Greg,

I was actually kind of looking forward to writing some code for this, and I
agree with you that I shouldn't reinvent the wheel. Thanks for the advice!

Matthew

On Tue, Jul 21, 2020 at 9:17 AM Greg Troxel <g...@lexort.com> wrote:

> Matthew <redin...@gmail.com> 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!
>


-- 
Matthew McCleary KF5VUD - www.kf5vud.org
voice 505.239.1044

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