14. Turning Moment Diagrams and Flywheel:14.1 Turning moment diagram: The turning moment diagram (also known as crankrepresentation of the turning moment Figure 1. The total area under the curve represents the work done by the crankshaft during the cycle.

If the resisting torque is constant, this is represented by the line the mean engine torque. Between points torque and the crank shaft accelerates, the area energy supplied during that time. Similarly, between than the resisting torque and the crankshaft decelerates, the area insufficiency in energy available during that time.

At the points of intersection, A, B, etc., the engine and load is no acceleration or deceleration of the flywheel; hence the speed is a maximum or minimum at these points.

14.2 Flywheel: A flywheel is a device to control the variations in speed during each cycle of an engineserves as a reservoir, which stores energy during the period when the supply of energy is more than the requirement, and releases it during the period when the requirement of energy is more than the supply. For the first period the flywheel speed increases and for the second period the flywheel speed decreases. Therefore, the flywheel keeps the speed of the engine within specified limits during each cycle.

14.3 Fluctuation of Energy: The variations of energy above and below the mean resisting torque line, Figure called fluctuations of energy.

14.3.1 Maximum Fluctuation of Energy:A turning moment diagram of a multi2. The horizontal line AG represents the mean torque line. Let aabove AG line and a2, a4, and a6 be the areas below quantity of energy which is either added or subtracted from the energy of the moving parts of the engine.

Figure 1

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. Turning Moment Diagrams and Flywheel:

The turning moment diagram (also known as crank-effort diagram) is the graphical representation of the turning moment or crank effort for various positions of the crank,

The total area under the curve represents the work done by the crankshaft during the cycle.

If the resisting torque is constant, this is represented by the line AE, which also represents n engine torque. Between points A and B, the engine torque exceeds the resisting

torque and the crank shaft accelerates, the area a of the loop representing the excess energy supplied during that time. Similarly, between B and C, the engine torque is less than the resisting torque and the crankshaft decelerates, the area b representing the insufficiency in energy available during that time.

, etc., the engine and load torques are equal, so that there is no acceleration or deceleration of the flywheel; hence the speed is a maximum or

A flywheel is a device to control the variations in speed during each cycle of an enginevoir, which stores energy during the period when the supply of energy is

more than the requirement, and releases it during the period when the requirement of For the first period the flywheel speed increases and for

nd period the flywheel speed decreases. Therefore, the flywheel keeps the speed of the engine within specified limits during each cycle.

The variations of energy above and below the mean resisting torque line, Figure

Maximum Fluctuation of Energy: A turning moment diagram of a multi-cylinder engine is shown by a wavy curve in Figure

represents the mean torque line. Let a1, a3, and abe the areas below AG line. These areas

quantity of energy which is either added or subtracted from the energy of the moving parts

Figure 2

effort diagram) is the graphical or crank effort for various positions of the crank,

The total area under the curve represents the work done by the crankshaft during the cycle.

, which also represents , the engine torque exceeds the resisting

of the loop representing the excess , the engine torque is less

representing the

equal, so that there is no acceleration or deceleration of the flywheel; hence the speed is a maximum or

A flywheel is a device to control the variations in speed during each cycle of an engine. It voir, which stores energy during the period when the supply of energy is

more than the requirement, and releases it during the period when the requirement of For the first period the flywheel speed increases and for

nd period the flywheel speed decreases. Therefore, the flywheel keeps the speed of

The variations of energy above and below the mean resisting torque line, Figure 1, are

cylinder engine is shown by a wavy curve in Figure , and a5 be the areas

represent some quantity of energy which is either added or subtracted from the energy of the moving parts

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If the energy of the flywheel at point A = U,

Energy at B = U + a1 Energy at C = U + a1 – a2 Energy at D = U + a1 – a2 + a3 Energy at E = U + a1 – a2 + a3 – a4 Energy at F = U + a1 – a2 + a3 – a4 + a5 Energy at G = U + a1 – a2 + a3 – a4 + a5 – a6 = Energy at A

Let us now suppose that the greatest of these energies is at B and least at E. Therefore,

Maximum energy in the flywheel = U + a1 Minimum energy in the flywheel = U + a1 – a2 + a3 – a4 ∴ Maximum fluctuation of energy = Max. energy – Min. energy = (U + a1) – (U + a1 – a2 + a3 – a4) = a2 – a3 + a4 14.3.2 Coefficient of Fluctuation of Energy: It may be defined as the ratio of the maximum fluctuation of energy to the workdone per cycle. =

The workdone per cycle may be obtained by using the following two relations:

1. Workdone per cycle = Tmean * θ

where Tmean = Mean torque, and θ = Angle turned (in radians), in one revolution.

