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# Van der Waals EOS

So for any new equation of state, we need to be able at the very least to calculate:

p(\rho, e) | e(\rho, p) | c_s(\rho, p) |

If we want to know the actual temperature - we also will need a thermal EOS

T(\rho, e) |

and if we want to do any roe averaging we will need

h(\rho, p) = \frac{e(\rho, p)+p}{\rho} | p(\rho, h) |

As an example, I looked at a Van der Waals gas

So for a Van der Waals gas the pressure takes the form P(\rho,T) = \frac{\rho RT}{1-\rho b}-a\rho^2 However to solve the euler equations we need a caloric equation of state

So we need the internal energy density (which can't be derived from the pressure function alone) e=\frac{f}{2} \rho RT-a\rho^2 where f is the degrees of freedom.

This allows us to write our caloric EOS e(p,\rho)=\frac{f(p+a\rho^2)(1-\rho b)}{2}-a\rho^2

as well as solve for the pressure given the internal energy density p(e,\rho)=\frac{2(e+a\rho^2)}{f(1-\rho b)}-a\rho^2

so we can calculate the specific enthalpy

h(p,\rho)=(e+p)/\rho=\frac{f(p+a\rho^2)(1-\rho b)}{2\rho}-a\rho+\frac{p}{\rho}

and then use the specific enthalpy to calculate the sound speed using various thermodynamic identities…

c^2=-\frac{\left (\frac{\partial h}{\partial \rho}\right )_p}{\left ( \frac{\partial h}{\partial p} \right) _\rho - \frac{1}{\rho}}

c^2=-\frac{f\left [ 2a\rho^2(1-\rho b)-b\rho(p+a\rho^2) -(p+a\rho^2)(1-\rho b) \right ] - 2\left(p+a\rho^2 \right)}{f\rho(1-\rho b)}

c^2=\frac{f+2}{f(1-\rho b)}\left(\frac{p}{\rho}+a\rho\right) - 2 a \rho

which when a = 0 reverts to

c^2=\frac{f+2}{f(1-\rho b)}\frac{p}{\rho}=\gamma\frac{p}{\rho}

and when b = 0 reverts to

c^2=\frac{f+2}{f}\frac{p}{\rho} - \frac{f-2}{f}a \rho

which gives damping sound waves when

p < \frac{f-2}{f+2}a\rho^2

and when a=b=0 reverts to

c^2=\frac{f+2}{f}\frac{p}{\rho}=\gamma \frac{p}{\rho}

and finally we need to invert the equation for the specific enthalpy

p(h,\rho)=\frac{2\rho h + 2a\rho^2-f(a\rho^2)(1-\rho b)}{(2+f(1-\rho b))}

So we ran the basic shock tube tests comparing ideal gas and VanDerWaals gas.

Note that as \rho \rightarrow \frac{1}{b} then pressure should begin to dramatically increase with density due to limited volume.

And if \rho \rightarrow \frac{RT}{a} then attractive Van der Waals forces begin to significantly modify the pressure.

The parameters in cgs were

a=1.38\times 10^6 |

b=3.258\times 10^{-2} |

f=5 |

R=231.11 \times 10^4 which corresponds to a mean molecular weight of 36 amu |

\rho_L=10 \mbox { and } \rho_R=1 |

T_L=5000 \mbox { and } T_R = 3000 |

Note since \rho_L=10 \mbox{ so } \rho_L b = .3258 \approx 1

And \rho_L << \frac{RT}{a} so molecular attraction is negligible.

In what follows, the red line is Van der Waals, and the blue line is ideal gas.

If we match the temperatures in the left and right states we get

and if we match the pressures

and if we match the internal energy

The energy match looks identical to the temperature match because the " *a* " constant is so small (although the bounds are different on the plots).

- Posted: 12 years ago (Updated: 12 years ago)
- Author: Jonathan
- Categories: (none)

### Attachments (3)

- VanDerWaals_TempMatch.png (26.8 KB) - added by Jonathan 12 years ago.
- VanDerWaals_EnergyMatch.png (26.5 KB) - added by Jonathan 12 years ago.
- VanDerWaals_PressMatch.png (27.0 KB) - added by Jonathan 12 years ago.

Download all attachments as: .zip

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