📘 Hard Relay Coordination Problem – Step by Step Solution
Three overcurrent relays R1 (downstream), R2 (middle), R3 (upstream) are protecting a radial feeder. Determine proper TMS settings to maintain coordination.
🔹 Given Data
- Fault current at end of feeder = 3000 A
- CT ratio for all relays = 400/5
- Plug setting = 100%
- Coordination time interval (CTI) = 0.3 sec
- Use IEC Standard Inverse characteristic
Operating Time T = TMS × (0.14 / (PSM0.02 − 1))
🔹 Step 1: Pickup Current
Secondary pickup = 5 A (100% setting) Primary pickup = 5 × (400/5) = 400 APickup Current = 400 A
🔹 Step 2: Calculate PSM
PSM = Fault Current / Pickup Current = 3000 / 400 = 7.5
Calculate: 7.50.02 ≈ 1.0410.14 / (1.041 − 1) = 0.14 / 0.041 ≈ 3.41
Thus: T = 3.41 × TMS🔹 Step 3: Downstream Relay R1
Assume R1 operating time = 0.5 sec (fast clearing) So: 0.5 = 3.41 × TMS₁TMS₁ = 0.5 / 3.41 = 0.147
TMS₁ ≈ 0.15
🔹 Step 4: Relay R2 Coordination
R2 must operate slower than R1 by CTI. Required time: 0.5 + 0.3 = 0.8 sec So: 0.8 = 3.41 × TMS₂TMS₂ = 0.8 / 3.41 = 0.235
TMS₂ ≈ 0.24
🔹 Step 5: Relay R3 Coordination
R3 must operate slower than R2 by CTI. Required time: 0.8 + 0.3 = 1.1 sec So: 1.1 = 3.41 × TMS₃TMS₃ = 1.1 / 3.41 = 0.322
TMS₃ ≈ 0.32
📊 Final Relay Settings
| Relay | Operating Time (sec) | TMS |
|---|---|---|
| R1 (Downstream) | 0.5 | 0.15 |
| R2 (Middle) | 0.8 | 0.24 |
| R3 (Upstream) | 1.1 | 0.32 |
🎯 Key Concepts
- Downstream relay operates fastest
- Upstream relay delayed by coordination time interval
- Proper TMS ensures selective tripping
- CT ratio affects pickup calculation
Selective Coordination = Only Faulty Section Isolated
.png)
No comments:
Post a Comment