| Franz J. Vesely > |
Power Wheel: A basic thermomechanical machineFranz J. Vesely franz.vesely@univie.ac.at Computational Physics Group, University of Vienna, Boltzmanngasse 5, A-1090 Vienna MotivationShort resume of simple dft:
1. Gas in constant grav field1.1 Ideal gasLet $N$ particles of mass $m$ be confined to a vertical tube of cross sectional area $F$ and height $L$. Let $\lambda_{DB}=\sqrt{2 \pi \hbar^{2}/mkT}$ (De Broglie w.l.) and $v(T)=\lambda_{DB}^{3}$ (De Broglie volume.) The reduced free energy of an ideal gas with local density $\rho(\vec{r})$ (in moles per $m^{3}$ and temperature $T$ is
$
A\{\rho\}/RT = \int \limits_{V} d\vec{r} \, \rho(\vec{r})
\left( \ln \rho(\vec{r}) v(T) -1 \right)
$
Assuming a field (and density) depending on $z$ only we have
$
A\{\rho\}/RT = F \int \limits_{0}^{L} dz \, \rho(z)
\left( \ln \rho(z) v(T) -1 \right)
\equiv F \int \limits_{0}^{L} dz \, f(z)
$ where $f(z) F dz= \rho(z) \left( \ln \rho(z) v(T) -1 \right) F dz $ is the total reduced free energy contained in a layer $(z,dz)$. The gravitational energy of such a layer is, of course, $u(z) F dz = \rho(z) m g z F dz$ (with molar mass $m$), such that the total reduced energy is $
A\{\rho\}/RT = F \int \limits_{0}^{L} dz \, \left(f(z)+u(z)/RT \right)
= F \int \limits_{0}^{L} dz \, \left\{ \rho(z) \left( \ln \rho(z) v(T) -1 \right)
- \rho(z) m g z/RT \right\}
$ This functional of $\rho$ must be minimized with the condition $F \int dz \rho(z) = N $. Dropping the arguments of $\rho$ and $v$ and introducing a Lagrange parameter $\lambda$ we have $
\begin{eqnarray}
\delta A/RT = 0 &=& F \int \limits_{0}^{L} dz \, \delta
\left\{ \rho \left( \ln \rho v -1 \right)
+ \rho m g z / RT - \lambda \rho \right\}
\\
&=& F \int \limits_{0}^{L} dz \,
\left\{ \delta \rho \left( \ln \rho v -1 \right)
+ \rho \, \delta \left( \ln \rho v -1 \right)
+ \delta \rho \, m g z / RT - \lambda \delta \rho \right\}
\\
&=& F \int \limits_{0}^{L} dz \, \delta \rho
\left\{ \ln \rho v + m g z /RT - \lambda \right\}
\end{eqnarray}
$ Thus, $
\ln \rho v + m g z / RT - \lambda = 0
\;\;\;\;
{\rm or}
\;\;\;\;
\rho(z) = \frac{\textstyle 1}{\textstyle v}e^{\textstyle \lambda}
e^{\textstyle - m g z/RT} = \rho_{0} e^{\textstyle - m g z/RT}
$ Using the normalization of $\rho$ we could calculate $\lambda$ and thence $\rho_{0}$, but in this simple example we immediately see that $
\rho_{0}= \frac{\textstyle N m g/RT}{\textstyle F(1-\exp[- m g L/RT])}
$ It is worth mentioning here that in case of a $z-$dependent tube profile $F(z)$ the value of $\lambda$, and thus $\rho_{0}$, will depend on $F(z)$. As a numerical example let us confine a xenon-like ideal gas in a long vertical tube of length $L=3 \cdot 10^{3} m$. Let the density be equal to the critical density of X, $\rho = 8.32 \, mol / m^{3}$. With molecular mass $m=131.3\, amu \;=2.2058 \cdot 10^{-25} kg$ we have, at a temperature of $300 K$, a de Broglie volume of $v(T)=4.386 \cdot 10^{-44} m^{3}$; the base density at $z=0$ is $ \rho(0)=16.482\, mol / m^{3}$, and the density decays exponentially to $\rho(L)=3.436\, mol/ m^{3}$; the center of mass of the gas column is at $z=1123.3 \, m$. Raising the temperature to $T=400 \, K$ we have a de Broglie volume of $v(T)=6.752 \cdot 10^{-44} m^{3}$; $\rho(z=0)=14.150 \, mol / m^{3}$, $\rho(L)=4.365 \, mol / m^{3}$; the center of mass of the gas column is now at $z=1212.6 \, m$. 1.2 Van der Waals gasThe reduced free energy in a VdW gas is, with $\rho = \rho(\vec{r})$ and de Broglie volume $v=v(T)$,$
