Transformer fault current -infinite bus method.

[mbrooke]

How do you go about finding the phase to ground and phase to phase fault current of a 25/33/45 MVA power transformer using the infinite buss method? Secondary side available fault current is my only concern.



Transformer is 115kv to 23kv, Dyn11 (Delta Grounded Y) 60Hz with an impedance of 8%. I did confirm 25 MVA is indeed the air cooled rating.

 

[JoeStillman]

The impedance will be based on the Ordinary Air cooled rating, not the fan-cooled. I am puzzled by the three different ratings you listed - 25/33/45? Are there two stages of fans? (ONAN/ONAF/ONAF) I would expect the lowest rating to be the ONAN rating. Just divide the secondary FLA by the impedance; 25 MVA @ 23 kV = 628 FLA. 628 / 0.08 = 7,844 SC Amps.

The impedance doesn't change when the fans come on.

 

[mbrooke]

The impedance will be based on the Ordinary Air cooled rating, not the fan-cooled. I am puzzled by the three different ratings you listed - 25/33/45? Are there two stages of fans? (ONAN/ONAF/ONAF) I would expect the lowest rating to be the ONAN rating. Just divide the secondary FLA by the impedance; 25 MVA @ 23 kV = 628 FLA. 628 / 0.08 = 7,844 SC Amps.

The impedance doesn't change when the fans come on.

The lowest is indeed the passive "air cooled rating". The other two are fan; and fan + oil recirculation through the convector fines. Basically 33 and 45MVA are increased ratings from cooling packages.

 

[mbrooke]

Heres what I ended up with on several units:

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?






The part that Im having difficulty with calculating the values for L-G faults, which compromise most faults in a power system.

 

[JoeStillman]

That looks about right. The L-G amps are typically about 1/3 of the L-L amps.

 

[lielec11]

That looks about right. The L-G amps are typically about 1/3 of the L-L amps.

Joe beat me to it. If you ask engineers who do this frequently they will give you the same answer. I recently did a study using the same method to obtain my L-G values.

 

[bob]

28/37/48 MVA @ 8Z= (702FLA/0.08)= 8,775 3 phase fault amps

Fault amps is based on the Air cooled rating and not the fanned rating.

 

[mbrooke]

That looks about right. The L-G amps are typically about 1/3 of the L-L amps.

Wouldn't they be more? From what I know the best way to do this is with positive, negative and zero sequnce currents. Trouble is applying it to fault currents on a transformer. Normally Im used to a line type deal and a breaker relying, but not transformers. If I remember right L-G faults are something around 1.25 times the 3 phase symmetrical fault current.

 

[jtester]

Heres what I ended up with on several units:

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?






The part that Im having difficulty with calculating the values for L-G faults, which compromise most faults in a power system.

The equation for three phase fault currents I3ph = E1/Z1, the equation for L-G faults is IL-G = 3E1/(2Z1+Z0). Since you have a delta high side, Z0 will only be the transformer impedance whereas Z1 will be transformer plus system high side Z1.

 

[mbrooke]

The equation for three phase fault currents I3ph = E1/Z1, the equation for L-G faults is IL-G = 3E1/(2Z1+Z0). Since you have a delta high side, Z0 will only be the transformer impedance whereas Z1 will be transformer plus system high side Z1.

Transmission fault limitations Im ignoring. I am assuming infinite source. While indeed it fault current be lower, its a 'strong' source so infinite will yield a good approximation. At least this is what Im assuming you mean by high side Z1.


Can you elaborate on 2Z1 and Z0? If it will be of help I have the transformer reactance and resistance based on a 100MVA base.

Thanks!

 

[jtester]

Transmission fault limitations Im ignoring. I am assuming infinite source. While indeed it fault current be lower, its a 'strong' source so infinite will yield a good approximation. At least this is what Im assuming you mean by high side Z1.


Can you elaborate on 2Z1 and Z0? If it will be of help I have the transformer reactance and resistance based on a 100MVA base.

Thanks!

I'm not the best to teach symmetrical components but Z1 is the transformer positive sequence impedance, Z0 is the zero sequence impedance. From what I've read, in transformers positive and zero sequence impedances are equal, or nearly so. If your calculations are on a 100 MVA base, follow the equations I posted, if not, refer to a text on converting to a different base and on changing from per unit to amperes.
Hope that helps
Jim Tester