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Fluid MechanicsQuestion about the Differential Energy Equation

  1. Nov 10, 2009 #1
    I am going through the derivation in my text and I have reached a line that makes absolutely no sense to me. I cannot see the correlation between one line and the next and it has led me to believe that it is either an error or just bad editing/wording.

    We are at the point where the energy equation has been completely derived, yielding:

    [tex]\rho\frac{d\hat{u}}{dt}+p(\nabla\cdot\vec{V}) = \nabla\cdot(k\nabla T) +\Phi \qquad (1)[/tex]​

    where [itex]\Phi[/itex] is the viscous work-dissipation function.

    He then says that
    Okay, that's great. Here, the next line is where I get all messed up:


    What?! How does [itex]d\hat{u}\approx c_vdT \qquad & \qqaud c_v,\mu,k,\rho\approx\text{constant}[/itex] imply that [itex]\nabla\cdot\vec{V}=0[/itex] ???

    Or are the two completely unrelated and the wording only makes it seem like the 2 statements follow logically.

    Is the [itex]\nabla\cdot\vec{V}=0[/itex] 'case' another condition that he is imposing on (1) in addition to the conditions [itex]d\hat{u}\approx c_vdT \qquad & \qqaud c_v,\mu,k,\rho\approx\text{constant}[/itex]?

    I am confused :confused:

    Any insight is appreciated as always :smile:
     
  2. jcsd
  3. Nov 10, 2009 #2

    Andy Resnick

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    The two statements are unrelated, AFAIK. Two separate simplifications.
     
  4. Nov 10, 2009 #3
    Yeah it's weird. He goes on a few lines later to talk about the special case where the fuid is at rest. But in the above, even though [itex]\nabla\cdot\vec{V}=0[/itex], that does not mean that the fluid is at rest since [itex]dT/dt \ne0[/itex], which contains velocity components.

    So I don't see when we would have [itex]\nabla\cdot\vec{V}=0[/itex] AND [itex]dT/dt \ne0[/itex].

    What special case is that? And why would he not explicitly give the details as to what that meant physically? Like when the fluid flows over kittens or something...what is the 'special case' ?
     
  5. Nov 10, 2009 #4
    They left out an equation.

    [itex] {\partial \rho \over \partial t} + \nabla \cdot (\rho \mathbf{V}) = 0[/itex]

    If you set \rho constant then you get [itex]\nabla\cdot\vec{V}=0[/itex]
     
  6. Nov 10, 2009 #5

    Oh wow. Nice catch twofish-quant :thumbsup: Maybe he did not 'leave it out;' maybe his intent was for the reader to really 'flex' that brain.

    This must be what he meant back in the beginning of the chapter on differential relations. He said something to the effect of "<paraphrasing> we must always return to the continuity equation else our results will be most likely be nonsensical...."

    It never occurred to me that just because the energy equation or momentum equation looks mathematically sound that it might not be physically sound.

    Thanks! :smile:
     
  7. Dec 28, 2010 #6
    Hi, I am trying to derive the full energy equation for a fluid (rho du/dt + P grad V=del dot (kgrad T) + the viscous dissipation function. I understand almost the entire derivation but my texts leave out a few steps. Instead of asking the specifics since I am not well versed with the equation editor, can someone post the entire step by step derivation? Thanks
     
  8. Jan 22, 2011 #7
    simply the the the divergence of the vector velocity (div(v)) THIS QUANTITY REPRESENT THE CONTINUITY EQUATION FOR A 3D INCOMPRESSIBLE FLOW.
    Go back to the continuity equation and you see that Prof Franl white he made a derivation in catesian coordinate and cylindrical coordinates system for an unsteady compressble conditions.
    but if we expande the the contnuity , and we assume that the flow is incompressble we get exactely the div(V)=nabla dot( V), nabla is a diferential operator when it operate on vector as grad(v) has no sense, but when it dot the result is a scalar function.
    *iF YOU NEDD A DETAILLED DERIVATION OF ALL THE EQUATION , AND THE SCALLING FACTORS, IN ORDER TO FIND THE CONTINUITY IN CYLINDRICAL AND SPHERICAL COORDINATES SEND ME A MESSAGE, AND I WILL SEND YOU THE FULL DETAILLED DERIVATIO. MY EMAIL=(london_maurice@hotmail.com)
     
  9. Jan 22, 2011 #8
    du=Cvdt this is the internal energy for a caloricaly, perfect gas. mu is the viscosity,k is the heat conduction,rho is the density.
     
  10. Jan 22, 2011 #9
    But i don't understand one things here Salds how did you get into this step without having problem with the steps before?
     
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