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

1. Jul 18, 2012

### 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 $\propto k_B T$)

2. Jul 19, 2012

### 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

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

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.

Last edited: Jul 19, 2012
3. Jul 19, 2012

### nonequilibrium

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

4. Jul 19, 2012

### Studiot

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

I just thought you'd appreciate some background.

5. Jul 19, 2012

### 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

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

6. Jul 19, 2012

### 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.

7. Jul 19, 2012

### 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?