NG Resistor and Parallel Sources

[billyzee]

There are two 34.5kV/4160 (delta/wye) transformers. Each transformer secondary has a 6 ohm neutral grounding resistor. The transformers feed switchgear which allows them to be paralleled. There are also 10 generators each with a 6 ohm NG resistor which also feeds switchgear that allows them to be paralled.

If only one transformer is feeding the load the ground fault current is roughly 400Amps (2400V/6ohm). If all the sources are in parallel and feeding the load the ground fault current is roughly 4800A (2400V/(6ohm x 15 parallel).

I am having difficulty grasping what sort of relay protection is used to protect MV cables in this circumstance. For instance at 400A the cable shield will not be damaged if the fault persist for 60 cycles but at 4800A the cable shield is damaged pretty quickly (4 cycles). Is a 51N relay the right device for this problem? Does a 51N relay come with a time versus current curve?

 

[JoeStillman]

Consider using concentric-neutral cable instead of tape shields.

You can get a time-current curve from the 51N relay manufacturer.

 

[Julius Right]

First, I think, the transformer current to Ground flows only through its resistor to Ground since the loop closed through generator [or the other transformer] passes through the generator winding -in order to reach the neutral resistor-and the total voltage in this circuit will be zero. So, never more than 400 A will flow through transformer winding. This could not damage the M.V. cable.
The shield is not connected directly with the secondary transformer winding, since there is not any galvanic contact between the two parts-only magnetic[at least, has not to be].
The shield could be damaged if an eddy current would flow through it. Check if it is not grounded at both ends-for instance. A short-circuit current at the M.V. transformer terminals can do it.
Another possibility is if the 34.5 KV is not grounded at all and then, a contact between a phase with the Ground will raise the Ground potential to the phase potential and then between grounded shield and the other phase will be 34.5 KV. In this case the insulation can be damaged- and the shield also.

 

[ron]

The resistance to ground is likely the same no matter how many transformers or generators are serving the load. The ground fault will be split across all resistors on its way back to the source neutral.

 

[Phil Corso]

Billyzee...

Paraleling both transformer and 10 generators conected together, with a ring-bus or a synchronous-bus, will produce a maximum ground-fault current, at the point-of-fault, of 12x400 or 4,800A.

If the above situation is true, that is, all sources simultaneously connected to the same bus, then current for a 3-phase fault could far-exceed switchgear SC-withstand ratings!

For clraification, can you provide a simple Single-Line-Diagram or a hand-drawn sketch of your system?

Regards, Phil Corso

 

[billyzee]

Ron, I do not think you are right. The idle machines which are not in-service will have an open circuit breaker so there is no fault return path through that particular NG resistor.

Julius, I agree that never more than 400 amps will flow through any transformer winding, but that doesn?t mean only 400 amps can flow through a cable shield. I think cable shields that are feeding loads down the line can experience ground fault currents in proportion to the number of sources that are parallel upstream line.

It seems to me that a zero sequence impedance diagram of a single line to ground fault for this system will be composed of 12 parallel legs if all machines are in service, 10 parallel legs if only 10 are in service etc. Each parallel leg is composed of the NG resistor. Zero sequence impedances of the transformer or generator and cables are small in comparison. I

Thank you for providing input with regards to this case. My thinking about the case, man, it has become uptight.

 

[Phil Corso]

Billzee,

I'm sorry, but you are wrong! Furthermore, you have provided exigent circumstances that affect proper response, which is why I requested the diagram!

Phil

 

[iceworm]

There are two 34.5kV/4160 (delta/wye) transformers. Each transformer secondary has a 6 ohm neutral grounding resistor. The transformers feed switchgear which allows them to be paralleled. There are also 10 generators each with a 6 ohm NG resistor which also feeds switchgear that allows them to be paralled.

If only one transformer is feeding the load the ground fault current is roughly 400Amps (2400V/6ohm). If all the sources are in parallel and feeding the load the ground fault current is roughly 4800A (2400V/(6ohm x 15 parallel). ...

