Backup Time of Single-Phase Battery

dear chandran

back of battery bank having 26 ah capacity is simply 26/5= 5.2 hour

s paliwal

That's right, chandran. the Amp Hour rating can be explained like this. Load factor in amperage will help you easily calculate length of battery life. 10AH is equal to 1 amp for 10 hours, 10 amps for one hour or 5amps for two hours...you get the idea. But also keep in mind that this is only theoretical. Without knowing the actual load (the higher the load, the greater the discharge by a percentage but not sure what percentage is). You also need to take into account the age and state of the batteries and you want to leave at least 25-30% because there are a lot of variables in what you are asking. Good luck!

Dear Mr.Chandran,

It can be 5 hrs if the battery spec says C5 rating. If the battery spec says C10 rating then 10 hrs. If it is C-5 then you would not more than 1 KVA and if it is C10 then not more than 500VA. For 5KVA UPS I think 26 Ah is not enough. Please visit http://hbl.in to download literature how to calculate battery voltage, Ah, etc for an application. Flooded batteries got C5 rating and most VRLA's got C10 rating.

The amount of backup time from a battery also is affected by the rate at which the battery is drained. Different manufacturers provide AH ratings on different standards- some use a 10 Amp discharge rate, others use a 20 Amp discharge rate. At 5 Amps, this should not be a problem. However, most batteries will only give you half the rated capacity before the voltage drops below a useful level. For planning purposes, I would suspect you can get only about 2.6 hours of reliable backup from a 26 AH battery that is fresh and has been properly maintained and not stored in a hot environment. Also, most batteries are rated at a particular operating temperature- increasing the storage/operating temperature above that specified by the manufacturer can impse significant limitations on how much time the battery is available...

Usually, in general calculation for the battery buck up time, I always use this formula to determine the hr of back up.

(KVA/220V)/(load consumption)

(5000/220)/(5A) = 4.55 hr or 4 Hour and 32 minute. The average time we will take about 4 hr of back up.

The battery cannot be discharged to 0 because of the chemical transformations inside it. Usually it has to have a reserve of about 25% of its capacity.

This reserve is also determined by the discharge rate, but 25% would be an acceptable limit.

In the described case, the available capacity would be 26 Ah * 0.75 = 19.5 Ah

The time would be determined by 19.5 Ah / 5A = 3.9 h = 234 min.

This assumes that the power backed up is 5A * 12 V = 60W.

If the current of 5A refers to a voltage of 220V, the power requirement is 5A * 220V = 1100W, which translates into 1100W/12V = 91.6A at the battery, which leads to a time of 19.5 Ah/91.6A = 0.212 h = 12.7 min.

Steady on there people. B.JEYACHANDRAN has only put on 1 post and has received some very dodgy advice.

5 amps relates to the output; 5 amps DC @ 12 v = 60 watts which would hardly light the front panel of a 5kVA UPS.

The output of a UPS is in kVA; let us assume a 230v output @ 5A = 1150 VA. We need to know the power factor of the load to work out the watts. If we assume say a 0.9 pf then the watts = 1035 W, but lets take 1 kW.

Now the UPS is not 100% efficient. At low loads the efficiency drops to about 90-95%, but this is end-to-end. On battery there are no input rectifier losses, but lets be nasty and take the output at 5% losses. This only equates to a 50W overhead, but as I said the UPS needs power to run itself (internal electronics, fans etc) and this is around 100W (more than the losses!). So we need 1150W or so from the batteries.

Now the Ah figure quoted by manufacturers is normally the C20 figure i.e. a drain over 20 hours, from nominal cell voltage down to 1.7 V (a 12V battery has 6 x 2V cells). See Vision batteries:

The exact discharge cut-off point used by UPS manufacturers varies, and in fact can be set in software. 1.65 V is common, which means you are squeezing more energy from the batteries. Lets use 1.7 however.

Note that when charging, a 12V cell is at 14.4V, quiescent at 13.2V (this allows internal volt drop to 12V under load), and discharged at 10.2V.

Now the big one: AH FIGURES ARE NOT BASED ON 1 HOUR.

Now the bigger one: AH FIGURES ARE NOT LINEAR.

Actually they are reasonably linear > 10 hours i.e. 26AH = 1.3A for 20 hours

≈2.6A for 10 hours

They are NOT linear at the "short end" < 5 hour i.e. 26AH ≈ 18A for 1 hour (not 26).

If you download the technical manual from the link above you'll see what I mean. They don't have a 26AH battery listed but 24AH is close enough:

5m 10m 15m 30m 1h 3h 5h 10h 20h time

98 69 47 28 16 7.1 4.6 2.4 1.2 amps

So you would expect a 24AH battery to give 96A for 15mins, but you only get 47A.

Now we have 16 batteries @ 12 V = 192V. Note that we could allow a higher voltage as at low loads the internal volt drop in the battery is low, so our battery voltage is closer to 13V. However 1150W @ 192V = 6amps. From the table above you will get somewhere around the 4 hour mark.

As mentioned by other posts, aging takes has an effect, as well as battery temperature.

Baby Guinness,

You assumed an output voltage of 230V, but when considering a string of 16 batteries the total voltage is 192 V. I think you should have selected 20 batteries in series ( 20 x 12 = 240V).

The rest of the explanation is OK. One question, since you determined the capacity required via load with power factor:

You indicated that for a 1150 VA load at a 0.9 power factor ( considered at 1000 W), you estimated losses at 150 W, which would translate as a total 1150 W at the DC part (battery), for which a certain time can be determined.

Do you think the same time would result if the output power requirement would be 1000 W at power factor 1 (considering the same losses as 150W, and assuming the same discharge type) ?

Hi Tomad

The 230V relates to the voltage on the output of the UPS rectifier. Drawing 5amps gives 1150 VA. If it was in the USA, the output would be 110V = 550 VA.

The battery voltage is independent of this. A range of battery voltages are available, the output inverter simply uses this energy to create the output.

The output pf has only a little effect on the power consumption; the reactive current simply flows in a loop around the circuit, performing no useful work. However, the increased current does cause increased losses in the cables and the SCRs on the output, as the current flows; but these losses are very small in comparison.

So basically yes; at pf 1 the power consumption is almost the same.

Incidentally, you cannot exceed either the kW or the kVA rating of the UPS. The kW is limited by the physical capacity of the device, the kVA is limited by the output current capacity.

Note also that this has nothing to do with the input power factor, which is a function of the design and the loading of the device.

Baby Guinness,

The 220V has been given only as an example, it only served to determine the output power. By the way, the standard voltage in the US is 120V not 110V.

The matter of where the reactive power for the load (UPS output) comes from has been the topic of another discussion earlier and it remained basically unclarified after probably 20-30 posts. The question at that thread was where does the reactive power needed by the load come from (capacitors ?, etc) and what limits it (Is the 0.8 pf to be considered a minimum or maximum).

One thing to keep in mind is that the above replys assume the batteries are new. As batteries age the lose their ability to hold a charge and the run time will not be as long when they are 3 years old compared to when they are new.