Transfomer short circuit current 1

[Mbrooke]

How do you calculate the phase to ground and phase to phase fault current of a 28/37/48 MVA power transformer using the infinite buss method? Secondary side available fault current is my main concern.



Transformer is 115kv to 23Ykv, Dyn11 (Delta Grounded Y) 60Hz with an impedance of 8%. I did confirm 28 MVA is indeed the air cooled rating.

 

[ZeroSeq]

28/37/48 is the ONAN (Oil Natural Air Natural), ONAF (Oil Natural Air Forced), OFAF (Oil Forced Air Forced) rating. The only rating that matters is the ONAN rating when calculating fault currents (There's only so much iron and copper in the transformer).

Maximum 3 phase fault current is the full load ampacity (28MVA/(1.732*23kV)) divided by the percent impedance.

To calculate the unbalanced LL and LG fault currents, you need to do symmetrical component analysis.

For LG:
I0 = I1 = I2 = V/(Z0 + Z1 + Z2)
Ia thus equals 3(I0)

for LL:
Iab = 1.732/(Z1 + Z2)

Positive and Negative sequence impedances are independent of transformer configuration, however the zero sequence impedance depends on connection.
In your case, since you have an infinite bus, you have no additional system impedances.

You may want to google symmetrical components, or check out Symmetrical Components for Power Systems Engineering by J. Lewis Blackburn.

 

[ZeroSeq]

If you want the numbers, they are as follows:
Infinite bus with X/R of 8.
28MVA Transformer with 8.0%Z and X/R of 25.2
LL fault current = 7582.20A
LG fault current = 8765.33A

Keep in mind as well these are the symmetrical fault currents and do not contain any asymmetrical components.

 

[davidbeach]

If you're transformer was tested per ANSI standards, that 8% impedance is on the 28MVA rating. BUT, if the transformer was tested per IEC standards that 8% impedance is on the 48MVA rating. Since it is 8% rather than nearly 14% it's probably safe to assume ANSI, but assumptions can always bite.

 

[Mbrooke]

Thanks! This unit is ANSI rated, its based on the self cooled rating. I will read up on applying sequence components to fault currents. Out of curiosity how did you come up with an X/R of 25.2?

 

[Mbrooke]

I found this online calculator, but its safe to say this only calculates 3 phase symmetrical? Something does not seem right about it?:

http://www.sandc.com/support/media/through-fault-current-calculator/

 

[ZeroSeq]

I just used a typical X/R value for a transformer of that size. X/R values only come into play when calculating asymmetrical values as it will dictate the amount of DC offset (transient AC decay).

What exactly are you requiring the through-fault current for?

 

[Mbrooke]

Ok I understand now. The fault current is for 2 reasons. One is determining what is the rough maximum amount of current lateral fused cutouts near the substation will see during a bolted fault. Even though the cutouts are rated 10ka interrupting, certain fuse combinations will not coordinate above certain fault levels. The other is to keep fault currents below the interrupting duty of the feeder breakers (currently rated 16ka).

If this unit is changed out with a larger one such as a 36/60MVA unit, the unit will have to be ordered with an Impedance that will keep fault currents below 12.5ka, or at least close to this unit.

 

[Mbrooke]

But mostly my concern is keeping fault current below the feeder breaker ratings. Ignoring transmission impedance is easier as well as the fact this particular line is considered "a strong source", so estimated fault current while higher, will not be that much far off.

 

[waross]

The available fault current calculated from the impedance is the steady state symmetrical current. The full offset current in the initial cycle of a fault may be much higher.
Fortunately the fault current rating of equipment is based on the symmetrical current with a suitable safety factor to allow for asymmetrical fault currents.

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

 

[Mbrooke]

So its safe to assume that if equipment is within its symmetrical fault rating the safety factor will also cover asymmetrical components? Even though the breakers are limited 16ka interrupting a 'rule' has been established to keep fault currents at or below 12,500amps. Past the substation short circuit currents are less of a concern since current falls off considerably down the line. Standard 10ka cutouts and recloser are employed down line.

 

[dpc]

So its safe to assume that if equipment is within its symmetrical fault rating the safety factor will also cover asymmetrical components?

No, not in general. The breaker or fuse will have been tested at some maximum X/R (or minimum pf). If the equipment is intended for use within a substation, you may be OK, but you can never make an assumption. If the test X/R is lower than your actual X/R, the calculated duty amps must be increased.

 

[ZeroSeq]

You can check the 1/2 cycle asymmetrical rating of the circuit breaker against the worst case calculation of the available to be sure.

 

[ScottyUK]

If you have local generation there's a good chance your X/R ratio will be high.

 

[Mbrooke]

No co-generation on the distribution. Customers are load only. The transmission does have strong generation, but IMO assuming infinite buss takes care of that.

 

[Mbrooke]

Heres what I came up with for several different transformer scenarios for 23kv GrY:

28/37/48 MVA @ 8Z= (702FLA/0.08)= 8,775 3 phase fault amps
28/37/48 MVA @ 12Z= (702FLA/0.12)= 5,850 3 phase fault amps
36/48/60 MVA @ 12Z= (903FLA/0.12)= 7,525 3 phase fault amps

Are these ok assuming symmetrical faults involving all phases?


My question comes into applying these values in finding LL and LG faults, particularly solving LG values since they comprise over 90% of faults:

"For LG:
I0 = I1 = I2 = V/(Z0 + Z1 + Z2)
Ia thus equals 3(I0)"

 

[mgtrp]

For a transfomrer:
Z1 = Z2 = nameplate impedance
If available, you're best to get Z0 from the transformer test report. If not, Z0 is typically approx 85-90% of Z1 for a delta-star three-limb core-form transformer, but can vary significantly depending on the actual construction of the transformer. In your case, I would assume at the lower end of the range (85% of Z1) if unknown to be conservative.
With this information, you should be able to solve the equation that you've already got.

 

[Mbrooke]

What do you mean by Z0, Z1 and Z2? I guess this is the part that is straying me off.

 

[davidbeach]

Zero-, Positive-, and Negative-Sequence Impedances.

 

[Mbrooke]

Ok, but how do you apply them to L-G faults? I just want to learn how to do this out by hand.