JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech III POWER SYSTEM PROTECTION Previous Question Paper (August/September - 2021 - R18)

Static relays offer several advantages over electromagnetic relays due to the absence of moving parts and the use of solid-state components:

• High Sensitivity and Precision: Static relays provide higher accuracy and sensitivity, allowing for faster response times and lower burden on instrument transformers.
• Low Power Consumption: They consume significantly less power, reducing the load on current and potential transformers, which improves overall system accuracy.
• Fast Operation and Reset: Due to the lack of mechanical inertia, static relays exhibit very fast operating and reset times, which is critical for clearing faults rapidly.
• Longer Life and Low Maintenance: The absence of mechanical contacts eliminates issues like contact wear, pitting, and friction, leading to longer service life and reduced maintenance requirements.
• Immunity to Mechanical Shocks: These relays are highly resistant to vibrations and mechanical shocks, making them suitable for environments where electromagnetic relays might malfunction.
• Compact Size: Solid-state components allow for a more compact design, facilitating easier installation and panel space optimization.

To determine the operating time, we first calculate the relay settings and the Plug Setting Multiplier (PSM):

1. Pickup Current: Pickup current = Rated current × Plug setting = 5A × 1.5 = 7.5A.
2. Secondary Fault Current: Secondary current = (Fault current / CT ratio) = 6000A / (400/5) = 6000A / 80 = 75A.
3. PSM Calculation: PSM = Secondary fault current / Pickup current = 75A / 7.5A = 10.
4. Operating Time (TMS=1): From the provided table, for a PSM of 10, the operating time at TMS=1 is 2.8 seconds.
5. Final Operating Time: Operating time = TMS × Time at TMS=1 = 0.4 × 2.8 seconds = 1.12 seconds.

The operating time of the relay for a fault current of 6000A is 1.12 seconds.