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--- electrical_engineering_1:task_f64r8g2jf4pdomfi_with_calculation [2023/04/03 11:48] – mexleadmin
+++ electrical_engineering_1:task_f64r8g2jf4pdomfi_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-{{tag>conversions energy power area chapter1_1}}
-{{include_n>100}}
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~  Conversion: Energy, Power and Area                    #@TaskText_HTML@#
-A Tesla Model 3 has an average power consumption of $ {{16~{\rm kWh}}\over{100~{\rm km}}}$ and an usable battery capacity of $60~{\rm kWh}$.
-Solar panels produces per $1~\rm m^2$ in average in December $0.2~{{{\rm kWh}}\over{{\rm m^{2}}}}$.
-The car is driven $50~{\rm km}$ per day.
-The size of a distinct Solar module with $460~{\rm Wp}$ (Wpeak) is $1.9~{\rm m} \times 1.1~{\rm m}$.
-. What is the average power consumption of the car per day?
-#@PathBegin_HTML~1~@#
-\begin{align*}
-{{W}\over{l}} &= {{16~{\rm kWh}}\over{100~{\rm km}}} = 0.16 {{~{\rm kWh}}\over{~{\rm km}}}  \\
-W             &= 50~{\rm km} \cdot 0.16 {{~{\rm kWh}}\over{~{\rm km}}} = 8~{\rm kWh}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~1~@#
-\begin{align*}
-W = 8~{\rm kWh}\end{align*}
-#@ResultEnd_HTML@#
-. How many square meters (=$m^2$) of solar panels are needed on average in December?
-#@PathBegin_HTML~2~@#
-\begin{align*}
-A = {{8~{\rm kWh}}\over{0.2 {{~{\rm kWh}}\over{~{\rm m^{2}}}}}} = 40~{\rm m^{2}}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~2~@#
-\begin{align*}
-A=40~{\rm m^{2}}
-\end{align*}
-#@ResultEnd_HTML@#
-. How many panels are at least needed to cover this surface?
-#@PathBegin_HTML~3~@#
-\begin{align*}
-A_{\rm panel}  &= 1.9~{\rm m} \cdot 1.1~{\rm m} = 2.1~{\rm m^{2}} \\
-n_{\rm panels} &= {{A             }\over{  A_{\rm panel      }}}
-                = {{40~{\rm m^{2}}}\over{2.1~{\rm m^{2}/panel}}} \\
-               &= 19.04~{\rm panels}  \rightarrow 20~{\rm panels}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~3~@#
-\begin{align*}
-n_{\rm panels} = 20~{\rm panels}
-\end{align*}
-#@ResultEnd_HTML@#
-.  What is the combined $\rm kWp$ of the panels you calculated in 3. ?
-#@PathBegin_HTML~4~@#
-\begin{align*}
-P &= 20~{\rm panels} \cdot 460~{\rm kWp / panel} \\
-  &= 9'200~{\rm kWp}
-\end{align*}
-#@PathEnd_HTML@#
-#@ResultBegin_HTML~4~@#
-\begin{align*}
-P = 9'200~{\rm kWp}
-\end{align*}
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#

conversions energy power area chapter1 1