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Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg. m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is______________?
  • a)
    595
  • b)
    350
  • c)
    656
  • d)
    136
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Maximum fluctuation of kinetic energy in an engine has been calculate...
∆E = Iω2Cs
⇒ 2600 =
⇒ I = 595
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Most Upvoted Answer
Maximum fluctuation of kinetic energy in an engine has been calculate...
Given data:
Maximum fluctuation of kinetic energy (ΔKE) = 2600 J
Average speed (ωavg) = 200 rpm

To keep the speed fluctuation within ±0.5% of the average speed, we need to calculate the polar mass moment of inertia (I) of the flywheel.

Step 1: Calculate the average kinetic energy (KEavg) of the flywheel.
The average kinetic energy can be calculated using the formula:
KEavg = (1/2) I ωavg^2

Step 2: Calculate the maximum kinetic energy (KEmax) of the flywheel.
The maximum kinetic energy can be calculated by adding the maximum fluctuation (ΔKE) to the average kinetic energy (KEavg):
KEmax = KEavg + ΔKE

Step 3: Calculate the maximum speed (ωmax) of the flywheel.
The maximum speed can be calculated using the formula:
ωmax = ωavg + (0.5% of ωavg)
= ωavg + (0.005 ωavg)
= 1.005 ωavg

Step 4: Calculate the maximum kinetic energy (KEmax) at the maximum speed (ωmax).
The maximum kinetic energy at the maximum speed can be calculated using the formula:
KEmax = (1/2) I ωmax^2

Step 5: Equate the maximum kinetic energy obtained from Step 2 and Step 4.
(1/2) I ωavg^2 + ΔKE = (1/2) I ωmax^2

Step 6: Substitute the value of ωmax from Step 3 in Step 5.
(1/2) I ωavg^2 + ΔKE = (1/2) I (1.005 ωavg)^2

Simplifying the equation:
I ωavg^2 + 2ΔKE = I (1.010025 ωavg^2)
I (1.010025 ωavg^2 - ωavg^2) = 2ΔKE

Step 7: Rearrange the equation to solve for I.
I = 2ΔKE / (1.010025 ωavg^2 - ωavg^2)

Step 8: Substitute the given values into the equation and calculate I.
I = 2(2600 J) / (1.010025 (200 rpm)^2 - (200 rpm)^2)
I ≈ 595 kg·m^2

Therefore, the polar mass moment of inertia of the flywheel to keep the speed fluctuation within ±0.5% of the average speed is approximately 595 kg·m^2. Hence, option A is the correct answer.
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Community Answer
Maximum fluctuation of kinetic energy in an engine has been calculate...
Apply delta E max = moment of inertia *( angular speed)2 * 0.01
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Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg. m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is______________?a)595b)350c)656d)136Correct answer is option 'A'. Can you explain this answer?
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Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg. m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is______________?a)595b)350c)656d)136Correct answer is option 'A'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg. m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is______________?a)595b)350c)656d)136Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Maximum fluctuation of kinetic energy in an engine has been calculated to be 2600 J. Assuming that the engine runs at an average speed of 200 rpm, the polar mass moment of inertia (in kg. m2) of a flywheel to keep the speed fluctuation within ±0.5% of the average speed is______________?a)595b)350c)656d)136Correct answer is option 'A'. Can you explain this answer?.
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