Exercise E6 Conversion: Energy, Power and Area

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 $460~{\rm W_p}$ (Watt peak) is $1.9~{\rm m} \times 1.1~{\rm m}$.

1. What is the average power consumption of the car per day?

Solution

\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*}

Result

\begin{align*} W = 8~{\rm kWh}\end{align*}

2. How many square meters (=$\rm m^2$) of solar panels are needed on average in December?

Solution

\begin{align*} A = {{8~{\rm kWh}}\over{0.2 {{~{\rm kWh}}\over{~{\rm m^{2}}}}}} = 40~{\rm m^{2}} \end{align*}

Result

\begin{align*} A=40~{\rm m^{2}} \end{align*}

3. How many panels are at least needed to cover this surface?

Solution

\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*}

Result

\begin{align*} n_{\rm panels} = 20~{\rm panels} \end{align*}

4. What is the combined $\rm kW_p$ of the panels you calculated in 3. ?

Solution

\begin{align*} P &= 20~{\rm panels} \cdot 460~{\rm kW_p / panel} \\ &= 9'200~{\rm kW_p} \end{align*}

Result

\begin{align*} P = 9'200~{\rm kW_p} \end{align*}