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\begin{LARGE}
Pipe flow problem: fluid comes in with pressure
$P_{\rm in}$ and exits with pressure $P_{\rm out}$.  
$P_{\rm in} - P_{\rm out}$ is what drives the fluid flow.

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\vspace{0.4cm}

{\bf How does one set up the pressure head $P_{\rm in} - P_{\rm out}$ in MD?}
\end{LARGE}

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\begin{LARGE}
{\bf Reflecting Particle Method to Generate Pipe Flow:}

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\flushleft$\bullet$ Conserve {\bf number of particles}
(efficient PBC setup)

\flushleft$\bullet$ Conserve {\bf total energy} 
(no need for temperature rescaling)

\flushleft$\bullet$ Faithful to Navier-Stokes eqn; reach
steady-state gracefully, etc.

\end{LARGE}

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\begin{LARGE}
\flushleft$\bullet\;$ {\bf Demon vs Heat -- Facts:}

-- Demons don't do work; system energy stays constant.
  
-- Heat {\em is} generated by viscous shearing inside fluid. 
Continuum solution predicts a dissipation rate of
 
$$\frac{dQ}{dt} =  \int \tau_{ij} v_{i,j} d\Omega =\frac{L_X L_Y w^3
  }{12\mu} \left ( \frac{\Delta P}{\Delta x} \right )^2.$$

-- Steady-state is observed to be reached; system does not heat up.

\vspace{0.3cm}
\flushleft$\bullet\;$ {\bf Then, where does the heat go?}

$\Delta Q = T\Delta S$. The demons reduce system entropy by
selectively reflecting atoms at $x=0$.

\end{LARGE}

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\begin{Large}

\flushleft$\bullet\;$ {\bf How to estimate $\Delta S$?}

Let $f_1/f_2$ be the rate of atoms hitting $x=0$ from left/right
when the system reaches steady-state. In unit time, the number 
of ways of crossing are

$$ \Omega_1 = \frac{(f_1+f_2)!}  {f_1!f_2!} 
\;\;\;\;\; \longrightarrow \;\;\;\;\;
\Omega_2 = \frac{(f_1+f_2)!} {(f_1+pf_2)!((1-p)f_2)!}$$ 

\vspace{0.1cm}
And, $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
\;\;\;\Delta S = k_{\rm B} \log\Omega_1 -   k_{\rm B} \log\Omega_2\;$,
\vspace{0.1cm}

therefore, there should be (?),

$$ \frac{dQ}{dt} = \frac{L_X L_Y w^3}{12\mu} \left (
\frac{\Delta P}{\Delta x} \right )^2 \;\;\;\;\;=^? $$

$$ k_{\rm B} T \log \frac{\Omega_1}{\Omega_2} \approx k_{\rm B}T \left \{
f_1 \log (1+\frac{pf_2}{f_1}) + pf_2 \log („+\frac{f_1}{f_2} ) +
f_2(1-p)\log(1-p) \right \}. $$
\end{Large}

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