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# Solar Thermal

As one example, solar thermal systems can be used for cooking.

Suppose we want to boil a kettle of water using sunlight. If we have a \(\SI{1}{\meter\squared}\) mirror that concentrates incident sunlight with an intensity of \(\SI[per-mode=symbol]{1000}{\watt\per\meter\squared}\) onto a \(\SI{1}{\liter}\) kettle of water, how long will it take (in seconds) to heat that water from \(\SI{20}{\celsius}\) to \(\SI{100}{\celsius}\)?

Details:

- Ignore the thermal mass of the kettle (only consider the water)
- Ignore heat losses from the kettle
- Assume all sunlight incident on the mirror ends up absorbed by the kettle
- Density of water is \(\SI[per-mode=symbol]{1}{\kilo\gram\per\liter}\) and heat capacity of water is \(\SI[per-mode=symbol]{4.18}{\joule\per\gram\per\kelvin}\)

If you put an absorber in sunlight, it continuously absorbs more energy. If the absorber constantly absorbs more energy, why doesn't the absorber's temperature increase indefinitely?

The reason is that as the absorber's temperature increases, it begins to lose heat to the ambient environment, which is at a lower temperature. The absorber stops heating up once it reaches a temperature where the absorbed sunlight is balanced by losses to the surroundings.

One mechanism for an absorber to lose heat to its surroundings is through **convection**. Convective heat loss occurs from a surface that's exposed to a fluid at lower temperature. As the fluid near the surface heats up, it carries thermal energy away from the surface and is replaced by fluid at the ambient temperature.

The effectiveness of a fluid at pulling heat away from a surface through this process is characterized by a convection coefficient, measured in \(\si[per-mode=symbol]{\watt\per\meter\squared\per\kelvin}\). The convection losses are proportional to the temperature difference between the surface and the fluid.

In the case of a solar absorber surface, which geometric parameter do convection losses depend on?

In the previous quiz, we considered a black absorber, which has an absorptance and emittance of \(1\). Under an intensity of \(\SI[per-mode=symbol]{50000}{\watt\per\meter\squared}\) and considering only radiation losses, we calculated the highest temperature the absorber could reach to be \(\SI{698}{\celsius}\).

If the black absorber was replaced with a gray absorber with an absorptance and emittance of \(0.3\), what is the maximum temperature it could reach (in Celsius)?

Only consider radiative losses and use \(T_\textrm{amb} = \SI{20}{\celsius}\).

A gray absorber has radiative properties that do not vary with wavelength. Its absorption and emittance are equal and are between \(1\) (corresponding to a black surface) and \(0\) (corresponding to a white surface).

**Remember** that radiative losses are given by \(A\epsilon\sigma(T_\textrm{abs}^4-T_\textrm{amb}^4)\), where the Stefan-Boltzmann constant \(\sigma = \SI[per-mode=symbol]{5.67e-8}{\watt\per\meter\squared\per\kelvin\tothe{4}}.\)

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