Short Answer Questions (2 Marks)

20
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.

Transcript of Short Answer Questions (2 Marks)

Page 1: Short Answer Questions (2 Marks)

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.

Page 2: Short Answer Questions (2 Marks)

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.

Page 3: Short Answer Questions (2 Marks)

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)

Page 4: Short Answer Questions (2 Marks)

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.

Page 5: Short Answer Questions (2 Marks)

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.

Page 6: Short Answer Questions (2 Marks)

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.

Page 7: Short Answer Questions (2 Marks)

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)

Page 8: Short Answer Questions (2 Marks)

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

Page 9: Short Answer Questions (2 Marks)

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.

Page 10: Short Answer Questions (2 Marks)

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

Page 11: Short Answer Questions (2 Marks)

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

scale.

By measurement, ON = 38 mm = 0.038 m

∴Acceleration of the reciprocating parts,

𝑎𝑅 = 𝜔2 x ON = (52.4)2 x 0.038 = 104.34 m/s

Inertia force, 𝐹𝑖 = 𝑚𝑅 𝑎𝑅 = 180 x 104.34 = 18.78 kN

Page 12: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

(6) The turning moment diagram for a multi-cylinder engine has been drawn

to a scale of 1 mm = 600 N-m vertically and 1mm = 3° horizontally. The areas

above and below the mean torque line are +52, -124, +92, -140, +85, -72, and

+107 mm2, when engine is running at a speed of 600 rpm. If the total

fluctuation of speed is not to exceed ±1.5% of the mean speed. Find the

necessary mass of the flywheel of radius 0.5 m.

Solution:

Given data: N = 600 rpm or ω = 2 𝜋 𝑥 600

60

= 62.84 rad/s; R = 0.5 m,

Horizontal scale, 1 mm = 3° = 𝜋

60; Vertical scale, 1 mm = 600 N-m

The total fluctuation of speed is not to exceed ±1.5% of the mean speed,

𝜔1 - 𝜔2 = 3% ω = 0.03 ω

Coefficient of fluctuation of speed, 𝐶𝑆 = 𝜔1 − 𝜔2

𝜔

= 0.03

1 𝑚𝑚2 on turning moment diagram = 600 x 𝜋

60 = 31.42 N-m

Let the total energy at A = E

Energy at B = E + 52 (Maximum Energy)

Energy at C = E + 52 – 124 = E - 72

Energy at D = E – 72 + 92 = E + 20

Energy at E = E + 20 – 140 = E – 120 (Minimum Energy)

Page 13: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

Energy at F = E – 120 + 85 = E - 35

Energy at G = E – 35 – 72 = E - 107

Energy at H = E – 107 + 107 = E = Energy at A

Maximum fluctuation of energy,

ΔE = Maximum energy – Minimum energy

= (E+52) – (E-120) = 172 = 172 x 31.42 = 5404 N-m

Maximum fluctuation of energy (ΔE),

5404 = m 𝑅2 𝜔2 𝐶𝑆 = m x (0.5)2 x (62.84)2 x 0.03 = 29.6 m

m = 5404

29.6

= 183 kg

(7) The turning moment diagram for a petrol engine is drawn to the following

scales: Turning moment, 1 mm = 5 N-m; crank angle, 1 mm = 1°. The turning

moment diagram repeats itself at every half revolution of the engine and the

areas above and below the mean turning moment line taken in order are 295,

685, 40, 340, 960, 270 mm2. The rotating parts are equivalent to a mass of

36 kg at a radius of gyration of 150 mm. Determine the coefficient of

fluctuation of speed when the engine runs at 1800 rpm.

