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Watt's Governor - Equation

  1. Oct 29, 2011 #1
    I was reading up on the mathematical modelling of Watt's governor from Brennan's Book on Differential Equations.

    The book has 2 equations one for the governor itself which I understand.
    However, there is an equation for the steam engine flywall & Drive shaft which I don't understand. This is the equation

    J * dΩ/dt = -β* Ω + τ

    This is the text following the equation
    J is the moment of Inertia of the flywheel
    Ω - rotational speed of the engine
    First term on the right is the torque due to the load & the second term on the right is the steam generated torque referred to the drive shaft.

    What exactly is β in the equation? What's the unit of β?
     
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  3. Oct 29, 2011 #2

    AlephZero

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    All the terms in the equation must be in the same units.

    [itex]\tau[/itex] is obviously a torque. So is [itex]J\, d\Omega/dt[/itex].

    So [itex]\beta\, \Omega[/itex] must also be a torque.

    [itex]\Omega[/itex] is the rotational velocity, so [itex]\beta[/itex] is a measure of the "size" of the engine, in other words how much torque it produces at 1 rad/sec.

    If [itex]\beta[/itex] is a constant, your equation is assuming that the torque produced by the engine is proportional to its RPM. That seems a strange assumption to make for a steam engine, but then your book is about differential equations, not engineering.
     
  4. Oct 29, 2011 #3
    I don't think β is a measure of the size of the engine. This is because of 2 reasons
    1) βω part is preceded by -minus sign - if it was the torque produced by the engine, it would be positive in the equation, not negative
    2) Also the text I quoted in my original post says "First term on the right is the torque due to the load" - i.e. it's the negative torque which acts on the system because of the load driven by the engine. The quoted text also says that τ is the torque produced by the engine.

    EDIT:
    Looking a few more places, it seems that β is load damping co-efficient (unit N-m/rad/s).
    However, it's not clear what exactly this is.
     
    Last edited: Oct 29, 2011
  5. Oct 29, 2011 #4

    AlephZero

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    I took "torque due to the load" in your OP as meaning the torque required to drive the load. That is not the same as the "torque delivered by the engine" if the speed is changing.

    With that interpretation, one sign on the right hand side should be + and the other -, as they are in your equation, though I did think they were using a strange sign convention!

    But OK, if you are now saying [itex]\tau[/itex] is the NET torque delivered by the engine, (i.e. the torque acting on the flywheel, which is the difference between the total engine torque and the torque that is driving the load) then it would make sense if [itex]\beta\, \Omega[/itex] was a "damping" force proportional to speed.

    In "real life" there would be many causes of the energy lost from the system, for example friction viscous forces in the oil lubrication, air resistance to the moving parts, etc. Lumping all those things together and assuming they produce a "torque" proportional to linear or angular velocity is a good way to make the math fairly simple, but you can't easily relate it to the detailed physics of what is going on.

    But this is often used as a modelling technique, because (as you will probably see when your book discusses what the solution to the equation looks like) you can find the value of beta for a particular system by measuring how the system responds to a sudden change in load, for example, and that is much easier than trying to model all the "real physics" involved.

    This modelling technique of including some "general" parameters (like [itex]\beta[/itex] in your example) in the math model to represent the "global" effect of a lot of details is often used, and the process of finding numerical values for the parameters from the behaviour of the real-world system is called "system identification". You can also use this process the other way round: play with the math to find a "good" value of [itex]\beta[/itex] to get the response you want, and then go back to the physics and design something to produce that value of [itex]\beta[/itex] in real life.
     
    Last edited: Oct 29, 2011
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