2. Workdone / cycle = P * 60 / N

where P = Power in Watts, and N = Speed in rpm

14.3.3 Coefficient of Fluctuation of Speed: The difference between the maximum and minimum speeds during a cycle is called the maximum fluctuation of speed. The ratio of the maximum fluctuation of speed to the mean speed is called the coefficient of fluctuation of speed.

Let N1 and N2 = Maximum and minimum speeds in rpm during the cycle, and N = mean speed in rpm ∴ = − = −

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14.4 Energy Stored in a Flywheel:

We know that when a flywheel absorbs energy, its speed increases and when it gives up energy, its speed decreases.

Let m = Mass of the flywheel in kg. k = Radius of gyration of the flywheel, in meters. I = Mass moment of inertia of the flywheel about its axis of rotation in kg m2. = m * k2 N1 and N2 = Maximum and minimum speeds during the cycle in rpm. ω 1 and ω2 = Maximum and minimum angular speeds during the cycle in rad/s. ω = Mean angular speed during the cycle in rad/s. = (ω 1+ω2)/2

We know that the mean kinetic energy of the flywheel, = 12 ∗ = 12

As the speed of the flywheel changes from ω1 to ω2, the maximum fluctuation of energy, = . . .− . . = 12 ( − ) = ∗ ( − ) = ∗ ( − ) = = 2 ∗ Example 1: A flywheel of an engine has a mass 6.5 tons and the radius of gyration is 1.8 m. It is found from turning moment diagram that the fluctuation of energy is 56 kJ. If the mean speed of the engine is 120 rpm, find the maximum and minimum speeds.

Solution:

ω = 2π*120/60 = 12.566 rad / s

= ∴ = = 560006500 ∗ 1.8 ∗ 12.566 = 0.016839

CS = (ω1 – ω2) / ω

ω1 – ω2 = 0.016509 * ω = 0.2116 rad / s

and

ω1 + ω2 = 2 * ω = 25.132 rad / s

∴ ω1 = 12.672 rad / s and ω2 = 12.46 rad / s

Example 2: The flywheel of a steam engine has a radius of gyration of kg. The starting torque of the steam engine is Determine: (a) the angular acceleration of the flywheel, flywheel after 10 seconds from the start.

Solution:

I = m * k2 = 2500

a) Angular acceleration of the flywheel

Let α = angular acceleration of the flywheel

We know that T = I * α

∴ α = T / I = 1500

b) Kinetic energy of the flywheel

First of all, let us find out the angular speed of the flywheel after start (i.e. from rest), assuming uniform acceleration.

Let ω1 = Angular speed at rest = ω2 = Angular speed after t = Time in seconds ω2 = ω1 + α * t = ∴ ℎ ℎ Example 3: The turning moment diagram for a petrol engine is drawing to the following scales: turning moment 1 mm = 5diagram repeats itself at every half-revolution of the engine and the areas above and below the mean turning moment line, taking in order, are rotating parts are equivalent to a mass of Determine the coefficient of fluctuation of speed when the engine runs at

Page 4

The flywheel of a steam engine has a radius of gyration of 1 m and mass kg. The starting torque of the steam engine is 1500 Nm and may be assumed constant.

a) the angular acceleration of the flywheel, and (b) the kinetic energy of the seconds from the start.

2500 * 12 = 2500 kg m2

Angular acceleration of the flywheel

= angular acceleration of the flywheel

1500 / 2500 = 0.6 rad/s2

Kinetic energy of the flywheel

First of all, let us find out the angular speed of the flywheel after 10 seconds from the start (i.e. from rest), assuming uniform acceleration.

