- (Daily Watt Hour Usage x 5) / Battery Voltage
- (1000 wh x 5 days) / 12 volts = 416.66 Amp Hours
OK so let’s say you want more or less than 2.5 days autonomy (cloudy weather days) and live in a cold climate. What follows is a simplified IEEE method system designers will use.
- Determine how many Watt Hours you need in a single 24 hour day. WH = ___________.
- Determine the maximum number of consecutive cloudy days. The range is from 1 to 5. Caution here choosing a low number less than 2 does not give you ability to get through a cloudy spell and you will likely need a generator and a conventional battery charger in addition to a solar charger. Recommend minimum is 2.5 days or a range of 2.5 to 5 days. Days = __________.
- Multiply Line 1 by Line 2. WH = ______________
- Determine maximum amount of Depth of Discharge in a percentage. This determines a safety factor to extend battery life up to the maximum limits and leaves some emergency power. 50% is usually the default choice. You can go less like 40%, but going lower than 50% makes things very expensive. DoD % = ___________.
- Divide line 3 by line 4 WH = ___________.
- From the Table below find the correction factor multiplier for your area winter ambient temperature that the batteries will be exposed too. Correction Factor = ____________.
- Multiply line 5 by line 6 WH = ____________.
- Now to find the battery AMP HOUR capacity needed Divide line 7 by the system Battery Voltage. AH = __________________.
80 F / 26.7 C = 1.00
70 F / 21.2 C = 1.04
60 F / 15.6 C = 1.11
50 F / 10.0 C = 1.19
40 F / 4.4 C = 1.30
30 F / -1.1 C = 1.40
20 F / -6.7 C = 1.59
So let’s run through another example using the same as we did with the KISS method with the same inputs of 1000 WH per day, 12 Volt batteries, 2.5 cloudy days and now operating at 40 degrees F ambient temperature in winter:
- 1000 WH
- 2.5 days
- 1000 wh x 2.5 = 2500 wh
- 50 % = .5
- 2500 wh / .5 = 5000 wh
- 1.30
- 5000 wh x 1.30 = 6500 wh
- 6500 wh / 12 volts = 542 Amp Hours @ 12 volts.