Turning Moment Diagram and Flywheels Question Bank R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam Short Answer Questions (2 Marks) (1) Write the formula for velocity of the piston in reciprocating engine and explain how does velocity is related to crank angle θ. Ans: Velocity of piston, V P = ω r [sin Ɵ + 2 2 ] where, ω = Angular velocity of the crank, rad/sec r = radius of crank Ɵ = angle turned by the crank n = ratio of length of connecting rod to the radius of the crank = l / r When Ɵ = 0°, velocity of piston = 0, Ɵ = 180°, velocity of piston = 0 Ɵ = 90°, velocity of piston = ω r (2) Explain the term coefficient of fluctuation of energy. What are the parameters required to calculate coefficient of fluctuation of energy? Ans: Coefficient of fluctuation of energy: It is the ratio of maximum fluctuation of energy to the work done per cycle. = () Work done per cycle = T m . Ɵ T m = Mean torque, and Ɵ = Crank angle Parameters required to calculate coefficient of fluctuation of energy are, Mean angular speed of the flywheel and crank angle. (3) When and why is the correction couple applied while considering the inertia of the connecting rod of a reciprocating engine. Ans: The difference of the torques is known as correction couple. This couple must be applied, when the masses are placed arbitrarily to make the system dynamical equivalent.
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
Short Answer Questions (2 Marks)
(1) Write the formula for velocity of the piston in reciprocating engine
and explain how does velocity is related to crank angle θ.
Ans:
Velocity of piston, VP = ω r [sin Ɵ + 𝑆𝑖𝑛 2𝜃
2𝑛]
where, ω = Angular velocity of the crank, rad/sec
r = radius of crank Ɵ = angle turned by the crank
n = ratio of length of connecting rod to the radius of the crank = l / r
When Ɵ = 0°, velocity of piston = 0,
Ɵ = 180°, velocity of piston = 0
Ɵ = 90°, velocity of piston = ω r
(2) Explain the term coefficient of fluctuation of energy. What are the
parameters required to calculate coefficient of fluctuation of energy?
Ans:
Coefficient of fluctuation of energy: It is the ratio of maximum fluctuation
of energy to the work done per cycle.
𝐶𝐸 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 (𝛥𝐸)
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒
Work done per cycle = Tm. Ɵ
Tm = Mean torque, and Ɵ = Crank angle
Parameters required to calculate coefficient of fluctuation of energy are,
Mean angular speed of the flywheel and crank angle.
(3) When and why is the correction couple applied while considering the
inertia of the connecting rod of a reciprocating engine.
Ans:
The difference of the torques is known as correction couple. This couple
must be applied, when the masses are placed arbitrarily to make the system
dynamical equivalent.
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(4) What types of stresses are set up in the flywheel rims?
Ans:
1. Tensile stress due to the centrifugal force.
2. Tensile bending stress due to restraint of the arms.
3. Shrinkage stresses due to the unequal rate of cooling of casting.
(5) Explain the terms fluctuation of energy and fluctuation of speed as
applied to flywheels.
Ans:
Fluctuation of energy: The variations of energy above and below the mean
resisting torque line are called fluctuations of energy.
Fluctuation of speed: The variations of speed above and below the mean
resisting torque line are called fluctuations of speed.
(6) Draw the turning moment diagram of a single cylinder double acting
steam engine.
Ans:
Single cylinder double acting steam engine
(7) Explain the terms: Piston effort, Crank effort.
Ans:
Piston effort: It is the net force applied on the piston.
Crank effort: It is the net force applied to the crank pin perpendicular to
the crank which gives the required turning moment on the crank shaft.
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(8) Explain the turning moment diagram of a four-stroke cycle internal
combustion engine.
Ans: In a four-stroke cycle internal combustion engine, there is one working
stroke after the crank has turned through two revolutions, i.e 4π radians.
Since the pressure inside the engine cylinder is less than the atmospheric
pressure during the suction stroke, therefore a negative loop is formed.
During the compression stroke, the work is done on the gases, therefore a
higher negative loop is obtained. During the expansion stroke, the fuel burns
and the gas expand, therefore a large positive loop is obtained. In this stroke
the work is done by the gases. During the exhaust stroke, the work is done
on the gases, therefore a negative loop is formed.
