Energy Stored in a Flywheel :Ā A flywheel is shown in Fig. 1. We have already discussed that when a flywheel absorbsĀ energy its speed increases and when it gives up energy its speed decreases.

Let m = Mass of the flywheel inĀ kg,

k = Radius of gyration of theĀ flywheel in metres,

I = Mass moment of inertia ofĀ the flywheel about theĀ axis of rotation in kg-m2

= m.k2,

N1 and N2 = Maximum and minimumĀ speeds during the cycle inĀ r.p.m.,

Ļ1 and Ļ2 = Maximum and minimumĀ angular speeds during the cycle in rad / s,

N = Mean speed during the cycle in r.p.m. = N1 + N2 /2

Ļ = Mean angular speed during the cycle in rad / s =Ā Ļ1+Ļ2 / 2

Cs = Coefficient of fluctuation of speed = N1-N2/N orĀ Ļ1-Ļ2 / Ļ

We know that mean kinetic energy of the flywheel,

E =Ā 1/2Ā Ć IĀ Ļ^2 = 1/2Ā Ćm k^2 Ļ^2Ā (in N-m or joules)

As the speed of the flywheel changes from Ļ1 to Ļ2, the maximum fluctuation of energy,

ĪE = Maximum K.E. ā Minimum K.E. = (1/2Ā Ć IĀ Ļ1^2)-(1/2Ā Ć IĀ Ļ2^2)

=Ā 1/2Ā Ć I ( Ļ1^2 – Ļ2^2)

= 1/2Ā Ć I ( Ļ1 +Ā Ļ2)( Ļ1 – Ļ2)

=Ā I.Ļ (Ļ1 ā Ļ2) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā (‘.’ Ļ = Ļ1 +Ā Ļ2/2) ……(i)

=Ā I.Ļ^2Ā (Ļ1 ā Ļ2) /Ļ Ā Ā Ā Ā Ā Ā Ā Ā …[Multiplying and dividing by Ļ]

=Ā I.Ļ^2.Cs = m.k^2.Ļ^2.Cs Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā …( ‘.’Ā I = m.k^2) Ā Ā ……(ii)

=Ā 2 E.Cs Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā (‘.’ E = 1/2Ć I Ļ^2) Ā Ā Ā Ā Ā Ā Ā Ā Ā Ā ……(iii)

The radius of gyration (k) may be taken equal to the mean radius of the rim (R), because theĀ thickness of rim is very small as compared to the diameter of rim. Therefore substituting k = R inĀ equation (ii), we have

Ī E = m.R^2.Ļ^2.CS = m.v^2.CS Ā Ā Ā Ā Ā Ā Ā Ā Ā …(Q v = Ļ.R )

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 massĀ moment of inertia of the hub and arms is neglected. This is due to the fact that the major portion of weight of theĀ 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 ofĀ rotation, therefore the moment of inertia of the hub and arms is very small.

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

3. The mass of the flywheel rim is given by

m = Volume Ć Density = 2 Ļ R Ć A Ć Ļ

From this expression, we may find the value of the cross-sectional area of the rim. Assuming theĀ cross-section of the rim to be rectangular, then

A = b Ć t

where b = Width of the rim, and

t = Thickness of the rim.

Knowing the ratio of b / t which is usually taken as 2, we may find the width and thickness of rim.

4. When the flywheel is to be used as a pulley, then the width of rim should be taken 20 to 40 mm greaterĀ than the width of belt.

Reference A Textbook of Machine Design by R.S. Khurmi and J.K. Gupta

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