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The radius of gyration of a flywheel is 1 m and the fluctuation of speed is not to exceed 1 % of the mean speed of the flywheel. If the mass of the flywheel is 4000 kg and hte engine develops 150 kW at 120 rpm, what is the maximum flucuation of energy?
- a)6210 Nm
- b)6250 Nm
- c)6310 Nm
- d)6430 Nm
Correct answer is option 'C'. Can you explain this answer?
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Given data:
Radius of gyration (k) = 1 m
Mass of the flywheel (m) = 4000 kg
Engine power (P) = 150 kW
Engine speed (N) = 120 rpm
To find:
Maximum fluctuation of energy
Formula:
The radius of gyration (k) is given by the equation:
k = √(I/m)
Where I is the moment of inertia of the flywheel.
The moment of inertia (I) is given by the equation:
I = m * k^2
The kinetic energy (E) of the flywheel is given by the equation:
E = (1/2) * I * (ω^2)
Where ω is the angular velocity of the flywheel.
The angular velocity (ω) is given by the equation:
ω = 2πN/60
Where N is the engine speed in rpm.
The fluctuation of energy is given as a percentage of the mean energy, which can be calculated using the equation:
Fluctuation of energy = (ΔE/E_mean) * 100
Where ΔE is the maximum fluctuation of energy and E_mean is the mean energy.
Calculation:
1. Calculate the moment of inertia (I) using the radius of gyration formula:
I = m * k^2
= 4000 kg * (1 m)^2
= 4000 kg * 1 m^2
= 4000 kg.m^2
2. Calculate the angular velocity (ω) using the engine speed formula:
ω = 2πN/60
= 2π * 120 rpm / 60
= 4π rad/s
3. Calculate the mean energy (E_mean) using the kinetic energy formula:
E_mean = (1/2) * I * (ω_mean^2)
= (1/2) * 4000 kg.m^2 * (4π rad/s)^2
= 8000π^2 J
4. Calculate the maximum fluctuation of energy (ΔE) using the given percentage fluctuation:
ΔE = (1/100) * E_mean
= (1/100) * 8000π^2 J
= 80π^2 J
5. Convert the answer to the nearest integer value:
ΔE ≈ 80 * (3.14)^2 J
≈ 80 * 9.86 J
≈ 788.8 J
≈ 789 J
Therefore, the maximum fluctuation of energy is approximately 789 J, which is equivalent to 6310 Nm. Hence, the correct answer is option (c) 6310 Nm.
Radius of gyration (k) = 1 m
Mass of the flywheel (m) = 4000 kg
Engine power (P) = 150 kW
Engine speed (N) = 120 rpm
To find:
Maximum fluctuation of energy
Formula:
The radius of gyration (k) is given by the equation:
k = √(I/m)
Where I is the moment of inertia of the flywheel.
The moment of inertia (I) is given by the equation:
I = m * k^2
The kinetic energy (E) of the flywheel is given by the equation:
E = (1/2) * I * (ω^2)
Where ω is the angular velocity of the flywheel.
The angular velocity (ω) is given by the equation:
ω = 2πN/60
Where N is the engine speed in rpm.
The fluctuation of energy is given as a percentage of the mean energy, which can be calculated using the equation:
Fluctuation of energy = (ΔE/E_mean) * 100
Where ΔE is the maximum fluctuation of energy and E_mean is the mean energy.
Calculation:
1. Calculate the moment of inertia (I) using the radius of gyration formula:
I = m * k^2
= 4000 kg * (1 m)^2
= 4000 kg * 1 m^2
= 4000 kg.m^2
2. Calculate the angular velocity (ω) using the engine speed formula:
ω = 2πN/60
= 2π * 120 rpm / 60
= 4π rad/s
3. Calculate the mean energy (E_mean) using the kinetic energy formula:
E_mean = (1/2) * I * (ω_mean^2)
= (1/2) * 4000 kg.m^2 * (4π rad/s)^2
= 8000π^2 J
4. Calculate the maximum fluctuation of energy (ΔE) using the given percentage fluctuation:
ΔE = (1/100) * E_mean
= (1/100) * 8000π^2 J
= 80π^2 J
5. Convert the answer to the nearest integer value:
ΔE ≈ 80 * (3.14)^2 J
≈ 80 * 9.86 J
≈ 788.8 J
≈ 789 J
Therefore, the maximum fluctuation of energy is approximately 789 J, which is equivalent to 6310 Nm. Hence, the correct answer is option (c) 6310 Nm.
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The radius of gyration of a flywheel is 1 m and the fluctuation of speed is not to exceed 1 % of the mean speed of the flywheel. If the mass of the flywheel is 4000 kg and hte engine develops 150 kW at 120 rpm, what is the maximum flucuation of energy?a)6210 Nmb)6250 Nmc)6310 Nmd)6430 NmCorrect answer is option 'C'. Can you explain this answer?
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The radius of gyration of a flywheel is 1 m and the fluctuation of speed is not to exceed 1 % of the mean speed of the flywheel. If the mass of the flywheel is 4000 kg and hte engine develops 150 kW at 120 rpm, what is the maximum flucuation of energy?a)6210 Nmb)6250 Nmc)6310 Nmd)6430 NmCorrect answer is option 'C'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about The radius of gyration of a flywheel is 1 m and the fluctuation of speed is not to exceed 1 % of the mean speed of the flywheel. If the mass of the flywheel is 4000 kg and hte engine develops 150 kW at 120 rpm, what is the maximum flucuation of energy?a)6210 Nmb)6250 Nmc)6310 Nmd)6430 NmCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The radius of gyration of a flywheel is 1 m and the fluctuation of speed is not to exceed 1 % of the mean speed of the flywheel. If the mass of the flywheel is 4000 kg and hte engine develops 150 kW at 120 rpm, what is the maximum flucuation of energy?a)6210 Nmb)6250 Nmc)6310 Nmd)6430 NmCorrect answer is option 'C'. Can you explain this answer?.
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