Spurious Earth-Fault Trip on a Healthy 11 kV Feeder During a Through-Fault
An 11 kV, 50 Hz distribution switchboard is fed from a 33/11 kV, 20 MVA transformer with an earthing transformer limiting earth-fault current to 300 A. Each of the six outgoing feeders (maximum load 300 A) has three 400/1 A, 5P10, 10 VA, 5P CTs, Rct <= 4 ohm, connected in residual (star-point) formation to one earth-fault element of a numerical relay. That element is set to 40 A primary, definite time 0 s. Phase overcurrent is 480 A primary on an IDMT curve with 0.3 s grading.
During a phase-to-phase fault on Feeder 4 (9.4 kA at the 11 kV bus), Feeder 5's earth-fault element also operated and tripped a healthy chemical plant. Feeder 5's CT secondary leads are 45 m of 6 mm2 copper run in a multicore, and a 2.5 VA panel ammeter is connected in the residual path of one phase only. The relay event record for Feeder 5 shows an apparent earth-fault current of 1.9 A secondary, lasting 30 ms and starting 10 ms after fault inception. Feeder 4 tripped correctly and the fault was cleared in 120 ms.
1. Explain the mechanism that produced the apparent earth current on Feeder 5 and support it with an order-of-magnitude check of the CT performance.
Read the worked answer
Mechanism. A residual connection sums the three phase currents in the secondary; it only measures zero if all three CTs transform faithfully. During a heavy through-fault the CTs saturate, and they do not saturate identically because of remanence, manufacturing spread, unequal lead burden and the ammeter loading one phase only. The secondary sum then contains a spill current that the relay reads as earth-fault current on a healthy feeder.
Order-of-magnitude check.
- Through-fault secondary current = 9400 / 400 = 23.5 A, i.e. 23.5 x rated.
- The CT is guaranteed only up to its accuracy limit factor of 10, i.e. 4000 A primary; at 9.4 kA it is past the limit and is in partial saturation for much of each half cycle.
- With the fully offset first peak (X/R of the source plus 20 MVA transformer, asymmetric factor up to ~1.8), flux demand roughly doubles again.
- Required kneepoint for faithful reproduction: Vk >= 2 x 23.5 x (4 + 0.26) ≈ 200 V, using a lead resistance per leg of 0.0175 x 45 / 6 = 0.13 ohm. A 5P10, 10 VA unit with Rct = 4 ohm gives Vk ≈ 10 x 1 x (4 + 10) ≈ 140 V - materially short.
The measured 1.9 A secondary (≈760 A primary equivalent) is 19 times the 40 A setting, so the element operates.
2. Recommend corrective measures and state which of the two elements of the scheme (connection method or setting) is the primary design error.
Read the worked answer
Primary error is the connection method, not the setting value. 40 A primary is a defensible sensitivity against a 300 A earth-fault limit; the error is asking a residual connection of three saturating phase CTs to discriminate against a spill current that, during a through-fault, can exceed the setting by more than an order of magnitude.
Corrective measures, in order of preference.
- Replace the residual scheme with a dedicated core-balance CT around the three feeder cores. It measures true residual flux and is unaffected by phase-CT saturation. Retain 40 A primary, but see point 4.
- If residual must be kept, specify class PX CTs sized for the through-fault: Vk >= K x If_secondary x (Rct + Rl) with an offset factor K ≈ 2, and remove the ammeter from the residual path (or use a 3-phase ammeter with matched burden on all three phases).
- Match the lead burden: equal-length, equal-cross-section leads, common return point at the relay, single-point earth.
- Add a definite time delay of 0.3 s or more so the element cannot race a through-fault, and grade it against downstream equipment.
- Add 2-out-of-3 phase-current supervision to block the earth element when a phase overcurrent element is picked up, which prevents most through-fault maloperation without losing sensitivity to genuine earth faults.
3. The client asks why a 40 A instantaneous earth-fault setting cannot simply be kept because 'the plant needs fast earth-fault clearance'. Construct the engineering response, including grading.
Read the worked answer
Trade-off to explain. Speed and sensitivity compete with stability. An instantaneous element set at 10% of CT rating has no time to discriminate the genuine 40–300 A earth fault from a 760 A spill transient created by a through-fault nineteen times larger than the setting.
Grading argument.
- Downstream of each feeder there are LV fuses and MCCBs; the feeder earth element must clear faster than they do for a fault in its own zone, but slower than any upstream element for a fault outside it.
- Typical electromagnetic/mechanical grading margin is 0.3–0.4 s; numerical relays permit 0.25–0.3 s. A 0.3 s definite-time earth element therefore still clears a feeder earth fault well inside the 1 s damage limit for 11 kV cable and switchgear, and inside typical personnel-protection targets on the LV side.
- Where the plant genuinely needs faster clearance, the answer is a unit-protection or CBCT-based scheme with a high-set instantaneous element set above the maximum through-fault spill, not a lower setting on the residual scheme.
Evidence to present. The event record (1.9 A for 30 ms starting 10 ms after inception), the CT accuracy-limit calculation, and a replayed fault record demonstrating that the revised settings still clear a 40 A resistive earth fault in the feeder zone.