This set of Farm Tractor Multiple Choice Questions & Answers (MCQs) focuses on “Flywheel”.
1. Find the mean kinetic energy of the flywheel having a radius of the gyration 1.5 m, the mass of the flywheel 500 kg, maximum and minimum angular speed as 10 rad/s, 8 rad/s respectively.
a) 45 joules
b) 47 joules
c) 48 joules
d) 49 joules
View Answer
Explanation: Kinetic energy = \(\frac {1}{2}\) * m * k2 * ω2
= \(\frac {1}{2}\) * 500 * (1.5)2 * ω2
Where ω = ω1 + ω2 / 2
ω = (10 + 8) / 2
ω = 9 rad/s
K.E = \(\frac {1}{2}\) * 500 * 2.25 * 81
= 45562.5 kN – m
= 45.56 joules
2. What is the purpose of a flywheel in an engine?
a) For energy – storing
b) For steering
c) For power – producing
d) For governing speed
View Answer
Explanation: The flywheel is a metal wheel – like structure attached to the engine. It is used for storing energy during the power stroke and for releasing the stored energy during non – power strokes for an even rotation of the crankshaft.
3. What happens to the speed of the flywheel while absorbing energy?
a) Increases
b) Decreases
c) Double
d) Quadruple
View Answer
Explanation: The flywheel absorbs excess energy generated during the power stroke and it will increase the rotational speed of the flywheel. It is the only power – producing stroke and it is more than the engine load.
4. What is the coefficient of fluctuation of speed in a flywheel?
a) N1 – N2 / N
b) N / N1 – N2
c) N / N1 + N2
d) N1 + N2 / N
View Answer
Explanation: The coefficient of fluctuation of speed is the ration of maximum fluctuation of speed (N1 – N2) to the mean speed (N) of the flywheel. It is also represented in the linear and angular speeds.
5. Find the coefficient of fluctuation of speed of a flywheel with a maximum speed of 180 rpm and a minimum speed of 160 rpm.
a) 0.1176
b) 0.2332
c) 0.3569
d) 0.6325
View Answer
Explanation: Cs = N1 – N2 / N
Where Mean speed, N = N1 + N2 / 2
= 2 (N1 – N2) / N1 + N2
= 2(180 – 160) / 180 + 160
= 0.1176
6. What is known as the reciprocal of the coefficient of fluctuation of speed?
a) Coefficient of steadiness
b) Coefficient of speed
c) Coefficient of energy
d) Coefficient of torque
View Answer
Explanation: The coefficient of steadiness of the flywheel is the reciprocal of coefficient of fluctuation of speed i.e. m = 1 / Cs where m – coefficient of steadiness. Also m = N / N1 – N2.
7. Find the mass of the flywheel having a radius of gyration 1.5 m, angular velocity 10 rad/s, coefficient of fluctuation of speed 0.10 and maximum fluctuation of energy 15 kN – m.
a) 650 kg
b) 666 kg
c) 645 kg
d) 630 kg
View Answer
Explanation: ∆E = M . k2 . ω2 . Cs
15000 = M * (1.5)2 * (10)2 * 0.10
M = 666 kg
8. Find the energy in the flywheel having maximum fluctuation of energy 20 kN – m and coefficient of fluctuation of speed 0.20.
a) 50 kN – m
b) 55 kN – m
c) 60 kN – m
d) 65 kN – m
View Answer
Explanation: ∆E = 2 E Cs
E = ∆E / 2 Cs
= 20000 / 2 * 0.20
= 50000
= 50 kN – m
9. What is the maximum fluctuation of speed in a flywheel?
a) N1 – N2
b) N1 + N2
c) N2 – N1
d) N1 / N
View Answer
Explanation: The maximum fluctuation of speed is the difference between maximum speed and the minimum speed of the flywheel during a cycle. Where N1 is the maximum speed and N2 is the minimum speed.
10. What happens to the speed of the flywheel while releasing energy?
a) Increases
b) Doubles
c) Decreases
d) Quadruples
View Answer
Explanation: During other strokes than a power stroke, the engine does not produce power. So the energy from the flywheel is released for the uniform rotation of the crankshaft. Flywheel is different from governor as it controls the speed of the engine.
11. Find the coefficient of steadiness of a flywheel with maximum and minimum speed of the flywheel is 200 rpm and 150 rpm respectively.
a) 3.5
b) 3.01
c) 3.99
d) 3.10
View Answer
Explanation: m = 1 / Cs
= N / N1 – N2
= N1 + N2 / 2(N1 – N2)
= 200 + 150 / 2(200 – 150)
= 3.5
12. Find the rim mean radius of the flywheel, given mass of the flywheel – 200 kg, angular velocity – 40 rad/s, coefficient of fluctuation of speed – 0.5 and maximum fluctuation of energy – 10 kN – m.
a) 0.25 m
b) 0.30 m
c) 0.50 m
d) 0.45 m
View Answer
Explanation: ∆E = M.R2.ω2.Cs
R2 = ∆E / M.ω2.Cs = 10000 / (200 * 402 * 0.5)
R = 0.25 m
13. Find the maximum fluctuation of energy of the flywheel with mass moment of inertia 2000 kg – m2, maximum and minimum angular speed are 15 rad/s, 10 rad/s respectively.
a) 125 kN – m
b) 120 kN – m
c) 123 kN – m
d) 124 kN – m
View Answer
Explanation: ∆E = I ω (ω1 – ω2)
ω = ω1 + ω2 / 2 = 15 + 10 / 2 = 12.5
∆E = 2000 * 12.5 (15 – 10) = 125 kN – m
Sanfoundry Global Education & Learning Series – Farm Tractor.
To practice all areas of Farm Tractor, here is complete set of Multiple Choice Questions and Answers.
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