4 ms·

> Solar averages ~10W/m^2 long term (days/weeks) That figure surprises me. What I’ve heard before is that full sun is roughly 1kW/m², and that over time you sh

> Solar averages ~10W/m^2 long term (days/weeks) That figure surprises me. What I’ve heard before is that full sun is roughly 1kW/m², and that over time you should expect to get 4–5 useful hours (depending on latitude and weather conditions of where you are—some will be outside this range on either side), so if you take 21% efficient solar panels, the total would be 1000 (watts) × 4 (hours) / 24 (hours) × 0.21 (efficiency) = 35W. 10W is a lot less than 35W. I presume I have some incorrect assumption or am overlooking something. (I’m interested in this because I’m planning to build a velomobile and live out of it while going round Australia for a year. Power systems are something I’ve been putting a fair bit of theoretical research into, but I have no practical experience whatsoever. I’ve been planning to build solar panelling into the body.)

ctdonath 6y ago
35 is pretty close to 10 when you start at 1300 and prune obstructions. Now address clouds, angles, dust, snow, opacity, malfunctions, storage inefficiencies, panel aging, and chronic optimism. 10W.
- chrismorgan 6y ago
  Most of that is what the 4–5 useful hours per day figure is supposed to take care of (that’s a year-round thing, too, averaging summer, which may be as much as double that, and winter, which may be half). Hmm… storage inefficiencies? Are you including battery storage and perhaps an inverter in your calculations? I guess that’d account for some loss. I’m guessing also from “snow” that you may be in a place that gets comparatively little sunlight?
  - gibspaulding 6y ago
    I wonder if "4-5 useful hours per day" is per sunny day, so you can cut your number in half again for cloudy days.
    - chrismorgan 6y ago
      No, this is from averaging it across the entire year, based mostly on locations in the south of Australia (Gippsland, Melbourne, Adelaide, Ballarat, those sorts of areas), but I looked briefly at other parts of the world also. The idea is then that a panel rated at 250W can produce an average of about 1kWh per day when pointed in the optimal direction and never shaded (but without tracking the sun); the average winter day may get you only 500W, and the average summer day over 1500W, with plenty of variation within those seasons day by day also.
  - adrianN 6y ago
    42,339 MW of PV installations produced 39,401 GWh of energy in 2017 in Germany[1], so about 38 days of peak generation, or roughly 10.5% of peak. A square meter of PV panels produces around 150W. So 15W average for a panel in a good location in Germany. 10W/m^2 seems like a good estimate if you want your stuff to work reliably year round. Better locations than Germany would do better of course, and restricting yourself to just summer probably helps quite a bit too (winters are dark and cloudy in Germany). [1] https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statistics https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statist...
    - Retric 6y ago
      Germany is north of Maine, it’s a terrible location for solar power and heavily subsidized. At best they hit a capacity factor of 11.6% in 2018. By comparison the US had a capacity factor of 28% for utility-sized PV plants in 2015, but it’s dropped as less efficient locations have been added. A square meter of PV panels produces around 150W, that’s some low efficiency panels, you can easily hit 200+W/M2 using a high efficiency panel like SunPower Maxeon 3. Solar isn’t a great fit everywhere. So, people really can see wildly different outputs from the same setup based on location. At 28% capacity factor averaging 60+W/M is very possible.
      - adrianN 6y ago
        Thanks for the reference to the 200W panel!
hinkley 6y ago
The most shocking thing I ever learned about solar panels is that they behave as a rectifying diode when they are not receiving sufficient photons. That means you can screw up the current from an entire bank of solar panels by having just one of them in deep shade. So if you want maximum generation, you either have to site them in a very clear area, or wire them all in parallel instead of serial, which means much fatter wires to reduce transmission losses, and then larger conversion losses?
- organsnyder 6y ago
  I'm not sure on the details, but according to my solar installer newer panels (or inverters or some other component) are no longer susceptible to this.
  - chrismorgan 6y ago
    The key term with this is bypass diodes. Not all panels will be equipped with them (which puzzles me, given the substantial benefit and what I imagine to be near-trivial BOM cost).
- senkora 6y ago
  I learned this at the Exploratorium in SF. They had an exhibit consisting of a miniature solar panel with an output reading and a small token that could cover a single panel. Very fun trip!
- yardie 6y ago
  I don't think this is the case with newer panels. Our boat stern mounted panels atleast had one panel shaded at anytime. The output drop was proportional to the area shaded. The bigger losses came from the angle of the sun. Output was mostly logarithmic, almost nothing at sunrise/sunset, full output at 12pm, 50% by 3pm.
  - PaulDavisThe1st 6y ago
    It's still true of the cells in the panels. But it's typical these days to add a device between the panels and the DC circuit than can improve the situation (e.g. by cutting the panel out of the circuit if it isn't productively contributing the array's generating capacity at the moment).