For instance at 400A the cable shield will not be damaged if the fault persist for 60 cycles but at 4800A the cable shield is damaged pretty quickly (4 cycles). Is a 51N relay the right device for this problem? Does a 51N relay come with a time versus current curve? ...

Is this a new installation design issue, or an existing install needing additional protection? I'm looking for the constrainsts you are working under.

ice

 

[billyzee]

The system is hypothetical. Estimation of line-line fault currents and their impact on switchgear ratings is not within the scope of my inquiry. I am attempting to validate my understanding of one thing and one thing only.

Given a system (4160V) with X paralleled sources. Each source as a 6 ohm NG resistor. All sources are 3 wire, wye connected (no neutral load). My question comes down to this. if there is a single line to ground fault on a downstream cable then is the single line to ground fault current that the cable shield sees equal to a, b or c.

a. VLN/RNG = (4160/1.73/6) = 400A; constant and independent of the number of sources.

b. VLN/(RNG of all the available sources) = 4160/1.73/0.4ohm = 6000A; constant and independent of the number of sources

c. VLN/(RNG of only the sources which are on-line and in parallel) = 4160/1.73/RNG parallel:
400A for 1 on-line source
800A for 2 on-line sources
1200A for 3 etc.

 

[ron]

Ron, I do not think you are right. The idle machines which are not in-service will have an open circuit breaker so there is no fault return path through that particular NG resistor.

billyzee, Although the breaker will be open to a source not in use, the breaker doesn't disconnect the equipment ground conductor or the N-G resistive bond. It is still in tact.

 

[billyzee]

Ron,

Thanks for the reply. I take it that you are choosing answer ?b? in my last post. I should have added that for my example the 6000A answer was for single line to ground fault for number of sources X = 15.

That still seems wrong to me though because the NG resistor is the only connection to each wye connected source. Any fault current going through that source (and its NG resistor) would enter the neutral and then need to go through the phase windings to connect the circuit back to the fault. Therefore that is why I?m stuck on C as being correct.

 

[Phil Corso]

BillyZee...

Consider the case of n-sources connected to the point-of-fault. Neglecting interconnection impedances, then the symmetrical component impedances are:
o Z1 = 1 / [ (1/ Z1a)+ (1/Z1b) + . . . ( 1/Z1n) ]
o Z2 = 1 / [ (1/ Z2a)+ (1/Z2b) + . . . ( 1/Z2n) ]
o Z0 = 1 / [ (1/ Z0a)+ (1/Z0b) + . . . (1/ {Z0n+3xR/n} ) ] when all grounding resistors are equal.

Phil Corso

 

[Phil Corso]

BillyZee...

Correction to my earlier post:

o Z0 = 1 / [ (1/ Z0a)+ (1/Z0b) + . . . (1/ Z0n) + (n/3R) ] when all grounding resistors are equal.

Also, I've presumed all source kVA capacities are equal!

Phiil

 

[Julius Right]

I meant always 34.5 KV cable shield. I forgot also 4.16 KV cable has to be shielded now [NEC 2011 art.310.10(E)]. However, does not have to be grounded both ends.
If the single-core cable shield build-up induced voltage is too large you may ground it both ends, but you have to derate the ampacity .Three-cores [three phases]cable, individual core shielded and all 3 shields are in permanent contact can be both ends grounded without derating. However, also, will be no expected build-up voltage.
All the energy sources ?transformers and generators-will participate to share the short-circuit current at the fault point but only its part will return to the source.
Let's say the shield is grounded both end then will carry a part of the return path short-circuit current [parallel with the NG]. Only if the fault point will be on the shield-between live conductor through insulation to the shield-then the shield will carry the entire short-circuit. The fault loop will be: common bus bar ,the faulted cable conductor up to the "failure point", the shield portion from the "point" up to its Ground connection, this Ground connection, spreaded again through each individual NG up to each source. But in this case, the cable is entirely damaged. No such rapid protection, which could save it.
I think the direction protection of this "source" has to work if 4800 A is not enough to put into operation the short-circuit protection relay-and to limit ?in a way-the damage.

 

[Julius Right]

"Let's say the shield is grounded both end then will carry a part of the return path short-circuit current [parallel with the NG]".

