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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...
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.
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.
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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?.
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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