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PV array size vs battery bank

Wyoming Ed

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Feb 20, 2021
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I'm trying to figure out PV array size vs battery bank. Currently I have 7.5 kw of panels , 48 volt system , 10 kw inverter, 16 gel batteries ( Renogy 200 ah) in series and parallel, maybe 2.5 peak sun in winter. I'm likely using 20 kwh to run furnace and all on a 1600 SQ ft house. The goal is to have enough battery and panels to make it through the night without back up generator..Any thoughts? I saw there is a spreadsheet, I'm not likely skilled enough to use that. Thanks
 
Assuming you have four sets of four 12v batteries you are setting at 38 kWH of DC storage. (200 ah x 48 v = 9,600 watt hours per set of four x 4 sets = 38.4 kWH). Discharging batteries isn't 100% efficient. Exactly how inefficient depends on a lot of factors that I honestly don't understand myself but lets use 75% to be safe. 38 x .75 means you should be able to get almost 28 AC kWH's out of your battery bank before it's 100% depleted which is too far for GEL batteries to do on a regular basis. Now you've got an empty battery bank so assuming charging batteries is 100% efficient which it's not your 7.5 kW array will need at least 5 hours to recharge your battery bank. (38/7.5 = 5) This back of the napkin math indicates you need at least 2x the array that you have in the winter which is why its pretty common for people to rely on generators in the winter. Hope that helped!
 
I'm trying to figure out PV array size vs battery bank.
The size of the battery bank is based on how much power you consume, and that can be determined from an Energy Audit, click the link and check it out - if you still don't understand it please let me know what's throwing you for a loop and I'll try to tighten it up - that way it'll help others too.

The math

7.5 kW solar array - From the insolation table, if the tilt is optimized for
winter you should be able generate an average of 7.5 x 3.99 = 30 kWh per day in December, more in every other month.

...maybe 2.5 peak sun in winter...
You don't have units on that number, but your tilt might not be optimized for winter.

...using 20 kwh...The goal is to have enough battery and panels to make it through the night without back up generator.
Since you're capable of making 30 kWh/d and you need 20 kWh/d, all you need is 20 kWh of battery (plus some reserve most likely, how much is up to you).

As "Wyoming" is in your Moniker, let's use Cheyenne insolation data

1615206120648.png

Tilt at 34°, optimized for winter, use PVWatts for more accurate numbers. Image is a link to the data source.

How many batteries do I need?
You need some key information to calculate this. First is the efficiency of the inverter in converting DC to AC and it's standby power consumption. You can find it on your inverter's datasheet, let's assume it's 92% efficiency and 20 watts consumption for these example calculations.

Depth of Discharge (DoD): Typically you limit this with a voltage cutoff to extend the battery life, discharging too deeply can radically reduce their life, more on this in the battery FAQ. For these example calculations we'll assume 50% for lead acid and 80% for LiFePO4.

Peukert Effect: Lead acid batteries have less than their stated capacity if you pull power faster than their 20-hour rate. To know how much, you need to know your current draw and see the battery's datasheet.

Temperature effects: Lead acid batteries lose a lot of power if they're cold. LiFePO4 can be damaged if charged when below freezing. For this example we'll assume the batteries are warm and operating with no temperature losses. If you know they'll be cold or hot at times, look at the datasheet for them and factor this in.

Round Trip losses (RTL): Some energy is lost in the act of charging and discharging. For this example we'll assume 80% efficiency for lead acid and 96% efficiency for LiFePO4.

Lead acid: number of watts per hour /.5 DoD x .92 Inverter Efficiency / .8 RTL
Example: 20,000 Wh AC needed / .5 / .92 / .8 = 54,348 Wh DC + 20W standby x 24 hrs/d = 54828. Wh / V = Ah, so at 48V, 1143 Ah​
LiFePO4: number of watts per hour /.8 DoD x .92 Inverter Efficiency / .96 RTL
Example: 20,000 Wh AC needed / .8 / .92 / .96 = 27,420 Wh DC + 20W standby x 24 hrs/d =27900. Wh / V = Ah, so at 48V, 582 Ah​
Caveats / Assumptions
Be sure to lookup the insolation data for your actual location and redo the math.

One of the things to double-check, is the the 20 kWh for the furnace. The above assumes you need a battery to run the full 20 kWh. But, if that 20 kWh is over the course of the day, then the panels could lift the day-time load and charge the batteries. For example, let say of the 20 kWh it's coldest at night and 70% of it is consumed at night and 30% is consumed when the sun is up and panels generating power...then you'd only need 70% of 20,000 Wh, or 70% of the 582 Ah for LiFePO4 batteries.

Hope that helps some!
 
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