- 2076 Bhadra · 6 marks
Operating time of two relays (250/1, 3000 A)
Short answer
Relay 1: pickup 312.5 A, PSM=3000/312.5=9.6, interpolated t(TMS=1)=2.852 s, t1=0.18×2.852=0.513 s. Relay 2: pickup 375 A, PSM=8, t(TMS=1)=3.191 s, required t2=0.513+0.4=0.913 s, TMS2=0.913/3.191=0.286.
Determine the time of operation of two relays assuming that both the relays have the characteristics as shown in the following table. The time multiplier setting of relay 1 is 0.18 and the time grading margin between the relays is 0.4 second. [Figure: CB – CT 250/1 with Relay 2 (plug setting = 150%) – CB – CT 250/1 with Relay 1 (plug setting = 125%) – fault, If = 3000 A.]
| PSM | 2 | 3.6 | 5 | 9.23 | 10 | 15 | 20 |
|---|---|---|---|---|---|---|---|
| Time in sec | 10 | 6 | 3.9 | 2.9 | 2.8 | 2.2 | 2.1 |
Relay 1 is nearer the fault, so it trips first; relay 2 must wait 0.4 s more. Times between table points are found by linear interpolation.
Relay 1
Pick-up = 1.25 × 250 = 312.5 A
PSM = 3000/312.5 = 9.6
Between 9.23 (2.9 s) and 10 (2.8 s):
t = 2.9 − (9.6 − 9.23)/(10 − 9.23) × 0.1 = 2.852 s
t1 = 0.18 × 2.852 = 0.513 s
Relay 2
Pick-up = 1.5 × 250 = 375 A
PSM = 3000/375 = 8
Between 5 (3.9 s) and 9.23 (2.9 s):
t = 3.9 − (8 − 5)/(9.23 − 5) × 1.0 = 3.191 s
t2 = t1 + 0.4 = 0.913 s
TMS2 = 0.913 / 3.191 = 0.286
| Relay | Pick-up | PSM | t at TMS=1 | TMS | Operating time |
|---|---|---|---|---|---|
| 1 | 312.5 A | 9.6 | 2.852 s | 0.18 | 0.513 s |
| 2 | 375 A | 8 | 3.191 s | 0.286 | 0.913 s |
Answer: t1 ≈ 0.513 s, t2 ≈ 0.913 s (TMS of relay 2 ≈ 0.286).
✓ Answer checked by Claude Opus 5.5 (high effort)
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