= Angular speed at rest = 0 = Angular speed after 10 sec, and

= Time in seconds t = 0 + 0.6 * 10 = 6 rad/s = 12 = 12 ∗ 2500 ∗ 36 = 45000

The turning moment diagram for a petrol engine is drawing to the following 5 N m; crank angle 1 mm = 1°. The turning moment revolution of the engine and the areas above and below

turning moment line, taking in order, are 295, 685, 340, 960, rotating parts are equivalent to a mass of 36 kg at a radius of gyration of

fluctuation of speed when the engine runs at 1800

Figure

m and mass 2500 Nm and may be assumed constant.

and (b) the kinetic energy of the

seconds from the

= 45

The turning moment diagram for a petrol engine is drawing to the following . The turning moment

revolution of the engine and the areas above and below , 270 mm2. The

kg at a radius of gyration of 150 mm. 1800 rev/min.

Figure 3

Solution: I = m * k2 = 36 * 0.152

ω = 2π*1800/60 = 188.5

Since the turning moment scale is 1π/180 rad, therefore

1 mm2 = on turning moment diagram =

Let the total energy at A = U, then referring to Figure

Energy at B = U + 295 Energy at C = U + 295 – 685 = U – Energy at D = U + 295 – 685 + 40 = U Energy at E = U + 295 – 685 + 40– Energy at F = U + 295 – 685 + 40– Energy at G = U + 295 – 685 + 40–

From above, the total energy is greatest at point B and least at point E.Maximum fluctuation of energy = . − . =Coefficient of fluctuation of speed, = ∗ ∴ = = 860.81 ∗ 188.5 = 0.00299 Example 4: The variation of crankshaft torque of a approximately represented by taking the torque as zero for crank angles 260 N m for crank angles 20° and 45°straight lines. The average speed is machine requiring a constant torque, find the mass of the flywheel, of radius of gyration 250 mm, which must be fitted in order that the total variation of speed shall be

Solution:

The crankshaft torque diagram is shown in Figure 4.

Page 5

2 = 0.81 kg m2

188.5 rad/s

1 mm = 5 N m and crank angle scale is

= on turning moment diagram = 5 * π/180 = π/36 Nm

Let the total energy at A = U, then referring to Figure 3,

390 = U – 350 340 = U – 690 340 + 960 =U +270 340 + 960– 270 = Energy at A

the total energy is greatest at point B and least at point E. = 985 = 985 ∗ 36 = 86

00299 = 0.299%

The variation of crankshaft torque of a 4-cylinder petrol engine may be approximately represented by taking the torque as zero for crank angles 0° and

45°, the intermediate portions of the torque graph being aight lines. The average speed is 600 rev/min. Supposing that the engine drives a

machine requiring a constant torque, find the mass of the flywheel, of radius of gyration mm, which must be fitted in order that the total variation of speed shall be

The crankshaft torque diagram is shown

Figure 4

N m and crank angle scale is 1 mm = 1° =

cylinder petrol engine may be and 180° and as

the intermediate portions of the torque graph being rev/min. Supposing that the engine drives a

machine requiring a constant torque, find the mass of the flywheel, of radius of gyration mm, which must be fitted in order that the total variation of speed shall be 1 per cent.

Work done in 1/2 revolution = area OABC = 12 ∗ 260 ∗ 20 ∗ 180 +∴ Tmean = Work done / θ = 465

From triangle ABC, 260 − 148 = 26020 → ∴ = 8And 260 − 148 = 260135 → ∴ = 58∴ x + y = 66.76° ∴ Fluctuation of energy represented by shaded area = 112 ∗ 25 ∗ 180 + 1122 ∗ 66 = 12 ∗ ( − ) = ∗ ∗ ∴ m = 46.2 kg

Example 5: An engine working on the two120°. The turning moment for any cylinder is assumed to increase uniformly from zero to a maximum while the crank turns 90°remain zero over the remainder of the revolutionIf the engine develops 15 kW per cylinder when running at a mean speed of the turning moment diagram for one cylinder and from it construct the combiDetermine the variation in the kinetic energy of the flywheel and its required mass for a radius of gyration of 0.3 m to limit the total speed variation to

Page 6

revolution = area OABC + 12 ∗ 260 ∗ 135 ∗ 180 + 260 ∗ 25 ∗ 180465 / π = 148 N m

8.61° 58.15°

represented by shaded area

66.76 ∗ 180 = 114 ∗ = ∗ 0.25 ∗ (2 60 ∗ 600) ∗ 0.01An engine working on the two-stroke cycle has three cylinders with cranks at

. The turning moment for any cylinder is assumed to increase uniformly from zero to 90°, to fall uniformly to zero over the next

of the revolution. kW per cylinder when running at a mean speed of

the turning moment diagram for one cylinder and from it construct the combiDetermine the variation in the kinetic energy of the flywheel and its required mass for a

m to limit the total speed variation to 2 rpm.

