Machine Dynamics Interview Questions and Answers for Experienced people on “Inertia Forces in a Reciprocating Engine Considering the Weight of Connecting Rod”.
1. Force which does not act on the connecting rod is ______
a) Weight of connecting rod
b) Inertia force of connecting rod
c) Radial force
d) Coriolis force
Answer: d
Clarification: Since there is no accelerated frame of reference having different velocities, therefore coriolis force does not act on the connecting rod.
2. Inertia forces on the reciprocating parts acts along the line of stroke.
a) True
b) False
Answer: a
Clarification: The reciprocating parts of the engine experience an Inertia force of magnitude m.ω2.r(cosθ + cos2θ/n), this inertia force acts along the line of stroke.
3. When mass of the reciprocating parts is neglected then the inertia force is _____
a) Maximum
b) Minimum
c) 0
d) Not defined
Answer: c
Clarification: Inertia force for neglected mass of reciprocating parts is 0 as it depends on the mass, since it has a value, it is defined.
4. For a steam engine, the following data is given:
Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
calculate inertia force at θ=30 degrees from IDC.
a) 19000 N
b) 19064 N
c) 19032 N
d) 20064 N
Answer: b
Clarification: l1 = l – G.C = 1.5 – 0.5 = 1m
Fi = (Mr + G.C,Mc/l).ω2.r(cosθ + cos2θ/n)
Mr=300 Kg
Mc=250 kg
ω=13.1 rad/s r=0.3m
substituting these values will give Fi = 19064 N.
5. Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
Find the equivalent length L of a simple pendulum swung about an axis.
a) 1.35 m
b) 1.42 m
c) 1.48 m
d) 1.50 m
Answer: b
Clarification: We know that equivalent length L is given by the expression
L = (Kg2 + l12)/l1
Kg = 0.65m l1 = 1m
therefore L = 1.42 m.
6. From the data given:
Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m ; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
Find the correcting couple in N-m?
a) 52.7
b) 49.5
c) 59.5
d)56.5
Answer: c
Clarification: The correction couple depends on equivalent length, l1 mass of connecting rod and angular position and velocity
Tc = Mc.l1.(l-L).(ω2.sin2θ/2n2)
substituting the values into the equation results in
Tc = 59.5 N-m.
7. Piston diameter = 0.24 m, length of stroke = 0.6 m, length of connecting rod = 1.5 m, mass of reciprocating parts = 300 kg, mass of connecting rod = 250 kg; speed of rotation = 125 r.p.m; centre of gravity of connecting rod from crank pin = 0.5 m ; Kg of the connecting rod about an axis through the centre of gravity = 0.65 m
Find angular acceleration of connecting rod in rad/s2.
a) 16.782
b) 17.824
c) 15.142
d) 17.161