A 20 MVA transformer, which may be called upon to operate at 30% overload, feeds 11-kV busbars through a circuit breaker. Other circuit breakers supply outgoing feeders. The transformer circuit breaker is equipped with 1000/5 A CTs, and the feeder circuit breakers with 400/5 A CTs. All sets of CTs feed inverse definite minimum time (IDMT) relays for overcurrent protection. The operating characteristic of relays on the feeder circuit breakers is shown in Fig. Q.2 below. The plug is set to 125% and the time multiplier is set to 0.3.

If a 3-phase fault current of 5000 A flows from the transformer to one of the feeders, determine:

1. The operating time of the feeder relay.
2. The minimum plug setting of the transformer relay.
3. The time setting, assuming a discriminative time margin of 0.5 seconds.

\[ \text{Relay Operating Time} = \frac{1}{0.14 / (\text{PSM} - 1)} \]

Note: The figure referenced in the problem is necessary for completing the calculations.

(i) To determine the operating time of the feeder relay, we need to refer to the operating characteristic curve in Fig.Q.2. Since the fault current is 5000A, we find the corresponding time on the curve, which is approximately 0.17 seconds.

(ii) The minimum plug setting of the transformer relay is determined by multiplying the rated current (20MVA) by the overload percentage (30%). This gives us 6MVA. Since the transformer is operating at 11kV, we can use the formula P = V x I to find the current: 6MVA / 11kV = 545.45A. Therefore, the minimum plug setting of the transformer relay is 6.8 (125% of 545.45A).

Assuming a discriminative time margin of 0.5 seconds means that the time setting of the transformer relay should be at least 0.5 seconds higher than the operating time of the feeder relay. Therefore, the time setting for the transformer relay would be approximately 0.67 seconds.

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