Dynamics of Machines Questions and Answers Part-13

1. Consider a three cylinder engine, the resultant turning moment diagram is the _____ of three cylinders.
a) Sum
b) Difference
c) Product
d) Independent

Answer: a
Explanation: For a multicylinder engine, the turning moment diagram is formed by combining the turning moment diagrams of individual cylinders.

2. For a multicylinder engine, the coefficient of fluctuation of speed would vary with _________
a) Number of cylinders
b) Remains unaffected
c) Length of connecting rod
d) Input temperature

Answer: b
Explanation: The coefficient of fluctuation depends only on the speed of the engine and is unaffected by the no. of cylinders.

3. For a 4 cylinder engine, if the minimum speed of the engine is half the maximum speed, then coefficient of fluctuation is _________
a) 0.5
b) 1.5
c) 2
d) 0.66

Answer: d
Explanation: Let N be the maximum speed, then minimum speed is N/2
mean speed = (N+N/2)/2 = 3N/4
Maximum fluctuation = N – N/2 = N/2
therefore coefficient of fluctuation = 2/3.

4. In the turning moment diagram of a multicylinder engine, the work done during exhaust stroke is by ______
a) The gases
b) On the gases
c) Piston wall
d) Valve

Answer: b
Explanation: During exhaust stroke, the area under the loop is negative. This indicates that the work is done on the gases and power has not been generated during the process.

5. For a 4 cylinder engine, when the pressure inside the cylinders exceeds the atmospheric pressure then.
a) Work is done by the gases
b) Work is done on the gases
c) Work is done on the piston wall
d) Work is done by the piston wall

Answer: a
Explanation: When the pressure inside the cylinders exceeds the atmospheric pressure then the area under the loop is positive. This indicates that the work is done by the gases.

6. The flywheel of a steam engine has a mass moment of inertia of 2500 Kg-m2. The starting torque of the steam engine is 1500 N-m and may be assumed constant, using this data find the angular acceleration of the flywheel in rad/s2.
a) 0.4
b) 0.6
c) 0.3
d) 1.2

Answer: a
Explanation: We know that T = I.α
T = 1500 Nm (constant )
I = 2500 kg-m2
therefore, \[\alpha\] = 15/25 = 0.6 rad/s2

7. The maximum fluctuation of speed is the
a) difference of minimum fluctuation of speed and the mean speed
b) difference of the maximum and minimum speeds
c) sum of the maximum and minimum speeds
d) variations of speed above and below the mean resisting torque line

Answer: b
Explanation: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

8. The coefficient of fluctuation of speed is the _____________ of maximum fluctuation of speed and the mean speed.
a) product
b) ratio
c) sum
d) difference

Answer: b
Explanation: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called coefficient of fluctuation of speed.

9. In a turning moment diagram, the variations of energy above and below the mean resisting torque line is called
a) fluctuation of energy
b) maximum fluctuation of energy
c) coefficient of fluctuation of energy
d) none of the mentioned

Answer: a
Explanation: The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation.The variations of energy above and below the mean resisting torque line are called fluctuation of energy.

10. If E = Mean kinetic energy of the flywheel, CS = Coefficient of fluctuation of speed and Δ E = Maximum fluctuation of energy, then
a) ΔE = E / CS
b) ΔE = E2 × CS
c) ΔE = E × CS
d) ΔE = 2 E × CS

Answer: d
Explanation: ΔE = Maximum K.E. — Minimum K.E. = 2 E × CS