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How to calculate equivalent inertia of an engine?

  1. Mar 26, 2015 #1
    Hello, is there a formula or technique to calculate the equivalent inertia of an engine?

    I think I will have to add up all the inertias of everything else rotating inside it but it has so many components like cams, followers, pulleys, water pump, crank, flywheel etc. What sort of effect will the reciprocating pistons have on the equivalent inertia?

    Is there no way of estimating a reasonable equivalent inertia of an engine without getting into all the complication of finding out the inertia for each and every rotating part? it does not have to be very accurate.
     
  2. jcsd
  3. Mar 26, 2015 #2

    SteamKing

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    It depends on what you ar planning to use this "equivalent inertia" for.
     
  4. Mar 26, 2015 #3
  5. Mar 26, 2015 #4
    I need to model linear acceleration of a vehicle which involves the use of this equation:

    Engine Torque - Clutch Torque = Equivalent inertia of engine*Angular acceleration of engine
     
  6. Mar 26, 2015 #5
    I am having a hard time understanding this. I am sorry if I am asking a dumb question but what is M and N, one of them seem to be the mass and the other M seems to be a moment ? I understand up till equation 4 but then equation 5 makes no sense to me.
     
  7. Mar 27, 2015 #6
    Of course you don't ask for a dumb question but for a known question which is explained in the text books.
     
  8. Mar 27, 2015 #7
    Where 60b0dfa02027be086cd7c576d05abd74_1b.gif is the force of the gas on the piston and is towards the crank.



    Inertia-Forces-and-Couples-0001.png
    magnify-clip.png


    It can be seen from the diagram that 43660fcd96ed949751b26c9be3390b74_05.gif is accompanied by a force in the connecting-rod of eb0356703289743ce82f268d488167d5_93d.gif , and the useful turning moment on the crankshaft during the power or outstroke is,


    e417df66d631f5bb9fd574667c7b898d_b901.gif


    "Actually it is calculating equivalent torque which is given by force*distance where P/cos is force and OM is the perpendicular distance same is with P x ON"
     
    Last edited by a moderator: Apr 17, 2017
  9. Mar 28, 2015 #8
    So the equivalent inertia = Torque/Angular acceleration of flywheel right?
     
    Last edited by a moderator: Apr 17, 2017
  10. Mar 30, 2015 #9
  11. Mar 30, 2015 #10
  12. Mar 30, 2015 #11

    OldEngr63

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    This whole problem is much easier if you formulate it using energy (such as the Lagrange method). In particular, the cylinder gas pressure is converted into crankshaft torque by means of a virtual work calculation with no trig involved. What I see here is the hard way to do the problem.
     
  13. Apr 4, 2015 #12
    I hvnt looked into the Lagrange method yet but would it work for the whole vehicle inertia too?
     
  14. Apr 4, 2015 #13

    OldEngr63

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    I'm not sure what you mean when you ask, "would it work for the whole vehicle inertia too?" Until you learn about the Lagrange formulation, I think it is premature to attempt to answer such a question.
     
  15. Apr 5, 2015 #14
    I am still trying to understand Lagrange method. What I have learnt till now is

    L = Kinetic Energy - Potential Energy

    Then: d/dt (dL/dx(dot)) - dL/dx = 0

    So I resolve for this and get an equation of motion but how do I relate this to inertia ? :oldconfused:

    EDIT: I found K.E also equals to 0.5*Inertia*angular velocity. So would this angular velocity be the RPM of the engine?
     
  16. Apr 5, 2015 #15

    OldEngr63

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    The equivalent inertia depends upon what you define as your generalized coordinate. The engine rotation angle would be a good choice, in which case, engine rpm, expressed in rad/s, is the associated generalized velocity.

    Be aware, however, that the generalized inertia, the I in your equation, is not a constant but rather varies with crank angle. Consider a one cylinder engine. When the piston is a TDC, a slight crank rotation does not involve significant piston motion. Thus, in this position, the inertia is determined by the crank and the connecting rod (mostly the crank); the same is true when the piston is at BDC. There are other points in between these two positions where piston motion is very significant for a small crank motion, and thus the generalized inertia is greater in these positions. With a multi-cylinder engine, there is some smoothing of the overall inertia, but it is still not a constant. Also, with a multi-cylinder engine, there is the possibility - a virtual certainty - that there will be torsional flexure between the crank throws, making this a many DOF problem.
     
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