TRANSFORMER FAULT CURRENT
Calculating the short Current when there is a Transformer within the circuit. Every transformer has “ %” impedance value stamped on the nameplate. Why is it stamped? it's stamped because it is a tested value after the transformer has been manufactured. The test is as follows: A voltmeter is connected to the primary of the transformer and thus the secondary 3-Phase windings are bolted in conjunction with an ampere meter to read the worth of current flowing within the 3-Phase bolted fault on the secondary. The voltage is mentioned in steps until the secondary full load current is reached on the ampere meter connected on the transformer secondary.
So what does this mean for a 1000KVA 13.8KV – 480Y/277V. First you'll need to know the transformer Full Load Amps Full Load Ampere = KVA / 1.73 x L-L KV
FLA = 1000 / 1.732 x 0.48
FLA = 1,202.85
The 1000KVA 480V secondary full load ampere is 1,202A.
When the secondary ampere meter reads 1,202A and thus the first Voltage Meter reads 793.5V. The percent of impedance value is 793.5 / 13800 = 0.0575. Therefore;
% Z = 0.0575 x 100 = 5.75%
This shows that if there was a 3-Phase Bolted fault on the secondary of the transformer then the utmost fault current that might flow through the transformer would be the ratio of 100 / 5.75 times the FLA of the transformer, or 17.39 x the FLA = 20,903A
Based on the infinite source method at the primary of the transformer. a fast calculation for the utmost Fault Current at the transformer secondary terminals is FC = FLA / %PU Z FC = 1202 / 0.0575 = 20,904A
This quick calculation can assist you identify the fault current on the secondary of a transformer for the aim of selecting the proper overcurrent protective devices which can interrupt the available fault current. the most breaker that's to be installed within the circuit on the secondary of the transformer possesses to possess a KA Interrupting Rating greater then 21,000A. remember that feeder breakers should include the estimated motor contribution too. If the actual connected motors aren't known, then assume the contribution to be 4 x FLA of the transformer. Therefore, during this case the feeders would be sized at 20.904 + (4 x 1202 = 25,712 Amps
Utility MVA at the first of the Transformer
MVAsc = 500MVA
Transformer Data
13.8KV - 480Y/277V
1000KVA Transformer Z = 5.75% MVA Value
1000KVA / 1000 = 1 MVA
MVA Value = 1MVA / Zpu = 1MVA / .0575 = 17.39 MVA
Use the admittance method to calculate Fault Current 1 / Utility MVA + 1 / Trans MVA = 1 / MVAsc
1 / 500 + 1 / 17.39 = 1 / MVAsc
0.002 + 0.06 = 1/ MVAsc MVAsc = 1 / (0.002 + 0.06) MVAsc = 16.129
FC at 480V = MVAsc / (1.73 x 0.48) FC = 16.129 / 0.8304
FC = 19.423KA FC = 19, 423 A
Reference: