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Matteo Cantiello edited Convective Efficiency.tex
over 9 years ago
Commit id: 33a5a325e5cf080e07bc7772bd32e5fcffce5dfb
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\end{equation}
The energy excess content carried by the convective element before dissolving is $\cp \rho \Delta T^{*} V$ with $\Delta T^{*}$ here being the temperature excess of the element over its surroundings at the end of its path. Obviously $\Delta T^{*} \simeq \Delta T$, and in the classic Bohm-Vitense work the choice $\Delta T^{*} = 2 \Delta T$ is made. Using these quantities one can define a convective efficiency $\Gamma$ as
\begin{equation}\label{eq:gamma}
\Gamma = \frac{2 \cp \rho \Delta T V}{\frac{4ac}{3}\frac{T^3}{\kappa \rho}\frac{\Delta T}{\lambda/2} \frac{\lambda A}{\bar{v}}} = \frac{\cp}{6ac}\frac{\kappa \rho^2 \bar{v}
\lamda}{T^3} \lambda}{T^3}
\end{equation}
where the second identity used the fact that
$V/A=\lamda/6$ $V/A=\lambda/6$ for a sphere of diameter $\lambda$.