Solution:

Given Data: m = 36 kg; k = 150 mm = 0.15 m;

N = 1800 rpm or ω = 2 𝜋 𝑥 1800

60 = 188.52 rad/s

Horizontal scale, 1 mm = 1° = 𝜋

180; Vertical scale, 1 mm = 5 N-m

1 𝑚𝑚2 on turning moment diagram = 5 x 𝜋

180 = 𝜋

36 N-m

Page 14: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

Let the total energy at A = E

Energy at B = E + 295 (Maximum Energy)

Energy at C = E + 295 – 685 = E - 390

Energy at D = E – 390 + 40 = E - 350

Energy at E = E - 350 – 340 = E – 690 (Minimum Energy)

Energy at F = E – 690 + 960 = E + 270

Energy at G = E + 270 – 270 = E = Energy at A

Maximum fluctuation of energy,

ΔE = Maximum energy – Minimum energy

= (E+295) – (E-690) = 985 = 985 x 𝜋

36 = 86 N-m

Maximum fluctuation of energy (ΔE),

86 = m 𝑅2 𝜔2 𝐶𝑆 = 36 x (0.15)2 x (188.52)2 x 𝐶𝑆 = 28787 𝐶𝑆

𝐶𝑆 = 86

28787

= 0.003 or 0.3%

Coefficient of fluctuation of speed = 0.3% Ans.

Page 15: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

(8) The turning moment diagram for a multi-cylinder has been drawn to a

scale of 1 mm = 325 Nm vertically and 1 mm = 30 horizontally. The areas above

and below the mean torque line are -26, +378, -256, +306, -302, +244, -380,

+261 and -225 sq.mm. The engine is running at a mean speed of 600 rpm. The

total fluctuation of speed is not to exceed ±1.8% of the mean speed. If the

radius of flywheel is 0.7 m, find the mass of the flywheel.

Solution:

Given Data: N = 600 rpm (or) ω = 2 𝜋 𝑥 600

60 = 62.84 rad/s; R = 0.7 m

Total fluctuation of speed is not to exceed ±1.8% of the mean speed.

∴ 𝜔1 - 𝜔2 = 3.6% ω = 0.036 ω

Coefficient of fluctuation of speed, 𝐶𝑆 = 𝜔1 − 𝜔2

𝜔

= 0.036

Horizontal scale, 1 mm = 3° = 𝜋

60; Vertical scale, 1 mm = 325 N-m

1 𝑚𝑚2 on turning moment diagram = 325 x 𝜋

60 = 17 N-m

Let the total energy at A = E

Energy at B = E - 26

Energy at C = E - 26 + 378 = E + 352

Energy at D = E + 352 - 256 = E + 96

Energy at E = E + 96 + 306 = E + 402 (Maximum Energy)

Page 16: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

Energy at F = E + 402 - 302 = E + 100

Energy at G = E + 100 + 244 = E + 344

Energy at H = E + 344 – 380 = E – 36 (Minimum Energy)

Energy at J = E – 36 + 261 = E + 225

Energy at K = E + 225 – 225 = E = Energy at A

Maximum fluctuation of energy,

ΔE = Maximum energy – Minimum energy

= (E+402) – (E-36) = 438 = 438 x 17 = 7446 N-m

Maximum fluctuation of energy (ΔE),

7446 = m 𝑅2 𝜔2 𝐶𝑆 = m x (0.7)2 x (62.84)2 x 0.036 = 69.66 m

m = 7446

69.66

= 107 kg

(9) In a turning moment diagram, the areas above and below the mean torque

line taken in order are: 5.81, 3.23, 3.87, 5.16, 1.94, 3.87, 2.58 and 1.94 cm2

respectively. The scales of the diagram are: Turning moment. 1 cm = 7 kN-m;

crank, 1 cm = 60°. The mean speed of the engine is 120 rpm and the variation

of speed must not exceed ±3 % of the mean speed. Assuming the radius of

gyration of the flywheel to be 106.67 cm, find the weight of the flywheel to

keep the speed within the given limits.

Solution:

Given Data: N = 120 rpm (or) ω = 2 𝜋 𝑥 120

60 = 12.56 rpm;

R = 106.67 cm = 1.0667 m

Total fluctuation of speed is not to exceed ±3% of the mean speed.

