Behold the Power of the Sun!

This post originally appeared on the Brilliant blog on 2/10/2013.

In this last problem of the week we confronted the basic physics of one of the more interesting methods of generating renewable energy, solar collector power generation. Before we launch into its solution, it is time to recognize the winners of our t-shirt raffles for our previous problems of the week.

The lucky solver of the problem Three Quarks for You!, was Norman Vincent C.

Additionally, probability smiled fondly upon Joel K, who solved Space Oddity Radio.

Congratulations Norman and Joel! Both gentlemen will be receiving a free Brilliant.org t-shirt in the mail.

Now onto solar collector energy. To recap, the problem was:

Behold the Power of the Sun!

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 $Watts/m^{2}$ 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 1m on a side, heating the cube to a steady $100^{o}$ Celsius so I can start generating steam and running a turbine. What is the necessary effective mirror surface area A in $m^{2}$ 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.

1. The cube absorbs and emits as a perfect blackbody. It's well insulated so you can ignore any conductive or convective heat transfer.
2. The air around the cube is $20^{o}$ Celsius.
3. Effective area takes into account the angle of the mirrors with respect to the sun, so when calculating the effective area you can ignore the orientation of the mirrors. If the mirrors are angled, the true surface area is larger than the effective area.

Solution

Since we can ignore convective or conductive heat transfer, the only way the cube will lose energy is due to radiation into the surrounding area. The power loss due to radiation is given by $P=A_c \sigma \epsilon(T^4-T_0^4)$, where $A_c$ is the surface area of the cube, $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the emissivity, T is the temperature of the cube and $T_0$ is the temperature of the air. Note that you cannot neglect the temperature of the air in this problem! Also, as a fun aside, the energy loss rate scaling as the temperature of the object to the fourth power is one of the reasons you feel 'cold' when you have a fever. Our body is actually sensitive to <strong>the rate</strong> of energy loss, so when you have a fever you lose energy more quickly and feel cold.Returning to our problem, since the cube is a perfect blackbody, $\epsilon=1$ and we can then calculate a total power loss of $4083$ Watts. This loss must be compensated for by the energy coming in from the mirrors, and so the total area of the mirrors is 4.083 $m^2$.