Figure 5

180 = 465

01

stroke cycle has three cylinders with cranks at . The turning moment for any cylinder is assumed to increase uniformly from zero to

uniformly to zero over the next 90° and to

kW per cylinder when running at a mean speed of 400 rpm, draw the turning moment diagram for one cylinder and from it construct the combined diagram. Determine the variation in the kinetic energy of the flywheel and its required mass for a

Page 7

Solution:

OABC is the diagram for one cylinder, Figure 5. ∴ Work done per cycle per cylinder = Power * 60/ N = (15000 * 60) / 400 = 2250 J ∴ 2250 = ∗ = 12 ∗ ∗ ∴ = 1433 ℎ = 3 ∗ 22502 = 1075

From Figure 5, ∴ = 1433 − 10751433 ∗ = 4 ∴ , = (1433 − 1075)2 ∗ 4 = 140.6 ∴ 140.6 = 12 ∗ ∗ ( − ) = ∗ ∗ ( − ) = ∗ 0.3 ∗ 2 60 ∗ 400 2 60 ∗ 2 ∴ = 178

Example 6: A single cylinder internal-combustion engine working on the 4-stroke cycle develops 75 kW at 360 rpm. The fluctuation of energy can be assumed to be 0.9 times the energy developed per cycle. If the coefficient of fluctuation of speed is not to exceed 1 per cent and the maximum centrifugal stress in the flywheel is to be 5.5 MN/m2, estimate the mean diameter and the cross-sectional area of the rim. Cast-iron has a density of 7.2 Mg/m3.

Solution: = 2 ∗ 36060 = 12 / = 75000 ∗ 60180 = 25000 ∴ = 0.9 ∗ 25000 = 22500 ∴ 22500 = 12 ∗ ∗ ( − ) = ∗ ∗ − = ∗ (12 ) ∗ 0.01 ∴ = 1584

Centrifugal stress, σ = ρ * v2 = ρ * ω2 * R2

where R is the mean rim radius,

i.e. 5.5* 106 = 7200 * 144π2 * R2

∴ R = 0.732 m, i.e. mean diameter, D = 2R = 1.464 m

I = mk2 = ρ * A * 2πR * R2

where A is the cross-sectional area of the rim,

i.e. 1584 = 7200 * A * 2π * 0.732 ∴ A = 0.0892 m2

Example 7: A machine shaft running at a mean speed of varies uniformly from 1200 Nm to constant for the next one revolution, decreases uniformly to revolution and then remains constant for the next of operations. It is driven by a constant speed motor and a flywheel of radius of gyration 0.6 m is fitted to the shaft. If the fluctuation of speed is

1. The power of the motor, and2. The mass of the flywheel required.

Solution:

Since the fluctuation of speed is ±2%speed, therefore total fluctuation of speed, − = 4% = 0.04

and coefficient of fluctuation of speed, = − = 0.04 The turning moment diagram for the complete cycle is shown in Figure 6.

ω = 2π*200/60 = 20.95 rad/s

We know that the torque required for one complete cycle

= Area OAEF + Area ABG + Area BCHG + Area DCH = 9π * 1200 + 2400π/2 = 19200π Nm ∴ Mean torque, Tmean = 19200π / 9π =

1. Power of the motor, Power = 2π*N*Tmean/60 = 2π *

2. Mass of the flywheel requiredLet m = mass of the flywheel required,First of all, let us find the values of LMFrom similar triangles ABG and BLM / = / → ∴ =

Page 8

sectional area of the rim,

0.7322

A machine shaft running at a mean speed of 200 rpm requires a torque which Nm to 3600 Nm during the first half revolution, remains

constant for the next one revolution, decreases uniformly to 1200 Nm during the next one on and then remains constant for the next two revolutions, thus completing a cycle

of operations. It is driven by a constant speed motor and a flywheel of radius of gyration m is fitted to the shaft. If the fluctuation of speed is ±2% of mean speed, fi

The power of the motor, and The mass of the flywheel required.

