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Putting It Together

Putting It Together

CHOOSING COMPONENTS

This is not a linear process due to the number of permutations possible, but this is one approach. We will use an example of 3000 WH per day average load:

Avg. daily load X No-Sun Days = Needed Battery Capacity

3000 WH X 3 = 9000 WH Needed Battery Capacity

This is also known as "days of autonomy." Now you can refer to our Ratings section and consider a battery:

If you want to use a Surrette S-460, for example, each battery has a usable capacity of:

460 AH X 6V X 25% = 690 WH

9000 WH / 690 WH = 13.04 Batteries Needed

The 25% is somewhat arbitrary and conservative. If you want to cycle to a 50% depth of discharge you can use 50% instead. Remember cold temperatures can reduce battery capacity considerably.

Of course we need an even number of batteries AND we need to have a battery bank at the voltage we want (12, 24 or 48). The S-460 is a 6V battery so if we had 12 of them we could choose 12 or 24 V for our battery bank, depending how we wire them.

12 Batteries X 690 WH = 8280 WH Usable Capacity

This is a little shy of our desired 9000 WH, but since we are being fairly conservative we could go with that. Or we could consider choosing the S-530 instead, which is rated:

530 AH X 6V X 25% = 795 WH

12 Batteries X 795 WH = 9540 WH Usable Capacity

Which meets our requirement. Now let's size the array:

To provide a healthy charge, Lead-Acid batteries can be charged at around a C/20 rate. This can be approximated by taking the total capacity of the battery bank, in amps or watts, and dividing by 20:

530 AH X 6V X 12 Batteries = 38,160 WH

38,160 WH / 20 H = 1908W

Add 20% inefficiency factor:

1908 W / 80% = 2385 W

This is the size of PV array which would charge the batteries very well indeed. We can confirm that this should work by noting that:

5 Hours Sun X 1908 W output = 9540 WH

Which means we can replace our usable battery capacity in one sunny day in the summer. The only problem is that achieving a C/20 rate gives us an array that is quite a bit larger than we really need to meet our average usage. Therefore, we will not make achieving the C/20 charge rate a necessity. Let's say we choose a 1000W-output array to keep up with our 3000 WH/day needs. What charge rate can we acheive that is it sufficient?

38,160WH / 1000W = C/38.16 charge rate

C/38 is not bad at all. Rates of C/60 or less may be acceptable if the system is left unused at times and has plenty of time to recharge. Note that if you are running loads during the day, much of the solar output will be diverted to that and not be available for recharging. You would want to either increase the size of the array or run a generator occasionally during cloudy periods to make sure batteries regularly get fully charged.

An inverter can be chosen based on features and the maximum surge/simultaneous load you need.

Exact components are also affected by the choice of system voltage. This is in turn affected by the length of wire runs and system size. The example system above could use 12, 24 or 48 Volts with the right components. Smaller systems would generally be 12 or 24 V, and larger would generally be 24 or 48 V.


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