Overcurrent relay grading on an 11 kV motor feeder
A 33/11 kV substation feeds an 11 kV industrial feeder through a 400/5 A CT and a microprocessor overcurrent relay set on the IEC standard inverse curve, plug setting 100% and TMS 0.10. The feeder supplies a 2.0 MW, 11 kV, four-pole induction motor with a rated current of about 123 A and direct-on-line starting current of 6 times full load, accelerating in 9 s. The downstream 11 kV/415 V transformer fuse clears a fault in 0.15 s. A three-phase fault at the downstream 11 kV bus is 4 kA. The relay has tripped three times during motor starting, although the motor starter shows no fault. The upstream 33 kV relay is graded at 0.3 s above this relay. The plant wants to raise the feeder load and asks whether the relay can simply be set to a higher TMS.
1. Calculate the relay operating time for the 4 kA fault using the IEC standard inverse formula. Does TMS 0.10 coordinate with the 0.15 s downstream fuse and the 0.3 s upstream grading margin?
Read the worked answer
Use the IEC standard inverse curve:
t = 0.14 × TMS / ((I/Is)^0.02 - 1)
With CT 400/5 and plug setting 100%, Is = 400 A. At I = 4000 A, PSM = 10.
- 10^0.02 = 1.0471
- Denominator = 1.0471 - 1 = 0.0471
- t = 0.14 × 0.10 / 0.0471 ≈ 0.297 s
The downstream fuse clears in 0.15 s, leaving only about 0.15 s margin. That is below the usual 0.3 s grading margin for electromechanical and fuse-to-relay coordination, so TMS 0.10 is too fast on the downstream side. The upstream relay is graded 0.3 s above this relay, so the upstream margin is about 0.3 s and is acceptable at this fault level. Raising TMS to 0.15 would give about 0.45 s, restoring roughly 0.30 s downstream margin, but the upstream relay would then also need to be re-graded to about 0.75 s. A higher TMS alone cannot be chosen without checking both directions.
2. Calculate the relay operating time at the motor starting current and explain why the relay trips during a normal start. What does this reveal about using a standard inverse overcurrent relay for motor feeders?
Read the worked answer
Motor starting current = 6 × 123 = 738 A.
PSM = 738/400 = 1.845.
- 1.845^0.02 = 1.0123
- Denominator = 0.0123
- t = 0.14 × 0.10 / 0.0123 ≈ 1.14 s
The motor needs 9 s to accelerate, but the relay reaches its operating time after about 1.1 s. The relay is not maloperating; it is doing exactly what the curve and settings tell it to do. The pickup of 400 A is only 3.25 times motor full load and 0.54 times starting current, so the start looks like a long-duration overload. Increasing TMS enough to ride through 9 s would require a TMS of about 0.8, which would make the 4 kA fault operate in about 2.4 s - far too slow for the downstream fuse and the upstream grading. A standard inverse overcurrent relay with pickup near load is therefore unsuitable as the only motor-start protection. The correct approach is a higher pickup above starting current, a definite-time start block, or a dedicated motor protection relay with thermal and starting supervision.
3. Recommend a coordinated solution that allows starting and preserves fault clearance, stating one check you would make on the CT.
Read the worked answer
Raise the relay pickup to about 200% of CT rating, i.e. 800 A, so the 738 A start is below pickup. At 4 kA the PSM becomes 5, giving:
t = 0.14 × 0.10 / (5^0.02 - 1) ≈ 0.43 s.
That restores about 0.28 s margin over the 0.15 s downstream fuse, and the upstream relay can be graded at about 0.73 s. Verify that the relay accepts a 200% plug setting and that the CT can carry the starting current and the 4 kA fault without severe saturation. A 400/5 CT on a 123 A load is also poorly utilised, so a 200/5 or 300/5 CT with a 5P20 accuracy limit factor may be a better choice; check the CT knee-point voltage and secondary burden. Add a definite-time start inhibit or motor thermal model so that a genuine locked-rotor condition still trips after a safe starting window. Do not solve the problem only by increasing TMS, because that delays fault clearance.