Medium Voltage Switch Gear / Circuit Breaker Application 1

[Powerengineer_84]

This question pertains to application of 3000A outdoor metal clad 15kV class ANSI switch gear. The available three fault current is calculated to be 38kA with an X/R = 37. As this substation has a resistor grounded medium voltage system and the SLG fault current is 4kA with an X/R<1.

The substation has 4 transformers run in parallel and the bus ties CANNOT be run normally open to reduce fault current.

Given the amount of three phase fault current and the large offset, there is one school of thought that is proposing an ANSI 63kA gear installation the rationale being that for a bolted three phase feeder fault (all UG feeders from this substation) right outside the station (the worst case) the amount of fault current and offset suggests 63kA gear as appropriate.

There is another school of thought proposing an ANSI 40kA gear. The rationale for the 40kA proposal is that each transformer will contribute <10 kA to a fault and that SLG faults are the most prevalent.

I am not sure how the < 10 kA / transformer secondary rationalizes this; for a bolted three phase fault on a feeder just outside the station the 38kA fault current is going to go thru the feeder breaker and that is what the breaker will need to interrupt.

Thoughts?

 

[waross]

Do the proponents of the 40 kA gear propose posting signs to the effect that
"LINE TO GROUND FAULTS ARE PROHIBITED ON THIS SYSTEM
UNDER NO CIRCUMSTANCES WILL LINE TO GROUND FAULTS BE ALLOWED TO DEVELOP INTO LINE TO LINE FAULTS"?
Sorry for the sarcasm, but really.-

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

 

[Zogzog]

Post a sign saying "This is a line to ground fault free zone"

 

[dpc]

1. Breaker short circuit ratings and tests (ANSI) are based on three-phase faults and should be evaluated against three-phase faults.
2. You're correct in looking at the maximum fault current that can flow through each breaker, not simply the total bus fault current
3. With the high X/R ratio, the close & latch rating becomes an issue, as well as the interrupting rating.

Simplest approach is evaluate breaker duties using commercial software such as EasyPower or SKM PowerTools. If you don't have access to that, send your fault data to the breaker manufacturers and get their recommendations.

Dave

 

[LionelHutz]

The 4 transformers in parallel total to 38kA with an X/R = 37, right?

 

[Powerengineer_84]

Yes -- all 4 transformers in parallel provide 38kA with X/R = 37 for the three phase fault..

 

[LionelHutz]

I should have asked where the breakers were in the system as well since your post didn't share that info either. You write about a fault outside the substation which implies the breaker is a single one on the output bus. But then you write that each transformer only contributes so much current which implies there is a breaker for each transformer.

If the breaker in question is a single breaker on the output bus then it does have to be capable of interrupting the 38kA fault current.

Now, for the rest of this post I'm going to write about a breaker on the output of each transformer with the output of all these breakers paralleled together to create the bus with the 38kA of fault current.

There are 2 possible fault cases for these breakers.

The first would be a short after the breaker. In this case, the breaker would only see the phase to phase fault current that a single transformer could provide. I'm thinking this is the case where the claim that each breaker would never see over 10kA of fault current came from. At any rate, it appears a 40kA breaker would easily handle this type of fault.

The second would be a short between the TX and the breaker. In this case, the breaker would see the phase to phase fault current that the other 3 transformers can provide as they back fed through the breaker into the fault. This case phase to phase fault current is more than 10kA but less than 38kA. Most likely around 30kA. It's the worst case.

So, figure out the worst case fault current and then determine if a 40kA breaker is suitable for that fault current.

 

[jghrist]

There are 2 possible fault cases for these breakers.

How about a fault on the load side of a feeder breaker. All 4 transformers will contribute.

 

[LionelHutz]

jghrist - I already posted that answer, assuming feeder breaker means what I think it means. So, try actually reading my whole post again. It works just fine as a whole.

 

[waross]

While a 40 kA rated breaker may be suitable, my comment was in response to this statement.

and that SLG faults are the most prevalent.

This is wishful thinking, not engineering.
To this mind set could be added;
"We don't need the full interrupting rating because most faults are less than the worst case."
More wishful thinking.
While this type of rationale may not make a difference in this instance, it may make a BIG difference on a future project.

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

 

[LionelHutz]

Bill,

Well, L-G faults are the most common types or at least how most of the faults start.

A breaker on the output side of each transformer would not see the full fault current. There are 4 transformers so from what I'm reading it's likely the theory was that a breaker on the output of each transformer could only see 1/4 of the fault current. Each transformer contributes 9.5kA through it's output breaker to sum up to 38kA total on the parallel bus. But, I believe they are neglecting the case of a short occurring between the transformer and the breaker.

A breaker on the output of the paralleled bus will see the full fault current. Someone could try to say 40>38, but the X/R ratio doesn't make it likely the breaker would be correct.

The OP posted attempting to attach a single line. Knowing where in the system all the breakers in question are would certainly clear up the confusion.

On another note, should the sign say that ONLY line to ground faults are allowed and that line to line faults are prohibited?

 

[Powerengineer_84]

I have attempted to attach a simplified single line diagram. There are four 13.8kV busses arranged in an "H" congfiguration. Each bus is supplied by a transformer. From each bus there are six underground feeders, I have shown one for simplicity. A three phase fault on the load side of any feeder breaker will be supplied by the four transformers in the station and the feeder breaker will need to interrupt the 38kA, X/R=37. This of course assumes negligible contribution from the other (unfaulted) feeders.

 

[dpc]

There are two separate breaker ratings to evaluate: interrupting and momentary (aka 1/2 cycle, aka close & latch). The primary rating is the interrupting rating and the Close and Latch rating is calculated from this for the ANSI ratings.

For ANSI tests and ratings, the 15 kV class breakers are tested at an X/R ratio of 17 (IIRC). Since your X/R is 37, some adjustments to the calculated duties are required to compare a fault at X/R of 37 to a fault at X/R of 17.

Without getting into the weeds, just keep in mind that you cannot simply calculate the symmetrical SC current and compare with the breaker SC rating if you are this close to the breaker rating (38 kA vs 40 kA). It is more complicated than that, unfortunately.

A good reference is "A Practical Guide to Short--Circuit Calculations" by Conrad St. Pierre. My guess is that the 40 kA breakers may not be adequate. Again, you can probably get help on this question from prospective breaker suppliers if you provide them with your SC data.

Good luck,

Dave

 

[LionelHutz]

Well, the breakers for the transformers and ties should not see the full fault current. But, the breakers on each end of bus3 would see more fault current then what Tc can provide, so anyone claiming those breakers would only see <10kA are wrong. The 6 feeders off of bus3 would certainly see the full fault current, so anyone claiming they would only see <10kA are also wrong.

If you were interested in what fault rating would be suitable for each breaker in the H bus configuration, then you'd have to calculate the fault current possible with a short at each point and also calculate which part of that current flows through which breakers. I can't see you mixing fault ratings though, so once you pick the breaker needed for the 6 feeders off each bus then you've got the rating needed for all of them.