any harmonic currents from the SCR would have to go thru the transformer to get to the load.
RE: Unbalanced Three Phase Load
Here are some new measurements. They are much worse than the last batch regarding load balance.
X-former primary amps
I ph.1 = 216A
I ph.2 = 224A
I ph.3 = 246A
Seems okay.
///The seems o.k. however; they confirm that there is an unbalanced load downstream. It is a good observation.\\\
X-former secondary amps
Conductor Gang 1 (Ph. X2 to shwave 1) = 866A
///Conductor gang 1 and conductor gang 4 shall have combined currents, i.e. vectorially summed, to obtain Ix2 current. It would be good if you could provide the phasor angle of the combined currents in conductor gang 1 and gang 4, with reference to, for example, conductor gang 2.\\\
Conductor Gang 2 (Ph. X1 to sheave 2) = 1300A
///It would be good if you could provide the phasor angle with reference to for example conductor gang 3.\\\
Conductor Gang 3 (Ph. X3 to sheave 3) = 1153A
///It would be good if you could provide the phasor angle with reference to for example conductor gang 1.\\\
Conductor Gang 4 (Ph. X2 to sheave 4) = 606A
///Conductor gang 4 and conductor gang 1 shall have combined currents, i.e. vectorially summed, to obtain Ix2 current.\\\
X-former secondary volts
X1 to gnd - 56VAC
X2 to gnd - 0V
X3 to gnd - 48VAC
Seems like a problem?
///The voltage analysis could take place after the current analysis is done. Incidentally, the current issue is your original concern.\\\
Load amps (delta)
Currents:
I leg1 = 836A
I leg2 = 673A
I leg3 = 529A
Seems very unbalanced?
///Yes. It is nice to know for the purpose of your questions in the original posting.\\\
X2 grounding conductor is carrying 46A
///It is a ground current which leaks to the ground and uses ground path to complete its circuit according to Kirchhoff Laws. It is called common-mode current and it is carrying zero sequence current too. The rest of load unbalance is done over the negative sequence currents in the main conductors.\\\
Comments?
///It is nothing new to those who are analyzing asymmetries including fault current values caused by various faults in power distribution systems.\\\
RE: Unbalanced Three Phase Load
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 as in the present case or delta windings, with one phase grounded?
Symmetrical components are very suitable for fault analysis, but I have found that trying to analyze unbalanced load situations with symmetrical components is difficult. There are some equivalent diagrams in the Westinghouse T&D Manual (Fig. 21 in the 4th Edition) but only with two phase loads identical and the third different.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
It is impossible to have a shorted SCR without other abnormal manifestations in the electrical system.
Have you measured the current in the annealed-wire segment leaving sheave 4 enroute to the quench tank?
Are you able to make voltage measurements between sheaves? As well as the voltage from each sheave to "ground?"
RE: Unbalanced Three Phase Load
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.\\\
RE: Unbalanced Three Phase Load
Io=(Ia + Ib + 0)/3=Ig/3
since the ground current Ig:
Ig=Ia + Ib + 0
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
Alternatively, can you provide an e-mail address? Or contact me at EpsiconInc@aol.com!
RE: Unbalanced Three Phase Load
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
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
I have v7.0 and v6.0.
RE: Unbalanced Three Phase Load
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!
RE: Unbalanced Three Phase Load
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:"
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
1st paragraph, kVAR should have been kVA!
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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
RE: Unbalanced Three Phase Load
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.}
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
RE: Unbalanced Three Phase Load
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!
RE: Unbalanced Three Phase Load
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?
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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.
RE: Unbalanced Three Phase Load
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:
"""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.//""""
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 "almost".\\\
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?\\\
RE: Unbalanced Three Phase Load
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 "almost" comment.
2) The use of measured vs calculated phase-angles values, is unwarranted. The "anticipated" 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.
RE: Unbalanced Three Phase Load