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 THE FLYWHEEL
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ron_o
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United Kingdom
1052 Posts

Posted - April 23 2014 :  17:29:37  Show Profile Send ron_o a Private Message  Reply with Quote
All pulse motors that have a rotor already have a flywheel , obvious really.

A flywheel stores energy and this stored energy can be released when required.

The main advantage of a flywheel as an energy storage device is that it never
looses it's efficiency ( obviously mechanical wear takes place over time )

The equation for the energy stored in a flywheel is :-

Energy stored ( joules ) = 1/2 x I x w^2

where
I = moment of inertia ( see below for more details )

w = rotational speed ( measured as angular velocity in radians per second )

MOMENT OF INERTIA :- I

I = k x m x r^2

Where

K :- is the Inertia Constant ...... please note that despite being called a "constant" it's
value actually depends upon the shape/form of the flywheel
for example :-

bicycle wheel , k = 1 ( the mass is at the circumference )

solid disk , k = 0.5 ( the mass is evenly distributed )

m :- is the total mass of the flywheel in Kg

r :- is the radius of the flywheel in metres

ANGULAR VELOCITY :- w

w = 2 x Pi x rpm / 60 radians/second

Where

Pi = 3.14159

rpm = revolutions per minute

EXAMPLE OF FLYWHEEL ENERGY STORAGE

flywheel mass = 0.420kg , flywheel diameter = 0.2 metres so r = 0.1 metres , rpm = 2000

First step is to convert rpm to angular velocity

w = 2 x Pi x rpm / 60

w = 2 x 3.14159 x 2000 /60

w = 209.44 rads/sec

Second step is to calculate the Moment of Inertia

We now have our first problem ! most pulse motors flywheels have the magnets
inset into the rotor plate so the actual value for the rotor constant is not known.
So initially a guess can be made and then the actual value can be "reverse engineered"

K = 0.7

I = k x m x r^2

I = 0.7 x 0.42 x 0.1^2

I = 0.00294

Finally step is to calculate the energy stored in the flywheel ( joules )

E = 0.5 x I x w^2

E = 0.5 x 0.00294 x 209.44^2

E = 64.48 joules

------------------------------------------

ron











Edited by - ron_o on April 23 2014 17:33:43

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ron_o
Moderator



United Kingdom
1052 Posts

Posted - April 23 2014 :  17:55:25  Show Profile Send ron_o a Private Message  Reply with Quote
THE FLYWHEEL part 2

Using the information and equations shown in part one it is possible to establish how much energy is held by the pulse motors rotor , this is important because it allows us to investigate ( and measure ) the change in system energy when for example a generator coil is loaded or shorted.

This is possible because the energy stored within the flywheel is measured in joules and has have said many times before 1 joule = 1 watt second.

Since we can measure rotational speed (rpm) , power supplied to drive the pulse motor(watts),
load duration etc . This information can be used to optimise coil design ( turns , shape , resistance , distance to rotor etc )

Looking forward to any comments

ron
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - May 08 2014 :  16:47:15  Show Profile Send ron_o a Private Message  Reply with Quote
THE FLYWHEEL part 3

A flywheel is as previously stated an energy storage device so now lets look at extracting energy from the flywheel.

Using the example given in part one let us see how much energy is released when the flywheel speed is allowed to fall from 2000 rpm to 1500 rpm.

The first step is to calculate the new value of angular velocity at 1500 rpm

w = 2 x Pi x rpm / 60

w = 2 x Pi x 1500 / 60

w = 3000 x Pi / 60

w = 157

The modified equation :-

E = 0.5 x I x ( w1^2 - w2^2 )

where w1 is the maximum speed and w2 is the minimum speed

E = 0.5 x 0.00294 x ( 209.44^2 - 157^2 )

E = 0.5 x 0.00294 x ( 43865 - 24649 )

E = 0.5 x 0.00294 x ( 19216 )

E = 28.24 joules ( approx )

NOTE .... the speed ( rpm ) has slowed down by ( 2000 - 1500 ) / 2000 = 0.25 or 25%
but the energy released is 28.24 / 64.48 = 43.79 % of its original value


This "released" energy ( if large enough ) could be used to say power a generator however conversion losses will be introduced so a conversion efficiency loss factor would need to be introduced to amend the final available energy value.

ron
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ron_o
Moderator



United Kingdom
1052 Posts

Posted - May 30 2014 :  16:07:16  Show Profile Send ron_o a Private Message  Reply with Quote
THE FLYWHEEL part 4

To calculate the torque required to accelerate a solid disk flywheel from rest to 2000 rpm in 5 seconds

Flywheel type :- solid disk , mass ( m ) = 0.64 Kg , diameter 0.250 metres ( r = 0.125 )

Initial angular velocity ( w1 ) is zero ( since it is at rest )

Final angular velocity ( w2 ) = rpm x 2 x Pi / 60 = 2000 x 2 x Pi / 60 = 209.4395 rad/second

Acceleration = ( w2 - w1 )/ ramp up time = ( 209.44 - 0 ) / 5 = 41.888 rad/second

Moment of Inertia I ( for a plain disk ) = m x r^2 / 2

I = 0.64 x 0.125^2 / 2 = 5 exp -3 Kg m^2

Torque = I x acceleration = 5 exp -3 x 41.888 = 0.20944 Nm

ron
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