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Thread: Engaged/Disengaged flywheel's effect on the shaft's power

  1. #1 Engaged/Disengaged flywheel's effect on the shaft's power 
    Junior Member
    Join Date
    May 2013
    Posts
    1
    Hi,

    I have this question and need your help please..

    If you have a crank shaft (lets say of a bike), rotates with variable speed, and you have a flywheel that is automatically engaged/disengaged to/from the shaft.

    How the flywheel, in this case, affects the power of the shaft ? if I want to include the effect of the flywheel on the overall shaft`s power what formula should I use ?

    I know the Inertia of the flywheel, and also I can calculate the energy of the flywheel using 1/2 * I * w^2. Can I say that: during engagement with the shaft, the work done on the flywheel by the shaft is the resultant flywheel energy which is (E=1/2 * I * W2^2 - 1/2 * I * W1^2), then take the derivative of the Energy to get the power(P=dWork/dt) and add it to the overall shaft's power as the flywheel's effect on the system ? or this is not correct ?

    Any idea would be appreciated..
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  2. #2  
    Senior Member
    Join Date
    Jan 2013
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    504
    Quote Originally Posted by shune2001 View Post
    Hi,

    I have this question and need your help please..

    If you have a crank shaft (lets say of a bike), rotates with variable speed, and you have a flywheel that is automatically engaged/disengaged to/from the shaft.

    How the flywheel, in this case, affects the power of the shaft ? if I want to include the effect of the flywheel on the overall shaft`s power what formula should I use ?

    I know the Inertia of the flywheel, and also I can calculate the energy of the flywheel using 1/2 * I * w^2. Can I say that: during engagement with the shaft, the work done on the flywheel by the shaft is the resultant flywheel energy which is (E=1/2 * I * W2^2 - 1/2 * I * W1^2), then take the derivative of the Energy to get the power(P=dWork/dt) and add it to the overall shaft's power as the flywheel's effect on the system ? or this is not correct ?

    Any idea would be appreciated..
    I replied to this very question posed in a PM. Look to your messages for a response.
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  3. #3  
    Junior Member
    Join Date
    Jun 2013
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    9
    Quote Originally Posted by shune2001 View Post
    Hi,

    then take the derivative of the Energy to get the power(P=dWork/dt) and add it to the overall shaft's power as the flywheel's effect on the system ? or this is not correct ?

    Any idea would be appreciated..
    Correct. The derivative (d/dt) of the kinetic energy of the fly wheel is the power -
    the power used by the shaft to speed up the flywheel , or the power the flywheel delivers to the shaft.

    Add/subtract that power to any other power taken from or delivered to the shaft.

    Eg the motor providing 10 kW at one end of the shaft, or the bike's wheels taking 5kW from it, or whatever.
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