Question

A wye-delta connected power transformer is protected by a percentage differential relay with 25% slope and 2 A minimum pick-up. A 3-phase fault occurs on the primary (bushing) terminals of the transformer. The resulting fault current infeed from the primary side is 4 kA and that from the secondary side is 24 kA. Assuming CT ratios of 30/5 A and 1 200/5 A on the primary and secondary sides, respectively, determine whether the differential relay will operate to trip the

Answer 1

To determine if the differential relay will operate to trip the transformer, calculate the fault currents and compare with the pickup current.

Step-by-step explanation:

In order to determine whether the differential relay will operate to trip the transformer, we need to compare the primary and secondary fault currents and the CT ratios.

1. Calculate the primary and secondary fault currents using the CT ratios. Primary fault current = (4 kA * 30/5 A) = 24 kA, Secondary fault current = (24 kA * 1200/5 A) = 57.6 kA.
2. Calculate the pickup current for the differential relay using the 25% slope. Pickup current = 25% of the maximum fault current = 0.25 * 57.6 kA = 14.4 kA.
3. Compare the pickup current with the minimum pickup. Since the pickup current (14.4 kA) is greater than the minimum pickup (2 A), the differential relay will operate to trip the transformer.

Answer 2

The differential relay will operate to trip the transformer due to a 3-phase fault. The current difference detected by the relay exceeds both the minimum pick-up value and the allowable percentage difference set by the relay's slope characteristic.

Step-by-step explanation:

To determine whether the differential relay will operate to trip due to a 3-phase fault, we must compare the secondary fault current transformed to the primary side with the actual primary fault current. The current transformer (CT) ratios transform the currents to the relay level. With a primary CT ratio of 30/5 A and a secondary CT ratio of 1,200/5 A, the transformed secondary fault current to the primary side becomes (24 kA * 5/1,200) which equals 0.1 kA or 100 A.

The actual primary fault current is 4 kA, which, when passed through the CT, becomes (4 kA * 5/30) or 0.667 kA, which is 667 A at the relay level. The relay will detect the difference between 667 A and 100 A, which is 567 A.

Now, we need to check if this difference is above the minimum pick-up current of the relay. Since 567 A is much higher than 2 A, the minimum pick-up is easily exceeded. Next, we check the percentage restraint characteristic. The slope of 25% means that the relay allows a 25% difference between the two sides (due to CT errors, etc.) without tripping, which is (0.25 * 667 A = 166.75 A). Because the actual difference (567 A) is greater than the allowed difference by the slope characteristic (166.75 A), the differential relay will operate and trip the transformer protection.