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IDMT relays across 132/33 kV transformer (2700 A)

Short answer

R1 operating time ≈ 0.37 s, R2 ≈ 0.97 s, TMS of R2 ≈ 0.255. From graph, R1 PSM=13.5 gives t=2.475 s at TMS=1; R2 PSM=6.75 gives t=3.81 s. Fault current on 132 kV side is 675 A.

It is given that fault current level at 33 kV side is 2700 A, CT ratio at 33 kV side is 200:1 and 132 kV side is 100:1 (figure below). If both the relays R1 and R2 are set for 100% plug setting, determine the operating time for both the relays when time grading margin of 0.6 second is given and TMS for relay R1 is 0.15. (PSM/Operation Time graph is given below.) [Figure: 132 kV bus – CB – CT 100/1 with relay R2 – 132/33 kV transformer – 33 kV bus – CB – CT 200/1 with relay R1 – fault, IF = 2700 A.] [Graph: operating time (seconds, 0–12) against PSM (0–20) for TMS = 1, a falling inverse curve; read approximately as PSM 2 → about 11.5 s, 4 → about 5.5 s, 6 → about 4 s, 10 → 3 s, 14 → about 2.4 s, 20 → about 2.1 s.]

R1 (33 kV) is the primary relay; R2 (132 kV) backs it up with a grading margin of 0.6 s.

R1: pick-up = 1.0 × 200 = 200 A
    PSM = 2700/200 = 13.5
    From graph (between 10→3 s, 14→2.4 s): t ≈ 2.475 s
    t_R1 = 0.15 × 2.475 = 0.371 s

R2: I(132 kV) = 2700 × 33/132 = 675 A
    pick-up = 1.0 × 100 = 100 A
    PSM = 675/100 = 6.75
    From graph (between 6→4 s, 10→3 s): t ≈ 3.81 s
    t_R2 = 0.371 + 0.6 = 0.971 s
    TMS_R2 = 0.971 / 3.81 = 0.255

Answer: t_R1 ≈ 0.37 s and t_R2 ≈ 0.97 s, with TMS of R2 ≈ 0.255. These values depend on how the graph is read, so they are approximate (±0.05 s).

✓ Answer checked by Claude Opus 5.5 (high effort)

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