Transformer Secondary Fault Current 1

[tmaly1]

Dear Folks,

We've calculated the available fault current at our substations (points of delivery) using the Per Unit method. I got a bit confused on how to calculate the SC current at the Sub TX secondary knowing the available SC current at the Sub TX Primary (where we used the MVA/KV base).

should I use the same MVA/KV base I used for the initial calculations, how to integrate the TX impedance with the calc.s?

Thanks in advance for your help

 

[Skogsgurra]

First, decide if you want to work in p.u. or engineering units.

Then, calculate SC impedance on the primary (Using U^2/S).

Then transform this impedance to the secondary by dividing with turns ratio squared.

Calculate secondary SC impedance from transformer's short circuit voltage (usually around 5 percent).

Add the two impedances to get total SC impedance and calculate SC current from nominal secondary voltage and total impedance.

If there is a big difference in phase angles in primary and secondary impedance, you have to vector add the two. If you ignore phase angles, you will get the lowest possible SC current. The real current seldom is more than 10 or 20 percent larger than this.

Always transform to WYE and work line-neutral for easiest calculations (and safest).

 

[rbulsara]

Z of source in p.u. = base MVA/fault MVA

where base MVA = your transformer rating will make life simpler.

fault MVA= 1.732* available sc in KA * rated kV (primary)

Now you can use the source Z in pu and xfmr Z in pu in series .


(YOu can use different base MVA but then you need to convert correctly.)

 

[GamezBeCJ]

Assuming your source is infinite, the worst case scenario is:
Data you need:
Transformer impedance: %Z (the nameplate usually states this value in %)

Transform your % impedance to pu's: Zpu = %Z / 100

The calculate you faul current in pu's: Isc = 1 / Zpu

Finally you just multiply this value to the transformer's full load current on the side you are interested (primary or secondary).

Carlos Gamez, P.E.
Industrial Consultant
Transformer Specialist

 

[cuky2000]

GamezBeCJ,

I concur with you for the Isc calc. for unloaded transformer. However, this is only a fraction of the total short circuit available that not includes contribution from other sources such as rotating machines.
For total SC see the enclose thread thread238-102304 and graphic below.

 

[GamezBeCJ]

From what I understood tmaly1 wanted to know the current on the secondary side of the transformer, so you can see the transformer as a source on the secondary voltage side. If we look at the diagram, not any of the devices contributing to the fault push a current TROUGH the transformer, so I beleive the calculation I posted is correct for giving you the available current at the transformer secondary side.
What do you think?

Carlos Gamez, P.E.
Industrial Consultant
Transformer Specialist

 

[tmaly1]

Folks,

Thanks for your feedback and help, I appreciate you all taking the time to contribute and offer me tips. I would go ahead and use the method recommended by Gamez and verify it with the Primary Analysis Dist. Software we're using.

THANK YOU ALL

 

[cuky2000]

Carlos, I believe that our “amigo” Tmaly1 need our help. Calculating the SC only at the secondary of the transformer bushing could mislead him and others if the intention is to determine the proper secondary device rating. Let’s review his questions.

QUESTION 1: We've calculated the available fault current at our substations (points of delivery)… [Transformer bushings, main breaker, feeder breaker, others

QUESTION 2: I got a bit confused on how to calculate the SC current at the Sub TX secondary…. What you mean by transformer secondary?


COMMENTS:
One of the main purpose of a short circuit study is to determine whether or not electrical equipment is rated properly for the maximum available fault current (ex. Interrupting & withstand) the electrical devices (i.e., protective device, buses, switchgear, etc) may see.
“Point of delivery” could be the main breaker, main bus, feeder breaker, etc. Calculating the SC at the transformer bushing only give him a partial picture of the problem.

Tmaly1, probably is good idea if you describe the main components of the system one-line diagram and what is the purpose of the SC calc.

 

[tmaly1]

cuky2000, thanks for your post
in fact we're using 'Circuit Switcher' as the main disconnecting and isolating mean for our Subs.

The available SC Current at the Circuit Switcher is 16,091 based on 100 MVA base and 230 KV base. Sub TX is 230/13.2 kv (Delta/Wye) with only 24 MVA capacity and 8.8%Z.

My question was how to calculate the SC Current on the 13.2 kv side of the TX based on the 16,091 value on the 230 KV side.

 

[rbulsara]

16091A is 16091A if that is the absoulte value (arrived at multiplying pu value with base I), regardless of the base MVA used for calculation.

Using the 'base' is a convenince and should result in same absolute value (pu values may vary by base).

Review my post again..

If you use 24MVA as base, the source Z in pu for 16901A at 230kV will be

24/(1.732*16.091kA*230kV) =0.003744 pu

add this to .088 pu of the xmr Z and you have total z at secondary of the tx.

This is without other contributions, but that is simple arithmatic add if you know them.


At 16KA at 230kV the fault MVA is 6410MVA, that is as good as infinity bus..as obvious by the calcs..

 

[rbulsara]

Refer to IEEE Buff book or Red book for reference...