I need to choose a suitable PMdc motor for a flywheel (Diametre = 0.38m, m =10kg). The flywheel need to utlimately rotate at 3000rpm over 3 hours.

The PMdc that I have is 24V, 3000rpm, 0.16N.m, 3.3A, 1.2kg and its shaft diametre is 7mm.

So, I need to know whether this motor can drive the flywheel described above using proper bearings? Also the starting torque of the flywheel?

I will appreciate your inputs.

 

Sure it will spin the flywheel, with proper bearings. What you haven't mentioned is what power you may be taking out. The size of the motor will control how fast you can bring it up to speed, but it takes very little power to keep it spinning if that's all it is doing. The starting torque of the flywheel will be the locked rotor torque of the motor multiplied or divided by any gearing, etc. Inertia works that way. All of this should be intuitive if you are going to play with stored energy.

Regards
cww

 

The same recommendations you need to take with amplifier choice, i.e. high peak and low contunie current (ratio app.5:1). Shunt regulator need for such flywheel application as a rule.

 

Thank you Curt,

Sorry to have replied late.

I want to ultimately to take 30 watts over 5min out of the flywheel. So hopefully it should do that? Can you confirm it can get this power out of the flywheel during the discharge?

Also, there is one thing that is bothering a lot: How can I simulate the whole flywheel system especially the discharge mode so that I am sure to get what to expect?

Any ideas on how to do that?

 

Hi Leonid,

How can I properly implement a shunt regulator?
Any ideas on how I can simulate the flywheel system in especially the discharge mode? I do not know which computer program or software can help me do that or any other method. I hope to get an answer.

 

It may be quite difficult to take out 30 watts for 5 minutes, and here is why: Assuming the best possible distribution of mass in the flywheel (all the mass out at .19 m from the center), the flywheel rotating at 3000 RPM will have 17814.6 Joules of kinetic energy stored. This is equivalent to 4.95 watt-hours, or 59 watts over 5 minutes. Therefore taking 30 watts over 5 minutes is taking over half the energy it has. This will bring the speed down to 2121 RPM. Since the flywheel cannot have such an optimal mass distribution, the actual situation must be worse than this. If the actual radius for most of the mass was only .13 m, then taking 30 watts over 5 minutes would be taking all it had. And that assumes no losses at all in the bearings.

Robert Scott
Real-Time Specialties

 

Perhaps one of the physicists on the list will do the calculus for total kinetic energy and the remaining value after 5 minutes. I suspect one would need the wheel profile to get the moments right. At least that's my thinking after midnight.

Regards

cww

 

Hi Robert,

I am not really sure of your approach. Can you explain to me how you got 30 watts over 5 min to bring the speed down to 2121 RPM?

I have calculated the stored energy using E=0.5*I*w2 which should be 8907.3 instead of 17814.6. May be I am wrong? Can you clarify it more?

If you have also any ideas on how to simulate the flywheel discharge rate analytically or by software program, I will be very glad to know it.
I am finding it very hard to get all the things right in my design.

 

Some servo amp manufacturers have power supply with build-in shunt regulator (Elmo, AMC, Copley, etc). Calculation of shunt regulator you may find inside Galil website.

Simulation of flywheel system may performed by MATLAB/Simulink as system with very high mismatch - you need to calculate it. Usually mismatch is more than 1:20.

 

I have tried the Matlab approach but it's just that I am noy getting right or may be I am not implementing it properly. I don't even the mismatch that you've talked about. My problem is I don't have any experience about simulating mechanical moving parts. I have very little background on mechanical engineering. I am not really sure of the MAtlab/Simulink method.

 

Here is how I figured it: If all the mass is at .19 m from the center, then one rev moves that mass 1.1938 m (the circumference). 3000 RPM means 50 revs per second. So in one second, the mass moves 50 x 1.1938 m = 59.69 m/sec. The energy can be calculated as if the mass were moving at 56.69 m/sec in a straight line. The formula for linear kinetic energy is:

1/2 m v^2

which is .5 x (10 kg) x 3562.8961 = 17814.48 Joules, which is 0.004948 kw-hr, or 4.948 watt hours. If we reduce the time from 1 hour to 5 minutes, then the power goes up from 4.948 to 59.376 watts. Therefore if you take out 30 watts over that time period, you will be taking out half the energy that is there. Since the energy is proportional to the square of the velocity, then cutting the energy in half cuts the velocity down by the square root of 2, which means 3000 x .7071 = 2121 RPM. You don't need to simulate the flywheel. You just have to assume conservation of energy.

Robert Scott
Real-Time Specialties

 

Ok, Robert.
From your first post, you said it will be difficult to take out 30 watt over 5min.And from your calculations, you show I should be able to get 30 Watt over 5min and brings the speed down to 2121rpm. Does this mean that I should ideally get the 30 watt over 5min plus another 30watt over 5min? Is there any catch involved?

 

In theory, you are right. It can be done. However that assumes there is no loss due to friction and that all the mass of the flywheel was out at the edge (.19 m from the center). If there is friction loss and if the mass is distributed over different radii, then you would have to do a more precise calculation of the moment of inertia. Finally, the process of extracting exactly 30 watts from a gradually slowing generator is complicated by the fact that the voltage output will generally decrease as the RPM decreases. If the load is resistive, then it will take less power as the speed decreases. Of course you could manage the power using devices that regulate the voltage. But any such regulation device has some inefficiencies of its own. Taking all these factors into consideration, I just think the job of getting 30 watts for 5 minutes is not trival.

Robert Scott
Real-Time Specialties

 

Hello,

It depends on the shape of the flywheel: for a solid disk, it's 8907J; for a spoked flywheel with negligible spokes, it's 17815J. Either way, getting 9000J of usable energy out of it seems implausible.

Jiri
--
Jiri Baum <[email protected]> http://www.baum.com.au/~jiri

 

I now understand what you’re trying to explain. You’re right about the voltage decreasing with the speed and I was exactly thinking of using a regulator circuit using a boost converter. It will keep on boosting the voltage when it decreases from a reference value due to speed. This involves closed –loop control of the system. For now that is my ides about. If there is a better way or any parts that can do that, you may let me know, please. As you know dealing with flywheel may be dangerous, if there is any measure of precaution that I can do, you may please advise me to do so.

 

...and how about that 3 hour requirement in the first posting? If the flywheel has 5 watt-hours of energy totally, then a friction loss of only .83 watts will use up half of that total energy in 3 hours. Is this flywheel spinning in a vacuum? I think it is going to take some really exceptional bearings to make considerably less than .83 watts of friction loss.

Robert Scott
Real-Time Specialties

 

In reply to Robert Scott: For a 12VDC PMDC automotive motor, typical efficiency is about 60% to 65%. Friction losses with the brushes, bearings, and windage are a significant proportion of the total losses.

You could probably quite safely raise your estimate of friction losses in this application by at least an order of magnitude.

 

No, the flywheel will not be running in vacuum. And the bearing that I would like to use is a ball bearing that can handle up to 38000 RPM (from catalogue). So, hopefully the friction loss will be less than 0.83. However, at this point I am not sure of it. I am still trying to find a better way.