(9) Write the formula for energy stored in a flywheel when it has
fluctuation in its speed. (or)
Prove that the maximum fluctuation of energy, ΔE = E x 2 𝐂𝐒, where E
= Mean kinetic energy of the flywheel and 𝐂𝐒 = Coefficient of fluctuation
of speed.
Ans: Energy stored in flywheel is given by,
E = ½ I 𝜔2 = ½ m 𝑘2 𝜔2
As the speed of the flywheel changes from ω1 to ω2 , the maximum
fluctuation of energy,
ΔE = Maximum K.E – Minimum K.E = ½. I. ω12 – ½. I. 𝜔2
2
= ½ I (ω1 + ω2)(ω1 − ω2) = I. ω(ω1 − ω2)
= I. ω2 [ω1− ω2
ω] = I. ω2. CS = 2. E. CS (in N-m or joules)
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(10) What is the function of flywheel? How does it differ from that of
a governor?
Ans: The function of a flywheel is:
a. To absorb energy when demand of energy id less than the supply
b. To give out energy when demand of energy is more than the supply.
Differences between Governor and flywheel:
Governor Flywheel
The function of a governor is to
regulate the mean speed of an
engine, when there are variations in
the load.
The function of a flywheel is to
reduce the fluctuations of speed
caused by the engine turning moment
during each cycle of operation.
It is provided or, prime movers
such as engines and turbines.
It is provided on engine and
fabricating machines viz., rolling
mills, punching machines, shear
machines, presses, etc.
It works intermittently, i.e., only
when there is change in load.
It works continuously from cycle to
cycle.
It has no influence over cyclic
speed fluctuation
It has no influence on mean speed of
the prime mover
(11) Explain the term maximum fluctuation of energy and maximum
fluctuation of speed in flywheel?
Ans: The different between the maximum and the minimum energies is known
as maximum fluctuation of energy.
∆E = Maximum energy – Minimum energy.
The difference between the maximum and the minimum speeds is known as
maximum fluctuation of speed.
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(12) Discuss the turning moment diagram of a multi-cylinder engine.
Ans: The turning moment diagram for a compound steam engine having three
cylinders is shown in the following figure.
The resultant turning moment diagram is the sum of the turning moment
diagrams for the three cylinders. The first cylinder is the high-pressure
cylinder, second cylinder is the intermediate cylinder and the third cylinder
is the low-pressure cylinder. The cranks, in case of three cylinders are
usually placed at 120° to each other.
(13) What are turning moment diagrams? Explain the uses of turning
moment diagram of reciprocating engines.
Ans: Turning moment diagram is the graphical representation of the turning
moment or crank effort for various position of the crank.
In turning moment diagram, the turning moment is taken as the ordinate (Y-
axis) and crank angle as abscissa (X-axis).
Uses of turning moment diagrams: 1) The area under the turning moment
diagram represents work done per cycle. This area multiplied by number of
cycles per second gives the power developed by the engine.
2) Dividing the area of the turning moment diagram with the length of the
base gives the mean turning moment. This enables to find out the fluctuation
of energy,
3) The maximum ordinate of the turning moment diagram gives the maximum
torque to which the crankshaft is subjected to. This enables to find the
diameter of the crank shaft.
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(14) Define the terms coefficient of fluctuation of energy and coefficient
of fluctuation of speed in case of flywheels.
Ans:
Coefficient of fluctuation of energy: It is the ratio of maximum fluctuation
of energy to the work done per cycle.
𝐶𝐸 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑐𝑡𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑛𝑒𝑟𝑔𝑦 (𝛥𝐸)
𝑊𝑜𝑟𝑘𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒
Coefficient of fluctuation of speed: The ratio of the maximum fluctuation of
speed to the mean speed is called the coefficient of fluctuation of speed.
𝐶𝑆 = 𝑁1− 𝑁2
𝑁
where N1 = Maximum speed
N2 = Minimum speed and
N = Mean speed = 𝑁1+ 𝑁2
2
(15) List out the few machines in which flywheel is used.
