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A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated pf = 0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H, in sec) will be _____. (Answer up to two decimal places)
    Correct answer is '2.44'. Can you explain this answer?
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    Calculating the Inertia Constant (H) of a Synchronous Generator

    In order to calculate the inertia constant (H) of a synchronous generator, we can use the formula:
    \[ H = \frac{2 \times \text{moment of inertia}}{P_{\text{rated}} \times (\frac{2\pi}{\text{frequency}})^2} \]

    Given Data
    - Moment of inertia = 27.5 x 10^3 kg-m^2
    - Rated power (P) = 500 MW = 500 x 10^6 W
    - Rated voltage (V) = 21 kV = 21 x 10^3 V
    - Rated frequency = 50 Hz
    - Power factor (pf) = 0.9

    Calculations
    1. Convert rated power to watts:
    \[ P_{\text{rated}} = 500 \times 10^6 \text{ W} \]

    2. Calculate the synchronous speed:
    \[ \text{Synchronous speed (N)} = \frac{120 \times \text{frequency}}{\text{number of poles}} \]

    3. Calculate the angular velocity:
    \[ \omega = \frac{2\pi \times \text{frequency}}{60} \]

    4. Substitute the given values into the formula for H:
    \[ H = \frac{2 \times 27.5 \times 10^3}{500 \times 10^6 \times (\frac{2\pi}{50})^2} \]

    5. Solve for H:
    \[ H = \frac{55,000}{500,000,000 \times (\frac{2\pi}{50})^2} \]

    \[ H = \frac{55,000}{500,000,000 \times (\frac{2\pi}{50} \times \frac{2\pi}{50})} \]

    \[ H = \frac{55,000}{500,000,000 \times (\frac{4\pi^2}{2500})} \]

    \[ H = \frac{55,000}{500,000,000 \times \frac{4\pi^2}{2500}} \]

    \[ H = \frac{55,000 \times 2500}{500,000,000 \times 4\pi^2} \]

    \[ H = \frac{137,500,000}{2,000,000,000\pi^2} \]

    \[ H ≈ 2.44 \text{ seconds} \]

    Therefore, the inertia constant (H) of the synchronous generator is approximately 2.44 seconds.
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    A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated pf = 0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H, in sec) will be _____. (Answer up to two decimal places)Correct answer is '2.44'. Can you explain this answer?
    Question Description
    A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated pf = 0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H, in sec) will be _____. (Answer up to two decimal places)Correct answer is '2.44'. Can you explain this answer? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated pf = 0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H, in sec) will be _____. (Answer up to two decimal places)Correct answer is '2.44'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 500 MW, 21 kV, 50 Hz, 3-phase, 2-pole synchronous generator having a rated pf = 0.9, has a moment of inertia of 27.5 x 103 kg-m2. The inertia constant (H, in sec) will be _____. (Answer up to two decimal places)Correct answer is '2.44'. Can you explain this answer?.
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