A three-phase fault occurs at the end of a transmission line connected to a bus. The systemconsists of a generator (1000 MVA, 22 kV) with a sub-transient reactance of 0.2 per unit,and a step-up transformer (1000 MVA, 22/220 kV) with a per-unit impedance of 0.1 perunit on its own base. The transmission line has an impedance of 0.1 + j0.5 per unit on a 220kV, 1000 MVA base.a. Convert all impedances to a common base of 1000 MVA, 22 kV.
Question
A three-phase fault occurs at the end of a transmission line connected to a bus. The systemconsists of a generator (1000 MVA, 22 kV) with a sub-transient reactance of 0.2 per unit,and a step-up transformer (1000 MVA, 22/220 kV) with a per-unit impedance of 0.1 perunit on its own base. The transmission line has an impedance of 0.1 + j0.5 per unit on a 220kV, 1000 MVA base.a. Convert all impedances to a common base of 1000 MVA, 22 kV.
Solution
To convert all impedances to a common base of 1000 MVA, 22 kV, we need to follow these steps:
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Generator impedance: The generator is already on the base of 1000 MVA, 22 kV, so its sub-transient reactance remains 0.2 per unit.
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Transformer impedance: The transformer is also on the base of 1000 MVA, but its voltage base is different. We need to convert it to the new voltage base. The per-unit impedance is directly proportional to the square of the base voltage. So, the new per unit impedance of the transformer is (22/220)^2 * 0.1 = 0.01 per unit.
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Transmission line impedance: The transmission line impedance is given on a base of 220 kV, 1000 MVA. We need to convert it to the new base of 22 kV, 1000 MVA. The per-unit impedance is directly proportional to the square of the base voltage. So, the new per unit impedance of the transmission line is (22/220)^2 * (0.1 + j0.5) = 0.01 + j0.05 per unit.
So, the impedances on a common base of 1000 MVA, 22 kV are:
- Generator: 0.2 per unit
- Transformer: 0.01 per unit
- Transmission line: 0.01 + j0.05 per unit.
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