kewo
Electrical
- Jan 17, 2009
- 5
What is the interrupting current rating required for a circuit breaker to protect this transformer?
Single phase transformer:
OA, 55C: 8333 kVA
OA, 65C: 9333 kVA
FA, 55C: 10000 kVA
%Impedance: 10.2% (at 75C, OA Base)
Is it:
1. FLA = 8333kV / 5.04kV = 1653A; IC = 1653A/.102 = 16206A,
2. FLA = 9333kV / 5.04kV = 1852A; IC = 1852A/.102 = 18157A, or
3. FLA = 10000kV / 5.04kV = 1984A; IC = 1984A/.102 = 19451A?
The %Impedance does say "75C, OA Base" so would IC be either (1) or (2)?
How is the 75C taken into account? Would the impedance actually be lower, thus a higher IC, since the ratings are listed at a lower temperature (55C and 65C instead of 75C)?
Thanks.
Single phase transformer:
OA, 55C: 8333 kVA
OA, 65C: 9333 kVA
FA, 55C: 10000 kVA
%Impedance: 10.2% (at 75C, OA Base)
Is it:
1. FLA = 8333kV / 5.04kV = 1653A; IC = 1653A/.102 = 16206A,
2. FLA = 9333kV / 5.04kV = 1852A; IC = 1852A/.102 = 18157A, or
3. FLA = 10000kV / 5.04kV = 1984A; IC = 1984A/.102 = 19451A?
The %Impedance does say "75C, OA Base" so would IC be either (1) or (2)?
How is the 75C taken into account? Would the impedance actually be lower, thus a higher IC, since the ratings are listed at a lower temperature (55C and 65C instead of 75C)?
Thanks.