∴ 𝜔1 - 𝜔2 = 6% ω = 0.06 ω

Coefficient of fluctuation of speed, 𝐶𝑆 = 𝜔1 − 𝜔2

𝜔

= 0.06

Horizontal scale, 1 cm = 60° = 𝜋

3; Vertical scale, 1 cm = 7 kN-m

Page 17: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

1 𝑐𝑚2 on turning moment diagram = 7 x 𝜋

3 = 7.33 kN-m

Let the total energy at A = E

Energy at B = E + 5.81

Energy at C = E + 5.81 – 3.23 = E + 2.58

Energy at D = E + 2.58 + 3.87 = E + 6.45 (Maximum Energy)

Energy at E = E + 6.45 – 5.16 = E + 1.29

Energy at F = E + 1.29 + 1.94 = E + 3.23

Energy at G = E + 3.23 – 3.87 = E – 0.64 (Minimum Energy)

Energy at H = E – 0.64 + 2.58 = E + 1.94

Energy at J = E + 1.94 – 1.94 = E = Energy at A

Maximum fluctuation of energy,

ΔE = Maximum energy – Minimum energy

= (E + 6.45) – (E - 0.64) = 7.09 = 7.09 x 7.33 = 52 kN-m

Maximum fluctuation of energy (ΔE),

52 x 1000 = m 𝑅2 𝜔2 𝐶𝑆 = m x (1.0667)2 x (12.56)2 x 0.06 = 10.77 m

m = 52 𝑥 1000

10.77

= 4828 kg

weight of the flywheel = 4828 x 9.81 = 47365 N

Page 18: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

(10) With reference to a reciprocating engine mechanism, derive the

relations for: a) The angular velocity and angular acceleration of the

connecting rod, and b) Turning moment on the crank shaft.

Ans:

Let l = Length of connecting rod between the centres,

r = Radius of crank or crank pin circle,

φ = Inclination of connecting rod to the line of stroke PO, and

n = Ratio of length of connecting rod to the radius of crank = l/r.

Page 19: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

The negative sign shows that the sense of the acceleration of the

connecting rod is such that it tends to reduce the angle φ.

Turning moment on the crankshaft,

T = 𝐹𝑇 x r = 𝐹𝑝 𝑠𝑖𝑛 (𝜃 + 𝜙)

𝑐𝑜𝑠 𝜙

x r

= 𝐹𝑝 (𝑠𝑖𝑛 𝜃 𝑐𝑜𝑠𝜙 + 𝑐𝑜𝑠 𝜃 𝑠𝑖𝑛 𝜙)

𝑐𝑜𝑠 𝜙

x r

Page 20: Short Answer Questions (2 Marks)

Turning Moment Diagram and Flywheels Question Bank

R. Syam Sudhakar Rao, Associate Professor, Guru Nanak Institutions Technical Campus, Ibrahimpatnam

= 𝐹𝑝 (𝑠𝑖𝑛 𝜃 + 𝑐𝑜𝑠 𝜃 𝑥 𝑠𝑖𝑛 𝜙

𝑐𝑜𝑠 𝜙 ) x r

= 𝐹𝑝 (𝑠𝑖𝑛 𝜃 + 𝑐𝑜𝑠 𝜃 𝑡𝑎𝑛 𝜙 ) x r

But l sin ϕ = r sin θ

sin ϕ = 𝑟

𝑙 sin θ = 𝑠𝑖𝑛 𝜃

𝑛

cos ϕ = √1 − 𝑠𝑖𝑛2 𝜙 = √1 −

sin2 θ

n2

= 1

𝑛 √𝑛2 − 𝑠𝑖𝑛2 𝜃

∴ tan ϕ = 𝑠𝑖𝑛 𝜙

𝑐𝑜𝑠 𝜙 = 𝑠𝑖𝑛 𝜃

𝑛

x 𝑛

√𝑛2− 𝑠𝑖𝑛2 𝜃 = 𝑠𝑖𝑛 𝜃

√𝑛2− 𝑠𝑖𝑛2 𝜃

Substituting the value of tan ϕ in T equation,

T = 𝐹𝑝 (𝑠𝑖𝑛 𝜃 + 𝑐𝑜𝑠 𝜃 𝑠𝑖𝑛 𝜃

√𝑛2− 𝑠𝑖𝑛2 𝜃) x r

= 𝐹𝑝 x r (𝑠𝑖𝑛 𝜃 + 𝑠𝑖𝑛 2𝜃

2√𝑛2− 𝑠𝑖𝑛2 𝜃)