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Author Topic: How do fly-wheels work?  (Read 9300 times)

Offline McQueen

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How do fly-wheels work?
« on: 08/05/2008 09:16:18 »
Fly-wheel power

A fly-wheel is in effect a mechanical battery. In the  early 1950s the Swiss company Oerlikon introduced the Gyrobus simultaneously in Switzerland, Leopoldville in the Congo and Belgium. The flywheel had a diameter of 1.5 metres (5ft.approx.), weighed 1.6 metric tones and was charged to 3000 rpm.  The fly-wheel was charged electrically by overhead electric booms.  Once fully charged the gyrobus could travel from 6 Km 12 Kms at speeds of 50 Kms/hr. 60 Kms/hr. (on flat ground).  Unfortunately none of the buses lasted more than a few years, they were extremely difficult to drive due to precession of the fly-wheel and so were almost impossible to steer, also they were extremely heavy about 10 tonnes. Braking was done electrically and was fed back to power the fly-wheel. You can see the details of these gyrobuses here :
The kinetic energy of such a flywheel can be calculated using the equation (at least  Ill give it a try):
E k    = ( * I * w)
Where I = the inertial momentum of the disc which can be calculated by I = * mr2
and w = the angular velocity in radians. r = radius.
Taking the figures which have been given above the Kinetic energy of the fly-wheel on the gyrobus can be worked out :
M = 3,520 lbs.
3000 rpm = 50 rps
r = 2.46 ft.
Then I =  mr^^2/2 =  3520 * 6.05 / 2 =  10,650.82
The Kinetic energy would then be :  (1/2 * 10,650.82 * (50  * 6.28)^^2
= (5,325.41 * 98596 )  = 525,023,700  ft. poundal/sec = 16,406,990 ft lbs/sec
= 22,247 Kw
= 6.18 kwh.
In metric :
M= 1600
R = 1.5
w = 50 x 6.28 =  314 rads/s
I =  mr^^2/2 =  1600 * 0.56 /2  = 448 Kg m^^2.
K.E = I * w = 448 * 314^^2/2 = 22,085,504 Joules = 6434 Kw
= 6 Kwh and assuming it ran for 10 minutes it would have an output of  50 hp.
So the fly-wheel in the gyrobus had an energy of  6.1 Kwh approx.  which equals an output of 50 hp for 10 mins.

If you take a smaller fly-wheel :
M =  110 lbs
Rpm = 30,000 = 500 rps
R = 0.75 ft
Then I = (1/2 * 110 * 0.56) = 30.8
E k = (1/2 * 30.8 * (500 * 6.28)^^2 = 151,837,840 ft. poundal/sec = 4,744,932 ft lb/sec
=  6434 Kw = . 1.78 Kwh which would give 14.38 hp over 10 minutes.

In metric:
M =  50
R = 0.23m
w = 3140 rad/sec
I = 50 * 0.25^^2 =  50 * 0.05 = 2.64/2 = 1.32 Kg m^^2.
K.E = (1.32 * 9859600)/2 = 6,519,660 w/sec
= 6519.7 Kw =  1.8Kwh

If you increase r to 1ft.
M= 110 lbs
w = 3140 rad/sec
r = 1
I = 55 lb ft sq
K.E = (55 x 9859600) /2 = 271,139,000 poundal/sec. = 8,473,093 ft lb/sec =
11,489 Kw/sec = 3.1 Kwh.

In metric :
M= 50
R = 0.3
w = 3140 rad/sec
I =  2.25 Kg sec sq.
K.E = (2.25 * 9859600)/2 =  11,092,050 w/sec =  3.09 Kwh. Or 24 hp over 10 minutes.
   30,000 rpm is a relatively low speed for a flywheel, and precession can be overcome by the use of gimbals.  The Rotary Pulse Jet would be the ideal solution to power such a system since it can be small enough to place the whole system on gimbals, with leads from the generator,  going to electric motors.   The Auragen electrical generator would be ideal for such an application used in multiples since it has very small physical dimensions and has already been tested extensively by the military.


