- 2070 Bhadra · 8 marks
Design IDMT grading for relays A and B
Short answer
Both relays set PS=100%. Relay B (200/5 CT) set TDS=0.1, giving 0.22s at its 3000A fault. Relay A graded at 3000A: PSM=10, required 0.72s (0.22+0.5), giving TDS=0.72/2.8=0.26. Actual times: Relay A operates at 0.624s for own 4000A fault and 0.728s for backup at B (margin 0.508s).
Design a time-current graded protection scheme based on two identical IDMT relays at points A and B for the following radial feeder. Select appropriate current tap setting and time dial settings for the relays. Find out actual time of operation of relay at A for fault in its zone and as back up. Relay at A should also back up for faults at protection zone of B. [Figure: Source – relay point A – feeder section – relay point B – feeder – far end C.]
| Relay point | CT ratio | Fault current, A |
|---|---|---|
| A | 300/5 | 4000 |
| B | 200/5 | 3000 |
| Far end of feeder at point C | – | 2000 |
IDMT characteristics (for TDS = 1):
| PSM | 2 | 3.6 | 5 | 8 | 10 | 15 | 20 |
|---|---|---|---|---|---|---|---|
| T.op. sec | 10 | 6 | 3.9 | 3.15 | 2.8 | 2.2 | 2.2 |
In a time-graded scheme the relay farthest from the source (B) is set fastest, and relay A behind it is set slower by a discrimination margin for every fault in B's zone, so that A acts as back-up.
Assumptions: relay rated current 5 A; load current is below the CT primary rating, so PS = 100% is used on both relays (pick-up = CT primary current); discrimination margin = 0.5 s (breaker time + relay overshoot + safety); TDS steps of 0.01; the curve between table points is interpolated linearly.
Source ─[A]──── AB ────[B]──── BC ──── C
300/5 200/5
4000 A 3000 A 2000 A
Relay B (last relay, set fastest)
PS = 100% → pick-up = 200 A primary
Fault at B (3000 A): PSM = 15 → t = 2.2 s (TDS 1)
Fault at C (2000 A): PSM = 10 → t = 2.8 s (TDS 1)
Choose minimum TDS = 0.1
t_B (fault at B) = 0.1 × 2.2 = 0.22 s
t_B (fault at C) = 0.1 × 2.8 = 0.28 s
Relay A
PS = 100% → pick-up = 300 A primary
Grading point: fault at B, 3000 A
(maximum fault current in B's zone)
PSM_A = 3000/300 = 10 → t = 2.8 s (TDS 1)
Required t_A = 0.22 + 0.5 = 0.72 s
TDS_A = 0.72 / 2.8 = 0.257 → choose 0.26
Actual operating times of relay A (TDS = 0.26)
| Fault at | Current | PSM_A | t (TDS 1) | t_A | Relay B time | Margin |
|---|---|---|---|---|---|---|
| A (own zone) | 4000 A | 13.33 | 2.40 s | 0.624 s | – | – |
| B (back-up) | 3000 A | 10.0 | 2.80 s | 0.728 s | 0.22 s | 0.508 s |
| C (back-up) | 2000 A | 6.67 | 3.48 s | 0.906 s | 0.28 s | 0.626 s |
Interpolations used: PSM 13.33 → 2.8 − (3.33/5)(0.6) = 2.40 s; PSM 6.67 → 3.9 − (1.667/3)(0.75) = 3.483 s.
Settings
| Relay | CT | PS | TDS |
|---|---|---|---|
| A | 300/5 | 100% (5 A) | 0.26 |
| B | 200/5 | 100% (5 A) | 0.10 |
Answer: relay B: PS 100%, TDS 0.1; relay A: PS 100%, TDS 0.26. Relay A operates in 0.624 s for a fault at A in its own zone; as back-up it operates in 0.728 s for a fault at B and 0.906 s for a fault at C, always at least 0.5 s after relay B.
✓ Answer checked by Claude Opus 5.5 (high effort)
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