- 2078 Chaitra · 8 marks
Operating time of two relays (300/1, 6000 A)
Short answer
Relay 1: pickup 375 A, PSM=6000/375=16, t(TMS=1)=2.2 s, t1=0.2×2.2=0.44 s. Relay 2: pickup 450 A, PSM=6000/450=13.33, t(TMS=1)=2.8 s, required t2=0.44+0.3=0.74 s, TMS2=0.74/2.8=0.264.
The time multiplier setting of relay 1 is 0.2 and the time grading margin between the relays is 0.3 second. Determine the time of operation of two relays. [Figure: CB – CT 300/1 with Relay 2 (plug setting = 150%) – CB – CT 300/1 with Relay 1 (plug setting = 125%) – fault, IF = 6000 A.]
| PSM | 2 | 3.6 | 5 | 9.23 | 13.33 | 16 | 20 |
|---|---|---|---|---|---|---|---|
| Time in sec | 10 | 6 | 3.9 | 2.9 | 2.8 | 2.2 | 2.1 |
Relay 1 (nearest the fault) operates first; relay 2 is the back-up and must operate 0.3 s after relay 1.
Data: CT 300/1 for both, plug setting R1 = 125%, R2 = 150%, TMS of R1 = 0.2, IF = 6000 A.
Relay 1
Pick-up current = 1.25 × 300 = 375 A
PSM = 6000/375 = 16
t(TMS=1) = 2.2 s (from table)
t1 = 0.2 × 2.2 = 0.44 s
Relay 2
Pick-up current = 1.5 × 300 = 450 A
PSM = 6000/450 = 13.33
t(TMS=1) = 2.8 s (from table)
t2 = t1 + margin = 0.44 + 0.3 = 0.74 s
TMS2 = 0.74 / 2.8 = 0.264
| Relay | Pick-up | PSM | t at TMS=1 | TMS | Operating time |
|---|---|---|---|---|---|
| 1 | 375 A | 16 | 2.2 s | 0.2 | 0.44 s |
| 2 | 450 A | 13.33 | 2.8 s | 0.264 | 0.74 s |
Answer: relay 1 operates in 0.44 s and relay 2 in 0.74 s (TMS of relay 2 ≈ 0.264).
✓ Answer checked by Claude Opus 5.5 (high effort)
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