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:
- Is my math right?
- Will the 12W panel (at 48Wh) cover my 42.47Wh power consumption?
Answer
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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