Junior Member
Posted July 3, 2018
Can someone help me with these two questions:
What is the theoretical maximum symmetrical fault current that would be available at the secondary terminals of a 3-phase 12,470-480Y/277V, 1000kva, 7.0% impedance transformer?
A given 1000kva wye-wye transformer has a phase voltage of 277; what would you expect the full load line current to measure?
Member
Posted July 3, 2018
i got the 1st question on my practice exam, and got it wrong also. waiting for answer too.....Originally Posted by adamstraub
Junior Member
Posted January 17, 2019
This quick calculation can help you determine the fault current on the secondary of a transformer for the purpose of selecting the correct overcurrent protective devices that can interrupt the available fault current.
Fault Current = Full Load Amps / % Impedance (Z)
Example: Calculate the maximum fault current for a transformer rated 13.8kV-480Y/277V 1000kVA 5.75%Z
Step 1: Determine Full Load Amps (FLA)
You can determine the Full Load Amps of a transformer with the following formula: FLA = VA / L-L Voltage x 1.732, so using the example above we get FLA = 1000000 / 480 x 1.732 or 1000000 / 831.36 = 1202.84 (note the conversion from kVA to VA, 1000 x 1000 = 1000000).
FLA = 1203 amperes
Step 2: Divide FLA by Impedance
The next step is to simply divide the number obtained in step 1 by the transformer nameplate impedance. Using the example above we get: 1203 / 0.0575 = 20921.73 (REMEMBER TO CONVERT PERCENT TO DECIMAL, if 100% = 1.00 then 5.75% = 0.0575).
FC = 20,922 amperes
If a main breaker was to be installed in the circuit on the secondary of the transformer used in this example, it would have to have an Interrupting Rating greater than 21,000A.
