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Solar panels

Energy and power · level 6, lesson 4 of 6

A question to start

A Mars habitat needs an average of 10 kW of electricity, day and night. How many square metres of solar panels would that take, and how does that compare with the same habitat on Earth?

The idea

A solar panel's power is

\[P = I \times A \times \eta,\]

where \(I\) is the sunlight falling on it, in W/m², \(A\) its area and \(\eta\) its efficiency. But \(I\) changes all the time:

  • At night, it is zero.
  • The Sun's angle: a flat panel catches less when the Sun is low. With the Sun at angle \(\theta\) from overhead, it gets \(I\cos\theta\).
  • The air: dust (on Mars) or clouds (on Earth) take some of it away.

So engineers use the average sunlight over a whole day. For a flat panel near the equator on a clear day, the average over a day and night is roughly \(1/\pi\) of the noon sunlight above the air, less whatever the air takes. A rough guide, assumed here:

A flat panel near the equator, clear sky Mars Earth
Sunlight above the air, noon 586 W/m² 1,361 W/m²
Share reaching the ground about 70% (dust) about 75%
Average over a whole day, about 130 W/m² 325 W/m²
Panel efficiency assumed 25% 25%
Average power per m² of panel about 33 W about 81 W

These are estimates for clear skies; real places vary with latitude, season and weather.

nightnoonnightaveragepeak

Worked examples

1. The habitat on Mars. 10,000 W ÷ 33 W/m² \(\approx 300\) m² of panels, a little more than a tennis court. Plus batteries for the night, and more panels for the storms.

2. The same habitat on Earth. 10,000 ÷ 81 \(\approx 125\) m². Mars needs about 2.5 times the area.

3. Tilting the panel. A panel tilted to face the noon Sun catches more. At a latitude of 40°, with the Sun 40° from overhead at noon, a flat panel gets \(\cos 40° = 0.77\) of what a tilted one does at noon.

Try it

  1. A 2 m² panel, 25% efficient, has 500 W/m² falling on it. Find its power.
  2. Why is the average over a day much less than the noon value?
  3. Using the table, how much panel area gives an average of 1 kW on Mars?
  4. A Mars greenhouse needs 50 kW of lamps on average. Using the table, how much panel area is that?
Answers
  1. \(500 \times 2 \times 0.25 = 250\) W.
  2. There is no sunlight at night, and in the morning and evening the Sun is low.
  3. \(1000/33 \approx 30\) m².
  4. \(50{,}000/33 \approx 1{,}500\) m², which is why lamps for food are one of the biggest energy costs.

Common mistake

Using the noon value for a whole day. A panel that gives 150 W at noon gives far less on average, about a third of that on a clear day, and nothing at night.

Where it leads