[MOD: subject altered to make it a clear question. CS]
« Last Edit: 27/05/2008 09:23:08 by chris »


 

lyner

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Re: How do fly-wheels work?
« Reply #1 on: 08/05/2008 10:55:10 »
If you mount the flywheel with a vertical axis there is virtually no precession because there would be very little tilting on a slow bus. It would tend to steer left or right, depending in whether you are accelerating or decelerating, though.
I wonder how long they took to 'spin up' the gyro; quite a stressful operation.
The scenario of a serious collision is a scary one; the flywheel would do a lot of damage if it broke free or even if the mounting became a bit wobbly.
btw, your rotary pulse jet seems to be your solution to virtually everything. Have you considered it as a serious construction project? Is there a prototype? You might then see the shortcomings of a reaction engine at such low speeds. Looking at your web pages again, I wonder in which direction you would point the exhaust which is high speed and full of energy. Backwards to go and forwards to stop?
« Last Edit: 08/05/2008 10:58:27 by sophiecentaur »
 

Offline McQueen

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Re: How do fly-wheels work?
« Reply #2 on: 08/05/2008 11:05:47 »
The gyro took anything between 30 secs and 3 minutes to charge up to speed. One of the reasons that I feel that the RPJ can fit into a lot of solutions is because it does offer pure rotary motion and thus has a lot more flexibility in its application I might just go forward with that thread which will once and for all demonstrate that the RPJ will work and what's more you can do the demonstration at home. How's that.  I think there is still a lot of confusion on what exactly I  expect the engine to do so this should clinch it.  As for the exhaust if you have the whole thing on gimbals the exhaust would have to be able to move with it, no great problem just needs a bit of thought. Your suggestion of a vertically mounted flywheel is excellent, probably that was how it was done in the original gyrobus. There used to be an article about a gyrobike but i can't seem to find it.
« Last Edit: 08/05/2008 11:10:53 by McQueen »
 

lyner

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Re: How do fly-wheels work?
« Reply #3 on: 08/05/2008 15:51:58 »
Quote
no great problem just needs a bit of thought
I think it is quite a big prob, actually.
I don't think you, personally, will see the problems until you actually try to design and build a working model.
The nearest thing to your idea is  gas turbine and to make one work efficiently, you need multiple pressure stages and several sets of blades at each pressure. There was one installed in a Rover(?) car in the 60s(?) and it worked fine on the open road but you couldn't use it at slow speed.  Your system only gets one go at the energy of combustion and then lets it all out. How can that be efficient or even effective?
I have no doubt that it would rotate but how much useful power will it deliver from 1litre of fuel every 5 minutes (which is what a modest car engine does, at high speed)?
 

Offline McQueen

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Re: How do fly-wheels work?
« Reply #4 on: 17/05/2008 15:45:37 »
Continuing the discussion on fly-wheel power. Hitherto using linear to rotary conversion engines such as the IC piston engine, it was impossible to use fly-wheels effectively for obvious reasons. Even the Wankel Engine which is often referred to as a rotary engine would not be suited for applications with fly-wheels because of its eccentric action and output. Having determined the amount of power that would be available from a 50 Kg flywheel at 30,000 rpm, it is time to determine how much power would be needed to bring the flywheel upto its rotational speed, in order to do this we have to multiply it moment of inertia by its angular acceleration .
Angular acceleration is represented by the greek symbol alpha. Taking w1 as 0 and w2 as 3140 rad/sec we have w1 w2 / t . If the time is equal to one minute  then we have 3140/60 =  52.3 rad /sec sq. I = 2.25 kg m sq I * alpha = 2.25 * 52.3 = 1.175 Kw/sec, which makes the scheme highly practical and also means that the engine has to be fired for only one minute in every ten which should make it extremely economical.
« Last Edit: 17/05/2008 18:19:41 by McQueen »
 

lyner

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Re: How do fly-wheels work?
« Reply #5 on: 18/05/2008 00:42:48 »
Quote
1.175 Kw/sec
What sort of unit is that? Rate of change of power?
You can get any 'power' you like from a flywheel with the appropriate gearing - as long as you only want it for a short time. It's surely the energy that counts - as usual.

