Behold The Power Of The Sun

Electricity and Magnetism Level 5

Solar collector power generation plants, such as this one in Spain, work by concentrating solar energy into a small region to drive a steam turbine.

Direct sunlight delivers about $1000~\mbox{Watts/m}^2$ of energy in the infrared, visible, and ultraviolet parts of the spectrum. I want to use mirrors with a total effective area $A$ to direct this solar energy into a cube $1~\mbox{m}$ on a side, heating the cube to a steady $100^\circ$ Celsius so I can start generating steam and running a turbine. What is the necessary effective mirror surface area $A$ in $\mbox{m}^2$ to do this?

Note that your $A$ will be smaller than you might expect. This is because we have used components that are 100% efficient - all the solar energy gets directed onto the cube. In a real life situation the amount of solar energy redirected by the mirrors is much, much less.

Assumptions

The temperature of the air is 20 $^\circ$ C

Image credit: Abengoa

The answer is 4.083.

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Soham Dibyachintan
Jan 28, 2014

Since we can ignore conductive and convective heat transfer, the only way the cube will lose energy is through radiation into the surrounding area. The power loss due to radiation is given by $P=A_c\sigma\epsilon((T_{Cube})^{4}-(T_{Air})^{4})$ where, $A_c$ is the surface area of the cube. $\sigma$ is the Stefan-Boltzmann constant. $\epsilon$ is the emissivity. $T_{Cube}$ is the temperature of the cube. $T_{Air}$ is the temperature of the air. Note that we can't negect the temperature of the air in this problem. As, the temperature of air isn't given, I have taken the temperature of the air to be $20^{\circ}C$ Since the cube is a perfectly black body, it's emissivity is $1$ . Now we can calculate total power loss to be equal to $4083 W$ . This loss must be compensated for by the energy coming in from the mirrors which are reflecting the sun rays. As the mirrors are $100 \%$ efficient, the effective area of the mirrors is $\frac{4083}{1000} m^{2}=\boxed{4.083 m^{2}}$

Nice solution but the question should have mentioned the temperature of air.

Sanat Anand - 7 years, 4 months ago

well room temperature is usually taken to be 300 K or 27 C, so could you please explain why air temperature was taken to be 20? the data provided is really inadequate if elaborations about the surroundings are not provided.

Kabir Chakravarti - 7 years, 3 months ago

I am pretty sure this answer is incorrect .. the Stefan-Boltzmann law depends on the absolute temperature of the black body .. it is independent of the temperature of the surrounding medium. I guess the point is supposed to be that the incoming power is less than it would be if the cube were surrounded by vacuum, but that really doesn't make much sense. Vacuum is a perfect insulator, so in such a case the ONLY losses would be radiative .. having the cube surrounded by air at a lower temperature INCREASES the thermal losses, because air is a better thermal conductor than vacuum. The way the calculation is done above assumes that there is an equivalent RADIANT source of energy that would have maintained the temperature of the cube at ambient in the ABSENCE of the air ... this is almost certainly not the case. The air temperature is maintained over huge areas by solar heating .. that energy was transferred to the cube under ambient conditions by thermal conduction, and does not factor into this calculation.

Thus, I am pretty sure that the inclusion of the air temperature correction in the Stefan-Bolzman law is wrong ... I did the problem using only the final temperature of the cube, and I got 6.59 sq. meters.

David Moore - 6 years, 5 months ago

Very nice solution.

Lily Luna - 7 years, 4 months ago

how is the cube a perfectly black body ?

Vijay Chauhan - 7 years, 4 months ago

also the power loss is not coming equal to 4083 W

Vijay Chauhan - 7 years, 4 months ago

Behold The Power Of The Sun

The answer is 4.083.

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