Matthew <[email protected]> writes: > I'm thinking of installing a solar electric system on my house, and am in > the planning phases. One of the things that you use in planning these > systems is "peak sun hours per day" for your location -- the number of > average hours each day during which the sun is shining at 100% of maximum > for a given location.
average hours at 100% of theoretical is very different from calculating KWh/m^2 per day. When the sun is low in the sky, it might only deliver 250 W/m^2 to horizontal, but still be "at 100%". But I don't see why people would care about this, rather than light arriving at the panel at the angle it's really at, perhaps combined with the W/m^2 ==> W(electricty) transfer function. Look at the graph of solar radiation. On blue-sky days, you will see zero at night, very slowly building to maybe 50 in the morning, and an abrupt transition up when the sun hits your sensor (when it gets over the tall pines trees, and yes I see you are NM not MA so that's a joke but the concept reamins). Then it traces a curve which is sort of sin(elevatio) until the sun goes behind an obstacle, and then it fades out. On cloudy days, you get less. On partly sunny days, there is sort of the curve I described, sometimes at it, sometimes less. And, there are often values above the blue-sky line, which I think is lensing from cloud edges. > You can look up these values for a nearby city (in my case, the closest is > Albuquerque, 6.77 peak sun hours per day or 6.77 kWh/m^2/day), but I've > been running a Davis Vantage Pro 2 Plus at my location (with a watt meter > and a UV sensor) for about fifteen months now, collecting the data with > WeeWX, so I should have pretty reliable data by now. But I don't know how > to turn the wattage data (instantaneous watts per meter) into what I need, > which is KWh / m^2 / day. Probably it's possible with a lot of math and/or > a spreadsheet, but I was wondering if perhaps WeeWX has some built-in > method. I would suggest writing a program and reading the weewx database and calculating, rather than extending weewx, because you want to calculate over your historical data, rather than have an online today's energy report. The Davis sensor tries to measure W/m^2, and I believe that's arriving on a horizonal square meter. So if you take a day, and for each value (assuming a 5 minute archive interval), multiply the value by 5m/60m (= 1/12) and add them all, you will get the total energy (in kWh, which is like Joules, but 3.6E6 bigger). Is your sensor level (bubble level)? You also need to check calibration of your sensor. I haven't done that yet for mine, but on a blue-sky day you should be able to calculate the standard solar irradiance for your latittude (and maybe humidity/temp??) and compare that to the curve you measure. Keep in mind that for solar, your panels will be angled. So the energy is different, and generally more. There are various libraries out there for doing this sort of thing. Before you write much code, see https://pvpmc.sandia.gov/applications/pv_lib-toolbox/ > This isn't at all critical to have, but it should be more accurate than > Albuquerque's data (Albuquerque being about 70 straight-line miles north of > me). Thanks! It's more accurate locally, but it's 1 year vs (probably) an average of multiple years, and you have the calibration issue to deal with. But I encourage you to do this anyway! -- You received this message because you are subscribed to the Google Groups "weewx-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. To view this discussion on the web visit https://groups.google.com/d/msgid/weewx-user/rmituy0ucbm.fsf%40s1.lexort.com.
