- Asked 2/16
- 2072 Asoj · 8 marks
- 2079 Chaitra · 8 marks
IDMT settings for 20 MVA transformer and feeder
Short answer
Feeder: 62.5 A, PSM=10, time=0.3×3=0.9 s. Transformer: 130% overload gives 6.82 A secondary, so PS=150% (7.5 A). Fault current 25 A, PSM=3.33, time at TMS=1 is 5.68 s. Required 1.4 s, TMS=0.25. Feeder time 0.9 s, transformer PS=150%, TMS=0.25.
A 20 MVA transformer which is used to operate at 30% overload feeds an 11 kV busbar through a circuit breaker. The transformer circuit breaker is equipped with a 1000/5 CT and the feeder circuit breaker with a 400/5 CT, and both the CTs feed IDMT relays having following characteristics:
| Plug Setting Multiplier | 2 | 3 | 5 | 10 | 15 | 20 |
|---|---|---|---|---|---|---|
| Time in Second | 10 | 6 | 4.1 | 3 | 2.5 | 2.2 |
The relay on the feeder CB has 125% plug setting and a 0.3 time multiplier setting. If a fault current of 5000 A flows from the transformer to the feeder, determine: i) Operating time of feeder relay ii) Suggest suitable plug setting and time multiplier setting of the transformer relay to ensure adequate discrimination of 0.5 s between transformer and feeder relay. (The 2072 Asoj paper omits the 400/5 feeder CT ratio.)
The feeder relay time is found from its PSM; the transformer relay plug setting must be above the transformer overload current, and its TMS gives 0.5 s more time than the feeder relay for the same fault.
Data: 20 MVA, 11 kV, 30% overload; transformer CT 1000/5; feeder CT 400/5; feeder relay PS 125%, TMS 0.3; fault current 5000 A; margin 0.5 s; relay rated current 5 A.
i) Operating time of feeder relay
CT ratio = 400/5 = 80
Relay current = 5000 / 80 = 62.5 A
Pick-up = 125% × 5 = 6.25 A
PSM = 62.5 / 6.25 = 10
Time at TMS 1 (table, PSM 10) = 3 s
Feeder relay time = 0.3 × 3 = 0.9 s
ii) Settings of transformer relay
Plug setting: the relay must not trip on the allowed 30% overload.
Full-load current = 20 × 10⁶ / (√3 × 11 × 10³) = 1049.73 A
Max load (130%) = 1.3 × 1049.73 = 1364.65 A
CT ratio = 1000/5 = 200
Secondary current = 1364.65 / 200 = 6.82 A
Required PS > 6.82 / 5 = 136.5%
Choose the next standard plug setting: PS = 150% (pick-up 7.5 A secondary = 1500 A primary).
Time multiplier setting:
Relay current = 5000 / 200 = 25 A
PSM = 25 / 7.5 = 3.333
From table (linear interpolation between
PSM 3 → 6 s and PSM 5 → 4.1 s):
t = 6 − (3.333 − 3)/(5 − 3) × (6 − 4.1)
= 6 − 0.1667 × 1.9
= 5.683 s (at TMS = 1)
Required time = 0.9 + 0.5 = 1.4 s
TMS = 1.4 / 5.683 = 0.2463
Choose TMS = 0.25 (next practical setting). Then actual time = 0.25 × 5.683 = 1.42 s, giving a margin of 0.52 s ≥ 0.5 s.
Results
| Relay | CT | PS | PSM | t (TMS 1) | TMS | Time |
|---|---|---|---|---|---|---|
| Feeder | 400/5 | 125% | 10 | 3 s | 0.3 | 0.9 s |
| Transformer | 1000/5 | 150% | 3.33 | 5.68 s | 0.25 | 1.42 s |
Answer: feeder relay time = 0.9 s; transformer relay PS = 150% and TMS ≈ 0.25 (calculated 0.246), operating in about 1.42 s. (If the feeder CT ratio is not given, as in the 2072 paper, 400/5 is assumed.)
✓ Answer checked by Claude Opus 5.5 (high effort)
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