Machine Dynamics Questions and Answers – Turning Moment Diagram for a Multicylinder Engine

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This set of Machine Dynamics Multiple Choice Questions & Answers (MCQs) focuses on “Turning Moment Diagram for a Multicylinder Engine”.

1. For a multicylinder engine, the moment of inertia of a flywheel is 6500 kg-m2, from the turning moment diagram, it is found that fluctuation of energy is 56000 N-m, if mean speed is 120 rpm, then find maximum speed.
a) 121
b) 119
c) 122
d) 118
View Answer

Answer: a
Explanation: We know that ΔE= I ω2. Cs
= I ω. (ω12)
= 6500. 120x(2π/60)2(N1-N2)
N1-N2 = 2
N=120
Therefore, N1=121.
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2. Multi cylinder engines are usually placed at an equal inclination to each other.
a) True
b) False
View Answer

Answer: a
Explanation: In case of multi cylinder engines, the cylinders are placed in such a way that the performance of the engine is optimum hence placing them at equal inclination improves the performance.

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

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

4. 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
View Answer

Answer: b
Explanation: The coefficient of fluctuation depends only on the speed of the engine and is unaffected by the no. of cylinders.
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5. 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
View Answer

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.

6. 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
View Answer

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.

7. 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
View Answer

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.
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8. 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
View Answer

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

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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn