Question

In the system shown below, the generator was initially supplying power to the grid. A temporary LLLG bolted fault occurs at \( F \) very close to circuit breaker 1. The circuit breakers open to isolate the line. The fault self-clears. The circuit breakers reclose and restore the line. Which one of the following diagrams best indicates the rotor accelerating and decelerating areas? 

• A. Fig. (i)
• B. Fig. (ii)
• C. Fig. (iii)
• D. Fig. (iv)

Hint:

In a fault-disturbed synchronous machine, use the {Equal Area Criterion} to assess transient stability: - Area between mechanical and electrical power curves during fault = {accelerating area} - Area after fault clearance = {decelerating area} The system remains stable if accelerating area = decelerating area.

Answer 1

The Correct Option is B

This is a classic Equal Area Criterion problem from power system stability.

- During the fault, electrical power output drops significantly, leading to rotor acceleration.
- After the fault is cleared and breakers reclose, the electrical power recovers, causing rotor deceleration.
- The accelerating area \( A_1 \) is the region between mechanical input power and the reduced electrical power curve during the fault.
- The decelerating area \( A_2 \) is between mechanical input and post-fault electrical power.

Equal Area Criterion: For system stability, the area under the acceleration (before fault clearance) and deceleration (after clearance) curves must be equal.

Among the diagrams:

- Fig. (ii) shows proper demarcation of accelerating and decelerating areas with correct rotor angle progression up to \( \delta_{\max} \), consistent with system dynamics and power-angle characteristics.

\[ \boxed{\text{Correct match: Fig. (ii)}} \]