Question

The single line diagram of a lossless system is shown in the figure. The system is operating in steady-state at a stable equilibrium point with the power output of the generator being \( P_{max} \sin \delta \), where \( \delta \) is the load angle and the mechanical power input is \( 0.5 P_{max} \). A fault occurs on line 2 such that the power output of the generator is less than \( 0.5 P_{max} \) during the fault. After the fault is cleared by opening line 2, the power output of the generator is \( \frac{P_{max}}{\sqrt{2}} \sin \delta \). If the critical fault clearing angle is \( \pi/2 \) radians, the accelerating area on the power angle curve is \_\_\_ times \( P_{max} \) (rounded off to 2 decimal places).
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Hint:

For fault analysis, integrate the power difference over the fault clearing angle to calculate the accelerating area.

Answer 1

Step 1: Define the accelerating area. The accelerating area is the integral of the difference between mechanical power and electrical power over the angle \( \delta \). For \( \delta \) from \( 0 \) to \( \pi/2 \): \[ A_{acc} = \int_{0}^{\pi/2} \big(P_m - P_e\big) \, d\delta. \] Step 2: Substitute the given powers. During the fault: \[ P_m = 0.5 P_{max}, \quad P_e = \frac{P_{max}}{\sqrt{2}} \sin \delta. \] Simplify: \[ A_{acc} = 0.12 \, P_{max}. \]