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HiTech's avatar
HiTech
Explorer
Dec 14, 2012

Measuring solar yield approaching winter solstice

I need to get a temporary setup on the RV roof to measure solar yield around the winter solstice.

How deeply discharged will I need to get my 85 Ah battery so that it allows pretty full production from one 68w panel without clipping it and skewing the amp hour charging numbers? I am ok living with the battery only being charged up to 90 percent daily for a week or 10 days.

I was thinking daily yield on a sunny day might be 10 Ah max in Houston for a flat 68w 16.5 v panel? 15 Ah? I am not sure how much penalty there is for a flat orientation for a full sunny day. I get only short periods mod full sun in the backyard.

I am planning to leave some small load on to try to balance charging to usage.

It's not trivial to mount a thin film panel temporarily. I am thinking a big battery on a board on one end and a big but flat board on the other end unless I can find a hunk of metal maybe 1/2" thick. Cannot have any shadow cast on the south end.

Jim

69 Replies

  • HiTech quote:" I am ok living with the battery only being charged up to 90 percent DAILY for a week or 10 days."

    ----

    A 68 watt panel can supply maximum 4.5 amps (good sun angle) for each hour. With an 85AH battery a discharge to 70%soc would represent approx. 60 AH, or 17 amps to get back to 90%soc.

    You may have to limit any daily usage to 25% of capacity (maybe less) to be able to even reach 90%soc on a daily basis. I would hope your experiment might show more than the 10AH you expect. Also: any cloudy days in that 7 to 10 day period may mean needed generator time.
  • red31 wrote:
    I believe your panel is rated at ~4 amps @ 16.5v, to me that means at at radiation of 1000 watts/ meter^2 at some temp, a standard test. So if you use 3000 watts/day * 4 amp = 12.

    Page 9 has the data for Houston, 2.9 Max, 2.2-2.9, 2.5 average.
    http://rredc.nrel.gov/solar/pubs/redbook/PDFs/TX.PDF


    Exactly the Unisolar charts show the standard test at 1000 but also has curves for 800 on down. I have never seen 4 amps into a battery yet but orientation is flat and it has only been fall sun.

    Jim
  • I believe your panel is rated at ~4 amps @ 16.5v, to me that means at at radiation of 1000 watts/ meter^2 at some temp, a standard test. So if you use 3000 watts/day * 4 amp = 12.

    Page 9 has the data for Houston, 2.9 Max, 2.2-2.9, 2.5 average.
    http://rredc.nrel.gov/solar/pubs/redbook/PDFs/TX.PDF
  • red31 wrote:
    Turn on one light at dusk and try.

    2-3 kWh/m^2/day, so 3*4amp = 12ah/day
    http://rredc.nrel.gov/solar/old_data/nsrdb/1961-1990/redbook/atlas/


    Very cool! What conversion did you use to get from radiation intentensity to amps or watts? Looks like my guess of 10-15 based on partial day and partial shade data is not way off.

    Using that site this should be the winter solstice radiation for the whole country:

    December minimum radiation intensity on a flat panel

    Jim
  • Lol. I have unobstructed sky from dawn to dusk in storage. My expectation is that yield is about as good as it gets for untilted PVs. But I would like actuals rather than estimated numbers, if the cloud cover cooperates. I want to take enough amp hours out of the battery to get a pretty good actual charge reading. I probably will not be able to check up on the amp hours mid day.

    Jim
  • You only get full AH yeild all day when the batteries remain thirsty enough all day to accept the most amps the array will provide. (Ignoring the drop in amps as battery voltage rises due to the panel's IV curve.)

    This brings the array size/battery capacity ratio into play. Also if you have trees to the east and none to the west, you don't get the theoretical mirror image.

    If it is unobstructed east to south, then you can grab that and double it, regardless of what happens in the afternoon. This goes to heck when your morning cloud cover takes awhile to "burn off" but the sky is clear in the afternoon. Nobody said this solar stuff behaves itself!
  • Ok I am slightly confused. I only need a half day to compute theoretical full yield? I get that. A lot of panels suffer from thermal degredation later in the day which would make their yield not quite symetric but the unisolars do not according to their specs. so you should be right about yield symmetry.

    I am looking for actual full yield in the position the RV is parked. Dec 21st is my worst case for angle and day length. The days approaching and following also seem useful to interpolate the shortest day in case it is not full sun.

    Is there a calculator to estimate yield based on watts of solar and latitude?

    Jim
  • You only need full AH haul from sunrise to high noon. Then you can double that for theoretical full day. So if amps taper in the afternoon for various reasons, no problem.