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

[nonequilibrium]

In my course they're using the equality U = \frac{p}{\alpha \rho} 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 p = \rho \beta T for some constant beta, and that U = \gamma T (note that U is energy per mass, i.e. up to a constant energy per particle \propto k_B T)

 

[Studiot]

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


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

For liquids in particular, Engineers commonly tabulate two quantities thus

The Volume Expansivity


\beta = \frac{1}{V}{\left( {\frac{{\partial V}}{{\partial T}}} \right)_P}


Isothermal Compressibility


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


Putting these definitions into the above equation leads to


\frac{{dV}}{V} = \beta dT - \kappa dP

For an incompressible fluid both β and κ are zero.

Now to link to ordinary thermodynamics


dH = TdS + VdP

and

{\left( {\frac{{\partial H}}{{\partial T}}} \right)_P} = {C_P} = T{\left( {\frac{{\partial S}}{{\partial T}}} \right)_P}

and (Maxwell)


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


Combining


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

Insert engineering definions


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

Also


U = H - PV


differentiate at constant temp


{\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

Thus inserting engineering definitions


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


There is more if you want it.

 

[nonequilibrium]

Sorry I seem to be missing your point. How does this answer my question?

 

[Studiot]

Are we not talking abou the same quantities, beta and kappa?

I just thought you'd appreciate some background.

 

[Studiot]

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


\ln \left( {\frac{{{V_2}}}{{{V_1}}}} \right) = \beta \left( {{T_2} - {T_1}} \right) - \kappa \left( {{P_2} - {P_1}} \right)

 

[nonequilibrium]

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 U \propto \frac{p}{\rho} in a liquid.

 

[Studiot]

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


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

if u is the internal energy per unit mass


u = U\left( {\rho dV} \right)


\frac{{DU}}{{Dt}} = \rho V\frac{{du}}{{dt}}

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?