- 2073 Magh · 8 marks
IDMT relays across 132/11 kV transformer (3500 A)
Short answer
R1 operating time ≈ 0.554 s, R2 ≈ 1.304 s, TMS of R2 ≈ 0.13. R1 PSM=14 gives t=2.214 s at TMS=1; R2 PSM≈2 gives t≈10 s. Transformer reduces fault current to 291.67 A on 132 kV side.
Two relays R1 and R2 are connected in two sections of a feeder shown in figure below. Fault current level at 11 kV side is 3500 A, CT ratio at 11 kV side is 200:1 and 132 kV side is 100:1. If R1 is set on 125% and R2 set on 150% plug setting, determine the operating time for both the relays when time grading margin of 0.75 second is given and TMS for relay R1 is 0.25. Make use of the following characteristics. [Figure: 132 kV – CB – CT 100/1 with relay R2 – 132/11 kV transformer – 11 kV bus – CB – CT 200/1 with relay R1 – fault, If = 3500 A.]
IDMT Characteristic:
| PSM | 2 | 4.6 | 5.33 | 6.73 | 8.2 | 10 | 12.3 | 13.6 | 15 |
|---|---|---|---|---|---|---|---|---|---|
| Operating time (sec) | 10 | 5 | 4.5 | 4 | 3.15 | 2.8 | 2.5 | 2.3 | 2 |
R1 on the 11 kV feeder is the primary relay and R2 on the 132 kV side is the back-up relay that must wait for the grading margin of 0.75 s.
Data: If = 3500 A at 11 kV; CT of R1 = 200/1, plug 125%; CT of R2 = 100/1, plug 150%; TMS of R1 = 0.25. Times between table points are found by linear interpolation.
Relay R1
Pick-up = 1.25 × 200 = 250 A
PSM = 3500 / 250 = 14
Between PSM 13.6 (2.3 s) and 15 (2.0 s):
t = 2.3 − (14 − 13.6)/(15 − 13.6) × (2.3 − 2.0)
= 2.3 − 0.2857 × 0.3 = 2.214 s (TMS = 1)
t_R1 = 0.25 × 2.214 = 0.554 s
Relay R2
I(132 kV) = 3500 × 11/132 = 291.67 A
Pick-up = 1.5 × 100 = 150 A
PSM = 291.67 / 150 = 1.94 ≈ 2
t (TMS=1) ≈ 10 s (start of the given curve)
Required t_R2 = 0.554 + 0.75 = 1.304 s
TMS_R2 = 1.304 / 10 = 0.13
Note: PSM of R2 is just below 2, so the relay works at the very steep start of its curve; the curve is taken as 10 s here. In practice the 150% setting is a little high for this fault level and a lower plug setting on R2 would give more reliable back-up.
| Quantity | R1 | R2 |
|---|---|---|
| Current seen | 3500 A | 291.67 A |
| Pick-up | 250 A | 150 A |
| PSM | 14 | 1.94 (≈ 2) |
| t at TMS = 1 | 2.214 s | ≈ 10 s |
| TMS | 0.25 | 0.13 |
| Operating time | 0.554 s | 1.304 s |
Answer: t_R1 ≈ 0.554 s, t_R2 ≈ 1.304 s (TMS of R2 ≈ 0.13).
✓ Answer checked by Claude Opus 5.5 (high effort)
Related questions
- IDMT relays across 132/33 kV transformer (2000 A) · Switchgear and Protection, chapter 7
- IDMT relays across 132/33 kV transformer (2700 A) · Switchgear and Protection, chapter 7
- IDMT grading: R1, R2 with 1000/5 CTs · Switchgear and Protection, chapter 7 · asked 2×
- CT ratio and connection for 11/132 kV · Switchgear and Protection, chapter 7
- IDMT settings for 20 MVA transformer and feeder · Switchgear and Protection, chapter 7 · asked 2×
- IDMT settings for sectionalised radial feeder A, B, C · Switchgear and Protection, chapter 7
- Design: 100 kVA 11/132 kV transformer with winding data · Electric Machine Design, chapter 3 · asked 3×
- Fault level at end of two 200 km lines · Power System Analysis II, chapter 3 · asked 2×
- Fault level: two 40 MVA generators, 2 Ω feeder · Power Plant Equipment, chapter 4 · asked 5×
- Fault current: two 50 MVA generators, feeder reactor · Power Plant Equipment, chapter 4 · asked 3×