Maybe this explanation below will help.
A. A simple diagram:
I assume now we all agree that a UPS is made up of two basic components that can be purchased as a unit called a UPS.
But we can achieve the same functionality if we were to use an inverter and a separate battery. (This is the scenario applicable to outdoor people who install inverters in the vehicles driven by the vehicle battery to supply AC power).
And it does not matter what the charging side looks like . It could be the vehicle alternator, or a Solar panel or an AC input to the inverter. Sure there are power management issues and control circuitry requirements that are different in these scenarios but that is NOT what OPs question is about.
Everything to the right of line A or B is the "load". So when looking at the battery, the load must include the inverter and the load that is to be provided with power.
B. The formulas and the units used:
Now to the circles on the right. Everyone remember these very basic formulas about electricity?
Volts X Amps = Resistance in Ohms, and Volts x Amps = Watts ( in DC circuits) and VA (in AC circuits) and to convert AC VA to Watts, the formula is V x A x p where p is the power factor which is a value between 0 and 1.
So now to convert into power over time, one multiples both sides of the equation by time.
V x A x Time (in hours) = VA-hr or W-hr
C. The battery:
Firstly, one has to careful not confuse things with batteries and ensure what you put on paper is not open to interpretation.
Two batteries placed in series is NOT the same as two batteries in parallel. In this example the numbers are unfortunate because those of you quoting the formulas for batteries in parallel get the same answers and therefore leave a nagging feeling that there is not something quite right here.
Two 12 Volt batteries in series = a total of 24 Volts, which is the correct value to plug into whatever formula you want to use.
Next, Yes there is an efficiency issue about any electrical device, which will result in the full capacity of the battery NOT being available ( OR, as I have seen many times, a battery able to deliver more than its rated capacity! i.e. better than 100% efficient.)
So in this example we can say the capacity in watts of the 24 volt battery is 24 Volts x 102 Amps = 2 448 w,
OR,
we can quote it in w-hr as 24 Volts x 102 A-hr = 2 448 W-hr.
THEN we can introduce an efficiency factor if we want to by reducing the available power by a factor. (Some have suggested 0.85 others have proposed to use 100 A-hr as an a reference value, whatever floats your boat).
Then one has to decide on the level of discharge. A simple (and by no means the only way) is simply cut the capacity by half for a 50% discharge. So the capacity is now
1224 w OR 1224 w-hr.
Now to determine the standby time (at point A) will be Time in hours = Capacity of the battery in A-hr x Volts /load, which must include the load plus the inverter load.
So
assume that the
load is 70 watts and the inverter uses 50w just to work, the load will be 120 w.
That means the standby time at the battery for a 50% discharge will be (102 x 24)0.5 /120 = 10.2 hours.
D: Refine the calculations:
To refine the calculations some more, one wants to calculate the load more accurately and also calculate the efficiency of the inverter.
