Matteo Cantiello edited Convective Efficiency.tex  over 9 years ago

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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 $\lambda$.