Transformer Exam Questions, NETA Level 2

[adamstraub]

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?

[hy3533]

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?

i got the 1st question on my practice exam, and got it wrong also. waiting for answer too.....

[herbertadiaz1]

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?

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.