2% of mean speed, therefore total fluctuation of speed,

and coefficient of fluctuation of speed,

The turning moment diagram for the complete

We know that the torque required for one complete cycle = Area of OALBCDEF

= Area OAEF + Area ABG + Area BCHG + Area DCH 2 + 2π * 2400 + 2π * 2400/2

π = 2133.3 Nm

π * 200 * 2133.3 / 60 = 44685 W = 44.685

Mass of the flywheel required = mass of the flywheel required,

LM and NQ BLM, = 0.61

Figure 6

rpm requires a torque which Nm during the first half revolution, remains

Nm during the next one two revolutions, thus completing a cycle

of operations. It is driven by a constant speed motor and a flywheel of radius of gyration of mean speed, find

= Area of OALBCDEF

44.685 kW

6

Page 9

Now from similar triangles CHD and CNQ, / = / → ∴ = 1.22 Since the fluctuation of energy is equal to the area above the mean torque line, therefore,

e = Area LBCQ = Area LBM + Area MBCN + Area NCQ

= 0.61π * 1466.7/2 + 2π *1466.7 + 1.22π * 1466.7/2 = 13435 Nm

And e = I * ω2 * CS = mk2 * ω2 * CS

∴ m = 13534 / (0.62 * 20.952 * 0.04) = 2126 kg

14.5 Dimensions of the Flywheel Rim:

Consider a rim of the flywheel shown in Figure 7;

Let D = Mean diameter of rim = 2R R = Mean radius of rim A = Cross-sectional area of rim ρ = Density of rim ω = Angular velocity of the flywheel σ = Centrifugal or hoop stress v = Linear velocity of the mean radius = ω * R b = Width of rim t = thickness of rim

D

t

b

σ = ρ * v2 = ρ * ω2* R2 = = ∗ ∗ 60 = ∗

mrim = Volume * density = 2π * R * A * ρ

Figure 7

Page 10

A = mrim / (2π * R * ρ) = b * t

Example 8: The turning moment diagram for a multi-cylinder engine has been drawn to a scale of 1 cm of 5000 Nm torque and 1 cm to 60° of crank displacement. The intercepted areas between output torque curve and mean resistance line taken in order from one end in square cm are: -0.3, +4.1, -2.8, +3.2, -3.3, +2.5, -3.6, +2.8, -2.6 cm2, when the engine is running at 800 rpm. The fluctuation of speed is not to exceed 2% of the mean speed. Determine a suitable mean diameter and cross-section of the flywheel rim for a limiting value of the safe centrifugal stress of 7 MN/m2. The material density may be assumed as 7.2 Mg/m3. The width of the rim is to be 5 times the thickness.

Solution: CS = 2% = 0.02 ω = 2π*800/60 = 83.8 rad/s

From the figure above we find that the maximum energy is at point E = U + 4.2 and the minimum is at point H = U – 0.2 ∴ e = U + 4.2 – U + 0.2 = 4.4 cm2

1 cm2 = 5000 * 60π/180 = 5235.987 ≅ 5236 N m ∴ e = 5236 * 4.4 = 23038.4 N m

Also e = 2E * CS ∴ E =23038.4 / 0.04 = 575960 N m σ = ρ * v2 → v = (7*106/7200)0.5 = 31.18 m/s

and v = π*D*N/60 → D = 0.74438 m = 12 ∗ ∴ mrim = 2E / v2 = 1184.868 kg

-0.3

+3.2

-2.6

Crank angle

Turn

ing

mom

ent

Page 11

A = mrim / (π * D * ρ) = 0.07037 m2

A = b * t, b = 5t, ∴ A = 5t2, ∴ t = 0.1186 m ≅ 119 mm ∴ b = 5 * 119 = 595 mm

Page 12

Problems (Flywheel):

Q1/ A double-acting steam engine runs at 100 rev/min. A curve of the turning-moment plotted on a crank angle base showed the following areas alternately above and below the mean turning-moment line: 780, 400, 520, 620, 260, 460, 340, and 420 mm2. The scales used were 1 mm = 400 N m and 1 mm = 1° crank angle. If the total fluctuation in speed is limited to 1.5 per cent of the mean speed, determine the mass of the flywheel necessary if the radius of gyration is 1.05 m. (Ans.: 3464 kg)

