📘 SF₆ Circuit Breaker – Numerical Problems with Detailed Solutions
These problems focus on breaking capacity, making capacity, short circuit MVA, and breaker selection in high voltage systems using SF₆ circuit breakers.
🔹 Problem 1: Breaking & Making Capacity
An SF₆ circuit breaker is installed in a 220 kV system. The symmetrical short circuit current at the bus is 40 kA. Calculate:
- 1) Breaking Capacity (MVA)
- 2) Making Capacity (kA)
Step 1: Breaking Capacity
Breaking Capacity = √3 × V × Isc
= 1.732 × 220 × 40
= 15241.6 MVA
Breaking Capacity ≈ 15242 MVA
Step 2: Making Capacity
Making Capacity = 2.55 × Isc
= 2.55 × 40 = 102 kA
Making Capacity = 102 kA
🔹 Problem 2: Short Circuit MVA Method
A 400 kV system has Thevenin impedance of 0.15 pu on 100 MVA base. Find:
- 1) Short Circuit MVA
- 2) Fault Current
Step 1: Short Circuit MVA
SC MVA = Base MVA / Zth
= 100 / 0.15 = 666.7 MVA
Short Circuit MVA = 666.7 MVA
Step 2: Fault Current
Isc = MVA / (√3 × V)
= 666.7 / (1.732 × 400)
= 666.7 / 692.8 = 0.96 kA
Fault Current ≈ 0.96 kA
🔹 Problem 3: Breaker Selection
If calculated fault current is 35 kA at 132 kV bus, which breaker rating should be selected?
Standard ratings available: 31.5 kA, 40 kA, 50 kA
Breaker rating must be greater than calculated fault current.
Select 40 kA breaker.
🎯 Exam Tips
- SF₆ breakers used in EHV systems
- Making capacity ≈ 2.55 × RMS breaking current
- Breaker rating must exceed system fault level
- Higher system MVA → stronger breaker required
Correct Breaker Selection = Reliable High Voltage Protection
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