Setting an 11 kV Feeder Overcurrent Relay Against a Downstream Transformer
A 33/11 kV substation feeds a 6 km, 11 kV express feeder with a peak load of 250 A. The feeder CT is 400/5 A and the relay is a numerical overcurrent unit with IEC 60255 standard inverse (SI) characteristic. The three-phase fault level at the 11 kV bus is 6.5 kA and at the far end of the feeder it falls to 4.2 kA. At the far end, a 1000 kVA, 11 kV/433 V distribution transformer is protected by its own HV relay set to operate in 0.10 s for faults on its terminals. The utility's protection policy requires a grading margin of 0.30 s between successive devices. A junior engineer proposes a plug setting of 1.0 and a TMS of 0.05 so that 'the feeder trips fast and the 33 kV breaker is never disturbed'.
1. Select the plug setting for the feeder relay and justify it against both load and fault current.
Read the worked answer
Pickup must ride through the maximum load with margin, but sit well below the minimum fault.
- Maximum load = 250 A. Using the usual 1.2–1.3 safety factor: 250 × 1.2 = 300 A.
- CT ratio 400/5 means one plug step = 400/n A primary. A setting of 300 A is exactly 0.75 × CT rated (75 % plug), and 300/80 = 3.75 A on the secondary side - comfortably above the relay's 0.1 A minimum and well within the 400/5 CT's accuracy range.
Sensitivity check at the far end: 4200 A / 300 A = 14 × pickup. That is ample - even a 2 × pickup threshold for dependable operation is exceeded many times over, so the far-end fault will be cleared dependably.
Why not 1.0 plug (400 A)? It still detects the far-end fault (10.5 ×), but it leaves only 1.6 × margin above peak load. A cold-load pickup after an outage, or a motor restart on the feeder, can exceed 300 A momentarily; 400 A gives less security against nuisance tripping, while 300 A is the better balance.
2. Calculate the TMS needed to grade with the downstream transformer relay, and state the actual operating time you would expect at the far-end fault.
Read the worked answer
Target time at the far-end fault: t = downstream time + grading margin = 0.10 + 0.30 = 0.40 s.
IEC SI characteristic: t = TMS × 0.14 / [(I/I_s)^0.02 − 1]
- I/I_s = 4200 / 300 = 14
- 14^0.02 = e^(0.02 × ln 14) = e^(0.02 × 2.639) = 1.0542
- Denominator = 1.0542 − 1 = 0.0542
- Without TMS: t = 0.14 / 0.0542 = 2.58 s
TMS = 0.40 / 2.58 = 0.155 → select 0.15 (numerical relays allow 0.01 steps).
Achieved time: 0.15 × 2.58 = 0.387 s at 4.2 kA - a grading margin of 0.287 s against the downstream relay, marginally short of the 0.30 s policy. Selecting TMS = 0.20 gives 0.517 s and a 0.417 s margin, which is the safer choice unless the downstream time can be tightened.
Check at the bus fault: I/I_s = 6500/300 = 21.7 → 21.7^0.02 = 1.0634 → t = 0.15 × 0.14/0.0634 = 0.33 s. Coordinated, and fast enough for the switchgear rating.
3. Evaluate the junior engineer's proposal and check whether the CT can support the settings you have chosen.
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The proposal is unsafe. A TMS of 0.05 gives 0.05 × 2.58 = 0.13 s at the far-end fault - faster than the 0.10 s downstream relay plus margin, so the feeder would trip before the transformer's own protection for faults inside the transformer. That destroys discrimination: a fault in a single consumer's transformer would black out the entire 6 km feeder. A TMS of 0.05 also falls below many relays' useful range and makes coordination with the upstream 33 kV relay impossible.
CT check: at the 6.5 kA bus fault the primary multiple is 6500/400 = 16.25 × rated. The CT secondary would try to deliver 16.25 × 5 = 81 A. The CT's accuracy-limit factor must therefore be at least 16.25 at the actual connected burden - specify ALF ≥ 20 (5P20) with the burden measured, not assumed. A 5P10 CT, common in older installations, would saturate here and could slow the relay enough to break grading.
Also verify: the relay burden plus pilot lead resistance must not exceed the CT rated VA; and confirm the 0.10 s downstream time is itself graded with the LV side protection before locking in TMS = 0.15.