Notices
Results 1 to 23 of 23

Thread: flywheel

  1. #1 flywheel 
    New Member
    Join Date
    Sep 2006
    Posts
    2
    We where given a problem in class yesterday (sort of a keep you thinking while out of class, not worth any marks, but just a fun challenge).

    The teacher asked us to calculate the theoretical energy (in watts) of a given flywheel,

    these are the specs,

    I = Moment of enertial
    W = Rotational velocity (RPM) = 6000 (RPM)
    I = k*M*(R*R) = 5

    k=Inertial constant = 0.8
    M= Mass = 25KG
    R= Radious= 0.5m

    the formula is quite simple

    Kenetic Energy = 1/2 * I * (W*W)

    so once you plug in the numbers you get

    Kenetic Energy = 90 000 000

    I'd love to convert this to watts but I have no clue what the unit for that 90 million is.

    please help me.

    Thanks Tony


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    Energy is measured in joules, power is measured in watts.

    power=energy/time

    (you might want to convert your time to units of second as opposed to minutes)


    Reply With Quote  
     

  4. #3  
    New Member
    Join Date
    Sep 2006
    Posts
    2
    J/s to W/s is 1:1

    so 90 000 000J/s = 90 000 000 watts/second,
    Reply With Quote  
     

  5. #4  
    Forum Freshman
    Join Date
    Dec 2005
    Location
    India
    Posts
    65
    Billards and BigT both of you got the whole concept wrong.

    The body is in rotational equilibrium since it has constant angular velocity.
    Threfore torque acting on the flywheel is zero.
    Power=torue*angular velocity
    =0 watts (since torque=0)
    Reply With Quote  
     

  6. #5  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    J/s=W

    So: 90 000 000 J/s = 90 000 000 W = 90 MW (megawatts)
    Reply With Quote  
     

  7. #6  
    Guest
    That does sound very efficient......

    Sort of like Perpetual Motion with a lot left over.... like 89.9xx.xxxMW
    Reply With Quote  
     

  8. #7 Re: flywheel 
    Forum Freshman
    Join Date
    Dec 2005
    Location
    India
    Posts
    65
    Quote Originally Posted by BigT
    the formula is quite simple
    Kenetic Energy = 1/2 * I * (W*W)
    so once you plug in the numbers you get
    Kenetic Energy = 90 000 000

    Thanks Tony
    Our target variable is power not KE.
    For a rotational body,
    Power=Torque*angular velocity
    =0 (since the torque=0)
    Torque is equal to 0 since the body is in rotational equilibrium.
    The body is in rotational equilibrium since the its angular velocity is constant.
    Reply With Quote  
     

  9. #8  
    Forum Freshman
    Join Date
    May 2006
    Location
    maryland
    Posts
    18
    First, what you need to do is convert rev/min into radians per second to give a consistent set of units.

    i.e. multiply your answer by (2*pi) radians/rev * (1/60) min/sec = 987000

    Second, this gives the rotational kinetic energy in (Kg m/s^2)*m or N*m or Joules

    Finally, since the flywheel is creating this energy in a constant manner it is creating 987 kJ per every second. kJ/sec = kW
    Reply With Quote  
     

  10. #9  
    Forum Freshman
    Join Date
    Dec 2005
    Location
    India
    Posts
    65
    Quote Originally Posted by lance
    First, what you need to do is convert rev/min into radians per second to give a consistent set of units.

    i.e. multiply your answer by (2*pi) radians/rev * (1/60) min/sec = 987000

    Second, this gives the rotational kinetic energy in (Kg m/s^2)*m or N*m or Joules

    Finally, since the flywheel is creating this energy in a constant manner it is creating 987 kJ per every second. kJ/sec = kW
    Lance if you feel you are right tell me whats went wrong in my interpretation of this problem. I got power=0.Since,

    Our target variable is power not KE.
    For a rotational body,
    Power=Torque*angular velocity
    =0 (since the torque=0)
    Torque is equal to 0 since the body is in rotational equilibrium.
    The body is in rotational equilibrium since the its angular velocity is constant.
    Reply With Quote  
     

  11. #10  
    Guest
    Guy's

    What your teacher is aking is; "what is the Kinetic energy of the flywgheel in these conditions". If left on its own, it will slow down due to friction, the energy would 'bleed away' - he/she wants you simply to calculate the instananeous energy in watts. It is not 90 Megawatts.

    A horsepower is around 750 Watts, how many horses do you think it would take to spin a 25KG flywheel up to around 100 Revs, second?

    If you think it's 120 THOUSAND then the answer 90MW is right. If you think 1 horse could do it in about 5 seconds then the answer would be 3.5KW - of course you cannot arrive at the answer using this method (well not easily anyway) but you CAN get a picture of the order of magnitude of the answer. You could go down to the local hardware store and look at a few generators, compare there size/weight with output power - again you will NOT get an answer but there comes a point where you need to start matching theory and reality.
    Reply With Quote  
     

  12. #11  
    Forum Freshman
    Join Date
    Sep 2006
    Location
    B.C., Canada
    Posts
    88
    untill you start to draw power from the system no watts are being produced it remains 90000000J of kninetic energy untill you start to draw power from it (or put it in) if you start removing 10000J every second then your running a 10000 watt system of whatever if this system where 90 MegaWatts as some put it and it didn't dissipate in 1 second you would be creating energy which would be breaking the laws of physics

    ps anybody got any good ideas for a flywheel battery
    Reply With Quote  
     

  13. #12  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    At this point I'd like to point out that I don't actually think the answer is 90MWatts, I was simply trying to explain that 90 million joules/sec was equivalent to 90MWatts.