ctdonath 6y ago

35 is pretty close to 10 when you start at 1300 and prune obstructions. Now address clouds, angles, dust, snow, opacity, malfunctions, storage inefficiencies, panel aging, and chronic optimism. 10W.

chrismorgan 6y ago
Most of that is what the 4–5 useful hours per day figure is supposed to take care of (that’s a year-round thing, too, averaging summer, which may be as much as double that, and winter, which may be half). Hmm… storage inefficiencies? Are you including battery storage and perhaps an inverter in your calculations? I guess that’d account for some loss. I’m guessing also from “snow” that you may be in a place that gets comparatively little sunlight?
- gibspaulding 6y ago
  I wonder if "4-5 useful hours per day" is per sunny day, so you can cut your number in half again for cloudy days.
  - chrismorgan 6y ago
    No, this is from averaging it across the entire year, based mostly on locations in the south of Australia (Gippsland, Melbourne, Adelaide, Ballarat, those sorts of areas), but I looked briefly at other parts of the world also. The idea is then that a panel rated at 250W can produce an average of about 1kWh per day when pointed in the optimal direction and never shaded (but without tracking the sun); the average winter day may get you only 500W, and the average summer day over 1500W, with plenty of variation within those seasons day by day also.
- adrianN 6y ago
  42,339 MW of PV installations produced 39,401 GWh of energy in 2017 in Germany[1], so about 38 days of peak generation, or roughly 10.5% of peak. A square meter of PV panels produces around 150W. So 15W average for a panel in a good location in Germany. 10W/m^2 seems like a good estimate if you want your stuff to work reliably year round. Better locations than Germany would do better of course, and restricting yourself to just summer probably helps quite a bit too (winters are dark and cloudy in Germany). [1] https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statistics https://en.wikipedia.org/wiki/Solar_power_in_Germany#Statist...
  - Retric 6y ago
    Germany is north of Maine, it’s a terrible location for solar power and heavily subsidized. At best they hit a capacity factor of 11.6% in 2018. By comparison the US had a capacity factor of 28% for utility-sized PV plants in 2015, but it’s dropped as less efficient locations have been added. A square meter of PV panels produces around 150W, that’s some low efficiency panels, you can easily hit 200+W/M2 using a high efficiency panel like SunPower Maxeon 3. Solar isn’t a great fit everywhere. So, people really can see wildly different outputs from the same setup based on location. At 28% capacity factor averaging 60+W/M is very possible.
    - adrianN 6y ago
      Thanks for the reference to the 200W panel!

hinkley 6y ago

The most shocking thing I ever learned about solar panels is that they behave as a rectifying diode when they are not receiving sufficient photons. That means you can screw up the current from an entire bank of solar panels by having just one of them in deep shade. So if you want maximum generation, you either have to site them in a very clear area, or wire them all in parallel instead of serial, which means much fatter wires to reduce transmission losses, and then larger conversion losses?

organsnyder 6y ago
I'm not sure on the details, but according to my solar installer newer panels (or inverters or some other component) are no longer susceptible to this.
- chrismorgan 6y ago
  The key term with this is bypass diodes. Not all panels will be equipped with them (which puzzles me, given the substantial benefit and what I imagine to be near-trivial BOM cost).
senkora 6y ago
I learned this at the Exploratorium in SF. They had an exhibit consisting of a miniature solar panel with an output reading and a small token that could cover a single panel. Very fun trip!
yardie 6y ago
I don't think this is the case with newer panels. Our boat stern mounted panels atleast had one panel shaded at anytime. The output drop was proportional to the area shaded. The bigger losses came from the angle of the sun. Output was mostly logarithmic, almost nothing at sunrise/sunset, full output at 12pm, 50% by 3pm.
- PaulDavisThe1st 6y ago
  It's still true of the cells in the panels. But it's typical these days to add a device between the panels and the DC circuit than can improve the situation (e.g. by cutting the panel out of the circuit if it isn't productively contributing the array's generating capacity at the moment).