Feeder F-7: An Earth-Fault Relay That Cannot See the Fault
At the 132/11 kV auxiliary switchyard of a 412 MW run-of-the-river station, 11 kV feeder F-7 supplies the dewatering pump house and the gallery ventilation boards over 3.2 km of 3 × 95 mm² Cu/XLPE cable. The feeder CT is 200/1 A and the numerical relay uses the IEC Standard Inverse curve with a plug setting of 1.3 A and TMS 0.10; the instantaneous element is disabled because of the 11 kV/415 V transformer at the far end. A grading margin of 0.3 s is required against the 132 kV backup, which clears a 180 A primary fault in 1.5 s. In five weeks F-7 has tripped four times. Each fault was a flashover at the pump-house termination after heavy rain, with post-fault currents of 150–180 A primary recorded. On one occasion the feeder relay did not pick up at all and the 132 kV backup cleared the fault. Feeder load current is 76 A; the 250 kW pump motor starts direct-on-line at 5.5 times its running current of 44 A for 8 s. Single-shot auto-reclose with a 1 s dead time is enabled on F-7. Peak monsoon is two weeks away.
1. Explain why F-7's relay failed to operate for a 150–180 A earth fault, and why the plug setting cannot simply be lowered to catch it.
Read the worked answer
Pickup check. Primary pickup = plug × CT ratio = 1.3 × 200 = 260 A. A 150–180 A fault is only 0.58–0.69 of pickup, so on the one occasion the relay never started its timing and the 132 kV backup at 1.5 s had to clear a fault on an 11 kV cable. The plug was evidently chosen from load and starting considerations, never checked against the minimum earth-fault current.
Why lowering the plug fails. To see 180 A with any margin the pickup must fall below about 150 A, i.e. plug ≤ 0.75 A. Take plug 0.6 A (pickup 120 A). During a healthy start the relay sees 5.5 × 44 = 242 A, M = 242/120 = 2.02. With the SI curve, t = TMS × 0.14/(M^0.02 − 1) = TMS × 0.14/(1.01416 − 1) = 9.9 × TMS. Even at TMS 0.10 the relay trips in about 1 s, inside an 8 s start - the pump could never be started.
The real conclusion. The starting current (242 A) is larger than the fault current (180 A). A single phase overcurrent element therefore cannot distinguish a healthy motor start from a resistive earth fault. The design has no element that looks only at residual current, which is the actual defect. This is a protection-design error, not a setting nuisance, and it must be corrected before the monsoon rather than by another plug tweak.
2. Recommend the revised protection scheme and verify it numerically against the motor start, the 180 A earth fault, and a 4 kA phase fault.
Read the worked answer
1. Phase overcurrent (phase faults only). Plug 1.0 A (pickup 200 A), TMS 0.30, SI curve.
- Motor start, 242 A: M = 1.21, M^0.02 = 1.00382, t = 0.30 × 0.14/0.00382 = 11.0 s > 8 s start, margin ≈ 37 %.
- 4 kA phase fault: M = 20, M^0.02 = 1.0618, t = 0.30 × 0.14/0.0618 = 0.68 s; add 60 ms breaker → 0.74 s. Cable check: for Cu/XLPE k = 143, 1 s rating = 143 × 95 = 13,585 A; permissible time at 4 kA = (13585/4000)² = 11.5 s, so the cable is amply protected. Grading against the 132 kV backup at 4 kA (slower than 1.2 s) leaves more than 0.4 s.
2. Sensitive earth fault (the actual fix). Core-balance CT 50/1 A, setting 0.02 A = 1 A primary, definite time 0.5 s. At 180 A it clears in 0.5 s; backup at 1.5 s gives 1.0 − 0.06 = 0.94 s margin, far above the 0.3 s required. Balanced starting current produces negligible residual, so starting is unaffected.
3. Installation cautions. The CBCT must enclose the three phase cores only, not the armour or earth conductor, or sheath current will cause maloperation. Check the residual connection is not left in the old Holmgren arrangement, and test with primary injection before energising.
Result: earth faults are now cleared in 0.5 s instead of being handed to the 132 kV backup, and phase protection still rides through every pump start.
3. Is single-shot auto-reclose appropriate on F-7? State the action and the other post-fault steps you would order.
Read the worked answer
Auto-reclose must be switched off. Auto-reclose is justified on overhead lines where most faults are transient arcs that self-clear in the dead time. A cable fault is essentially always permanent - a punctured XLPE wall, a flashed stress cone, a water tree - and reclosing drives a second arc through already-damaged insulation, converts a repairable termination into a cable joint, and puts a second full fault current through the cable, the CT and the transformer. There is also a safety issue: the pump house termination is at working height and may be attended.
The 1 s dead time is worse than useless here because the fault path is wet and the arc restrikes readily.
Actions, in priority order:
- Disable auto-reclose on F-7 and audit every other 11 kV cable feeder in the station; keep auto-reclose only on overhead feeders where it was designed in.
- Replace the flashed termination with a cold-shrink kit, re-terminate with correct stress-cone dimensions, and prove with insulation resistance plus a VLF test before recharging.
- Fix the cause, not only the symptom: check gland plate sealing, cable trench drainage and the route's rainwater path - four flashovers in five weeks after rain is a moisture-ingress pattern, not bad luck.
- Log the events with recorded currents and give the 132 kV backup setting a review, since it has been carrying 11 kV faults it should never see.
Doing only the re-termination without the SEF element and without disabling auto-reclose guarantees a fifth event.