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📘 Generator Protection – Numerical Problems (Detailed Solutions)

These numerical problems cover reverse power, negative sequence, differential protection and loss of excitation conditions.

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🔹 Problem 1: Reverse Power Setting

A 100 MW generator operates at full load. Reverse power relay setting = 5% of rated power. Find pickup power.

Solution

Pickup power = 0.05 × 100 MW = 5 MW

Relay trips if generator absorbs more than 5 MW.

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🔹 Problem 2: Negative Sequence Protection

A generator rated current = 1000 A. Unbalanced current (negative sequence current) = 200 A. Thermal constant K = 10 (I₂² × t = K). Find allowable time before tripping.

Solution

I₂² × t = 10 (200)² × t = 10 40000 × t = 10 t = 10 / 40000 t = 0.00025 seconds

Very short time allowed – immediate trip required.

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🔹 Problem 3: Generator Differential Protection

CT ratio = 1000/1 Primary current at stator terminal = 900 A Neutral side current = 850 A Pickup setting = 0.2 A Check relay operation.

Step 1: Convert to Secondary

Terminal side = 900 / 1000 = 0.9 A Neutral side = 850 / 1000 = 0.85 A

Step 2: Differential Current

I_diff = |0.9 − 0.85| = 0.05 A

Since 0.05 < 0.2 → Relay does NOT operate

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🔹 Problem 4: Loss of Excitation

Generator rated voltage = 11 kV Current = 500 A Power factor becomes leading (0.8 leading). Calculate reactive power absorbed.

Solution

Q = √3 × V × I × sinφ φ = cos⁻¹(0.8) = 36.87° sinφ = 0.6 Q = 1.732 × 11000 × 500 × 0.6 = 5.72 MVAR (approx)

Generator absorbs 5.72 MVAR (Loss of excitation condition)

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🎯 Key Learning Points

• Reverse power protects turbine
• Negative sequence protects rotor heating
• Differential protection detects internal stator faults
• Loss of excitation causes generator to absorb reactive power

Generator Protection = Electrical + Mechanical Safety