Ans:
1. Punching machines.
2. Shearing machines,
3. Riveting machines, and
4. Crushing machines.
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
10 Marks Questions
(1) In a vertical petrol engine, the crank radius is 6 cm, and the connecting
rod is 22 cm long. The piston weighs 9.8 N. The connecting rod may be
regarded as being equivalent to a mass of 0.5 kg at the piston together with
a mass of 1 kg at the crank pin. Find the amount and the direction of the
force exerted on the crank pin when the crank has moved 30° from the top
dead centre. The engine speed is 2000 rpm, and in this position the force on
the piston due to gas pressure is 735 kN/𝑚2.
Solution:
Given Data: r = 6 cm = 0.06 m; l = 22 cm = 0.22 m; 𝑊𝑅 = 9.8 N (or) 𝑚𝑅 = 1 kg;
θ = 30°; N =2000 rpm (or) ω = 2 𝜋 𝑥 2000
60 = 209.44 rad/s; p = 735 kN/𝑚2;
Assume engine diameter, D = 100 mm = 0.1 m
n = 𝑙
𝑟 = 0.22
0.06
= 3.67
Force due to gas pressure, 𝐹𝑝 = p x 𝜋
4 x 𝐷2
= 735 x 1000 x 𝜋
4 x (0.1)2 = 5773 N
Inertia force on the piston,
𝐹𝑏 = 𝑚𝑅 𝜔2 𝑟 (cos 𝜃 + cos 2𝜃
𝑛)
= 1 x (209.44)2 x 0.06 (cos 30 + cos 60
3.67
) = 2632 N
Net force on the piston, F = 𝐹𝑝 + 𝑊𝑅 - 𝐹𝑏
= 5773 + 9.8 – 2632 = 3151 N
sin ϕ = sin 𝜃
𝑛
= sin 30
3.67
= 0.1362 ϕ = 7.83°
Force exerted on the crank pin, 𝐹𝑡 = 𝐹
cos 𝜙
sin (θ + ϕ)
= 3151
cos 7.83
sin (30 + 7.83) = 1951 N (perpendicular to crank)
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(2) The connecting rod of a gas engine weighs 700 N, and has a radius of
gyration of 400 mm about an axis through the center of gravity. The length
of the rod between centers is 1 m and the C.G. is 350 mm from the crank pin
center. If the crank is 250 mm long, and revolves at a uniform speed of 300
rpm, find the magnitude and direction of the inertia force on the rod, and of
the corresponding torque on the crankshaft when the inclination of the crank
to the IDC is 1350.
Solution:
Given data: 𝑊𝑐 = 700 N; 𝑚𝑐 = 700/9.81 = 71.35 kg; k = 400 mm = 0.4 m;
l = 1 m; CG = 350 mm = 0.35 m; 𝑙1 = l – CG = 1 – O.35 = 0.65 m; r = 250 mm =
0.25 m; θ = 135°; N = 300 rpm (or) ω = 2 𝜋 𝑥 300
60
= 31.42 rad/s.
n = 𝑙
𝑟 = 1
0.25
= 4
Acceleration of the piston,
𝑎𝑝 = 𝜔2𝑟 (cos 𝜃 + cos 2𝜃
𝑛)
= 31.422 x 0.25 (cos 135 + cos 270
4)
= -174.7 m/𝑠2
Mass of the connecting rod at P,
= 𝑙 − 𝑙1
𝑙
x 𝑚𝑐 = 0.35
1
x 71.35 = 25 m
∴ Inertia force on the rod, 𝐹𝑖 = 25 x (-174.7) = -4363 N
Torque exerted on the crank shaft, 𝑇𝑖
= 𝐹𝑖 x r (𝑠𝑖𝑛 𝜃 + 𝑠𝑖𝑛 2𝜃
2√𝑛2− 𝑠𝑖𝑛2 𝜃)
= 4363 x 0.25 (𝑠𝑖𝑛 135 + 𝑠𝑖𝑛 270
2√42− 𝑠𝑖𝑛2 135) = 633 Nm
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(3) The length of connecting rod of a gas engine is 500 mm, and its C.G. lies
at 165 mm from the crank pin centre. The rod has a mass of 80 kg and a
radius of gyration of 180 mm about an axis passing through the centre of
mass. The stroke of piston is 225 mm, and the crank speed is 300 rpm.
