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Expansion 1D Euler Eq.?

  1. Oct 27, 2011 #1
    Expansion 1D Euler Eq.??

    Trying to figure out an expansion step for 1D Euler Equations for unsteady gas flow.
    Continuity:
    [itex]\frac{\partial(\rho F)}{\partial t}[/itex]+[itex]\frac{\partial (\rho uF)}{\partial x}[/itex]=0

    After Expansion:
    [itex]\frac{\partial(\rho)}{\partial t}[/itex]+[itex]\frac{\partial (\rho u)}{\partial x}[/itex]+[itex]\frac{\rho u}{F}\frac{dF}{dx}[/itex]=0

    I understand to go from step 1 to step 2, you divide by F; however, for Step 2 where is the third term coming from and what does it mean?

    I have the same question for the momentum equation:
    Momentum:
    [itex]\frac{\partial (\rho uF)}{\partial t}[/itex]+[itex]\frac{\partial (\rho u^{2}+p)F}{\partial x}[/itex]-[itex]p\frac{dF}{dx}[/itex]+[itex]\frac{1}{2}\rho u^{2}f\pi[/itex] D=0

    After Expansion:
    [itex]\frac{\partial (\rho u)}{\partial t}[/itex]+[itex]\frac{\partial (\rho u^{2}+p)}{\partial x}[/itex]+[itex]\frac{\rho u^{2}}{F}\frac{dF}{dx}[/itex]+[itex]\frac{1}{2}\rho u^{2}f\pi[/itex] D=0

    Again, where is the [itex]\frac{\rho u^{2}}{F}\frac{dF}{dx}[/itex] term coming from and what does it mean?
     
  2. jcsd
  3. Oct 27, 2011 #2

    LCKurtz

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    Re: Expansion 1D Euler Eq.??

    It would be helpful to us who aren't familiar with gas flow problems and notation if you would tell us what variables ρ, u, and F depend on. I'm going to assume, from what I see, that F doesn't depend on t but does depend on x.
    [tex]\frac{\partial(\rho F)}{\partial t} = \frac {\partial \rho}{\partial t}F[/tex]
    because F doesn't depend on t. But [itex]\frac{\partial (\rho uF)}{\partial x}[/itex] requires the product rule since [itex]\rho u[/itex] and F depend on x:
    [tex]\frac{\partial (\rho uF)}{\partial x}=F\frac{\partial (\rho u)}{\partial x}
    +\rho u\frac{\partial (F)}{\partial x}[/tex]

    So if you differentiate the equation you get those three terms on the right above and you divide the equation by F. I didn't check your second question but I'm guessing it might be the same problem.
     
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