    The question asked to give the energy in watts, I was trying to explain that you don't measure energy in watts, you measure it in joules, and you measure power in watts. Whether this got through to the original threader remains uncertain. I was by no means trying to calculate anything.
    Reply With Quote  
     

  14. #13  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    Okay, I'm not sure if this is right because to be quite frank I haven't done classical mechanics for about 3 years now. But I get the kinetic energy to be equal to 1.3 million joules. How did I calculate it?

    I integrated the kinetic energy from r=0 to r=0.5, using a velocity equal to 200*pi*r. And then I multiplied by 2pi to correct for the geometry of the disc.

    K.E. = 0.5mv^2

    v(r)=[(6000*2pi)/60]r=200*pi*r
    thus v(r)^2= 40000*pi^2*r^2

    so: total K.E.=2pi*0.5m(integral between 0 and r)v^2 .dr

    or, subbing for v^2(and taking constants out of the integral):

    total K.E.=40,000*pi^3*m(integral between 0 and r)r^2 .dr

    then I did the integration and subbed in r=0.5

    Incidentally, one million joules is about the energy required to boil 3 litres of water. If you were to discharge all of this kinetic energy in one second, the resultant power would be 1.3 MW.
    Reply With Quote  
     

  15. #14  
    Forum Freshman
    Join Date
    Sep 2006
    Location
    In your room. Under your bed. Check it out.
    Posts
    9
    Then who is right?
    Reply With Quote  
     

  16. #15 Re: flywheel 
    Forum Freshman
    Join Date
    Dec 2005
    Location
    India
    Posts
    65
    Friends you are calculating the wrong target variable.

    Quote Originally Posted by BigT
    The teacher asked us to calculate the theoretical energy (in watts) of a given flywheel,
    We are supposed to calculate power not kinetic energy, since the units of power is watts.

    For a rotational body,
    Power=Torque*angular velocity=0 watts(since the torque=0)
    Torque is equal to 0 since the body is in rotational equilibrium.
    The body is in rotational equilibrium since the its angular velocity is constant.
    Reply With Quote  
     

  17. #16  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    I know, you've said that already. But as a matter of interest what is the kinetic energy of the system? To be honest, I don't think my above answer is correct because I failed to take into account the (non linear) change in mass from a ring of thickness r to a ring of thickness r+dr.
    Reply With Quote  
     

  18. #17  
    Forum Freshman
    Join Date
    Sep 2006
    Location
    B.C., Canada
    Posts
    88
    I = k*m*r^2 = .8*25*.5^2 = 5 (already known) no integration needed inertial constant gives the weight distribution info

    w = 6000*2*pi*r/60 = 314(apx)

    T(kinetic energy) = r*I*w^2 = .5*5*314^2 = 246740

    guy who posted this you squared the RPM which is not the right units that's it's speed in RPM you dont want to use that you need m/s
    Reply With Quote  
     

  19. #18  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    I revised the moment of inertia thing and I can report with confidence that the answer to this problem comes out in a calculator as 986,960 J. Or basically 1 MJ.
    Reply With Quote  
     

  20. #19  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    Actually, I've just realised that this flywheel must have an uneven weight distribution with denser material concentrated along the edge. I got bored (and stoned) last night and decided to go over all this moment of inertia jazz, I can't be bothered to type out all my working, but I derived an equation for the moment of inertia of a uniform disk spinning about its centre of mass.

    I=1/2(MR^2)

    So presumably then the "inertial constant" for such a disk is 1/2? In the original question the "inertial constant" was given to be 4/5, the fact that this constant is greater (than 1/2) implies that the weight distribution is not even, and that there is more mass concentrated towards the edge of the disk.

    Incidentally, if the disk did have an even weight distribution its kinetic energy would be slightly lower, I calculated it would then have 822kJ of kinetic energy. I must confess that the whole concept of an "inertial constant" was new to me before this thread, and it is thanks to this site that I have now (in my own mind) reached a new level of understanding in an area of mechanics that I never really had before.
    Reply With Quote  
     

  21. #20  
    Forum Freshman
    Join Date
    Sep 2006
    Location
    B.C., Canada
    Posts
    88
    Hey billiards how did you come up with that number I did it twice and it still comes out the same I showed all my math if you see something wrong let me know
    Reply With Quote  
     

  22. #21  
    Forum Freshman
    Join Date
    Sep 2006
    Location
    B.C., Canada
    Posts
    88
    I just did it on a calculator and still got the same answer and I can guarantee you my equations are just fine I would still love to know how you got to your answer
    Reply With Quote  
     

  23. #22  
    Forum Senior
    Join Date
    Sep 2006
    Posts
    399
    Quote Originally Posted by Beky
    I = k*m*r^2 = .8*25*.5^2 = 5 (already known) no integration needed inertial constant gives the weight distribution info

    w = 6000*2*pi*r/60 = 314(apx)

    T(kinetic energy) = r*I*w^2 = .5*5*314^2 = 246740

    guy who posted this you squared the RPM which is not the right units that's it's speed in RPM you dont want to use that you need m/s
    You need to convert the RPM into radians/second. This gives you the angular frequency w, which I see you have tried to calculate.

    To convert RPM into rad/sec you simply multiply by (2*pi)/60. This comes out as w=200*pi.

    Your mistake was to take this a step further by multiplying by R, this will effectively give you the speed of a particle at the very edge of the disk (not exactly what you want). The beauty of the moment of inertia I, is that it accounts for the variations in speed of all the particles as you go from the centre to the edge of the disk (as well as the weight distribution info). Remember, w (omega) is only a measure of how fast the object is spinning, and applys across the whole disk.
    Reply With Quote  
     

  24. #23  
    Forum Freshman
    Join Date
    Sep 2006
    Location
    B.C., Canada
    Posts
    88
    Ah yes I didn't notice I did that I wasn't thinkin straight, it must be the lack of THC
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •