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Deriving the Bernoulli EquationFluids

  1. Oct 23, 2009 #1
    I am trying to follow along with the text's derivation of the Bernoulli equation and I am a little confused with a particular line. The images show an elemental streamtube control volume (CV) of variable area A(s) and length ds.

    We assume that though the properties may change with time and position 's' they remain uniform over the cross section 'A.'

    So treating as a 1-dimensional inlet/outlet CV we can write conservation of mass as:

    [tex]\frac{d}{d\,t}(\int_{CV}\rho d\,V) +\dot{m}_{out} - \dot{m}_{in} = 0 \approx \frac{\partial{\rho}}{\partial{t}}d\,V + d\,\dot{m}[/tex]

    Okay. I am missing somethings here:

    (1) I am assuming that they set [itex]\dot{m}_{in} =\dot{m}[/itex]


    [itex]\dot{m}_{out} =\dot{m} + d\,\dot{m}[/itex]


    [itex]\dot{m}_{out} - \dot{m}_{in} =d\,\dot{m} [/itex].

    Does that sound right?

    (2) So the thing that is really bothering me is this. Where did the integral go? AND why did [itex]d/d\,t[/itex] turn into [itex]\partial/\partial{t}[/itex]
    I keep getting jammed up on these total derivatives turning into partials. What does it all mean?

    thanks :smile:

    Last edited: Oct 24, 2009
  2. jcsd
  3. Oct 24, 2009 #2
    I have made another assumption that seems to explain where the integral went:

    Since the control volume is of differential size, then upon integration we simply end up with 'dV.'

    But I don't really like that. I can't seem to work out the math of it.......maybe that's not right at all. I am just grasping at straws here...
  4. Oct 25, 2009 #3
    Any thoughts on this one? Or is Bernoulli like so 1986 or something...
  5. Oct 25, 2009 #4


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    It's a bit messy, saladsamurai, I will cough up my old notes and see if I can make a good post explaining this, on Reynolds' transport theorem and such.

    There is something strange about how they have set this up, so I need some time to think about it.
  6. Oct 25, 2009 #5
    Okay! I understand the RTT just fine.

    The left hand side of

    \frac{d}{d\,t}(\int_{CV}\rho d\,V) +\dot{m}_{out} - \dot{m}_{in} = 0 \approx \frac{\partial{\rho}}{\partial{t}}d\,V + d\,\dot{m}

    is the RTT with mass as its 'dummy variable.'

    It's whatever 'tricks' (simplifying assumptions) they are using to get the right hand side that are getting me jammed up.

    Thanks for looking :smile:
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