IEEE80 GPR on Grounded Y side of Transformer 2

[RSChinn]

IEEE80(-2000)page 78, third and fourth paragraphs: "For a grounded Y on the secondary (distribution) side of a substation transformer, ground faults at the transformer low-side terminals circulate through the grid with negligible leakage current to the earth and thus no effect on the substation's GPR. However, for distribution (or low-side) faults a remote distance away a large portion of the fault current will return to the transformer neutral via the substation grid, thus contributing to the GPR." Q1: Does that last sentence mean a small (and no longer neglibible) current now flows in the soil and will create a GPR that is no longer insignificant? Q2: By using the high side fault current values to calculate GPR, we will not need to worry about the remote low side GPR since the high-side always will be bigger (provided there are no other low-side sources or transformers connected in parallel)?

 

[cranky108]

Consiter what happens if a high side conductor drops into a ground lead, into the ground grid. Would there not be a GPR event? Yes part of the current would treverse through the grid into the soil.

 

[RSChinn]

To Cranky108: I'm OK with the high side faults. My questions focus on the low-side faults 1) close-in and 2)remote. For IEEE80, it appears we do not need to worry about low-side fault GPRs if the fault is at the low-side transformer terminals or close-in to the substation (like actually in the substation). IEEE80 appears to say if the low-side fault is far away (remote) then there will be a GPR value that is not negligible. Can we simply alway use the high side fault currents in our GPR calculations since the low-side is neglibile or generally lower than the high side on a non-multiple source low-side arrangement? Thanks for responding Cranky108!

 

[cranky108]

Actually there are many cases where the low side will have a larger available fault current than the high side, with a single transformer.

Consiter where a line leaves a substation, and the shield with should break and fall into the phase below. Both sides of the break are multi-grounded, but the two may not be connected. What would be the GPR at the border of the substation?

 

[waross]

Time for a reality check. You can't ignore or minimize remote faults because close in faults may cause negligible GPR.
Remote faults may cause considerable GPR.
Accept it and plan for it. This is an area of design where shortcuts and unwarranted assumptions get people killed.
You cannot assume that a low side fault may be equal to or less than a high side fault.
Do the calculations and determine what the low side fault current and GPR due to a remote fault will be.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[NAZ55]

RSChinn,

Fault always tries to return back to the source. I believe what the above paragraph is saying is that if there's a fault close to the transformer, since it has a transformer neutral connection grounded the fault will use that conductor, along with the portion of the grid conductor and as such the potential rise is minimal.

The remote fault will have to use the soil/shield wire since that's the only way current can travel back to the source and as such higher GPR.

 

[NAZ55]

GPR = I(fault) * Zg


Therefore Higher fault current value for a given system grounding impedance (Zg) will always give you higher GPR

 

[RSChinn]

To Cranky108/Waross: There is no question that low-side fault currents may be higher than the high-side fault currents. We have the high-side and low-side currents calculated for all of our substations and, in our system, the low side is higher than the high side about 10% of the time. But it is not the current values that concern me as much as it is the GPR that is produced by these fault currents in a substation. Remote faults absolutely can and do cause considerable GPR at the fault site, but the GPR at the substation is what we are after. We will not ignore the remote fault and its contribution to the substation GPR. See comments below to ZAZMAT. We do have a reality check at the fault site away from the substation and we are doing one now at a substation. I do not know of any injuries or fatalities in substations due to a fault event in the history of the electrical industry. If you know of any, please convey them to me as quite of few of my colleagues want this information to support their extensive efforts trying to meet IEEE80. Regardless of the statistics, however, we will comply with IEEE80. That is the reason for the GPR questions in this thread.

To ZAZMAT: Thanks for the two comments; I absolutely enjoy these technical blogs! I think that the current division factor Sf is 0 for a close-in fault because If = XXX and Ig=0 (page 75 of IEEE80-2000). Sf is 1 for a fault on our high side since it is ungrounded and ALL of the If goes through the earth and Ig=If (again page 75). The Sf of the remote fault is truly a current split since If will split into Ig +Ic (again page 75; c actually represents f1 in the picture). Please comment on this thought (zazmat, cranky108, waross and anyone else): When calculating GPRmax for the remote fault scenario using a fractional Sf times the low-side fault current value and the result is higher than the high-side fault current value, we will use that resultant number to calculate GPRmax. If the resultant is lower than the high-side value, we will then use the high-side value to get the maximum GPR. GPR max = I(fault)max times Zg.

 

[jghrist]

A fault on a distribution line close in to the substation will produce high fault currents, but they will return mostly through the neutral.

A fault at a distance from the substation will be lower because of the impedance of the line. Phase to ground faults drop off fairly fast because the zero-sequence impedance of distribution lines is high. A higher percentage of the fault current will return through the earth for a distant fault, however.

Software such as SES FDIST can calculate how much of the current returns to the grid through the earth for a fault on a distribution line. In the distribution substations where I have checked, the high side faults produced higher earth currents, but this isn't necessarily so.

 

[cranky108]

OK, but in the event the neutral wire breaks and contacts a phase, where does the ground fault current go?
Since the neutral wire is disconnected from the substation, but is still multi-grounded, the return path must be through the ground, and hence GPR. If the event happens close to the substation then the substation GPR could be of a sizeable value.

