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A belt-driven DC shunt generator running at 300 RPM delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 Q, field resistance is 50 Q, and brush drop is 2 V. Ignoring armature reaction, the speed of the motor i s ______ RPM. (Round off 2 decimal places)
    Correct answer is '275.18 (274.00 to 276.00)'. Can you explain this answer?
    Most Upvoted Answer
    A belt-driven DC shunt generator running at 300 RPM delivers 100 kW to...
    Solution:

    Given,

    Generator:

    Speed of the generator, N = 300 rpm

    Power generated, P = 100 kW

    Voltage of the grid, V = 200 V

    Motor:

    Power taken from the grid, P = 10 kW

    Armature resistance, Ra = 0.025 Ω

    Field resistance, Rf = 50 Ω

    Brush drop, Vb = 2 V

    To find: Speed of the motor, N.

    Step 1: Calculate the generated emf of the generator.

    From the formula of DC shunt generator,

    Eg = V + IaRa + Vb

    Where,

    Ia = (P/Eg)

    We know,

    P = EgIa

    Substitute the values,

    100000 = Eg(Ia)

    Ia = 500 A

    Substitute the values of V, Ra, Ia, Vb,

    Eg = V + IaRa + Vb

    Eg = 200 + (500 x 0.025) + 2

    Eg = 212.5 V

    Step 2: Calculate the speed of the motor.

    When the belt breaks, the generator continues to run as a motor taking 10 kW from the grid.

    Power input to the motor, P = 10 kW

    We know,

    P = EgIa - Ia^2Ra

    Substitute the values of Eg, Ra, P,

    10000 = 212.5Ia - Ia^2 x 0.025

    Ia^2 - 8500Ia + 400000 = 0

    Solving the quadratic equation,

    Ia = 46.01 A or Ia = 184.99 A

    Since the motor is a shunt motor, field current remains constant.

    Field current, If = Eg/Rf

    Substitute the values,

    If = 212.5/50

    If = 4.25 A

    At full-load,

    Ia = 184.99 A

    Total current, I = Ia + If

    I = 189.24 A

    Back emf, Eb = Eg - IaRa - Vb

    Eb = 212.5 - (184.99 x 0.025) - 2

    Eb = 207.13 V

    From the formula of DC shunt motor,

    N = ((Eb / Φ) - Ra) / ((1.5 x If) + Ia)

    Where,

    Φ = flux per pole

    Number of poles = 4

    We know,

    Φ = (Eg / N) x (60 / 2π) / 4

    Substitute the values of Eg and N,

    Φ = (212.5 / 300) x (60 / 2π) / 4

    Φ = 0.212 Wb

    Substitute the values in the formula of speed,

    N = ((207.13 / 0.212) - 0.025) / ((1.5 x 4.25) + 184.99)

    N = 275.18 rpm

    Therefore, the speed of the motor is 275.18 rpm.
    Free Test
    Community Answer
    A belt-driven DC shunt generator running at 300 RPM delivers 100 kW to...
    Ia = 504 A
    Eq =  V + Ia Ra + Brush drop
    = 200 + 504 (0.025) + 2V
    = 214.6 V
    In motoring case : VI = 10kW, V = 200 V
    If = 4A, Ia = IL - If
    = 46 A
    Eb = V - IaRa - Brush drop
    = 200 - 46 (0.025) - 2 =196.85 V
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    A belt-driven DC shunt generator running at 300 RPM delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 Q, field resistance is 50 Q, and brush drop is 2 V. Ignoring armature reaction, the speed of the motor i s ______ RPM. (Round off 2 decimal places)Correct answer is '275.18 (274.00 to 276.00)'. Can you explain this answer?
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    A belt-driven DC shunt generator running at 300 RPM delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 Q, field resistance is 50 Q, and brush drop is 2 V. Ignoring armature reaction, the speed of the motor i s ______ RPM. (Round off 2 decimal places)Correct answer is '275.18 (274.00 to 276.00)'. 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 belt-driven DC shunt generator running at 300 RPM delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 Q, field resistance is 50 Q, and brush drop is 2 V. Ignoring armature reaction, the speed of the motor i s ______ RPM. (Round off 2 decimal places)Correct answer is '275.18 (274.00 to 276.00)'. 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 belt-driven DC shunt generator running at 300 RPM delivers 100 kW to a 200 V DC grid. It continues to run as a motor when the belt breaks, taking 10 kW from the DC grid. The armature resistance is 0.025 Q, field resistance is 50 Q, and brush drop is 2 V. Ignoring armature reaction, the speed of the motor i s ______ RPM. (Round off 2 decimal places)Correct answer is '275.18 (274.00 to 276.00)'. Can you explain this answer?.
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