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I'm looking for confirmation to see if this battery and this solar panel will cover my needs for a remote timelapse project I'm setting up. The timelapse will run 6 days a week for 9 hours. My area gets 4 hours of good sun a day.

So far, based off all the reading I've been doing, my math looks like this:

The device's power consumption:

12V battery x 275mA device current = 3.3W x 9 hours a day = 29.7Wh + 10% solar controller losses = 32.67Wh

For a 1 day reserve (to try and get past any rain that might stop charging), this upsizes my power consumption to: 32Wh + 30% reserve = 42.47Wh

For battery capacity, I know I shouldn't drain the battery more than 50% to preserve it's health.

42.47Wh / 12V = 3.5Ah x 2 (for the 50%) = 7Ah battery

In order to size a panel to charge this system:

12W panel x 4 h sunlight = 48Wh of power

48Wh of power is within my 42Wh of consumption with upsize to charge.

So I'm looking to confirm:

  1. Is my math right?
  2. Will the 12W panel (at 48Wh) cover my 42.47Wh power consumption?
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  • \$\begingroup\$ Consider also that any solar panel rating is going to be ideal. To get close to ideal, you need to have the panel tracking the sun for maximum coverage. If your panel is in a fixed position, you'll need to derate it by some factor. \$\endgroup\$ – JYelton Jun 25 '14 at 18:40
  • \$\begingroup\$ @JYelton: Actually, for a panel mounted at a fixed angle (equal to the site's lattitude), insolation equivalent to 4 hours of "full sun" is about the right rule of thumb. \$\endgroup\$ – Dave Tweed Jun 25 '14 at 23:52
  • \$\begingroup\$ @Dave Good to know, I hadn't considered that was already in the figures. \$\endgroup\$ – JYelton Jun 26 '14 at 22:45
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Your approach is basically correct, but I question some of your numbers:

How does "1 day of reserve" only require 30% additional capacity, rather than 100%?

You've neglected to take into account the storage efficiency of the battery itself (Wh in does not equal Wh out), which is only about 70% for lead-acid.

For your 275 mA × 9 h = 2.5 A-h daily load, you need a 10 A-h battery, which accounts for your 100% functional reserve, plus 100% "battery reserve".

Your charger will need to supply 7 A-h per day, which is the functional capacity (including reserve) divided by the storage efficiency of the battery.

If you have a very good charge controller, it might achieve 90% power efficiency (but possibly much less). 7 A-h × 12 V = 84 W-h nominal. 84 W-h / 0.90 efficiency / 4 h (equivalent insolation) = 24 W panel.

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  • \$\begingroup\$ Thanks for confirming my math (and fixing it). I've been staring at this for a long time and reading many manuals, so I'm not sure where 30% came from :). I will look at a 30w panel. I have a 12v AGM battery now, but not sure of it's A-h capacity. Also, you mention fixed angle above. So if my latitude is 42, then I should mount the panel at 42 degrees? \$\endgroup\$ – Pat Jun 26 '14 at 2:22
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    \$\begingroup\$ 42 degress would just be an approximation, and it would mainly apply to a panel that's being used over one or more full years. This sounds like a less-permanent project than that, so you'll want to pick an angle that represents the average position of the sun for the time of year that the panel will be deployed. \$\endgroup\$ – Dave Tweed Jun 26 '14 at 3:05

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