Dr. Ahmed Nabhan Mahmoud

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Dr. Ahmed Nabhan Mahmoud

MINIA University

Faculty of Engineering

Production Engineering and Design Department

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DR. AHMED NABHAN MAHMOUD

PRODUCTION ENGINEERING AND DESIGN DEPARTMENT FACULTY OF ENGINEERING - MINIA UNIVERSITY

❑A flywheel used in machines serves 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 supply.

❑In case of steam engines, internal combustion engines, reciprocating

compressors and pumps, the energy is developed during one stroke and the engine is to run for the whole cycle on the energy produced during this one stroke.

Production Engineering and Mechanical Design Department

Faculty Of Engineering - Minia University Dr. Ahmed Nabhan Mahmoud

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PRODUCTION ENGINEERING AND DESIGN DEPARTMENT FACULTY OF ENGINEERING - MINIA UNIVERSITY Production Engineering and Mechanical Design Department

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 coefficient of fluctuation of speed.

Let,

N₁ = Maximum speed in rpm during the cycle,

N₂ = Minimum speed in rpm during the cycle, and N = Mean speed in rpm

N = ^N¹2⁺^N²

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Therefore, the coefficient of fluctuation of speed

❑ The coefficient of fluctuation of speed is a limiting factor in the design of flywheel.

❑ It varies depending upon the nature of service to which the flywheel is employed.

The following table shows the permissible values for coefficient of fluctuation of speed for some machines.

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Note:

The coefficient of fluctuation of speed is known as coefficient of steadiness and it is denoted by “m”.

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The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation.

Consider a turning moment diagram for a single cylinder double acting steam engine as shown in Fig.

The vertical ordinate represents the turning moment and the horizontal ordinate represents the crank angle.

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➢ It is defined as the ratio of the maximum fluctuation of energy to the work done per cycle. It is usually denoted by “C_E”.

➢ Mathematically, coefficient of fluctuation of energy can be expressed as follows:

➢ The work done per cycle may be obtained by using the following relations:

Work done per cycle = T_mean x 

 = Angle turned in radius per revolution

cycle per

done Work

energy of

n fluctuatio Maximum

C_E =

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Where

T_mean = Mean torque, and

 = Angle turned in radians per revolution

 = 2 π, in case of steam engines and two stroke internal combustion engines.

 = 4 , in case of four stroke internal combustion engines.

The mean torque (T_mean) in Nm may be obtained by using the following relation i.e.,

= 



=  P

N 2

60 T

_mean

P

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The work-done per cycle may also be obtained by using the following relation:

Where,

n = Number of working strokes per minute.

= N, in case of steam engines and two stroke internal combustion engines(ICE).

= N / 2, in case of four stroke (ICE).

n 60 cycle P

per done

Work 

=

(10)

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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 the axis of rotation in kg.m²

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I = m.k²

N₁ = Maximum speed during the cycle in r.p.m., N2 = Minimum speed during the cycle in r.p.m.,

₁ = Maximum angular speed during the cycle in rad. / s,

₂ = Minimum angular speed during the cycle in rad. / s,

2

1 N

rpm N in

cycle the

during speed

Mean

N = = +

. 2 sec /

. ¹ ²

 = Mean angular speed during the cycle in rad = ⁺

2

2 1

2

1 −  +

=

= or

N N speed N

of n fluctuatio of

t coefficien C_S

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The mean kinetic energy of the flywheel can be expressed as follow:

As the speed of the flywheel changes from ₁ to ₂ , the maximum fluctuation of energy:

joules or

m . N in

2 mk I 1

2

E = 1 ² =  ²²

2 2 2

1 I

2 I . 1 E . K Minimum .

E . K Maximum

E = − =   − 



( ) ( )

 

⁽ 1 2⁾⁽ 1 2⁾

2 2 2

1 I

2 I 1

2

E = 1  −  =   +  −



( )

(

₁ ₂

)

2 1

E 2 1



−



=



 +



=



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The radius of gyration (k) may be taken equal to the mean radius of the rim (R), because the thickness of rim is very small as compared to the diameter of rim.

S 2 2 S

2

2 2 2 1

C . . k . m C

. . I E

k . m I

, As

I E



=



=



=



−

 

=



S

2 S

2

C . E . 2 E

2 I E 1 ,

if and

C . E 2 E

2 I E 1 ,

If

=







=







=

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❑Therefore substituting k = R in equation (ii), we have:

❑From this expression, the mass of the flywheel rim may be determined.

Notes:

1. In the above expression, only the mass moment of inertia of the rim is considered and the mass moment of inertia of the hub and arms is neglected. This is due to the fact that the major portion of weight of the flywheel is in the rim and a small portion is in the hub and arms. Also the hub and arms are nearer to the axis of rotation, therefore the moment of inertia of the hub and arms is very small.

) R v

....(

...