A\{\rho\}/RT = \int \limits_{V} d\vec{r} \, \rho
\left( \ln \rho v -1 - \rho a/RT - \ln (1-\rho b) \right)
$ where $a$ and $b$ are the molar Van der Waals constants. In terms of a particle radius $r$ and a potential well depth $\epsilon$ we have $
a= N_{0}^{2} 2 \pi \epsilon r^{3}/3
\;\;\;\;\;\;\;\;\;
b=N_{0}2 \pi r^{3}/3
$ with $N_{0}= 6.022 \cdot 10^{26}$. On the other hand, $a$ and $b$ are related to the critical volume and temperature by $
b= V_{c}/3
\;\;\;\;\;\;\;\;\;
a=27 T_{c} R b / 8
$ The reduced free energy density in the VdW gas is thus $
f(\rho,T) \equiv A\{\rho,T\}/VRT =
\rho \left( \ln \rho v -1 \right) - \rho^{2} a/RT - \rho \ln (1-\rho b)
= f_{id}(\rho,T)- \rho^{2} a/RT - \rho \ln (1-\rho b)
$ which is, of course, consistent with $
p = - f + \rho \frac{\textstyle \partial f}{\textstyle \partial \rho }
= \rho \, RT - a \rho^{2} + \rho \, RT \frac{\textstyle b \rho}{\textstyle 1-b \rho }
$
Returning to our earlier example, we now treat the xenon gas as a Van der Waals fluid. The relevant parameters are: critical temperature $T_{c}=289.78\, K$; critical density $\rho_{c}=8.32 \, mol/m^{3}$; critical pressure $P_{c}=5.83 \cdot 10^{6} \, Pa$. The molar Van der Waals parameters are thus $b= V_{c}/3 = 4.006 \cdot 10^{-2} \, m^{3}$ and $a=27 b R T_{c}/8= 3.256 \cdot 10^{5} J m^{3}$. 
Figure 1: "Xenon" in a cylinder with height L=3000 m; (a) a) T=300 K, i.e. slightly above critical (289.78 K); red: ideal gas with xenon mass; green: Van der Waals gas with xenon parameters; density: critical; (b) T=400 K; magenta: ideal gas; light blue: VdW gas. For the VdW gas the center of mass moves from z= 894.1 m to 1136.1 m. [Codes: dft2.f, gp_dft2.gnu] Figure 1 shows the density profile in the cylinder. The c.o.m. of the fluid is at height $z_{m}=894.05 \, m$, the molal free energy is $A/n = 2.572 \cdot 10^{8} \, J/mol$ and the gravitational energy is $E/n= 0.2330 \cdot 10^{8} \, J/mol$. Heating the gas to $T=400 \, K$ we find: $z_{m}= 1136.1 \, m$; $A/n= 3.423 \cdot 10^{8} \, J/mol$; $E/n= 0.2961 \cdot 10^{8} \, J/mol$. 2. Gas in centrifugal field2.1 Ideal gasA cylindrical vessel is filled with gas and mounted on a disk rotating with angular frequency $\omega$. The inner and outer ends of the cylinder are at $r_{0}$ and $r_{1}=r_{0}+L$, the cross section area is $F$. The analysis proceeds in the same way as above, except that the potential centrifugal energy in a layer is now $\rho(r) F\, dr\, m \omega^{2} (r_{1}^{2}-r^{2})/2 $ instead of $ \rho(z) m g z F dz$. Minimizing the total energy (free plus centrifugal) we have$
\begin{eqnarray}
\delta A/RT = 0 &=& F \int \limits_{r_{0}}^{r_{1}} dr \, \delta
\left\{ \rho \left( \ln \rho v -1 \right)
+ \rho m \omega^{2} (r_{1}^{2}-r^{2}) / 2RT - \lambda \rho \right\}
\\
&=& F \int \limits_{r_{0}}^{r_{1}} dr \,
\left\{ \delta \rho \left( \ln \rho v -1 \right)
+ \rho \, \delta \left( \ln \rho v -1 \right)
+ \delta \rho \, m \omega^{2} (r_{1}^{2}-r^{2}) / 2RT
- \lambda \delta \rho \right\}
\\
&=& F \int \limits_{r_{0}}^{r_{1}} dr \, \delta \rho
\left\{ \ln \rho v + m \omega^{2} (r_{1}^{2}-r^{2}) / 2RT
- \lambda \right\}
\end{eqnarray}
$ Writing $\gamma \equiv m/RT$ we find that $
\rho(r) = \rho_{1} e^{\textstyle - m \omega^{2} (r_{1}^{2}-r^{2}) / 2RT}
$ with an appropriate $\rho_{1}$ such that the density profile is normalized to $\int = n$. |
franz vesely may-08