Hacker Newsnew | past | comments | ask | show | jobs | submitlogin

> "In terms of efficiency, the solar cells are currently showing around a 2-2.5 per cent power conversion rate" [1]

Since you can probably get 250-350 watts of sunlight (avg) per square meter, it means 1 sq m generates 5 to 8.75 watts. Which is 0.5 to 0.875 watts per dollar (or $1.14-$2 per watt).

I don't think that's cheaper than the current offerings, especially if you don't consider these printed panels take 10X more space. Also the lifespan isn't known.

But consider that the price of these new panels will likely fall, if mass-produced.

[1]: https://cleantechnica.com/2017/05/17/researchers-australias-...

EDIT:

Other people say it's $0.40 per watt. [2]

[2]: http://www.thefifthestate.com.au/energy-lead/energy/cheap-an...



I'm not sure where you are getting the 250-300W/m2 of irradiance. The standard value used in all solar industry calulations is 1000W/m2, but in practice, you should expect to see 800W/m2 at sea level and moderate lattitude.


Average capacity factor for solar cell is about 25% (night, clouds, seasons, etc).

1000W/m2 * 0.25 => ~250W/m2.


1000W/m^2 is average daily peak irradiance. That number is useful for designing system capacity, but doesn't reflect output throughout the day.


1000W/m2 is at noon at the equator without clouds/haze/dust.


1000W m^2 is actualy closer to high noon at Kansas on a fully clear day (this is equivalent to 2,000 umol/cm^2/s-1) Equatorial solar irradiance is actually higher (closer to 1,050W/m^2) because the equator is just a bit closer to the sun.


>Equatorial solar irradiance is actually higher (closer to 1,050W/m^2) because the equator is just a bit closer to the sun.

No, no, no. Equatorial is higher because the sun is more directly overhead on average than in Kansas.


That's not how the inverse square law works for photon flux density.


Maybe that's the 24-hour average?




Guidelines | FAQ | Lists | API | Security | Legal | Apply to YC | Contact

Search: