Requesting Available Fault Current

[Shujinko]

I have a new project on a campus that is fed by an existing Medium Voltage (MV) 12.470V/3-phase campus loop. We are putting a new high-rise building on the campus and tying to an existing MV pad mounted source switch. I ran the preliminary fault current calcs using the infinite bus method at the source. Now I need to run a more precise fault calculation. I need to ask the utility company for the available fault current. I usually ask the utility company for the available fault current and they give me this info at the primary or secondary terminals of the service transformer. In this case there is no utility transformer, the project is providing a 12470V-480Y/277V XFMR. Can I ask the utility for the available fault current at the MV pad mounted source switch? Is this standard practice? Is this info that they would typically be able to give me? Or will they need to calc this since it's not at a utility transformer?

 

[petersonra]

It is a customer owned xfmr so it seems to me that it is up to the utility to tell you what the short circuit current is at the line side of the transformer, or where ever else the line of demarcation is between what is under utility control and what is under customer control.

If they can't or won't do that I think it is necessary to use infinite bus method.

 

[JoeStillman]

They should be able to tell you what is available on the line side of a customer-owned transformer. The trick is getting past all the gatekeepers to an engineer who understands the question.

I would submit a service application form as if it was new service, and just call it a "load increase". That should get their attention.

 

[Hv&Lv]

I don’t know anyone that does the calculations any more.
System Modeling software gives you fault current at any node you click on.
All the cable impedances, lengths, etc are all in the model. An engineer should be able to do this in about 5 minutes or less.

Now will they? Depends. IOU or cooperative?

 

[ggunn]

If they can't or won't do that I think it is necessary to use infinite bus method.

The infinite bus model will be more conservative if it makes any difference at all.

 

[Bwas]

They should be able to give you a fault current at the service delivery point. You'll need to model it from there. You'll need to get the details from the owner or from a field survey for everything on the load side of the service delivery point. If you do the survey make sure you have the proper training to do it safely - I've seen a lot of dangerous installations in similar situations. I unknowingly took a lot of risks early in my career that I would never take today.

 

[ggunn]

I unknowingly took a lot of risks early in my career that I would never take today.

I think we can all say that.

 

[Julius Right]

According to IEC 600076-5 Standard Power Transformers Part 5: Ability to withstand short circuit
Table 2 – Short-circuit apparent power of the system
For voltage system of 7 to 24 kV the short-circuit apparent power is 500 MVA.

 

[wwhitney]

Beginner type question: to calculate AFC at various points on the customer owned wiring and equipment, don't you need two pieces of data from the utility, the AFC and the X/R ratio? I.e. you model the utility source as a perfect voltage source in series with an impedance Z, and AFC gives you the |Z|, but you need the X/R ratio to fully determine Z?

Thanks,
Wayne

 

[wwhitney]

According to IEC 600076-5 Standard Power Transformers Part 5: Ability to withstand short circuit
Table 2 – Short-circuit apparent power of the system
For voltage system of 7 to 24 kV the short-circuit apparent power is 500 MVA.

Are you saying that the IEC standard says that the worst case you need to consider absent any utility data isn't an infinite bus, but rather an AFC of 500 MVA / V*sqrt(3), for a 3-phase system with L-L voltage V between 7 and 24 kV?

Cheers, Wayne

 

[MyCleveland]

According to IEC 600076-5 Standard Power Transformers Part 5: Ability to withstand short circuit
Table 2 – Short-circuit apparent power of the system
For voltage system of 7 to 24 kV the short-circuit apparent power is 500 MVA.

Are you saying that the IEC standard says that the worst case you need to consider absent any utility data isn't an infinite bus, but rather an AFC of 500 MVA / V*sqrt(3), for a 3-phase system with L-L voltage V between 7 and 24 kV?

Cheers, Wayne
-----------------------------------------------------------------------

JR....would like to see your response to Wayne's question.

 

[Julius Right]

Sorry. As usually, I am late.
Indeed, in IEC 600076-5 Standard POWER TRANSFORMERS –Part 5: Ability to withstand short circuit is written:
cpt.3 Requirements with regard to ability to withstand short circuit
3.2.2.4 The short-circuit apparent power of the system at the transformer location should be specified by the purchaser in his enquiry in order to obtain the value of the symmetrical short circuit current to be used for the design and tests.
If the short-circuit apparent power of the system is not specified, the values given in table 2.
Table 2 – Short-circuit apparent power of the system shall be used..

