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.

 

[mgtrp]

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.

 

[Mbrooke]

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.

 

[mgtrp]

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.

 

[Mbrooke]

Thanks I will read up one those. 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?

 

[7anoter4]

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

 

[7anoter4]

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

 

[Mbrooke]

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