math :: finding battery SOC

Knowing how much energy is left in the batteries is the most important bit of day-to-day information needed to run the off-grid system. The first calculation needed is the maximum capacity of the batteries. Because information was limited on the Energizer (Johnson Control) 6V golf cart battery, I used the specifications from a comparable Trojan battery to make my calculations.

Trojan also has a wonderful guide to battery maintenance: http://www.trojanbattery.com/Tech-Support/BatteryMaintenance.aspx. I assembled the information from their battery maintenance section into a pdf titled Batteries Care that is kept at the cabin in a binder. I started a binder originally so the most important parts of the inverter manual could be quickly referenced, and eventually it grew into an all-encompassing reference manual for all the components of the off-grid system, including maintenance.

CALCULATIONS: Total battery bank capacity:

225 Ah (at 20 hr discharge rate) x 14 batteries = 3150 Amphours

3150 Amphours x 6 Volts = 18900 Watts (DC)

18900 Watts x 85% efficiency of inverter (DC to AC conversion) = 16065 useable 120 VAC Watts

Since it is undesirable to drain a battery 100% it was decided that we should recharge at 50% SOC (state of charge) ideally and 30% if needed.

16065 AC Watts x 50% discharge = 8032.5 useable Watts

16065 AC Watts x 70% discharge (30% SOC remaining) = 11245.5 useable Watts

Assumptions: in order for the estimates to be exact (which is impossible) the following would need to be true:

  • batteries have exactly 225 AH capacity
  • 6V from each battery is constant throughout the entire discharge of the battery
  • the inverter always operates at 85% efficiency
  • there are no internal changes in resistance under different discharge rates of the batteries (a 1000 watt load would have to use exactly the same amount of energy in 1 hour as a 100 watt load in 10 hours).

All these assumptions will vary. At best, these calculations are a little better than a best guess.

TEMPERATURE AND BATTERIES

Temperature does some funny things to lead acid batteries. At 80°F (the temperature most batteries are rated at) a fully charged battery my read 12.7 V under no load conditions. At 30°F that same battery may read the same, or lower. Not a lot of difference. The real fun begins when we try to estimate the SOC of the battery bank when a load is applied.

Most of the time we use about 300 Watts in the cabin – primarily for lighting, fans, 26″ LCD TV, laptop charger, and any appliances that are plugged in drawing phantom loads. When our system is 95% charged under a 300 Watt load at 80°F the control panel will read 12.5V. Under the same conditions at 30°F it will read 12.1V. Normally we would recharge the bank as soon as we drop to 12V. However, at 12V @ 30°F the bank is actually SOC 90%. Why does this happen?

To understand why the voltage drops disproportionally at different temperatures under the same 300 Watt load I’ll use oil as an analogy. At 80°F oil will flow through a funnel quite quickly, but at 30°F the oil will be much thicker and flow much slower. Pouring one quart of oil will take more time at 30°F. Note that it takes longer for the oil to flow through the funnel but the amount of oil is the same – temperature does not lower the capacity of the batteries, it only slows down the chemical reaction that produces the flow of electricity. This decreased flow shows as a drop in voltage. This can be verified by taking specific gravity readings of the SOC at different temperatures while under a 300 Watt load.

In short – if a system has a voltage readout, it is important to account for the load (Watts used at a given point in time) that is most commonly used by the system. While a no-load voltage readout will almost always be more accurate, it is often not practical to turn off the system for 2 hours (ideal) and then take a voltage reading.

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2 thoughts on “math :: finding battery SOC

  1. Ian McDonald

    Are you wiring your batteries in series / parallel? I ask because the only way to increase you AH is to run them in parallel, then to increase from 6v to 12v (48v would be better but YMMV) you would place them in series. So I am curious how you calculated the 3150AH, in other words I would like to see the math.

    I am in the process of designing my own array for off grid and I am trying to determine the right battery for my needs while build a system with sufficient power for a couple days without charging.

    Reply
    1. offgridcabin Post author

      Ian,

      Thanks for visiting the blog! 3150 AH is based solely from taking 14 batteries x 225 AH per 6V battery. If you are thinking the AH of the battery bank when wired at 12V then it will be 1/2 of 3150 AH or 1575 AH (I have used 220 AH for the batteries in other posts which would result in 1540 AH @ 12V).

      If I were to do it over again (if I knew then what I know now) I would have wired in series-parallel to achieve 24V for the battery bank (4 batteries per string for 3 strings, total 12 batteries). But 12V has its advantages and suites this system quite well (if I wanted a larger solar array then I could begin to encounter problems).

      Keep reading, if you find anything else confusing or you want to know more let me know. I can almost guarantee your knowledge base for off-grid will double shortly after you install a system. The biggest strength of any set up is the ability to be flexible (easy to add or change things).

      Reply

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