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[fluid dynamics] are they trying to use the ideal gas law for LIQUIDS?

  1. Jul 18, 2012 #1
    In my course they're using the equality [itex]U = \frac{p}{\alpha \rho}[/itex] with alpha some constant (U = internal energy per mass, p = pressure, rho = density). They explicitly derive it for an ideal gas yet later apply it to a liquid (in the context of deriving the Navier-Stokes energy equation). Seems pretty unfounded... However, is there perhaps a reason we should expect such an equation to hold in more general cases?

    NB: to see it follows from the ideal gas law, note that [itex]p = \rho \beta T[/itex] for some constant beta, and that [itex]U = \gamma T[/itex] (note that U is energy per mass, i.e. up to a constant energy per particle [itex]\propto k_B T[/itex])
     
  2. jcsd
  3. Jul 19, 2012 #2
    OK for any fluid you need an equation of state connecting P, V & T, which you can solve and differentiate for one in terms of the other two eg


    [tex]dV = {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}dT + {\left( {\frac{{\partial V}}{{\partial P}}} \right)_T}dP[/tex]

    For liquids in particular, Engineers commonly tabulate two quantities thus

    The Volume Expansivity


    [tex]\beta = \frac{1}{V}{\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}[/tex]


    Isothermal Compressibility


    [tex]\kappa = - \frac{1}{V}{\left( {\frac{{\partial V}}{{\partial P}}} \right)_T}[/tex]


    Putting these definitions into the above equation leads to


    [tex]\frac{{dV}}{V} = \beta dT - \kappa dP[/tex]

    For an incompressible fluid both β and κ are zero.

    Now to link to ordinary thermodynamics


    [tex]dH = TdS + VdP[/tex]

    and

    [tex]{\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} = {C_P} = T{\left( {\frac{{\partial S}}{{\partial T}}} \right)_P}[/tex]

    and (Maxwell)


    [tex]{\left( {\frac{{\partial S}}{{\partial P}}} \right)_T} = - {\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}[/tex]


    Combining


    [tex]{\left( {\frac{{\partial H}}{{\partial P}}} \right)_T} = V - T{\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}[/tex]

    Insert engineering definions


    [tex]\begin{array}{l}
    {\left( {\frac{{\partial S}}{{\partial P}}} \right)_T} = - \beta V \\
    {\left( {\frac{{\partial H}}{{\partial P}}} \right)_T} = \left( {1 - \beta T)V} \right) \\
    \end{array}[/tex]

    Also


    [tex]U = H - PV[/tex]


    differentiate at constant temp


    [tex]{\left( {\frac{{\partial U}}{{\partial P}}} \right)_T} = {\left( {\frac{{\partial H}}{{\partial P}}} \right)_T} - P{\left( {\frac{{\partial V}}{{\partial P}}} \right)_T} - V[/tex]

    Thus inserting engineering definitions


    [tex]{\left( {\frac{{\partial U}}{{\partial P}}} \right)_T} = \left( {\kappa P - \beta T} \right)V[/tex]


    There is more if you want it.
     
    Last edited: Jul 19, 2012
  4. Jul 19, 2012 #3
    Sorry I seem to be missing your point. How does this answer my question?
     
  5. Jul 19, 2012 #4
    Are we not talking abou the same quantities, beta and kappa?

    I just thought you'd appreciate some background.
     
  6. Jul 19, 2012 #5
    For an incompressible fluid there is no equation of state connecting P, V & T since V is constant.

    For small compressibility it is common to integrate the fourth equation in my first post to yield


    [tex]\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) = \beta \left( {{T_2} - {T_1}} \right) - \kappa \left( {{P_2} - {P_1}} \right)[/tex]
     
  7. Jul 19, 2012 #6
    The alpha's, beta's, gamma's I'm using are just symbols I used since I didn't want to specify what constants they were.

    My question was if there is a justification for [itex]U \propto \frac{p}{\rho}[/itex] in a liquid.
     
  8. Jul 19, 2012 #7
    Well if you think about it, if the liquid is incompressible then density = a constant.

    However the energy changes must go somewhere and the basic equation of energy balance in a flowing fluid is


    [tex]\frac{D}{{Dt}}\left( {U + KE} \right) = P + Q[/tex]

    if u is the internal energy per unit mass


    [tex]u = U\left( {\rho dV} \right)[/tex]


    [tex]\frac{{DU}}{{Dt}} = \rho V\frac{{du}}{{dt}}[/tex]

    depending upon conditions you can use this in the energy balance to obtain a relationship between U, P and T

    Is this what you are after?
     
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