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.
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
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
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
RE: Transfomer short circuit current
RE: Transfomer short circuit current
http://www.sandc.com/support/media/through-fault-c...
RE: Transfomer short circuit current
What exactly are you requiring the through-fault current for?
RE: Transfomer short circuit current
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
RE: Transfomer short circuit current
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
RE: Transfomer short circuit current
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
RE: Transfomer short circuit current
RE: Transfomer short circuit current
RE: Transfomer short circuit current
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
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
RE: Transfomer short circuit current
RE: Transfomer short circuit current
RE: Transfomer short circuit current
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
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
RE: Transfomer short circuit current
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
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
RE: Transfomer short circuit current