Determine the inertia force on the crankshaft when the crank has turned
through 125° from the inner dead centre.
Solution:
Given Data: l = 500 mm = 0.5 m; 𝑚𝑐 = 80 kg; Distance of centre of mass = 165
mm from the crank pin centre; k = 180 mm; L = 225 mm; N = 300 rpm; 𝜃 =
125°; r = 225/2 = 112.5 mm = 0.1125 m; n = 500/112.5 = 4.45
ω = 2 𝜋 𝑥 300
60
= 31.42 rad/s
Mass at crank pin, 𝑚𝑎
= 80 (500 −165
500) = 53.6 kg
Mass at gudgeon pin, 𝑚𝑏 = 80 – 53.6 = 26.4 kg
∴ Mass of reciprocating parts = 26.4 kg
Acceleration of the reciprocating parts,
a = r 𝜔2 (cos 𝜃 + cos 2𝜃
𝑛)
As θ is greater than 90°, it is towards the left, and the inertia force is
towards right.
Inertia force, 𝐹𝑏
= m a = m r 𝜔2 (cos 𝜃 + cos 2𝜃
𝑛)
= 26.4 x 0.1125 x (31.42)2 (cos 125 + cos 250
4.45)
= 1907 N Ans.
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(4) In a vertical double-acting steam engine, the connecting rod is 4.5 times
the crank. The weight of the reciprocating parts is 120 kg and the stroke of
the piston is 440 mm. The engine runs at 250 rpm. If the net load on the
piston due to steam pressure is 25 kN when the crank has turned through an
angle of 120° from the top dead centre, determine the (i) thrust in the
connecting rod (ii) pressure on slide bars (iii) tangential force on the crank
pin (iv) thrust on the bearings (v) turning moment on the crank shaft.
Solution:
Given Data: r = 440/2 = 220 mm = 0.22 m; F = 25 kN; θ = 120°; N = 250 rpm;
m = 120 kg; n = 4.5; ω = 2 𝜋 𝑥 250
60 = 26.18 rad/s
sin β = sin 𝜃
𝑛 = sin 120
4.5
= 0.1925 (or) β = 11.1°
Accelerating force, 𝐹𝑏 = m r 𝜔2 (cos 𝜃 + cos 2𝜃
𝑛)
= 120 x 0.22 x (26.18)2 (cos 120 + cos 240
4.5) = - 11058 N
Force on the piston,F = 𝐹𝑝 + mg - 𝐹𝑏 = 25000 +120x9.81 –(-11058) = 37235 N
(i) Thrust in the connecting rod,
𝐹𝑐 = 𝐹
cos 𝛽 = 37235
cos 11.1
= 37945 N
(ii) Pressure on slide bars,
𝐹𝑛 = F tan β = 37235 tan 11.1 = 7305 N
(iii) Tangential force on the crank pin,
𝐹𝑡 = 𝐹𝑐 sin (θ + β) = 37945 x sin (120 + 11.1) = 28594 N
(iv) Thrust on the bearings,
𝐹𝑟 = 𝐹𝑐 sin (θ + β) = 37945 x cos (120 + 11.1) = - 24944 N
(v) Turning moment on the crank shaft,
T = 𝐹𝑡 x r = 28594 x 0.22 = 6290.7 Nm
Turning Moment Diagram and Flywheels Question Bank
R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam
(5) By using graphical method, find the inertia force for the following data
of an I.C. engine.
Bore = 175 mm, stroke = 200 mm, engine speed = 500 rpm, length of
connecting rod = 400 mm, crank angle = 60° from T.D.C. and mass of
reciprocating parts = 180 kg.
Solution:
Given Data: D = 175 mm; L = 200 mm = 0.2 m (or) r = L/2 = 0.1 m; N = 500 rpm
(or) ω = 2 𝜋 𝑥 500
60 = 52.4 rad/s; l = 400 mm = 0.4 m; 𝑚𝑅 = 180 kg.
Graphical method:
Draw the klein’s construction acceleration diagram OCQN to some suitable