The rotational energy will surely be 'I omega squared/2' where I is moment of inertia. You must get your basics right or your sums will not be reliable.
 

Offline McQueen

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Re: How do fly-wheels work?
« Reply #6 on: 18/05/2008 06:08:59 »
Quote
The rotational energy will surely be 'I omega squared/2' where I is moment of inertia. You must get your basics right or your sums will not be reliable.
I am pretty sure of the figures in the first post b'cos I checked and double checked them and did them in both metric and imperial so they should be correct. I'm not so sure about the calculation in the second post but it  should be approximately correct  and so should serve as a guideline on what to expect. Yes a fly-wheel does have quite a lot of power. In the article on gyrobuses I had referred to, it is plain that they could run for distances of up to 10 - 12 kms at a speed of 60 kmh, so it appears that a fly-wheel can let out its power slowly as well as all at once.  The main problem of course, is that we have no purely rotational output motors (except for electric ones) and so the question of what a flywheel can do has remained more or less moot. P.S. I have a more accurate answer to this now:
It turns out that this is simpler than it looks, the fly-wheel is essentially an energy storage device so the amount of power that  can be taken from it is the same as the amount of power that is put into it. So Energy = I x 0.5 x w ^^2.  If you are going to take 3.4Kwh out of the flywheel over a time of 10 minutes and you want to bring it back to speed in 1 min, it is necessary to put in 3.4K Joules of energy in 1 min. Which works out to about 204 Kilojoules/sec. Which works out to about 4.5 hp over a minute. This is still very reasonable when you think about it.
« Last Edit: 18/05/2008 13:55:02 by McQueen »
 

lyner

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Re: How do fly-wheels work?
« Reply #7 on: 18/05/2008 16:44:47 »
But what do your figures mean?
If the units are meaningless then the numbers with them are meaningless, too.
Kilowatt hours is a measure of energy - kilowatts is a measure or power.

3.4kW, running for 10 minutes would be (3.4/6 )kWhr = 0.59kWhr
OR
3.4kWhr taken in 10 minutes would mean a power of (3.4 X 6)kW = 20.4kW.

to put this 'back' in 1minute would involve ten times the power =204kW. That is quite high (Nearly 100A at mains volts, for a motor, for instance)
In horsepower it is 204/0.76 = 268hp. Quite a lot but quite possible.

I don't see why you think a 'purely rotational' motor would, per se, have real advantages. The KE lost in a reciprocating engine is far less than you seem to imply - particularly in a multi cylinder engine, where the ke of one piston is transferred to the KE of another, not all lost. The pistons of short stroke engines don't go all that fast, really.
Stick to SI units - imperial doesn't help, really and it's a bit like sticking to Farenheit temperatures just for comfort.
« Last Edit: 19/05/2008 13:24:55 by sophiecentaur »
 

Offline McQueen

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Re: How do fly-wheels work?
« Reply #8 on: 20/05/2008 05:39:29 »
The fact that keeps this thread interesting so that it does not deteriorate in to a Yes it can !  .. No it cant ! type of argument is that  there is, at least,  some physical and mathematical basis to support the arguments. So far we have been able to get a fairly good idea of  how much power a fly-wheel can store under given conditions,  how much energy would be needed to store that energy in the fly-wheel in the first place , get an idea of what the power output from that fly-wheel would be over a given period of time and lastly to estimate whether it would be possible for an engine, either an IC piston engine or the rotary Pulse jet Engine to supply the needed power to the fly-wheel.  As regards the last point, the argument that a 12 cylinder engine, (since it gives almost continuous output would be as efficient as a Rotary Pulse Jet ) is spurious for the following reasons:- In a Rotary Pulse Jet engine with combustion chambers having a base( the combustion chamber is designed to direct all the resultant pressure from the combustion of gases to the back plate ) of 3 ins diameter the total force on the back plate from an internal pressure of 500 psi  would be 3532 lbs force. If both combustion chambers fire simultaneously the force generated would be twice that or 7064 lbs force, acting in the same direction. The advantage comes from the fact that the reactive force resulting from the expulsion of gases is acting perpendicular to the axis, (in the manner of a lever attached to the axle) so the force can be calculated directly in Newton metres or foot pounds  depending on the distance from the axle. Thus if the force is applied 1 ft. from the axle it results in 1 foot pound at the axle. Since in the RPJ the combustion chambers are situated 6 ins. from the axle the force would be (taking both combustion chambers into account ) 3532 ft lbs or roughly 20 times the torque of a 4 x 4 piston engine which delivers around 160 ft pounds of torque. (See . http://www.geocities.com/rotarypulsejet/page4.htm ) beginning at the fourth para from the bottom of the page.  The actual force might be less taking other losses into account but still an impressive improvement on the IC piston engine. (. P.S the example at the web-site is given in foot pounds.)
 

