\documentclass{../bkbp}
\begin{document}
    \heading{Goss}{Beleuchtungstechnik}{8.5.2024}
    \section{Aufgabe: Beleuchtung Turnraum}
    \paragraph{Gegeben}
    \begin{equation}
    \begin{split}
        \bar{E} &= 300lx (0.75m ü.O.F.) \\
        H &= 4.5m \\
        a = L &= 13.5m \\
        b = B &= 11.1m \\
        LVK &= A2 \\
        WF = F_W &= 0.6 \\
        \rho_{Decke} &= 60\% \\
        \rho_{Wand} &= 50\% \\
        \rho_{Boden} &= 40\% \\
    \end{split}
    \end{equation}
    \paragraph{Berechnung}
    \begin{equation}
    \begin{split}
        A &= 13.5m*11.1m \\
        A &= 149.85m^2 \\
        h = H - 0.75m &= 3.75m \\
        \\
        k &= \frac{A}{(L+B)*h} \\
        k &= \frac{149.85m^2}{(13.5m+11.1m)*3.75m} \\
        k = 1.62 &\approx 1.5 \\
        \\
        \eta_R &= 83\% \\
        \eta_{LB} &= 100\% \\
        \eta_B = \eta_{LB}*\eta_R &= 83\% \\
        \\
        \phi &= \frac{\bar{E}*A}{\eta_B*F_W} \\
        \phi &= \frac{300lx*149.85m^2}{83\%*0.6} \\
        \phi &= 90271.08lm \\
        \\
        n &= \frac{\phi}{\phi_{LE}} \\
        n &= \frac{90271.08lm}{12klm} \\
        n &= 7.52259 \approx 8
    \end{split}
    \end{equation}
    \section{Aufgabe: Wartezimmer}
    \paragraph{Gegeben}
    \begin{equation}
    \begin{split}
        U &= 230V \\
        f &= 50Hz \\
        \phi_{Le} &= 6.3klm \\
        P &= 50W \\
        \cos(\varphi) &= 0.94 \\
        \eta_{LB} &= 1 \\
        H &= 3m \\
        L &= 6m \\
        B &= 5m \\
        \bar{E} &= 300lx (0.75m ü.O.F.) \\
        \Delta_H &= 0.5m \\
        LVK &= A2 \\
        F_W &= 0.8 \\
        \rho_{Decke} &= 60\% \\
        \rho_{Wand} &= 35\% \\
        \rho_{Boden} &= 40\%
    \end{split}
    \end{equation}
    \paragraph{Berechnung}
    \begin{equation}
    \begin{split}
        A = L*B &= 6m*5m \\
        A &= 30m^2 \\
        h = H-\Delta_H-0.75m &= 3m-0.5m-0.75m\\
        h&=1.75m\\
        \\
        k &= \frac{A}{(L+B)*h} \\
        k &= \frac{30m^2}{(6m+5m)*1.75m} \\
        k = 1.56 &\approx 1.5 \\
        \\
        \eta_R &= 77\% \\
        \eta_{LB}&=1\\
        \eta_B=\eta_{LB}*\eta_R&=77\%\\
        \\
        \phi &= \frac{\bar{E}*A}{\eta_B*F_W} \\
        \phi &= \frac{300lx*30m^2}{0.77*0.8} \\
        \phi &= 14610.39lm \\
        \\
        n&=\frac{\phi}{\phi_{Le}}\\
        n&=\frac{14610.39lm}{6.3klm}\\
        n&=2.32\approx 3
    \end{split}
    \end{equation}
\end{document}