# Warning: Might be a stupid battery question

Solar Expert Posts: 116 ✭✭
If you have a weak battery bank (batteries have lost some of their capacities). They are monitored for Ah and KWh of discharge. When I look at the monitor (Mate and FNDC) it seems we are using more Ah during the night than we use to,even though we have made no changes or added anything running at night. My question is will/would weak batteries be the cause of the reason for the increased power drain/ usage during the night? Voltage just seems to drop more quickly than it use to.

Weak batteries, drop their output voltage faster than "good batteries"... A simple mode/set of assumptions for a 48 volt FLA battery bank:

Roughly 50.8 Volts full charge... 46.0 volts at 50% capacity (again very rough estimate for "good batteries" under medium load). And 42.0 volts for a "dead" FLA battery bank.

And you have to look at you major loads--For example, lets say that your loads are running off an AC inverter. AC inverters are "constant power" devices... I.e., no matter what the battery bank voltage is, the output is held at (for example) 120 VAC @ 60 Hz... From 50.8 volts to 42.0 Volts (other loads have different Current/Voltage curves... Filament bulbs are roughly constant current loads. DC motors running fans will slow down and use less power (Watts). Motors driving Water Pumps will use more current as the voltage falls. Electronics (TVs, LED lighting, Radio, Computers, etc. tend to be constant power).

So, just using the "constant power" AC inverter/loads assumption (i.e., you have a constant 500 Watt load in this example--And the inverter has to draw more current as the battery bus voltage falls):
• Power = Voltage * Current
• Current = Power / Voltage
• For example, assume 500 Watt load (losses, etc.)
• Battery charging Absorb set point: 500 Watts / 59.0 volts = 8.47. Amps to constant power load(s)
• Not charging, battery 100%: 500 Watts / 50.8 Volts = 9.84 Amps
• Good battery 50%: 500 Watts / 46 volts = 10.87 Amps
• Dead battery 0%: 500 Watts / 42 Volts = 11.90 Amps
• 100% battery / 50% battery = 9.84 Amps / 10.87 amps = 0.91 = 91% current drop from 50% to 100%
• 100% battery / 0% battery = 9.84 Amps / 11.90 amps = 0.83 = 83%
So--Very roughly, depending on your battery bank condition and state of charge, etc. you could see something like a reduction in current of 10-20% with a good bank, or an increase of 10% to 20% more AH for a "weak to very weak" battery bank under load.

10% increase or decrease in AH for that nights loads--That is a pretty "small" change in a day to day battery bank operation (your nightly loads probably vary by 10% or more--Unless your schedule is very consistent from night to night/loads on timer/etc.).

Because of the "constant power loads" typical for battery backed computers (uninteruptable power supply for computer/computer room/etc.), batteries designed for UPS use not only have the Amp*Hour ratings, they also have Watt*Hour ratings--Which is "more accurate) for constant power loads.

Make sense, or clear as mud?

-Bill
Near San Francisco California: 3.5kWatt Grid Tied Solar power system+small backup genset
• Solar Expert Posts: 116 ✭✭
Got it!