This set of Machine Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Coefficient of Fluctuation of Energy”.
1. In the figure given below, the areas BbC, CcD represent _______
a) Power generated
b) Power lost
c) Fluctuation of energy
d) Change in momentum
View Answer
Explanation: The variations of energy above and below the mean resisting torque line are called fluctuations of energy. The areas under the following curves: BbC, CcD, DdE and so on represent fluctuations of energy.
2. The fluctuation of energy is the variation of energy above the line of minimum torque.
a) True
b) False
View Answer
Explanation: The given statement is false because the variations of energy above and below the mean resisting torque line are called fluctuations of energy. It is represented by the enclosing curves above and below the mean line.
3. From the turning moment diagram shown below, the maximum speed will occur at point _____
a) p
b) b
c) q
d) c
View Answer
Explanation: The maximum engine speed occurs at point q because the flywheel absorbs energy when the crank moves from point p to q. This results in a gain of speed.
4. From the turning moment diagram shown below, the minimum speed will occur at point _____
a) p
b) b
c) q
d) c
View Answer
Explanation: The minimum engine speed occurs at point p because the flywheel gives out some of its energy when the crank moves from point a to p. This results in loss of speed.
5. Coefficient of fluctuation of energy is ratio of _______
a) Mean fluctuation energy to work done per cycle
b) Maximum fluctuation energy to work done per cycle
c) Minimum fluctuation energy to work done per cycle
d) Mean fluctuation energy to power generated per cycle
View Answer
Explanation: The ratio of the maximum fluctuation of energy to the work done per cycle is known as the coefficient of fluctuation of energy. Mathematically, coefficient of fluctuation of energy (CE) = Maximum fluctuation of energy/Work done per cycle.
6. Work done per cycle is calculated as _______
a) Tmean × θ
b) Tmin × θ
c) Tmax × θ
d) Tmax × θ/2
View Answer
Explanation: The work done per cycle (in N-m or joules) may be obtained by using the following relations: Work done per cycle = Tmean × θ, where Tmean = Mean torque, Mean torque is same as the mean resisting torque.
7. For a 4 stroke IC engine, the angle θ assumes a value equal to _______
a) π
b) 2π
c) 4π
d) π/2
View Answer
Explanation: For a 4 stroke internal combustion engine, a power stroke takes place every 2 revolution therefore the total angle turned is equal to 4π.
8. To calculate power generated, product of maximum torque and angular speed is taken.
a) True
b) False
View Answer
Explanation: To calculate power generated, product of mean effective torque and angular speed is taken.
P = Tmean × ω
9. For a 4 stroke IC engine, Tmean = 1875 N-m, then find the work done per cycle.
a) 23.56 kJ
b) 11.78 kJ
c) 5.89 kJ
d) 2.94 kJ
View Answer
Explanation: For a 4 stroke internal combustion engine, a power stroke takes place every 2 revolution therefore the total angle turned is equal to 4π. Hence θ=4π
therefore work done per cycle = 23.56 kJ.
10. A shaft fitted with a flywheel rotates at 250 r.p.m. and drives a machine, the torque fluctuated about 18750 N-m, find the power required to drive the machine.
a) 49.125 kW
b) 24.56 kW
c) 12.28 kW
d) 6.14 kW
View Answer
Explanation: ω = 2π × 250/60 = 26.2 rad/s,
P = Tmean ×ω
= 1875 x 26.2
= 49.125 kW.
Sanfoundry Global Education & Learning Series – Machine Dynamics.
To practice all areas of Machine Dynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
- Apply for Mechanical Engineering Internship
- Check Dynamics of Machinery Books
- Check Aeronautical Engineering Books
- Practice Aeronautical Engineering MCQs
- Practice Mechanical Engineering MCQs
