The Power Protection Blog

I’ve mentioned in other posts that runtime is battery. On our websites we use calculated values to determine the runtime based on a particular load. The purpose of this post is to give you an understanding of this process.

The first thing you need to know is that battery discharge is non-linear. For those of you who don’t understand the expression I’ll elaborate. A linear expression is one where, for example, you put two in, you get four out. So it follows that if you put three in, you get six out, or you put five in, you get ten out. EG. whatever you put in  you get twice out. In the non-linear world this doesn’t hold true, for example, you put two in, you get four out, but when you put three in, you get nine out etc.. This non-linearity makes the discharge characteristics very difficult to express mathematically.

Luckily, the battery manufacturers provide us with discharge tables that we can look up, but first we need to know some information about the UPS, the load and about the method.

End of Discharge Point

At what point will the UPS switch off? Your normal 12V lead acid battery contains 6 cells each of a nominal voltage of 2V (so you get 6x2V=12V). In practice the voltage is slightly higher than this and reduces as the battery is discharged. It is important not to allow the battery to become too discharged, so the UPS will monitor the cell voltage and cut off when it gets to a predetermined point. This is usually around 1.7V per cell or 10.2V for 12V battery.

UPS Efficiency?

Well, more precisely, the inverter efficiency. The inverter is used to convert the battery DC power into AC power. There will be losses associated with this. The better the inverter, the lower these losses are. If you’re unsure, use a worst case of say, 80% efficient. This means that for every 100W provided to the load, the batteries will need to provide 125W (simply 100/efficiency).

The Load Power Factor

Is the load purely resistive, or does it have a power factor? We’re only interested in the amount of WATTS that are needed.

Amps or Watts Method?

Firstly, there are two methods for calculating runtime, the Amps method, or the Watts per Cell Method. Generally, Watts per Cell is used for short term discharges and Amps is used for long term discharges.

Process

It’s easier to do this with an example, so let us take a standard server type load with a number of ancillary devices. We know from measurement  that the Ampere draw is 7Amps and we have mains voltage of 235V. Our Watts therefore (assuming unity power factor) is 1645W.

Our UPS has an inverter that is rated at 90% efficient, so the amount of power from the battery required to deliver 1645W is 1645/0.9 = 1828Watts.

Our UPS is a 3KVA, 2.1KW UPS that contains 8 batteries connected in series. Each battery is rated at 12V 7.2 Ah.

 I now need to look up the manufacturers data sheet and I find the following for a 7.2Ah battery:

Constant Power Discharge Method 

Our battery load is 1828W and we have 6×8=48 cells. Therefore our Watts per Cell is 38Wpc.

We know the FV (Final Value or End Of Discharge Point) is 1.7Vpc so looking along the Constant Power Discharge Table, we can see that 52Wpc would give 5 minutes, so we will get more than 5 minutes runtime. We can see that 35.16Wpc would give 10 minutes runtime, so we will get less than this.

So the calculated runtime for this example, based on constant power discharge is between 5 and 10 minutes.

 Constant Current Discharge Method

We have a total of 8x12V batteries in series, giving us a string voltage of 96V. We need to deliver 1828W so our Amperage is 1828/96 (from Power=VoltsxAmps, so Amps=Power/Volts). = 19Amps.

Now looking along the table above in the Constant Current Section, with our FV of 1.7, we see that a 26A discharge would give us 5 minutes, so we will get longer than this. A 17.6Amp discharge would give 10 minutes, so we will get less than this.

Therefore using the constant current discharge method we will get between 5 and 10 minutes runtime.

Working it out in reverse – I want 2 hours runtime – how many batteries do I need?

Using Watts Per Cell Method.

From the table under Constant Power Discharge, we can see that for our FV of 1.7, for a 2 hour runtime we need to have a WPC discharge of no more than 6.05. Our load is 1828W, so we need 1828/6.05 = 302 cells, which is 50.35 batteries. The battery requires a 96V string voltage, based on banks of 8, so we will require 6 banks to get close (that is 48 batteries), or 7 banks to be sure (that is 56 batteries).

Using Constant Current Method.

From the table above under Constant Current Discharge, you can see for our FV of 1.7V we need to have no more than a 3.05A discharge from each of our batteries to achieve a 2 hour runtime.

Our total current draw is 1828/96 = 19Amps (1828Watts load/Battery String Voltage = 96V)

Dividing the 19Amps total current by 3.05 gives us the number of strings needed to achieve 2 hours runtime which is 6.24. Obviously we cannot add in a quarter of a string so we need to round up. In this case we require 7 battery strings, or a total of 56 batteries to achieve a runtime of 2 hours.

Alternatively, you could of course opt for higher capacity batteries, and maintain the same number of batteries. The examples above were using 7.2Ah lead acid batteries but there are other choices available.