Q2/ The turning-moment diagram for an engine is drawn on a base of crank angle and the mean resisting torque line added. The areas above and below the mean line are +4400, -1150, +1300, -4550 mm2, the scale being 1 mm = 100 N m torque and 1 mm = 1° of crank angle. Find the mass of flywheel required to keep the speed between 297 and 303 rpm, if its radius of gyration is 0.525 m. (Ans.: 1460 kg)

Q3/ The turning-moment diagram for an engine, which has been drawn to scales of 1 mm to 50 N m and 1 mm to 1° of rotation of crankshaft, shows that the greatest amount of energy which has to be stored by the flywheel is represented by an area of 2250 mm2. The flywheel is to run at a mean speed of 240 rpm with a total speed variation of 2 per cent. If the mass of the flywheel is to be 450 kg, determine suitable dimensions for the rim, Cast iron has a density of 7.2 Mg/m3. (Ans.: 1.237 m external diameter, 274 mm width)

Q4/ A vertical diesel engine running at 350 rpm develops 600 kW and has 4 impulses per revolution. If the fluctuation of energy is 25 per cent of the work done during each impulse, estimate the cross-sectional area of the rim of the flywheel required to keep the speed within 2 rpm of the mean speed when the mean peripheral speed of the rim is 1350 m/min. Cast iron has a density of 7.2 Mg/m3. (Ans.: 0.04 m2)

Q5/ A machine press is worked by an electric motor, delivering 2.25 kW continuously. At the commencement of an operation, a flywheel of moment of inertia 50 kg m2 on the machine is rotating at 250 rpm. The pressing operation requires 4.75 kJ of energy and occupies 0.75 second. Find the maximum number of pressing that can be made in 1 hour and the reduction in speed of the flywheel after each pressing. Neglect friction losses. (Ans.: 1705, 23.5 rev/min)

Q6/ An engine has 3 single acting cylinders, the cranks being spaced 120° apart. For each cylinder, the crank effort diagram consists of a triangle:

Angle 0° 60° 180° 180° to 360° Torque (N m) 0 200 (max.) 0 0

Find the mean torque and the moment of inertia of the flywheel in kg m2 necessary to keep the speed within 180 ± 3 rpm. (Ans.: 150 N m; 2.21 kg m2)

Q7/ A shaft fitted with a flywheel rotates at 250 rpm and drives a machine the resisting torque of which varies in a cyclic manner over a period of three revolutions. The torque rises from 675 Nm to 2700 Nm in a uniform manner during 0.5 revolution and remains constant for 1 revolution, the cycle being then repeated.

Page 13

If the driving torque applied to the shaft is constant and the flywheel has a mass of 450 kg and a radius of gyration of 0.6 m, find the power necessary to drive the machine and the percentage fluctuation of speed. (Ans.: 44.2 kW, ± 3.58 per cent)

Q8/ A single cylinder four stroke internal combustion engine develops 30 kW at 300 rev/min. The turning-moment diagram for the expansion and compression strokes may be taken as two isosceles triangles, on bases 0 to π and 3π to 4π radians respectively, and the net work done during the exhaust and inlet strokes is zero. The work done during compression is negative and is one quarter of that during expansion. Sketch the turning moment diagram for one cycle and find the maximum value of the turning moment during expansion. If the load remains constant, mark on the diagram the points of maximum and minimum speed. Also find the moment of inertia, in kg m2, of a flywheel to keep the speed fluctuation within ±1.5 per cent of the mean speed. (Ans.: 10.186 kN m, 457.8 kg m2)

Q9/ Figure 8 shows the variation with time of the torque required on the driving shaft of a machine during one cycle of operations. The shaft is direct coupled to an electric motor which exerts a constant torque and runs at a mean speed of 1500 rpm. The rotating parts are equivalent to a flywheel of mass 18 kg with a radius of gyration of 250 mm. Determine (a) the power of the motor, neglecting friction; (b) the percentage fluctuation of speed. (Ans.: 2.075 kW; 6.275 per cent)