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A general question about Available Short Circuit Amps from the Utility 1

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bdn2004

Electrical
Jan 27, 2007
794
The Utility sent us the available short circuit for the Plant – appx 13 kA at 12.47 kV. X/R = 19

Since then, I actually looked at the nameplate on the Utility transformer. It has these ratings: 15/20/25 MVA ONAN/ONAF/ONAF. 13800-13090/7558 V and 9.10% impedance at that voltage and 15MVA.

I tried to model in our software this transformer and produce on the secondary side the same short circuit value the Utility had provided. But no matter how much primary short circuit MVA is available – the 15 MVA transformer will never produce 13kA and will produce like 7.5kA. The infinite bus method answer which makes some sense.

If I use 25 MVA as the maximum transformer size – it does give a much closer number as what they had sent us from the Utility. Is that the answer – you must use the maximum size listed on the nameplate? If that’s the case why don’t they give you the impedance at that maximum value on the nameplate and just call this a 25 MVA transformer?
 
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Sounds like you have fault current contributions from other 13kv sources, be it generation or normally closed bus tie/bus coupler.


Under the IEEE you use the minimum MVA rating, under IEC built units you typically use the max MVA.
 
Thanks.... I failed to mention there is another identical transformer in their switchyard connected in a ring configuration on both the primary and secondary side.

I assumed the one utility transformer will feed one side of our bus-tie-bus switchgear, the other utility transformer, the other. But the the utility transformers could certainly be paralleled.

That makes much more sense. Thanks !
 
Paralleling two transformers the SC virtually twice the SC produced by one unit.
 
I think you are looking at this all wrong.

You say the utility provided you with the short circuit information at 12.47kV.

You have a transformer that steps the voltage down with a ratio of 7558/(13800-13090).

You are trying to determine the fault current on the secondary side of the transformer and are surprised that it is not the same value that the utility gave you for at the high side.


You need to take their short circuit information and create the thevenin equivalent and then apply a fault on the secondary side of your transformer. Convert everything to PU and do the below.

Infinity bus ================= utility thevenin equivalent ========================== transformer impedance ========================fault

The fault current supplied at the high side terminals will not be the same in amps or pu as at the secondary side of the transformer.

------------------------------------------------------------------------------------------
If you can't explain it to a six year old, you don't understand it yourself.
 
No I don't think I'm looking at it wrong now.

The Utility did provide the short circuit information at 12.47kV - our utilization voltage. That in general is all we ever get. You have to wonder how did they come up with that number?

But I was able to get into their substation and looked at the nameplate on the large transformer. The transformer can be set to give us our 12.47kV. I originally wrongly assumed they do their transformers like our side is to be operated: bus - open transition normally open tie - bus.

But that's not right on their side - there are two identical transformers feeding our sub in parallel.
 
If the utility does not give you what you need, you should ask. Just say hay, what were your assumptions.

The one thing I think people don't do enough of is say something when you need more information.

Like when a consultant sends me 'the form' to fill out for fault studies. I turn around and call the consultant and ask,'what exactly are you needing?'.

And I would hope that if a customer needs more information that they will ask.

Just have your computer call (or e-mail) my computer.
 
Cranky,

Our Utility has been very helpful. This is a new installation. I've met with the Utility engineer and he is the one that gave me a tour of the inside of the new substation.

But there is another person at the Utility that sends us a dated standard Utility form with the short circuit values at the point of service.
As far I know that's all they will give us. It's their stuff, things change all the time.

I was just trying to simulate in our software the next voltage level up and still get the values they've given us - to see what would happen if for example that second transformer were taken out of service for maintenance or something. Does that effect anything on our end?
 
The second transformer could change motor starts, if you are using across the line starting. However, so many companies use soft starts, that that might not matter.

Yes the utility can change equipment, so use the minimum transformer impedance as the only source impedance, and then you know what the maximum fault current can be.
 
Several of our municipal utilities have direct-connected industrial customers for whom we must have two transformers on load for DOL/ATL large motor starts i/e a 30,000 horsepower motor driving a cryogenic liquid oxygen production plant; as well as having two transformers in service, we also must place the transformer tap changers on manual and drive the voltage up from its normal ~13.8 kV value to about 14.5 kV or so.

We leave the tapchangers on manual so they will not operate while the motor start is taking place. Upon successful completion of start we are so advised via normal comm channels & we return the t/c's to auto.

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
 
I've seen something like that years ago, except we inserted capacitors prior to starting the large motors.

Except our customer was not working with liquid O2, but was compressing gas to extract He, and taking out CO2 as a side product.
 
One more effect: When the utility goes from two transformers in parallel to a single transformer, the lower fault currently may actually cause the arc flash exposure to increase if the protection moves from the instaneous portion of the curve to the time curve.
 
I know, that’s one of the reasons I wanted to model the Utility side more accurately. At what point does this become an issue? Should we consider this as worst case?

One thing I’ve learned from them, our Utility, is the extensive use of bus differential protection. That’s how they protect their valuable assets. Protection not dependent on fault current sensing. The time - over current protective devices are backup.
 
Hi bdn2004,
Your question: If that’s the case why don’t they give you the impedance at that maximum value on the nameplate and just call this a 25 MVA transformer?
You should remember that,
1) The SC impedance given on the name plate of the transformer is for its ON rating for the transformers manufactured per IEEE C57.12 or CSA-C88.
2) The SC impedance given on the name plate of the transformer is for its FULL FAN RATING for the transformers manufactured per IEC 60076.
I believe that the above answers your question.
 
Under IEEE all transformer impedances are based on the self-cooled ratings of the transformers.

Besides fans and pumps, one can add soaker hoses, or fire hoses for additional cooling, but why should that matter?

If you don't know how to convert the impedance to another base of your choosing, then... well...
 
Where I am, all 25 kV s-c max contributions are 12 kA. Have the same letter from the utility, for different projects around the province.
 
For equipment sizing the utility is going to provide a number they're pretty sure won't be exceeded. For instance most of our 13kV (12.47kV) customers get told 20kA. That number won't be exceeded even if a new substation is built closer to the customer site and allows paralleling of transformers during switching without exceeding the equipment ratings.

Once the customer is ready to talk arc flash calculations and protection coordination they get a more refined number with the understanding that it is only for the present configuration of the system and could change at any time.
 
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