Sunday, 22 February 2026

📘 Advanced Multi-Source Relay Coordination – Hard Numerical

Two generators feed a common bus through two feeders. Overcurrent relays are installed at both feeders and at outgoing line. Design proper relay coordination.

🔹 Given Data

Fault at end of outgoing feeder = 5000 A
Generator 1 contribution = 3000 A
Generator 2 contribution = 2000 A
CT ratio = 600/5
Plug setting = 100%
Coordination Time Interval (CTI) = 0.3 sec
IEC Standard Inverse Curve

T = TMS × (0.14 / (PSM^0.02 − 1))

🔹 Step 1: Pickup Current

Secondary pickup = 5 A Primary pickup = 5 × (600/5) = 600 A

Pickup Current = 600 A

🔹 Step 2: Outgoing Feeder Relay (R1)

Total fault current = 5000 A

PSM₁ = 5000 / 600 = 8.33

Calculate: 8.33^0.02 ≈ 1.044

Constant = 0.14 / (1.044 − 1) = 0.14 / 0.044 ≈ 3.18

Assume fast clearing time = 0.4 sec

0.4 = 3.18 × TMS₁ TMS₁ = 0.4 / 3.18 = 0.126

TMS₁ ≈ 0.13

🔹 Step 3: Generator Relays R2 & R3

Generator 1 Fault Current = 3000 A

PSM₂ = 3000 / 600 = 5

5^0.02 ≈ 1.032

Constant = 0.14 / 0.032 ≈ 4.375

R2 must operate slower than R1 by CTI: Required time: 0.4 + 0.3 = 0.7 sec

0.7 = 4.375 × TMS₂ TMS₂ = 0.7 / 4.375 = 0.16

TMS₂ ≈ 0.16

Generator 2 Fault Current = 2000 A

PSM₃ = 2000 / 600 = 3.33

3.33^0.02 ≈ 1.025

Constant = 0.14 / 0.025 = 5.6

Required time = 0.7 sec

0.7 = 5.6 × TMS₃ TMS₃ = 0.7 / 5.6 = 0.125

TMS₃ ≈ 0.13

📊 Final Relay Settings

Relay	Fault Contribution	TMS	Operating Time
R1 (Feeder)	5000 A	0.13	0.4 sec
R2 (Gen 1)	3000 A	0.16	0.7 sec
R3 (Gen 2)	2000 A	0.13	0.7 sec

🎯 Important Concepts

Multi-source systems require contribution-based coordination
Each relay sees different fault current
Coordination interval must be maintained
PSM differs for each source

Multi-Source Protection = Intelligent Time Grading

Shaktimatha 369 Technologies

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