deletions | additions       diff --git a/Convective Efficiency.tex b/Convective Efficiency.tex  index 226e709..b14be06 100644  --- a/Convective Efficiency.tex  +++ b/Convective Efficiency.tex 
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E_{rad} = \frac{4ac}{3}\frac{T^3}{\kappa \rho}\frac{\Delta T}{\lambda/2} \frac{\lambda A}{\bar{v}}.   \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} \begin{equation}  \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} \lambda}{T^3}  \end{equation}  where the second identity used the fact that $V/A=\lambda/6$ for a sphere of diameter $\lambda$. The next step is substituting the calculated average velocity from \ref{eq:velocity} to obtain  \begin{equation}\label{eq:gamma}  \Gamma = \frac{\cp}{12\sqrt{2} a c}\frac{\kappa g Q^{1/2} \rho^{5/2} \lambda^2}{\P^{1/2} T^3}(\nabla - \nabla')^{1/2}.   \end{equation}