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Merge pull request #37 from thermotools/soret_error
Fix stupid Soret coefficient bug
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\section{Soret coefficients} | ||
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The Soret coefficient is a measure of the steady state separation in a mixture induced by a temperature gradient. Ortiz de Zárate \cite{ortiz2019definition} defines the Soret coefficients of a multicomponent mixture ($s$ components) through | ||
\begin{equation} | ||
\begin{bmatrix} | ||
x_1 (1 - x1) & x_1 x_2 & \hdots & x_1 x_{s - 1} \\ | ||
x_2 x_1 & x_2 (1 - x_2) & \hdots & x_2 x_{s - 1} \\ | ||
\vdots & & \ddots & \vdots \\ | ||
x_{s - 1} x_1 & & & x_{s - 1}(1 - x_{s - 1}) | ||
\end{bmatrix} | ||
\begin{pmatrix} | ||
S_{T, 1} \\ | ||
S_{T, 2} \\ | ||
\vdots \\ | ||
S_{T, s - 1} | ||
\end{pmatrix} | ||
= | ||
- \frac{1}{\nabla T} | ||
\begin{pmatrix} | ||
\nabla x_1 \\ | ||
\nabla x_2 \\ | ||
\vdots \\ | ||
\nabla x_{s - 1} | ||
\end{pmatrix}, | ||
\end{equation} | ||
or more compactly, | ||
\begin{equation} | ||
\Mat{X} \Vec{S}_T = - \frac{\nabla \Vec{x}}{\nabla T}. | ||
\end{equation} | ||
Similarly to the thermal diffusion coefficients, this definition carries the advantage that | ||
\begin{equation} | ||
\Mat{X} \Vec{S}_T = - \frac{\nabla \Vec{x}}{\nabla T} \quad \iff \quad \Mat{W} \Vec{S}_T = - \frac{\nabla \Vec{w}}{\nabla T}, | ||
\end{equation} | ||
such that mole- and mass fractions can be used interchangably with the same Soret coefficients. | ||
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In the state with vanishing mass fluxes ($\Vec{J} = \Vec{0}$). From the condition of vanishing mass fluxes, we find that we can compute the Soret coefficients as\footnote{See the memo on definitions of the diffusion coefficient for notes on $\Mat{D}^{(z)}$.} | ||
\begin{equation} | ||
\begin{split} | ||
\Vec{J} = - c \left( \Mat{X} \Vec{D}_{T}^{(z)} \nabla T + \Mat{X} \Mat{D}^{(z)} \Mat{X}^{-1} \nabla \Vec{x} \right) &= \Vec{0} \\ | ||
- \frac{\nabla \Vec{x}}{\nabla T} &= \Mat{X} \left(\Mat{D}^{(z)}\right)^{-1} \Vec{D}_{T}^{(z)} \\ | ||
\Vec{S}_T &= \left(\Mat{D}^{(z)}\right)^{-1} \Vec{D}_{T}^{(z)}. | ||
\end{split} | ||
\end{equation} | ||
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For a binary system (1, 2), this definition thus yields | ||
\begin{equation} | ||
S_{T, 1}^{(b)} = \frac{D_{T, 1}^{(z, b)}}{D_{11}^{(z, b)}}. | ||
\end{equation} | ||
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From the preceding relations it is possible to show that when using this definition of the Soret coefficient \cite{ortiz2019definition}, | ||
\begin{equation} | ||
\begin{split} | ||
\lim_{x_1 \to 0} S_{T, 2} = S_{T, 2}^{(b, 3)} \\ | ||
\lim_{x_2 \to 0} S_{T, 1} = S_{T, 1}^{(b, 3)} \\ | ||
\lim_{x_3 \to 0} S_{T, 1} - S_{T, 2} = S_{T, 1}^{(b, 2)}, | ||
\end{split} | ||
\label{eq:soret_binary_limit} | ||
\end{equation} | ||
where $S_{T, i}^{b, j}$ denotes the Soret coefficient of component $i$ in a binary mixture with $j$, with species $j$ being the dependent species. Figure \ref{fig:soret_binary_limit} shows how this convergence behaviour is obeyed. | ||
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\begin{figure} | ||
\centering | ||
\includegraphics[width=.85\textwidth]{soret_limit.pdf} | ||
\caption{The convergence of the ternary Soret coefficient to the corresponding binaries as indicated in Eq. \eqref{eq:soret_binary_limit}.} | ||
\label{fig:soret_binary_limit} | ||
\end{figure} |
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