📘 Busbar Protection – Numerical Problems (High Impedance Scheme)
These problems explain differential current calculation, stability check and stabilizing resistor selection in busbar protection.
🔹 Problem 1: Differential Current Calculation
A busbar has 3 feeders connected. Secondary CT currents are:
- I₁ = 2.5 A (incoming)
- I₂ = 1.8 A (outgoing)
- I₃ = 0.7 A (outgoing)
Solution
Sum of outgoing = 1.8 + 0.7 = 2.5 A Incoming = 2.5 A Differential Current = 2.5 − 2.5 = 0 A
Relay does NOT operate (External fault or normal condition).
🔹 Problem 2: Internal Fault Case
CT secondary currents:
- I₁ = 3 A
- I₂ = 1 A
- I₃ = 0.5 A
Solution
Sum outgoing = 1 + 0.5 = 1.5 A Incoming = 3 A I_diff = 3 − 1.5 = 1.5 A
If pickup = 0.5 A:Since 1.5 > 0.5 → Relay Operates.
🔹 Problem 3: Stabilizing Resistor Selection
Maximum external fault current = 5000 A CT ratio = 1000/1 CT secondary resistance = 1 Ω Relay pickup voltage = 100 V Find required stabilizing resistor.
Step 1: Convert Fault Current to Secondary
I_secondary = 5000 / 1000 = 5 A
Step 2: Calculate Total Required Resistance
V = I × R R_total = V / I = 100 / 5 = 20 Ω
Step 3: Stabilizing Resistor
R_stab = R_total − CT resistance = 20 − 1 = 19 Ω
Required Stabilizing Resistor ≈ 19 Ω
🔹 Problem 4: Percentage Differential Busbar
I₁ = 4 A I₂ = 3 A Slope setting = 40% Check operation.
Solution
I_diff = |4 − 3| = 1 A I_avg = (4 + 3)/2 = 3.5 A Bias limit = 0.4 × 3.5 = 1.4 A
Since 1 < 1.4 → Relay does NOT operate.
🎯 Key Learning Points
- Busbar differential protection is instantaneous
- Stabilizing resistor prevents mal-operation
- Internal faults create high differential current
- External faults should not trip relay
Busbar Protection = Fastest Substation Protection
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