Correction:
Shield grounded both ends it is not parallel to any NG. The shield is directly bonded to Ground-Grounding Grid or different grounding electrodes-and may transfer the potential from a point to another. The current flowing through the two ends grounding shield depends upon this potential difference.

 

[billyzee]

Phil Corso, I think you equation needs modified. Your equation says:

o Z0 = 1 / [ (1/ Z0a)+ (1/Z0b) + . . . (1/ Z0n) + (n/3R) ]

For a simple system where N=1 source this would reduce to:

Z0 = 1/[1/Z01 + 1/3R]

however IEEE 142 Figure 1.1 gives the zero sequence diagram for a single source with NG resistor as:

Z0 = Z0source + 3R

These results are not the same, so your equation must be incorrect. I think it is correctly modified by having a series/parallel circuit. Each NG resistor in series with its machines zero sequence reactance and then each of these parallel with those of the other sources:

Z0 tot = 1/{[1/Z01 + 3R] + [1/Z02 + 3R] ... + [1/Z0N + 3R]}

Then for a system were each machine has a Z0<<3R and each machines neut-gnd resistor the same then the equation reduces to:

Z0 tot = 3R/N.

Which, if correct, answers my original question. If this equation is correct then as the number of sources increases, the effective NG resistance decreases and the SLG fault current goes up.

 

[billyzee]

Julius,

Your last post was helpful to me. Can you elaborate on what you meant by "I think the direction protection of this "source" has to work if 4800 A is not enough to put into operation the short-circuit protection relay-and to limit ?in a way-the damage."

Thanks

 

[Phil Corso]

BillyZee... You are right!
I erred...Big Time! I failed to show the 3R component should have been added to the Zero-sequence for each source. I attempted to show that the equivalent impedance is equal to the parallel combination of individual source impedances!
Zo(System) = Zoa || Zob... ZoN, for N sources!
Thus, for the simplified case having ?N? grounding-resistors, neglecting all sequence-impedances, the equivalent grounding-resistor equals, Re = (3R)/N!
Regards, Phil

 

[rcwilson]

The change in ground fault current as additional sources are paralleled is why most practical systems use either a single ground point or disconnect switches on the generators' neutral resistors. For example, a system I just worked on had ten 2.5 MW generators in parallel at 11 kV. The 11 kV bus had two zig-zag grounding transformers with 400A fault limiting resistors in the neutral-to-ground connections. The zig-zags were interlocked so only one was on line at a time. Generation was inhibited when both were off. For protection prior to synchronizing, each generator had a 10 amp neutral grounding transformer/resistor that was disconnected when the generator breaker closed. System fault ground fault current was constant at 400A (except momentarily when swapping zig-zags).

In addition to limiting cable damage, high resistance grounding is used on MV systems to limit ground fault damage to motor and generators. 400A is enough fault current to quickly operate ground relays but low enough to minimize iron damage to motor and generator stator laminations. At higher currents the laminations will probably be damaged for a winding fault meaning the entire machine is scrapped instead of just rewinding.

Designing a system with variable ground fault current makes it difficult to apply the protective relaying and increases the chances of expensive fault damage.

In your three scenarios, answer B = constant fault current, is only true for a 3-phase, 4-wire system with the neutrals of all sources tied together. That is not common for industrial systems, unless the design uses a single grounding resistor connecting the common neutral bus to ground. Most MV systems are 3P3W with each transformer and generator neutral grounded independently. When a transformer or generator is off line it does not contribute any fault current. The fault level varies with the number of sources in parallel.

 

[billyzee]

Phil,

Thanks for reading my posts and replying. I think I am almost through with this thread and have beaten it to death. All the posts have been helpful.

My conclusion is that for a system with several resistance-grounded sources the SLG fault current is quite variable depending on the number of parallel sources that are on-line. Not being that familiar with these systems, I found this conclusion to be a surprise so I came here to air it out.

I may end up starting a different thread associated with exactly which relay type (51N?) is best for protecting a MV cable shield from overheating due to fault currents running in the shields, but I'm going to talk to some vendors first.