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Transfomer short circuit current

Transfomer short circuit current

Transfomer short circuit current

(OP)
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.

RE: Transfomer short circuit current

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.

RE: Transfomer short circuit current

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.

RE: Transfomer short circuit current

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.

RE: Transfomer short circuit current

(OP)
Thanks!smile 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?


RE: Transfomer short circuit current

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?

RE: Transfomer short circuit current

(OP)
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.

RE: Transfomer short circuit current

(OP)
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.

RE: Transfomer short circuit current

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

RE: Transfomer short circuit current

(OP)
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.

RE: Transfomer short circuit current

Quote:

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.

RE: Transfomer short circuit current

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

RE: Transfomer short circuit current

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

RE: Transfomer short circuit current

(OP)
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.

RE: Transfomer short circuit current

(OP)
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)"

RE: Transfomer short circuit current

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.

RE: Transfomer short circuit current

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

RE: Transfomer short circuit current

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

RE: Transfomer short circuit current

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

RE: Transfomer short circuit current

Try finding a basic power engineering text covering sequence components (there are plenty in the FAQ on this site) if you would like a proper understanding, but basically you are using the formula given above for line to ground faults (assuming a fault immediately on the transformer LV terminals and assuming infinite grid strength:
If = 3I0 = 3 x V / (Z0 + Z1 + Z2)
As you add in additional items of equipment to your system, the equation gets larger (each item of equipment has it's on Z0, Z1 and Z2 to include). Different fault types, system grounding, also affect the calculations - refer to an engineering text.

RE: Transfomer short circuit current

(OP)
Any links?

I like the answer Zeroseq gave, just wondering how to obtain same values as him manually. He included the X/R values in the equation which is correct but just not sure how to include them. FWIW I have the Resistance and reactance values of each unit based on a 100MVA base.


RE: Transfomer short circuit current

Have a look at the reference FAQ on this forum, What are good references for a Power Engineer? Most of the general, symmetrical components and protection references should have a section that covers fault calculations. The Alstom Network Protection and Automation Guide certainly does, and is available for free download online once you complete the request form.

RE: Transfomer short circuit current

(OP)
Thanks I will read up one those. smile Computer calculators while a blessing really do push manual arithmetic skills aside.

As a rough rule of thumb, is it good to assume that a line to ground fault is on average 1.25 times that of a 3 phase fault?

RE: Transfomer short circuit current

According to IEC 60909-0 Short-circuit currents in three-phase a.c. systems -Part 0: Calculation of currents:
The positive-sequence short-circuit impedances of two-winding transformers is:
ZT = RT + JXT where : ZT = uk/100*UrT^2/SrT ; RT=PrT/3/IrT; XT=SQRT(ZT^2-RT^2)
UrT is the rated voltage of the transformer on the high-voltage or low-voltage side;
IrT is the rated current of the transformer on the high-voltage or low-voltage side;
SrT is the rated apparent power of the transformer;
PkrT is the total loss of the transformer in the windings at rated current;
ukr is the short-circuit voltage at rated current in per cent
Z1=ZT ; Z2=ZT Zo =0.8-1 ZT[in your case].
For a network transformer- [A network transformer is a transformer connecting two or more networks at different voltages] for two-winding transformers with and without on-load tap-changer, an impedance correction factor KT is to be introduced in addition to the impedance evaluated according to above equations.
ZTk=KT*ZT
The following simple cases are [far from a power station]:
Three phase short-circuit I"k3=c*Un/sqrt(3)/ZTK
Line-to-line short circuit I"k2=c*Un/(Z1+Z2)=c*Un/(2*ZTK)
Line-to-line short circuit with earth connection:
I"kE2E=sqrt(3)*c*Un/(Z1+2*Zo) if Z2=Z1
Line-to-earth short circuit I"k1=c*Un*sqrt(3)/(Z1+Z2+Z3)
Voltage factor c=1.1 for maximum c=1 for minimum

RE: Transfomer short circuit current

Line-to-earth short circuit I"k1=c*Un*sqrt(3)/(Z1+Z2+Zo), of course!blush

RE: Transfomer short circuit current

(OP)
Thanks! This is a substation distribution transformer, but those equations are of help!

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