C . v . m C

. . R . m E

S 2 S

2

2  = = 

=



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2. The density of cast iron may be taken as 7260 kg/m³ and 7800 kg/m³ for cast steel,

3. The mass of the flywheel rim is given by:

m = Volume × Density = 2 π R × A × ρ

expression, we may find the value of the cross-sectional area of the rim.

Assuming the cross-section of the rim to be rectangular, then A = b × t

Where,

b = Width of the rim, and t = Thickness of the rim.

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Example 1.

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 = 1º. 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.

Determine the mass of 300 mm diameter flywheel rim when the coefficient of fluctuation of speed is 0.3% and the engine runs at 1800rpm. Also determine the cross-section of the rim when the width of the rim is twice of thickness. Assume density of rim material as 7250 kg / m³.

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Given : D = 300 mm or R = 150 mm = 0.15 m ; C_S = 0.3% = 0.003 ; N = 1800 rpm or ω = 2 π × 1800 / 60 = 188.5 rad/s ; ρ = 7250 kg / m³

Mass of the flywheel

The turning moment diagram is shown in Fig.

Since the scale of turning moment is 1 mm = 5 N-m, and scale of the crank angle is 1 mm = 1° = π / 180 rad,

therefore 1 mm² on the turning moment diagram.

= 5 × π / 180 = 0.087 N-m Let the total energy at A = E_A Energy at B = E_A+ 295

Energy at C = E_A+ 295 – 685 = E_A– 390

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Energy at D = E_A– 390 + 40 = E_A– 350 Energy at E = E_A– 350 – 340 = E_A – 690 Energy at F = E_A– 690 + 960 = E_A + 270 Energy at G = E_A+ 270 – 270 = E_A

∴ Maximum energy = E_A + 295 and minimum energy = E_A – 690

We know that maximum fluctuation of energy,

Δ E = Max energy - Min energy = (E_A + 295) – (E_A – 690) = 985 mm²

= 985 × 0.087 = 86 N-m

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We also know that maximum fluctuation of energy (Δ E), 86 = m.R².ω².C_S = m (0.15)² (188.5)² (0.003) = 2.4 m

∴ m = 86 / 2.4 = 35.8 kg.

Cross-section of the flywheel rim

b = 2 t ...(Given)

∴ Cross-sectional area of rim, A = b × t = 2 t × t = 2 t² We know that mass of the flywheel rim (m),

35.8 = A × 2πR × ρ = 2t2 × 2π × 0.15 × 7250 = 13 668 t2

∴ t2 = 35.8 / 13 668 = 0.0026 or t = 0.051 m = 51 mm Ans.

and b = 2 t = 2 × 51 = 102 mm

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Example 2.

A single cylinder double acting steam engine develops 150 kW at a mean speed of 80 rpm. The coefficient of fluctuation of energy is 0.1 and the fluctuation of speed is ± 2% of mean speed. If the mean diameter of the flywheel rim is 2 m and the hub and spokes provide 5 percent of the rotational inertia of the wheel, find the mass of the flywheel and cross-sectional area of the rim. Assume the density of the flywheel material (which is cast iron) as 7200 kg / m³.

Given : P = 150 kW = 150 × 10³ W ; N = 80 rpm ; C_E = 0.1; ω₁ – ω₂

= ± 2% ω ; D = 2 m or R = 1 m ; ρ = 7200 kg/m³

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Mass of the flywheel rim

Let m = Mass of the flywheel rim in kg.

We know that the mean angular speed, 𝜔 = ^{2 𝜋 𝑁}

60 = ^{2 𝜋 80}

60 = 8.4 𝑟𝑎𝑑/𝑠

Since the fluctuation of speed is ± 2% of mean speed (ω), therefore total fluctuation of speed,

ω1 – ω2 = 4 % ω = 0.04 ω

and coefficient of fluctuation of speed, 𝐶_𝑠 = ^𝜔¹^{− 𝜔}²

𝜔 = 0.04

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We know that the work done by the flywheel per cycle

= ^{𝑃× 60}

𝑁 = 150×1000×60

80 = 11250 𝑁. 𝑚

We also know that coefficient of fluctuation of energy, 𝐶_𝐸 = Maximum fluctuation of energy

Work done / cycle

∴ Maximum fluctuation of energy,

ΔE = C_E × Work done / cycle = 0.1 × 112500 = 11250 N-m

Since 5% of the rotational inertia is provided by hub and spokes, therefore the maximum fluctuation of energy of the flywheel rim will be 95% of the flywheel.

∴ Maximum fluctuation of energy of the rim, (Δ E)rim = 0.95 × 11250 = 10 687.5 N-m

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We know that maximum fluctuation of energy of the rim (Δ E)rim, 10 687.5 = m.R².ω².CS = m × 1² (8.4)² 0.04 = 2.82 m

∴ m = 10 687.5 / 2.82 = 3790 kg Cross-sectional area of the rim

Let A = Cross-sectional area of the rim.

We know that the mass of the flywheel rim (m),

3790 = A × 2πR × ρ = A × 2π × 1 × 7200 = 45 245 A

∴ A = 3790 / 45245 = 0.084 m²