Feeder Breaker Trips Before the Downstream ACB Clears
An 11 kV feeder from a 33/11 kV substation carries a peak load of 100 A and supplies five distribution transformers, one of which is a 630 kVA, 11/0.433 kV unit with 4.5% impedance. The feeder circuit breaker has a 200/1 A CT and a numerical relay set to IEC 60255 standard inverse, pick-up Is = 120 A primary, TMS = 0.15, and a high-set (instantaneous) element at 5 × Is = 600 A primary, definite time 0 s. During a fault on the LV side of the 630 kVA transformer, close to its terminals, the transformer's LV air circuit breaker clears the fault in 0.10 s, but the 11 kV feeder breaker has already tripped and blacked out the other four transformers, dropping about 1.8 MW of load. The fault was correctly cleared by the LV breaker; there was no equipment damage on the 11 kV side. Operators have logged three identical events in two months. The upstream 33 kV relay at the grid substation is set with TMS 0.30.
1. Explain, with a calculation, why the feeder breaker operated for a fault that the LV breaker cleared. Refer the LV fault current to the 11 kV side.
Read the worked answer
LV fault current at the transformer terminals, ignoring source impedance because the transformer dominates:
I_LV = 630,000 / (√3 × 433 × 0.045) = 630,000 / 33.75 ≈ 18,670 A.
Referred to 11 kV: I_11 = 18,670 × (433/11,000) ≈ 735 A.
The high-set element is set at 600 A primary (3.0 A secondary on the 200/1 CT). The referred fault current of 735 A is 1.22 times the high-set setting, so the instantaneous element operates in its definite time of 0 s - well before the LV ACB's 0.10 s. That is the miscoordination: the high-set setting is below the maximum through-fault current that the downstream transformer can deliver into an LV fault.
Checking the IDMT element confirms the same fault would have been handled correctly: M = 735/120 = 6.12, and t = 0.15 × 0.14/(6.12^0.02 − 1) = 0.021/0.037 = 0.57 s, which gives the LV breaker its 0.10 s plus a healthy grading margin. The problem is not the IDMT curve, the CT ratio or the TMS - it is solely the high-set threshold sitting inside the downstream protection zone.
2. Propose revised high-set and IDMT settings, and check them against transformer inrush, feeder load current and the minimum feeder-end fault level. State any assumptions.
Read the worked answer
Revised high-set: it must sit above the maximum LV-side through-fault referred to 11 kV with a margin for CT and calculation error. Using 1.3 × 735 ≈ 955 A, set 1000 A primary (5.0 A secondary).
Check against feeder-end faults: if the minimum single-phase or two-phase fault at the remotest point is, say, 1600 A, the high-set at 1000 A still trips there, but confirm the actual value from the fault study; if the far-end fault is below 1000 A, the high-set simply will not see it and the IDMT element will clear it, which is acceptable.
Check against inrush: a 630 kVA transformer draws roughly 8–12 times rated current at energisation for about 0.1 s. Rated HV current = 630,000/(√3×11,000) = 33 A, so inrush is roughly 260–400 A, well below the 120 A pick-up? No - it is above pick-up but the IDMT element at 400/120 = 3.3 gives t = 0.15 × 0.14/(3.3^0.02 − 1) = 0.021/0.024 = 0.88 s, so the 0.1 s inrush decays before operation. The high-set at 1000 A is immune to inrush.
Check load: peak 100 A is below Is = 120 A, so no normal-load trip, but the margin is thin; verify the peak is a true 15-minute maximum before leaving Is at 120 A.
Grading: the 33 kV relay at TMS 0.30 sees the same 735 A fault and must be slower than 0.57 s plus a 0.3–0.4 s margin; confirm its curve at the same current before commissioning.
3. Beyond changing the numbers, what else would you do before re-energising, and how would you prove the new settings are correct?
Read the worked answer
Re-energising on paperwork alone is how repeat trips happen. A practical sequence:
Verify the data, not the nameplate. Confirm the 630 kVA transformer's actual impedance from its test certificate (4.5% may be a design figure), the true fault level at the 11 kV bus, and the LV ACB's measured clearing time. The 735 A figure moves materially if the transformer is 5.5% instead of 4.5%.
Run a full coordination study across all five transformers, including the 33 kV upstream relay, and record the curves on one sheet so the grading margins (typically 0.3–0.4 s between devices) are visible.
Use primary injection testing on the feeder CT and relay, including the high-set element at 1000 A, to prove the secondary settings of 5.0 A and the CT ratio of 200/1 are actually what the relay applies. Secondary injection alone will not reveal a CT ratio or wiring error.
Check CT saturation and the dc offset for the close-in LV fault; a transient over-reach can still trip a high-set element set too close to the through-fault current, which is why the 1.3 margin matters.
Record the settings in the relay file with date and signature, log the event in the protection register, and arrange a follow-up review after the next LV fault to confirm the feeder holds.