 

[kingpb]

According to IEC 600076-5 Standard Power Transformers Part 5: Ability to withstand short circuit
Table 2 – Short-circuit apparent power of the system
For voltage system of 7 to 24 kV the short-circuit apparent power is 500 MVA.

Under no circumstances can you utilize IEC information for calculations in the US. The calculation methods are completely different and take into account different information. NEVER utilize IEC calcs to compare to ANSI/IEEE short circuit ratings. There is no cross reference.

 

[Julius Right]

For power transformers IEC 60076 is different from ANSI/IEEE C57.12.00, indeed.
In ANSI/IEEE C57.12.01 GENERAL REQUIREMENT FOR DRY-TYPE TRANSFORMER is written:
If in absence of system information from the user, the system symmetrical short-circuit current available at the transformer terminals shall be assumed to be 36 kA.
According to IEEE Std C57.12.00-2015 IEEE Standard for General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers
Table 14 —Short-circuit apparent power of the system to be used unless otherwise specified for less than 46 kV 63 kA.

 

[MyCleveland]

For power transformers IEC 60076 is different from ANSI/IEEE C57.12.00, indeed.
In ANSI/IEEE C57.12.01 GENERAL REQUIREMENT FOR DRY-TYPE TRANSFORMER is written:
If in absence of system information from the user, the system symmetrical short-circuit current available at the transformer terminals shall be assumed to be 36 kA.
According to IEEE Std C57.12.00-2015 IEEE Standard for General Requirements for Liquid-Immersed Distribution, Power, and Regulating Transformers
Table 14 —Short-circuit apparent power of the system to be used unless otherwise specified for less than 46 kV 63 kA.

JR
Can you expand on this some more...
Your first statement,
" If in absence of system information from the user, the system symmetrical short-circuit current available at the transformer terminals shall be assumed to be 36 kA. ".
This applies if your SERVICE xfmr is dry-type and/or any dry-type downstream of the service ?

Your second statement,
" Table 14 —Short-circuit apparent power of the system to be used unless otherwise specified for less than 46 kV 63 kA. ".
If you are given the apparent power by utility, could they not as easily provide Isc and the X/R, or are you saying if you have no information provided to use the apparent power and a table of values is provided in this IEEE document ?

 

[Julius Right]

If we compare IEC 60076-5 with ANSI/IEEE C57.12.01 and C57.12.00
for 13 kV a system short-circuit power at your transformer high-voltage terminals it could be 500, 800 or 1500 MVA. In my opinion, for short-circuit current calculations I should take 1500 MVA, but for voltage drop 500 MVA.

 

[MyCleveland]

If we compare IEC 60076-5 with ANSI/IEEE C57.12.01 and C57.12.00
for 13 kV a system short-circuit power at your transformer high-voltage terminals it could be 500, 800 or 1500 MVA. In my opinion, for short-circuit current calculations I should take 1500 MVA, but for voltage drop 500 MVA.

I am sorry JR…just not following you. Thanks…

 

[Revous]

Beginner type question: to calculate AFC at various points on the customer owned wiring and equipment, don't you need two pieces of data from the utility, the AFC and the X/R ratio? I.e. you model the utility source as a perfect voltage source in series with an impedance Z, and AFC gives you the |Z|, but you need the X/R ratio to fully determine Z?

Thanks,
Wayne

I remember taking a class from SKM on this and I think you are right. You need the AFC at the service point and the X/R Ratio but you are lucky if you get the AFC from the utility sometimes.

But I think one of the IEEE standards had some some suggested or placeholder values you could use... but I can't remember if its the purple book or the new IEEE 3000 series standards that had the table.

 

[kingpb]

Information needed from utility; the three phase short circuit MVA with the X/R ratio, and the single phase short circuit MVA with the X/R ratio. As an alternative to knowing the short circuit MVA and X/R, the system impedance (Z) and the X/R can be used to calculate the short circuit MVA. Also, the X and R components can be used.

 

[jim dungar]

I remember a recommendation of 12 as a value for unavailable utility data and 20 for generators.