Available short circuit at transformer

[cppoly]

Is the available short circuit current at the secondary of a transformer other than the utility transformer for a building calculated only from the kVA, voltage, and impedance of the transformer and not the impedances leading up to the primary? For instance, the further away from the utility transformer, the less available short circuit current a panelboard will see. Now if there's another transformer downstream of the utility, is the impedance up to it not used in calculating the available short circuit current at its secondary? Starting from "scratch"?

 

[mivey]

Is the available short circuit current at the secondary of a transformer other than the utility transformer for a building calculated only from the kVA, voltage, and impedance of the transformer and not the impedances leading up to the primary? For instance, the further away from the utility transformer, the less available short circuit current a panelboard will see. Now if there's another transformer downstream of the utility, is the impedance up to it not used in calculating the available short circuit current at its secondary? Starting from "scratch"?

To be accurate, use the source impedance as well. One problem is that the source impedance can change (bigger wire, bigger transformer, etc) so many times the source impedance is dropped to get a more conservative number for equipment specs (results in higher fault current). You can also substitute conservative values for the source impedance to take care of any normal changes that might occur.

For an arc-flash study you need the source data as accurate as you can get it.

 

[templdl]

This may generate some discussion:
I have recommended to my customers to simply go with 100% available fault current plus 100% motor contribution. That way you would expect that to be the maximum available fault current.
Then what are the OCPD kaic's available? Depending upon the device and the voltage they can be 10, 14, 18, 22, 25, 35, 50, 65, 100 etc.
Then I compare what the max. may be and see if it would pay off to do a further study where you would be able to reduce your costs. Depending upon the devices that you have to choose from going through the trouble of calculating a max fault current with unlimited available fault current of the primary of the transformer with 100% motor contribution which may be 95KA where you would need a 100KAIC device how much help can you get by doing an actual fault systems study to be able to get the KA down to 65KA or less in order to use a 65Kaic device? In other words are close enough to make it worth your while to allow you to go to the next lower kaic device? Consider the max KA and do what ever makes sense from there.
Is there going to be a cost to do a study and is the payback worth it?
I is a matter of finding out if you're going to gain anything.

 

[erickench]

The customer transformer will actually limit the KVA contributed by the utility. The maximum can be determined by dividing the rated KVA by the impedance of the customer transformer. Using the KVA method the calculation would be thus:

(KVA_U)(KVA_C)
(KVA_U)+(KVA_C)

The procedure is similar to calculating a parallel resistance or a series capacitor circuit. If the utility short circuit KVA is much higher i.e. 10X, then you could approximate by just using the customer transformer short circuit KVA.

 

[JoeStillman]

Whenever I find out the impedance of the utility feeding a transformer, I find that it's negligable compared to the transformer impedance. That's why a lot of people just divide the transformer secondary FLA by the impedance to get the available fault current (before motor contribution, naturally.) You'll always have a conservative estimate that way.

Here in PECO world, they'll tell you there is 150 MVA available wherever you are, whether you're 1 mile from the substation or 1 block. I had two customers on the same line that were a mile apart. The utility engineer wouldn't budge on the available fault current though, since they have all differnet ways to feed that line. He has to give you his worst case.

 

[cppoly]

Looks like taking the available short circuit at the customer transformer is the best way & conservative way to go for equipment specs. Thanks!

 

[kingpb]

This may generate some discussion:
I have recommended to my customers to simply go with 100% available fault current plus 100% motor contribution. That way you would expect that to be the maximum available fault current.
Then what are the OCPD kaic's available? Depending upon the device and the voltage they can be 10, 14, 18, 22, 25, 35, 50, 65, 100 etc.
Then I compare what the max. may be and see if it would pay off to do a further study where you would be able to reduce your costs. Depending upon the devices that you have to choose from going through the trouble of calculating a max fault current with unlimited available fault current of the primary of the transformer with 100% motor contribution which may be 95KA where you would need a 100KAIC device how much help can you get by doing an actual fault systems study to be able to get the KA down to 65KA or less in order to use a 65Kaic device? In other words are close enough to make it worth your while to allow you to go to the next lower kaic device? Consider the max KA and do what ever makes sense from there.
Is there going to be a cost to do a study and is the payback worth it?
I is a matter of finding out if you're going to gain anything.

What you are referring too is Duty Cycle, or Equipment Rating and works for purchasing the equipment only. If you have protective devices that have to be set, you have to do a short circuit analysis.

 

[kingpb]

Whenever I find out the impedance of the utility feeding a transformer, I find that it's negligable compared to the transformer impedance. That's why a lot of people just divide the transformer secondary FLA by the impedance to get the available fault current (before motor contribution, naturally.) You'll always have a conservative estimate that way.

Here in PECO world, they'll tell you there is 150 MVA available wherever you are, whether you're 1 mile from the substation or 1 block. I had two customers on the same line that were a mile apart. The utility engineer wouldn't budge on the available fault current though, since they have all differnet ways to feed that line. He has to give you his worst case.

Yes, that is conservative, and depending on the size of the transformer your talking about, in most cases, it would mean you have an infinte bus on the HV side of the transformer, thereby the fault current on the LV side of the transformer will be very close to the rated current of the Xfmr divided by the Xfmr impedance.

This is generally OK for specifying equipment, but not a good method for setting protection devices.

 

[templdl]

What you are referring too is Duty Cycle, or Equipment Rating and works for purchasing the equipment only. If you have protective devices that have to be set, you have to do a short circuit analysis.

Interesting. I alway understood that the OCPD kaic had to be rated for the fault current availble at the position that it was installed to assure that it woud be capable of clearing the available fault. I wasn't aware that it was a duty cycle. Also, I wasn't aware that one would have to do a short circuit analysis in order to set and coordinate OCPDs either which is something new to me.

 

[kingpb]

Interesting. I alway understood that the OCPD kaic had to be rated for the fault current availble at the position that it was installed to assure that it woud be capable of clearing the available fault. I wasn't aware that it was a duty cycle. Also, I wasn't aware that one would have to do a short circuit analysis in order to set and coordinate OCPDs either which is something new to me.

You are correct, it has to be rated to withstand the maximum available fault current. But, when you use an infinite bus situation, the maximum possible may be higher than the actual maximum encountered. Especially when the contribution from feeders or buses onto the bus will not be seen by say the main breaker.

Therefore, say the ultimate rating is 65kA of the equipment. The actual fault current may be 50kA. If you were to set the instantaneous setting above 50kA, the device wouldn't trip on the fault current until maybe some time later.

So, it is necessary to know minimum current for settings, maximum for equipment rating.

 

[mivey]

Whenever I find out the impedance of the utility feeding a transformer, I find that it's negligable compared to the transformer impedance.

That may just be the area you are in as that is certainly is not true for many locations. You should expect the more remote you get the more significant the source impedance.