It's unlikely, but what are you risking?

 

[apowerengr]

The low side faults would not be negligible as compared to the high side faults if the distribution system neutral is small in size, poorly grounded, becomes disconnected or severed in the fault path or because low side fault clearing times might be much longer than high-side faults (high side ground relays set for low pickup values and fast clearing times if all the T-D transfomrers are connected delta on the high-side). Typically you are correct that there is some point on the high side line from its source to the substation under study which results in the maximum GPR. Given that systems continue to be improved upon and that fault currents continue to rise, how do you plan to design for future changes or are you continually reevaluating all grids on the systems under your study?

 

[stevenal]

I am not understanding the claim that a higher percentage of distant distribution fault current returns through the earth. I am assuming a multigrounded system. The fault current from a downed conductor will initially flow through the earth. By Carson's formulas, the main earth current path will be directly below the distribution line. While traveling along this path, the current should encounter a pole ground in a quarter mile maximum (per NESC). At this point the current will divide by the relative impedance of the two available paths, one path being the metallic neutral going back. The more distant the fault, the more chances a particular coulomb will have to get back on the neutral, with all the pole grounds and service grounds.

If the conductor drops in the first span right outside the substation, the first available path to the neutral will be the grid.

I would expect a lower percentage earth return for the distant fault than for the close fault. Please explain why this is not so.

 

[jghrist]

If the conductor drops to the ground right outside the substation and the neutral is not involved, most of the current will flow to the grid. You would have to add some fault resistance to the calculation of Ø-grd fault current, however. The normal calculation would include negligible zero-seq line impedance because the length is close to zero. In any case, for this situation, is the danger to the public from the fallen conductor would seem to be higher than the danger to utility personnel who might happen to be in the substation at the time.

The calculation of Ø-grd fault currents using Carson's equations assume that there is both a metallic path and a ground return path for the current. A conductor falling on the ground that does not involve the neutral would be a high impedance fault with lower currents than normally calculated.

If the neutral is assumed to be involved in the fault, then clearly directly at the station, all of the current would flow into the grid and none would flow in the earth. J. Endrenyi showed in Fault-Current Analysis for Station Grounding Design, Ontario Hydro Research Quarterly, second quarter 1967, that the grid current increases up to a maximum value, and then decreases as the distance to the fault increases.

 

[RSChinn]

To jghrist: Can you help me understand why the grid current increases to a maximum value and then decreases as the distance to the fault increases in the Endrenyi paper? I can see why it decrease with distance, but why does it increase in the beginning? Thanks

 

[NAZ55]

kudos jghrist for your valuable responses!

apowerengr I agree with you completely that our design atleast of the ground grid should definitely inculde future fault current increase, but I think you lost me where you mentioned that high side fault current would be cleared much faster. I would argue that it depends on whether you are talking about generation or distribution side.

 

[jghrist]

RSChinn,

Total fault current decreases as the distance from the station increases. This is because the line impedance increases.

The impedance of the earth path doesn't change too much with distance from the station because the current can flow through a wider area when the fault is distant from the station. So, even though the return distance increases, the "conductor" area also increases.

The impedance of the metallic neutral path is pretty much proportional to the distance from the station.

Because the current splits according to the relative impedance of earth and neutral, close to the station, a higher percentage flows through the neutral. For a fault at a large distance from the station, a higher percentage flows through the earth.

After a point, the decrease in the total fault current becomes more of a factor than the increase in the percentage flowing in the earth, so the earth current reaches a maximum.

 

[RSChinn]

Thanks, JGhrist. New power engineers as well as those that have been around a while can always get something out of these discussions.

Question to be answered by anyone: The IEEE80 says that a transformer with an ungrounded high-side has the high-side fault current If = Ig (page 75, figure 29). The GPR is Ig times the grid resistance, but does that GPR go out along the paths of the substation feeder neutrals? The neutrals go out overhead, but are grounded at every-other pole. The high-side fault goes back to a grounded source that feeds the high-side, so wouldn't the ground around where the neutrals are installed at every-other pole get GPR (transferred Voltage)? Is this a danger to the public?

 

[jghrist]

Figure 29 does not include any transmission shield wires or distribution feeder neutrals. If these were present, then some of the fault current would flow through them.

The return current will flow more in paths that are coupled to the source fault current. More will flow in the transmission shield wire and in the feeder neutrals because it is better coupled with the phase wires.

There is a possibility of GPR at feeder pole grounds. One way to look at it is the current flowing in the neutral returning to earth and to the source through the pole grounds. Current flowing into the earth through the pole grounds will cause local GPR, touch- and step-voltages. Another way to look at it is from a transferred potential standpoint. The feeder pole grounds are metallically connected to the station grid, so if there were little or no current in the neutral, there would be little voltage drop and the pole grounds would be near the station grid potential.

Is this a danger to the public? It depends. It is possible to calculate the step- and touch-voltages, but there is no code requirement to do so. In one case that we reviewed, the GPR at a facility a mile from the substation was 1,500 volts for a 12 kV fault at the facility. The touch-voltage was 430 volts. For a 100 kV fault at the substation, the touch-voltage would be 3,300 volts. The area has high resistivity soil and the substation resistance is over 5 ohms. The problem is, there's not a whole lot you can do about the situation without a very large expense.