voltage transients in a resistance grounded system

[mull982]

Hello

I was wondering if anyone could give me an explanation as to why there are higher voltage transients for a line to ground fault in a resistance grounded system as opposed to a solidly grounded system. We have a 4160 V system, and I was told that for a L-G fault on a solidly grounded system the maximum voltage would only be 2400V where as a L-G ground fault on a resistance grounded system could produce voltage transients of up to 2x the L-L voltage. I was hoping that someone could give me an explanation as to why this is.

Thanks for the help

mull982

 

[steve066]

I think this can be explained (with a lot of effort) by a smith chart.

http://en.wikipedia.org/wiki/Smith_chart

Smith charts are normally used for radio frequency circuits, but they also apply to power systems where the size of the system is very long (i.e. several miles or more of wires connect the source to the load).

The basic theory (as best as I can remember) is that a voltage will travel from the source to the load if the impedence (Z) of the source equals the Z of the line, and the Z of the line equals the load Z.

But once the Z's are not all equal, a standing wave is set up. That is, the voltage travels from the source, to the load as we would expect (this happens very fast, but not instantly)

But the supprise is that some of the voltage is reflected, and travels from the load back toward the source. The reflected voltage may be inverted and subtract from the original voltage. Or it may add to the original voltage making it increase. (The reflected voltage can even be in phase at some points along the line, and out of phase at other points. So the voltage is not the same at all points on the line.)

If what you say about a 2x transient is true, then I assume a load Z of 0 causes a full reflected voltage that adds to the original voltage. And apparently a high load Z (your high resistance ground) causes a load reflection that opposes the applied voltage. The difference is that in the first case the Z of the load is less than the Z of the line, and that causes one type of reflection. In the other case, the Z of the load is greater than teh Z of the line, and that causes another type of reflection.

Again, I am trying to remember this from a class I took years ago, so if anyone can explain this better, I'm all ears too.

Steve

 

[rattus]

I don't think this can be explained by standing waves. Power transmission lines are not matched as are communication lines. The wavelength at 60Hz is 3160 miles which is much longer than any line I can imagine.

I would like to hear a few more details of the problem.

 

[jcormack]

It has always been my understanding that Reistance grounded systems "decreased" voltage transients when dealing with arcing faults. the problem of increased transients only becomes a problem when the HGR reistance is too high.....thus approaching open connection. As long as the resitance is chosen so that the fault current is larger than the system capacitive charging current by a safety factor the V-G levels should not exceed L-L levels. I have read several IEEE papers on this over the years & HGR is typically viewed as a good system when engineered properly & will mitigate several GF related issues.

From NEC Digest
High-resistance grounding
NEC Article 250.36 states that:

250.36 High-Impedance Grounded Neutral Systems, High-impedance grounded neutral systems in which a grounding impedance, usually a resistor, limits the ground fault current to a low value shall be permitted for 3-phase AC systems of 480 volts to 1000 volts where all of the following conditions are met:

The conditions of maintenance and supervision ensure that only qualified persons service the installation.
Continuity of power is required.
Ground detectors are installed on the system.
Line-to-neutral loads are not served.
A high-resistance grounded system is an alternative to the ungrounded system. Both provide improved service continuity by allowing continuous process systems to remain on line during a ground fault. However, the high-resistance grounded system has several additional advantages. High-resistance grounding requires selecting a resistor to limit the ground fault current to slightly higher than the capacitive charging current, but typically no more than 10 amps. The power system can usually continue to operate at this level of current since it is not in a damaging range. In some respects, the high-resistance ground system behaves like an ungrounded system since process continuity can be maintained. However, with a properly selected resistor, high-resistance grounding can also reduce the transient overvoltages that are characteristic of ungrounded systems. The connection to ground through a resistor provides an additional, more direct connection to ground rather than only through the weak capacitive coupling found in the ungrounded system. Similar to the ungrounded system’s requirements, the high-resistance grounded system requires a ground fault detection scheme, since a second ground fault on a different phase would produce a phase-to-phase fault.

 

[mull982]

The way this subject first came up around our plant was when we went to buy 5kv Shielded Ground Conductor cable or "dragline cable" The question came up as weather to buy 5kV shielded or 8kV shielded cable. We were told to look into buying 8kV shielded cable because we have a resistance grounded system here at the plant, and on a resistance gounded systems it is possible to get 2 times line to line voltage transients. Our system is a 4160V system, so we were told that to be on the safe side we should order 8kv shielded/insulated cable.

I really did not understand the explanation that the engineer gave for this so I was wondering if anyone could provide further information. Like I said in my first post, I was told that in a solidly grounded system there is only a possible 2400 V avaliable during a L-G fault, where in a resistance grounded system there is 2x the L-L voltage avaliable during a L-G fault.

Thanks

mull982