Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to calculate how much power required to turn a flywheel?

  1. Apr 10, 2015 #1
    How to calculate how much power required to turn a flywheel? Suppose I have a flywheel of 24 inch diameter, width 5 inches,50kg. How do I calculate the required power (HP) to turn this or any other with different values?
     
  2. jcsd
  3. Apr 10, 2015 #2
    The information given is not enough to find the power.
    First what do you mean by "to turn"? To accelerate from rest to a given speed? Or just to keep a given speed constant?
     
  4. Apr 10, 2015 #3
    To accelerate from rest to 2000rpm
     
  5. Apr 10, 2015 #4

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    In what time?
     
  6. Apr 10, 2015 #5
    In 10-12 seconds. :-)
     
  7. Apr 10, 2015 #6
    Then find the kinetic energy of the rotating flywheel at the required rotation rate and divide by the time.
    You can approximate the flywheel with a cylinder when looking up the moment of inertia.
     
  8. Apr 10, 2015 #7
    Ring's KE(joules): 101932.01
    Time is 10 second
    101932/10=10193.2
    Is it the power required to rotate that flywheel? Please elaborate, I really don't know how to calculate? I have attached flywheel's energy calculations.
     

    Attached Files:

  9. Apr 10, 2015 #8

    i got some values, please check
    values @ 2000rpm, 50 kg, 0.3048m radius
    rpm to rad/sec = rpm*2*pi/60; i.e. 209.33 rad/sec ;
    I = 0.5 m r^2 = 0.5*50*0.3048*0.3048 = 2.3225;
    E = 0.5 I ω ^2 = 0.5*2.3225*209.33*209.33 = 50884.8705 Joules; <==

    torque = mass moment of inertia * rotational acceleration rate = I*a= 2.3225*(20.933)=48.6168 Nm;
    instantaneous power consumption = torque * max rotation rate = 48.6168*209.33= 10176.9741 watt; <==

    is this the correct method to calculate power required to rotate flywheel?
     
  10. Apr 10, 2015 #9
    You don't need the torque. Once you have the kinetic energy, the average power will be that energy divided by the time.
     
  11. Apr 10, 2015 #10
    So in this case, power required to rotate 50 kg, 0.3048m radius flywheel @ 2000 rpm for 10 sec is
    101932/10=10193.2 watts. Is this correct?


    Power = KE/time;
    What does the time means? Is it the total time for which we r rotating the flywheel? What will be the value if I want to rotate that flywheel for hours? Please
     
    Last edited: Apr 10, 2015
  12. Apr 11, 2015 #11
    Maybe you should ignore power at first and use torque (T):
    (the torque is deemed to be constant)
    calculate the acceleration rate (α) in rad/sec/sec
    (ive attached a sheet to use)
    then calculate the torque T ( in N-m ) from:
    T = I * α
    ( I = flywheel moment of inertia (kg - m^2) , α = rotational acceleration rate (rad/sec/sec) )
    With the torque value (T) you can then calculate the power (Watts) at any rad/ sec ( ω ) from:
    Power = T * ω
     

    Attached Files:

    • rlin.gif
      rlin.gif
      File size:
      13 KB
      Views:
      1,888
  13. Apr 11, 2015 #12

    jbriggs444

    User Avatar
    Science Advisor

    No. This is not correct.

    The figure of 10 kilowatts is the startup power requirement. While the wheel is being spun up from rest to its desired rotation rate, you must provide "power". The faster you want to reach the steady-state RPM, the more power you must supply. Remember that in the physics vernacular, "power" is the rate at which energy is delivered. The amount of energy you must supply to spin the wheel up from rest is fixed. The rate at which you must supply that energy depends on how much time you are willing to spend getting the wheel up to speed. One Watt of power means you are delivering energy at a rate of one Joule per second.

    The steady state power requirement to maintain the flywheel at 2000 rpm is zero. In principle, it take no power at all to keep a flywheel spinning forever. In practice, the power requirement will depend on how good your bearings are and how much air resistance the wheel is subject to.

    In addition it is overly optimistic to quote the computed result to 6 significant figures when the inputs (mass and time interval) were known to only one significant figure.
     
  14. Apr 11, 2015 #13

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Did you forget what you were asking about?
     
  15. Apr 11, 2015 #14
    no, i was asking that time divided the KE is what? is it the total time or something else? i got the answer. thanks A.T. :-)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How to calculate how much power required to turn a flywheel?
Loading...