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Published in:
2022 | OriginalPaper | Chapter
7. Lösungen der Übungsbeispiele
Author : Herbert Sigloch
Published in: Technische Fluidmechanik
Publisher: Springer Berlin Heidelberg
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Zusammenfassung
Übung 1
Wasserverlust
$$\begin{aligned}\displaystyle\Updelta V&\displaystyle=V_{\text{R}}-V_{\text{D}}\\ \displaystyle V_{\text{D}}&\displaystyle=\dfrac{1}{2}\cdot l\cdot H\cdot B\\ \displaystyle l&\displaystyle=H/\tan\alpha\end{aligned}$$
$$\begin{aligned}\displaystyle\tan\alpha&\displaystyle=\dfrac{a}{g}=\dfrac{3}{{9{,}81}}={0{,}3058}\\ \displaystyle l&\displaystyle={2{,}5}/{0{,}3058}\,\mathrm{[m]}={8{,}175}\,{\text{m}}\\ \displaystyle V_{\text{D}}&\displaystyle=\frac{1}{2}\cdot{8{,}175}\cdot{2{,}5}\cdot{2{,}8}\,\mathrm{[m^{3}]}={28{,}6}\,\mathrm{m^{3}}\\ \displaystyle\boldsymbol{\Updelta V}&\displaystyle\boldsymbol{=50-{28{,}6}=21{,}4\,\mathrm{m}^{3}}\end{aligned}$$