Engineering Questions with Answers - Multiple Choice Questions
Machine Dynamics MCQ – Balancing of V-engines
1 - Question
A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 11.5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?
a) 736
b) 836
c) 936
d) 636
View Answer
Answer: b
Explanation: Maximum resultant secondary force is given by the equation:
2–√ (m/n)xω2.r
substituting the values we get
Fsmax = 836 N.
2 - Question
If the mass of the reciprocating parts is doubled, then the primary force is halved.
a) True
b) False
View Answer
Answer: b
Explanation: The primary force has a direct relation with the mass of the reciprocating parts thus doubling the mass will result in an increment of two times in the primary force.
3 - Question
A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 23 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in Newtons?
a) 7172
b) 1672
c) 1122
d) 1272
View Answer
Answer: b
Explanation: Maximum resultant secondary force is given by the equation:
2–√ (m/n)xω2.r
substituting the values we get
Fsmax = 1672 N.
Explanation: Maximum resultant secondary force is given by the equation:
2–√ (m/n)xω2.r
substituting the values we get
Fsmax = 1672 N.
4 - Question
A V-twin engine has the cylinder axes at 90 degrees and the connecting rods operate a common crank. The reciprocating mass per cylinder is 34.5 kg and the crank radius is 7.5 cm. The length of the connecting rod is 0.3 m. If the engine is rotating at the speed is 500 r.p.m. What is the value of maximum resultant secondary force in kN?
a) 2.238
b) 2.508
c) 2.754
d) 2.908
View Answer
Answer: b
Explanation: Maximum resultant secondary force is given by the equation:
2–√ (m/n)xω2.r
substituting the values we get
Fsmax = 2508 N.
5 - Question
From the following data of a 60 degree V-twin engine, determine the maximum value for secondary forces in newtons:
Reciprocating mass per cylinder = 1.5 Kg
Stroke length = 10 cm
Length of connecting rod = 25 cm
Engine speed = 2500 rpm
a) 890.3
b) 760.4
c) 580.3
d) 970.6
View Answer
Answer: a
Explanation: The maximum value for secondary force is given by the equation:
3–√ /2x(m/n).ω2.r
substituting the values we get
Fpmax = 890.3 N.
6 - Question
From the following data of a 60 degree V-twin engine, determine the maximum value for primary forces in newtons:
Reciprocating mass per cylinder = 1.5 Kg
Stroke length = 10 cm
Length of connecting rod = 25 cm
Engine speed = 2500 rpm
a) 7711
b) 4546
c) 2508
d) 8764
View Answer
Answer: a
Explanation: The primary force is maximum, when θ = θ°
therefore the equation for maximum primary force becomes
(m/2)ω2rx3
substituting the values we get
Fpmax = 7711 N.
7 - Question
From the following data of a 60 degree V-twin engine, determine the minimum value for primary forces in newtons:
Reciprocating mass per cylinder = 1.5 Kg
Stroke length = 10 cm
Length of connecting rod = 25 cm
Engine speed = 2500 rpm
a) 7711
b) 4546
c) 2570
d) 8764
View Answer
Answer: c
Explanation: The primary force is maximum, when θ = 90°
therefore the equation for maximum primary force becomes
(m/2)ω2r
substituting the values we get
Fpmax = 2570 N.
8 - Question
For a V-twin engine, which of the following means can be used to balance the primary forces?
a) Revolving balance mass
b) Rotating balance mass
c) Reciprocating balance mass
d) By the means of secondary forces
View Answer
Answer: a
Explanation: We know that primary resultant force depends on speed of revolution, mass and radius of rotation. Hence revolving mass can be used to balance the primary forces.