lyner

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Re: How do fly-wheels work?
« Reply #9 on: 23/05/2008 21:30:08 »
Do you not understand, McQueen, that Force is not Energy. It is Energy that takes you from A to B. I can use a car jack to lift a 10tonne vehicle by a few cm, which requires an enormous force (100,000N), the amount of actual Energy I put in would be a few hundred kJ.
The Energy wasted in the KE of your exhaust gases cannot be ignored. Unless they are 'slowed down' and transfer the KE to some mechanical component you have LOST that energy. During the 'reaction' in the pulse jet chamber some energy will go to the KE of the rotor, true, but unless you do something about the KE in the exhaust, the rest of the energy is wasted.
You really have to accept these simple basics of School Science.

 

Offline McQueen

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How do fly-wheels work?
« Reply #10 on: 02/06/2008 14:52:44 »
sophiecentaur:
Quote
Do you not understand, McQueen, that Force is not Energy. It is Energy that takes you from A to B. I can use a car jack to lift a 10tonne vehicle by a few cm, which requires an enormous force (100,000N), the amount of actual Energy I put in would be a few hundred kJ.
You are missing out on the whole argument, we are talking  about gases under pressure, like when a scuba tank is punctured and takes out a whole block of buildings!!! So, yes power is present and it can be harnessed usefully!!!
 

lyner

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How do fly-wheels work?
« Reply #11 on: 02/06/2008 18:28:06 »
We're in the land of definitions here McQ.
Power is rate of transfer of energy. If you let the air out of the cylinder very slowly then that is a low power. Let it out fast and it is high power. Same amount of energy transferred. You can't have a sensible conversation unless you either use the correct terms or define you own terms accurately and consistently.
There's a very high Power from a camera flash but not many Joules of energy - it's all over in a flash!
It doesn't matter what the context is, you still must use the right terms or it is gobbledegook.
 

Offline syhprum

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How do fly-wheels work?
« Reply #12 on: 24/08/2008 15:32:16 »
The construction regulations for F1 race cars allow for the use of KER (kinetic energy recover) systems of up to 400KJ from braking in the 2009 season.
Some teams are researching electrical systems while others contemplate vacuum enclosed flywheels running at up to 80,000 RPM.
It would be handy if correspondents who wish to use non SI units could provide a link to conversion tables
 
« Last Edit: 24/08/2008 15:34:53 by syhprum »
 

lyner

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How do fly-wheels work?
« Reply #13 on: 24/08/2008 17:56:44 »
 
sophiecentaur:
Quote
Do you not understand, McQueen, that Force is not Energy. It is Energy that takes you from A to B. I can use a car jack to lift a 10tonne vehicle by a few cm, which requires an enormous force (100,000N), the amount of actual Energy I put in would be a few hundred kJ.
You are missing out on the whole argument, we are talking  about gases under pressure, like when a scuba tank is punctured and takes out a whole block of buildings!!! So, yes power is present and it can be harnessed usefully!!!

Perhaps, McQueen, you could agree on some definitions of what we all mean by Power, Work, Energy, Force and a few other things. You can't have a scientific discussion if you keep using salesman's language.  I think you cannot be agreeing with my (generally accepted) definitions of the terms as I am using them.
'Power' is not a thing which can be 'present'; it is a rate of energy transfer. Energy can be present and you can use it at a certain rate.

Whether we are talking about gases under pressure, flywheels spinning or barrels of oil the technical  terms are all the same.
 

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How do fly-wheels work?
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