How to choose , use and calculations of the Flywheels !

 

 

 Load Equalization :

If the load fluctuates between wide limits in space of few seconds, then large peak demands of current will be taken from supply and produce

heavy voltage drops in the system. Large size of conductor is also required for this Process of smoothing out these fluctuating
loads is commonly referred to as load equalization and involves storage of energy during light load periods which can be given out during the peak load period, so that demand from supply is approximately constant. Tariff is also affected as it is based on M.D. (Maximum Demand) For example, in steel rolling mill, when the billet is in between the rolls it is a peak load period and when it comes out it is a light load period, when the motor has to supply only the friction and internal losses, as shown in figure

 

 Use Of Flywheels :

 

The method of Load Equalization most commonly employed is by means of a flywheel. During
peak load period, the flywheel decelerates and gives up its stored kinetic energy, thus reducing the
load demanded from the supply. During light load periods, energy is taken from supply to accelerate
flywheel, and replenish its stored energy ready for the next peak. Flywheel is mounted on the motor
shaft near the motor. The motor must have drooping speed characteristics, that is, there should be a
drop in speed as the load comes to enable flywheel to give up its stored energy. When the Ward -Leonard system is used with a flywheel, then it is called as Ward - Leonard Aligner control.

There are two choices left for selecting a flywheel to give up its maximum stored energy:

 

1.  Large drop in speed and small flywheel (But with this the quality of production will suffer, since a speed drop of 10 to 15% for maximum load is usually employed).
2.  Small drop in speed and large flywheel. (This is expensive and creates additional friction losses. Also design of shaft and bearing of motor is to be modified.) So compromise is made between the two and a proper flywheel is chosen.

 

Flywheel Calculations :

The behavior of flywheel may be determined as follows 

 Fly wheel Decelerating :- (or Load increasing)

Let :

TL : Load torque assumed constant during the time for which load is applied in (Kg.m)

Tf  : Torque supplied by freewheel in (Kg.m)

To : Torque required on no load to overcome friction internal losses in (Kg.m)

Tm : Torque supplied by the motor at any instant in (Kg.m)

ωo  : No load speed of the motor in (rad/sec)

ω : Speed of motor at any instant in (rad/sec)

s : Motor slip speed ( ωo-ω) in (rad/sec)

I : Moment of inertia of flywheel in (Kg.m2)

g : Acceleration due to gravity in (m/sec2)

t : Time in (sec)

When the flywheel decelerates, it gives up its stored energy. 

Tm = TL - Tf  ......... Equation (1)

Energy stored by flywheel when running at speed ‘ω’ is 1/2 Iω2/g.

If speed is reduced from ω0 to ω. 

The energy given up by flywheel is

= (I/2g)*[(ωo)² - (ω)²]   ......... Equation (2)

(ωo +ω)/2 : Mean speed . Asumming speed drop of not than 10% , this may be assumed equal to ω .
(ωo +ω)/2 = ω 
Also   (ωo - ω)= s 
From equation (2),energy given up = (I/g).ω.s
Power given up = (I/g).ω.(ds/dt)
But the torque = ( Power/ω )
Torque supplied by freewheel
 Tf = (I/g)*(ds/dt)
From equation (1), Tm = TL - (I/g)*(ds/dt)
For values of slip speed up to 10% of No-load speed,slip is proportional to torque or , 
S = K * Tm
Tm = TL - (I/g) * K *(dTm/dt)

This equation is similar to the equation for heating of the motor 

(TL - Tm) = (I/g) * K * (dTm/dt) 

g*(dt/IK) = dTm / (TL - Tm )

By integrating both sides 

ln (TL - Tm) = [(t*g)/(I*K)] + C1     ............ Equation (3)

At t = 0 , When load starts increasing from no load

 i.e. Tm = To

Hence, at t = 0       Tm = To 

C1 = - ln ( TL - To )

By substituting the value of C1 above, in equation (3) in ( TL - Tm ) = ( t*g)/(I*K) - ln ( TL - To )

ln [(TL - Tm)/(TL - To)] = - (t*g)/(I*K)

Tm = TL - (TL - To)* e^[( - t*g)/(I*K)]

If the load torque falls to zero between each rolling period, then Tm = TL - [1 - e^(- t*g)/(I*K)] , so  ( To = 0 )

Load Removed (Flywheel Accelerating)

Slip speed is decreasing and therefore ( ds/dt) is negative .

Tm = To + Tf  = To - [(I/g)*(ds/dt)]

To - Tm =(I/g)*K*(dTm/dt) 

(g*dt)/(I*K) = dTm / (To - Tm) 

After integrating both sides, 

ln (To - Tm) = [(t*g)/(I*K)] + C

At t = 0 , Tm = Tm'  motor torque at the instant, when load is removed .

C = - ln ( To - Tm' ) putting this value of C in the above equation 

ln (To - Tm) = [(t*g)/(I*K)] - ln(To - Tm')

ln [(To - Tm)/(To - Tm') = (- t*g)/(I*K)

To - Tm = (To - Tm')* e^[(-t*g)/(I*K)

Tm = To + (Tm' - To) * e^[(-t*g)/(I*K)]

Where Tm' = the motor torqur, at the instant the load is removed .

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