- 2077 Chaitra · 6 marks
Minimum TMS of relay A (16000 A fault)
Short answer
Relay B: PSM=16000/(1.25×800)=16, t=0.2×2=0.4 s. Required t_A includes breaker time 0.2 s and errors ±0.15 s each: t_A=0.4+0.2+0.15+0.15=0.9 s. Relay A pickup 1600 A, PSM=10, t(TMS=1)=3 s, so minimum TMS_A=0.9/3=0.3.
As shown in figure, the relay at B has a plug setting of 125% and TMS of 0.2. The circuit breakers take 0.20 sec to clear the fault and the relay error in each case is 0.15 sec. For a plug setting of 200% on the relay A, determine the minimum TMS on that relay for it not to operate before the circuit breaker at B has cleared the fault. A relay operating time curve is shown in table. [Figure: Bus A – CB – CT 800/5 with relay R2 – bus B – CB – CT 800/5 with relay R1 – fault F, If = 16000 A.]
IDMT characteristic:
| PSM | 2 | 3.5 | 5 | 6.5 | 8 | 10 | 12 | 14 | 16 |
|---|---|---|---|---|---|---|---|---|---|
| Operating Time (sec) | 10 | 6 | 5 | 4 | 3.15 | 3 | 2.9 | 2.5 | 2 |
Relay A is the back-up for relay B. It must not trip until relay B has operated and breaker B has cleared the fault, allowing for errors in both relays.
Data: CT 800/5 for both; relay B: PS 125%, TMS 0.2; relay A: PS 200%; If = 16000 A; CB time 0.2 s; relay error 0.15 s each.
Relay B
Pick-up = 1.25 × 800 = 1000 A
PSM = 16000/1000 = 16 → t(TMS=1) = 2 s
t_B = 0.2 × 2 = 0.4 s
Required time of relay A
Relay B may be 0.15 s slow and relay A may be 0.15 s fast, so:
t_A = t_B + CB time + error(B) + error(A)
= 0.4 + 0.2 + 0.15 + 0.15 = 0.9 s
TMS of relay A
Pick-up = 2.0 × 800 = 1600 A
PSM = 16000/1600 = 10 → t(TMS=1) = 3 s
TMS_A = 0.9 / 3 = 0.3
Answer: minimum TMS of relay A = 0.3 (operating time 0.9 s). Any smaller TMS could let relay A trip before breaker B has cleared the fault.
✓ Answer checked by Claude Opus 5.5 (high effort)
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