CT Knee-Point Voltage for an 11 kV Feeder and Nuisance Earth-Fault Tripping
An 11 kV feeder uses a 400/1 A, 5P20, 5 VA bar-primary CT with a secondary winding resistance of 3.5 Ω. The numerical relay sits 180 m away in the switchgear panel and is wired with 2.5 mm² copper cable whose loop resistance is 0.0148 Ω/m. Relay burden is 0.5 VA at 1 A and its decision time is 20 ms. The prospective three-phase fault level at the feeder origin is 12 kA and the source X/R ratio is 25. During primary injection at 6 kA the relay behaved correctly, so the protection engineer is being asked to sign off the scheme. A second problem has appeared: the feeder's earth-fault element has tripped twice during faults on an adjacent feeder. A spare 600/1 A, 5P20, 15 VA CT with Rct = 4.2 Ω is available in stores.
1. Compute the secondary fault current and the knee-point voltage required for the installed CT, and compare it with what a 5P20, 5 VA unit can deliver.
Read the worked answer
Secondary fault current: Is = 12 000/400 = 30 A.
Burden seen by the CT:
- Leads: 0.0148 Ω/m × 180 m = 2.66 Ω
- Relay: 0.5 VA at 1 A = 0.5 Ω
- CT secondary: Rct = 3.5 Ω
- Total: 6.66 Ω
Steady-state knee voltage needed: Vk = 30 × 6.66 = 200 V. With X/R = 25 the first-cycle flux can reach about twice the steady-state peak, so roughly 400 V is required for the CT to stay linear long enough for the 20 ms relay decision.
Available knee voltage for a 5P20, 5 VA unit: Vk ≥ 20 × 1 × (3.5 + 5) = 170 V, and that figure assumes a purely symmetrical current. The installed CT is therefore roughly 2.4 times short of what the transient case demands.
Consequence: on a close-in fault the core saturates within the first cycle, the secondary current collapses mid-wave and the relay sees a reduced, distorted current - the instantaneous element can under-reach, and the d.c. tail can operate sensitive elements spuriously. The 6 kA injection test, being below the worst case and free of offset, would not have revealed this.
2. Assess whether the spare 600/1 CT solves the problem, including the sensitivity and thermal trade-offs, and state any further change you would make to the wiring.
Read the worked answer
With 600/1: Is = 12 000/600 = 20 A; burden = 4.2 + 2.66 + 0.5 = 7.36 Ω; required Vk = 20 × 7.36 = 147 V, doubled for d.c. offset = 294 V.
Available Vk ≥ 20 × 1 × (4.2 + 15) = 384 V, so the spare has about 30% margin and is acceptable.
Trade-offs to check:
- Sensitivity: a 60 A earth fault now produces only 0.1 A secondary. The 1 A relay must set down to ~0.1 A in fine steps - most numerical relays can, but the setting must be re-verified.
- Thermal: 12 kA through a 600/1 CT is 20 times rated; confirm the short-time thermal rating (typically 25 kA/1 s for this class).
- Wiring: change the leads to 4 mm² (0.00461 Ω/m per conductor), giving a loop of 1.66 Ω instead of 2.66 Ω. Required Vk then falls to 20 × (4.2 + 1.66 + 0.5) × 2 = 254 V, increasing the margin to about 50% for negligible cost.
Also confirm the CT's class and remanence behaviour; a Class PS CT with a specified Vk and magnetising current would be preferable to a P-class unit on a critical feeder.
3. Explain why the earth-fault element trips during faults on an adjacent feeder, and set out the remedies in order of preference.
Read the worked answer
The earth-fault element is almost certainly connected residually (three phase-CT secondaries in parallel). Under an external fault the three phase currents are not exactly reproduced: the CTs differ in Rct, lead length and remanent flux, and the d.c. offset drives the cores into different parts of their magnetising curves. The result is a substantial false residual current that the sensitive earth-fault setting sees as an internal fault. The low knee point computed above makes this much worse.
Remedies, best first:
- Core-balance CT (CBCT) dedicated to the feeder earth fault. All three conductors pass through one core, so external phase currents cancel by construction and no residual is produced. This removes the problem at its source.
- If residual connection must be retained, use matched Class PS CTs with equal leads and a high-impedance (stabilising-resistor) scheme so that the external-fault spill current is diverted from the relay.
- Coordinate in time - add 200–300 ms of definite-time delay to the earth-fault element so it rides through external faults; this is a patch, not a cure.
- Check for a genuine second earth on the feeder before assuming the trips are spurious.
Simply raising the earth-fault pickup is the worst option: it hides the symptom and destroys sensitivity to genuine high-resistance earth faults, which is the whole reason the element exists.