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Relative to $Trec$


\begin{displaymath}
\frac{\partial Tcal}{\partial Trec} =
\frac{\partial Tcal}{...
...artial Tcal}{\partial Tau} \frac{\partial Tau}{\partial Trec}
\end{displaymath} (33)

From Eq. (18)
\begin{displaymath}
\frac{\partial Tcal}{\partial T_{emi}}
= \frac{Tcal}{T_{emi}-T_{load}}
\end{displaymath} (34)

and from Eq. (20)
\begin{displaymath}
\frac{\partial T_{emi}}{\partial Trec} = -1 + \frac{T_{emi}+Trec}
{T_{load}+Trec}
\end{displaymath} (35)

thus
\begin{displaymath}
\frac{\partial Tcal}{\partial T_{emi}} \frac{\partial T_{emi}}{\partial Trec}
= \frac{Tcal}{T_{load}+Trec}
\end{displaymath} (36)

For the second term, from Eq. (18)
\begin{displaymath}
\frac{\partial Tcal}{\partial Tau} = Air\_mass * Tcal
\end{displaymath} (37)

and from Eq. (27)
\begin{displaymath}
\frac{\partial Tau}{\partial Trec} =
\frac{\partial Tau}{\partial T_{emi}} \frac{\partial T_{emi}}{\partial Trec}
\end{displaymath} (38)


\begin{displaymath}
\frac{\partial Tau}{\partial Trec} =
\frac{1}{Air\_mass * T_{atm} * F_{eff}} \frac{\partial T_{emi}}{\partial Trec}
\end{displaymath} (39)


\begin{displaymath}
\frac{\partial Tcal}{\partial Tau} \frac{\partial Tau}{\part...
...cal}{T_{atm} * F_{eff}} \frac{T_{emi}-T_{load}}{T_{load}+Trec}
\end{displaymath} (40)

giving finally
\begin{displaymath}
\frac{1}{Tcal} \frac{\partial Tcal}{\partial Trec} =
\frac{...
...eff} - T_{load} + T_{emi}}
{F_{eff}*T_{atm}*(T_{load} + Trec)}
\end{displaymath} (41)

The typical numbers mentionned above, with $T_{load} = T_{cab}$, yield

\begin{displaymath}
\frac{1}{Tcal} \frac{\partial Tcal}{\partial Trec} =
\frac{F_{eff}-1}{F_{eff}} \frac{1}{T_{load}+Trec}
\end{displaymath} (42)

of the order of $3$ $10^{-4}$ per K. Note that this could be higher for higher opacities (hence higher $T_{emi}$).


next up previous contents
Next: Relative to Up: Derivatives Previous: Derivatives   Contents
Gildas manager 2018-06-21