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# Machine Dynamics MCQ – Flywheel

In a four stroke I.C. engine, the turning moment during the compression stroke is

a) positive throughout

b) negative throughout

c) positive during major portion of the stroke

d) negative during major portion of the stroke

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Answer: b

Explanation: Since the pressure inside the engine cylinder is less than the atmospheric pressure during the suction stroke, therefore a negative loop is formed. During the compression stroke, the work is done on the gases, therefore a higher negative loop is obtained.

The maximum fluctuation of energy is the

a) difference between the maximum and minimum energies

b) sum of the maximum and minimum energies

c) variations of energy above and below the mean resisting torque to the

d) ratio of the mean resisting torque to the workdone per cycle

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Answer: a

Explanation: The difference between the maximum and the minimum energies is known as maximum fluctuation of energy.

The co-efficient of fluctuation of energy is the ratio of maximum energy to the minimum energy.

a) True

b) False

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Answer: b

Explanation: The co-efficient of fluctuation of energy is the ratio of the maximum fluctuation of energy to the work done per cycle.

Which of the following statement is wrong?

a) The difference between the maximum and minimum energies is called maximum fluctuation of energy.

b) The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

c) The variations of energy above and below the mean resisting torque line is known as fluctuation of energy.

d) None of the mentioned

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Answer: d

Explanation: All the statements are correct.

The difference between the maximum and minimum energies is called maximum fluctuation of energy.

The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

The variations of energy above and below the mean resisting torque line is known as fluctuation of energy.

The ratio of maximum fluctuation of energy to the workdone per cycle is called

a) fluctuation of energy

b) maximum fluctuation of energy

c) coefficient of fluctuation of speed

d) none of the mentioned

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Answer: c

Explanation: The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

Maximum fluctuation of energy in a flywheel is equal to

a) Iω(ω1 – ω2)

b) Iω2CS

c) 2ECS

d) all of the mentioned

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Answer: d

Explanation: None

A flywheel is fitted to the crankshaft of an engine having W as the amount of indicated work per revolution and permissible limits of coefficient of fluctuation of energy and speed as CE and CS respectively. The kinetic energy of the flywheel is given by

a) 2WCE/CS

b) WCE/2CS

c) WCE/CS

d) WCS/2CE

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Answer: b

Explanation: None

If the rotating mass of a rim type flywheel is distributed on another rim type flywheel whose mean radius is half the mean radius of the former, then energy stored in the latter at the same speed will be

a) four times the first one

b) same as the first one

c) one fourth of the first one

d) one and a half times the first one

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Answer: c

Explanation: None

The ratio of maximum fluctuation of speed to the mean speed is called

a) fluctuation of energy

b) maximum fluctuation of energy

c) coefficient of fluctuation of speed

d) none of the mentioned

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Answer: c

Explanation: The co-efficient of fluctuation of speed is the ratio of maximum fluctuation of speed to the mean speed.

The flywheel of a machine having weight of 4500 N and radius of gyration of 2 m has cyclic fluctuation of speed from 125 r.p.m to 120 r.p.m. Assuming g = 10m/s2, the maximum fluctuation of energy is

a) 12822 N-m

b) 24200 N-m

c) 14822 N-m

d) 12100 N-m

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Answer: d

Explanation: Mass of flywheel = weight of flywheel/Acceleration due to gravity = 4500/10kg

Moment of Inertia = mk2

= 1800 kgm2

ω1 = 2π/60 x 125rad/sec

ω2 = 2π/60 x 120rad/sec

Emax = 1/2 I(?)2

= 12087.2 N-m

= 12100 Nm

A circular solid disc of uniform thickness 20 mm, radius 200 mm and mass 20 kg, is used as a flywheel. If it rotates at 600 rpm, the kinetic energy of the flywheel, in Joules is

a) 395

b) 790

c) 1580

d) 3160

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Answer: b

Explanation: For flywheel K.E = 1/2Iω2

ω = 2πN/60 = 62.83 rad/s

I (for solid circular disk) = 1/2mR2 = 0.4 kg m2

Hence, K.E = 790 Joules.

The speed of an engine varies from 210 rad/s to 190 rad/s. During the cycle the change in kinetic energy is found to be 400 Nm. The inertia of the flywheel in kg/m2 is

a) 0.10

b) 0.20

c) 0.30

d) 0.40

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Answer: a

Explanation: Given ω1 = 210 rad/ sec, ω2 = 190 rad/ sec, ΔE= 400 Nm

As the speed of flywheel changes from ω1 to ω2, the maximum fluctuation of energy,

ΔE = 1/2I [(ω1)2 (ω2)2] I = 0.10 kgm2

For a certain engine having an average speed of 1200 rpm, a flywheel approximated as a solid disc, is required for keeping the fluctuation of speed within 2% about the average speed. The fluctuation of kinetic energy per cycle is found to be 2 kJ. What is the least possible mass of the flywheel if its diameter is not to exceed 1 m ?

a) 40 kg

b) 51 kg

c) 62 kg

d) 73 kg

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Answer: b

Explanation: Given N = 1200 rpm, ΔE = 2kJ = 2000 J, D = 1m, Cs = 0.02

Mean angular speed of engine,

ω = 2πN/60 = 125.66 rad/ sec

Fluctuation of energy of the flywheel is given by,

ΔE = Iω2Cs = 1/2mR2ω2Cs For solid disc I = mR2/2

m = 51 kg