Skip to main content

Relay 1 time and relay 2 TMS (3000 A)

Short answer

For relay 1: pickup 1.25×300=375 A, PSM=3000/375=8, t at TMS=1=3.15 s, operating time=0.4×3.15=1.26 s. For relay 2: pickup 1.5×300=450 A, PSM=6.67, t at TMS=1=3.5 s, required time=1.26+0.5=1.76 s, TMS=1.76/3.5=0.503.

Calculate the time of operation of relay 1 and time setting multiplier for relay 2 for the conditions given below: (i) Time grading margin between the relays is 0.5 sec (ii) The time setting multiplier of relay 1 is 0.4 (iii) CT ratio of both CTs is 300/1 (iv) The fault occurs at point F and fault current is 3000 A (v) Time current characteristic of both relays is given in table. [Figure: CB – CT with Relay 2 (plug setting = 150%) – CB – CT with Relay 1 (plug setting = 125%) – fault point F.]

PSM23.656.678101520
Time in sec1063.93.53.152.82.22.1

Relay 1 (near fault F) is the primary relay; relay 2 backs it up with a 0.5 s margin.

Data: CT 300/1 for both, plug setting R1 = 125%, R2 = 150%, TMS of R1 = 0.4, If = 3000 A.

Relay 1

Pick-up = 1.25 × 300 = 375 A
PSM = 3000/375 = 8 → t(TMS=1) = 3.15 s
t1 = 0.4 × 3.15 = 1.26 s

Relay 2

Pick-up = 1.5 × 300 = 450 A
PSM = 3000/450 = 6.67 → t(TMS=1) = 3.5 s
Required t2 = 1.26 + 0.5 = 1.76 s
TMS2 = 1.76 / 3.5 = 0.503
RelayPick-upPSMt at TMS=1TMSTime
1375 A83.15 s0.41.26 s
2450 A6.673.5 s0.5031.76 s

Answer: operating time of relay 1 = 1.26 s; TMS of relay 2 ≈ 0.503 (≈ 0.5), giving t2 = 1.76 s.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Switchgear and Protection chapter 7 questions →