Worked exercise — Solar panel

Problem

A solar panel of area $S = 2\;\text{m}^2$ is struck by sunlight, of intensity $\abs{\vv I} = 1\;\text{kW/m}^2$. The direction of propagation of the light forms an angle of $60^\circ$ with the normal to the panel. The absorbed energy is used to heat $2\;\text{L}$ of water from $20\,^\circ\text{C}$ to $100\,^\circ\text{C}$. How much time is needed? (specific heat of water $c_s = 4186\;\text{J/(kg·K)}$)

Solution

Setup. The intensity $\abs{\vv I}$ is the power per unit surface carried by the wave (magnitude of the time-averaged Poynting vector). The power actually collected by the panel depends on the tilt: only the component of the flux along the normal counts, hence the $\cos 60^\circ$ factor.

Power incident on the panel.

$P = \abs{\vv I}\,S\,\cos 60^\circ = 1000\cdot 2\cdot 0{,}5 = 1000\;\text{W}$

Energy needed to heat the water. The mass of $2\;\text{L}$ of water is $m = 2\;\text{kg}$; the temperature rise is $\Delta T = 80\;\text{K}$:

$E = m\,c_s\,\Delta T = 2\cdot 4186\cdot 80 \approx 6{,}7\cdot 10^5\;\text{J}$

Time required. Assuming all the collected power goes into heating the water:

$t = \frac{E}{P} = \frac{6{,}7\cdot 10^5}{1000} \approx 670\;\text{s}$

$\ev{t \approx 670\;\text{s} \approx 11\;\text{min}}$

In practice more time would be needed, because the real efficiency is below $100\%$ and there are heat losses to the environment.

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Topics: Onde elettromagnetiche Concepts: Vettore di Poynting · Intensità di un’onda