Both relays trip: grading two IDMT overcurrent relays on an 11 kV feeder
A 33/11 kV substation feeds a depot distribution board through 11 kV feeder F1. Two overcurrent relays use the standard inverse curve t = 0.14 × TM / ((I/Is)^0.02 − 1). The incomer relay R1 has a 400/1 A CT, a plug setting of 75% (300 A primary pickup) and a time multiplier of 0.15. The feeder relay R2 has a 200/1 A CT, a plug setting of 100% (200 A primary) and a time multiplier of 0.10. A single source feeds the busbar, so for a fault on a consumer's cable downstream of R2 both relays see the same current, 2250 A. On three separate occasions that fault has tripped R1 as well as R2, blacking out the whole 11 kV board and the two 800 kW motors running on it. The board's maximum demand is about 185 A at 11 kV and the motors are direct-on-line started. A coordination time interval of 0.4 s is required to cover relay overshoot, CT error and breaker opening time.
1. Calculate the operating time of both relays at 2250 A and determine whether the present settings provide adequate discrimination.
Read the worked answer
R2 (feeder relay). Is = 200 A; PSM = 2250 / 200 = 11.25. 11.25^0.02 = exp(0.02 × ln 11.25) = exp(0.02 × 2.4204) = exp(0.04841) = 1.0496. t = 0.14 × 0.10 / 0.0496 = 0.014 / 0.0496 = 0.282 s.
R1 (incomer relay). Is = 300 A; PSM = 2250 / 300 = 7.5. 7.5^0.02 = exp(0.02 × 2.0149) = exp(0.04030) = 1.0411. t = 0.14 × 0.15 / 0.0411 = 0.021 / 0.0411 = 0.511 s.
Margin = 0.511 − 0.282 = 0.229 s.
Verdict. The required coordination interval is 0.4 s; the settings deliver 0.23 s. That is less than the relay's own repeatability plus breaker opening, so a heavy through-fault can be cleared by both relays - exactly the observed blackout. Note why the margin is worst here: at high multiples of pickup the standard inverse curve flattens towards its definite minimum time, so the absolute time gap shrinks precisely at the fault levels where grading matters most. The likely history is that R1's TM was reduced to 0.15 in an earlier exercise, possibly to speed up a slow upstream clearance, without re-checking this case.
2. Propose revised settings that restore discrimination, and verify that your change does not create a problem at lower fault currents or at the upstream 33 kV level.
Read the worked answer
Required R1 time = R2 time + 0.4 s = 0.282 + 0.4 = 0.682 s at 2250 A.
Required TM = t × (PSM^0.02 − 1) / 0.14 = 0.682 × 0.0411 / 0.14 = 0.200.
Set R1 TM = 0.20 (the next available tap). Check: t = 0.14 × 0.20 / 0.0411 = 0.681 s, giving a margin of 0.399 s ≈ 0.4 s. Correct.
Check at a lower fault current, say a 950 A fault on the distribution transformer LV side:
- R2: PSM = 4.75, 4.75^0.02 = 1.0317, t = 0.014 / 0.0317 = 0.44 s.
- R1: PSM = 3.17, 3.17^0.02 = 1.0233, t = 0.028 / 0.0233 = 1.20 s. Margin = 0.76 s - comfortable. The change therefore does not worsen grading in the low-current region; it only slows R1 where it was already far too fast.
Checks that must still be done:
- Re-grade against the 33 kV incomer relay so R1 is not now the slowest link in a chain that has its own CTI to meet.
- Confirm R1 at 1.20 s is still inside the cable and transformer short-time withstand and damage curves for the relevant currents.
- Verify R1's 300 A pickup sits above maximum demand (185 A) plus the larger motor's direct-on-line starting current for the several seconds of acceleration; at PSM near 1.5 the IDMT time is long, so this is normally acceptable but must be stated.
- Check CT performance at 2250 A - 5.6 times rated for the 400/1 CT, so saturation is not the limiting factor here.
3. A technician suggests 'just set the incomer instantaneous element at 2400 A and leave the time multipliers alone'. Evaluate that proposal.
Read the worked answer
It would not fix the reported problem. The through-fault that causes both relays to trip is 2250 A. A high-set instantaneous element set at 2400 A does not pick up at 2250 A, so discrimination at that fault level is unchanged and the board would still black out.
It is also set too close to the fault level to be safe. A setting only 6.7% above the maximum through-fault current is inside the combined relay accuracy, CT ratio error and DC offset transient. Any of these would let the element operate for a fault that R2 should clear, giving the same double trip, and with no time delay at all.
If an instantaneous element is wanted, set it at not less than about 1.3 times the maximum through-fault current seen by R1 - roughly 2900 A, so a 3000 A setting - and verify it lies below the switchgear and cable fault rating. Its purpose is to clear close-in busbar faults quickly, not to grade with a downstream feeder.
It also must clear the motor starting duty: the element must be above the largest direct-on-line starting current plus the standing load, and above transformer magnetising inrush if the transformer is switched from that busbar.
Conclusion: reject the proposal, implement TM 0.20 on R1, and treat any instantaneous element as a separate, correctly calculated addition - not a substitute for grading.