Calculating Solar Panel Area Needed for a 1MW Power Output

[Ethan_Tab]

Homework Statement

Hello everyone, I'm working on a (simplified) Mars energy model in which I need to calculate the area of land I need to cover in solar panels in order to achieve an particular power output.

Say the land I'm looking at receives a constant solar irradiance of 100W/m^2 from the sun. How would I determine how many panels I need to put up if each 1mX1m solar panel has a power output of 300W and the colony needs a total of 1MW of continuous power.

Homework Equations

The Attempt at a Solution

I understand that I could simply divide 1,000,000W/300W to obtain the amount of panels I need and then figure out the area which they take up. However I'm not sure where the 100W/m^2 solar irradiance comes into play.

 

[berkeman]

100W/m^2 from the sun. How would I determine how many panels I need to put up if each 1mX1m solar panel has a power output of 300W

Yes, those numbers do not match up, right?

However I'm not sure where the 100W/m^2 solar irradiance comes into play.

Check for errors in the problem statement?

Also, with a rotating planet, you don't get the peak insolation for the whole day/night period of the planet...

 

[Ethan_Tab]

So if I just knew that I had an solar irradiance of 100W/m^2 and I needed to power a 1,000,000 W colony how would I determine how large of a solar farm I would need?

 

[berkeman]

So if I just knew that I had an solar irradiance of 100W/m^2 and I needed to power a 1,000,000 W colony how would I determine how large of a solar farm I would need?

Model how that 100W/m^2 peak insolation translates into the insolation versus the full planetary rotation cycle. There has to be lots of info on the Internet about how to do that. Please show your work.

 

[haruspex]

a constant solar irradiance of 100W/m^2

Clearly there is no such patch of land, on Earth or on Mars. The equatorial tangential velocity is nearly 1000km/h, so we're not going to solve it by a mobile system.

According to NASA, the peak irradiance on Mars is about 580 W/m². That corresponds to an average of 145W/m² over the whole surface, so perhaps the 100W/m² figure is supposed to be the daily average. Whatever, batteries will need to be included.

If the peak panel output is 300W/m² then that suggests an efficiency of around 75%, so a daily average output of only 75W/m².

Admittedly, that's a lot of guesswork. Are you sure you have quoted the question exactly as given to you?