The fluctuation of energy may be determined by the turning moment diagram for one complete cycle of operation. Consider the turning moment diagram for a single cylinder double acting steam engine as shown in Fig. 16.1.

**FLUCTUATION OF ENERGY**

The fluctuation of energy may be determined by the turning moment
diagram for one complete cycle of operation. Consider the turning moment
diagram for a single cylinder double acting steam engine as shown in Fig. 16.1.
We see that the mean resisting torque line *AF*
cuts the turning moment diagram at points *B*,
*C*, *D* and *E*. When the crank
moves from *a* to *p*, the work done by the engine is equal to the area *aBp* , whereas the energy required is
represented by the area *aABp*. In
other words, the engine has done less work (equal to the area *a AB*) than the requirement. This amount
of energy is taken from the flywheel and hence the speed of the flywheel
decreases. Now the crank moves from *p*
to *q*, the work done by the engine is
equal to the area *pBbCq*, whereas the
requirement of energy is represented by the area *pBCq*. Therefore, the engine has done more work than the
requirement. This excess work (equal to the area *BbC*) is stored in the flywheel and hence the speed of the flywheel
increases while the crank moves from *p *to* q*.

Similarly, when the crank moves from *q*
to *r*, more work is taken from the
engine than is developed. This loss of work is represented by the area *C c D*. To supply this loss, the flywheel
gives up some of its energy and thus the speed decreases while the crank moves
from *q* to *r*. As the crank moves from *r*
to *s*, excess energy is again
developed given by the area *D d E* and
the speed again increases. As the piston moves from *s* to *e*, again there is a
loss of work and the speed decreases. The variations of energy above and below
the mean resisting torque line are called ** fluctuations of energy.** The areas

A little consideration will show that the engine has a maximum speed
either at *q* or at *s*. This is due to the fact that the
flywheel absorbs energy while the crank moves from *p* to *q* and from *r* to *s*.
On the other hand, the engine has a minimum speed either at *p* or at *r*. The reason is that the flywheel gives out some of its energy
when the crank moves from *a* to *p* and *q* to *r*. The difference
between the maximum and the minimum energies is known as *maximum fluctuation**of
energy*.

**1 Determination of Maximum
Fluctuation of Energy**

A turning
moment diagram for a multi-cylinder engine is shown by a wavy curve in Fig.
16.4. The horizontal line *A G*
represents the mean torque line. Let *a*1,
*a*3, *a*5 be the areas above the mean torque line and *a*2, *a*4 and *a*6 be the areas below the mean torque
line. These areas represent some quantity of energy which is either added or
subtracted from the energy of the moving parts of the engine

Let the energy in the flywheel at *A* = *E*,
then from Fig. 16.4, we have

Energy at *B*
= *E* + *a*1 Energy at *C* = *E* + *a*1–
*a*2

Energy at *D*
= *E* + *a*1 – *a*2 + *a*3 Energy at *E* = *E* + *a*1 – *a*2
+ *a*3 – *a*4

Energy at *F* = *E*
+ *a* 1 – *a*2 + *a*3 – *a*4 + *a*5
Energy at *G* = *E* + *a*1 – *a*2 + *a*3
– *a*4 + *a*5 – *a*6 = Energy at *A* (*i.e.*
cycle repeats after *G*)

Let us now suppose
that the greatest of these energies is at *B*
and least at *E*. Therefore, Maximum
energy in flywheel

= *E* + *a*1
Minimum energy in the flywheel

= *E *+* a*1* *–*
a*2* *+* a*3* *–* a*4

∴ Maximum fluctuation of energy,

∆ *E *=* *Maximum energy – Minimum energy

= (*E* + *a*_{1}) – (*E* + *a*_{1} – *a*_{2} + *a*_{3} – *a*_{4})
= *a*_{2} – *a*_{3} + *a*_{4}

**Fig. 16.4. **Determination of maximum fluctuation of energy.

**2 Coefficient of Fluctuation of
Energy**

It may be
defined as the **ratio of the maximum
fluctuation of energy to the work done** **per
cycle. **Mathematically, coefficient of fluctuation of energy,

*C*E* *= Maximum fluctuation of energy / Work done per cycle

The work done per cycle (in N-m or joules) may
be obtained by using the following two

Relations

**1. **Work done per cycle =** ***Tmean*** **×** **θ

The mean
torque (*Tmean*) in N-m may be obtained
by using the following relation :

where P = Power transmitted in watts,

N = Speed in r.p.m., and

ω = Angular speed in rad/s = 2 π N/60

2. 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,

= *N* /2, in case of four stroke internal
combustion engines.

The following table shows the values of coefficient of fluctuation of
energy for steam engines and internal combustion engines.

Coefficient
of fluctuation of energy (*C*_{E})
for steam and internal combustion engines.

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