Battery Help

A 12V 8.2Ah battery will sustain a 15W load for no longer than 6.4h, and the voltage will drop away towards the end of that period. The presence of the inverter introduces a load onto the circuit in addition to the 15W of lighting, therefore the battery-to-inverter-to-lamp circuit will stay energised for rather less than this time.

Why not discard the inverter and use 3 x 12V/5W lamps instead of the mains cfls? Further, 12V LED lighting is very efficient, and for the same light output one might get rather more time from this relatively small battery on the basis of it producing more lumens per watt, or in other words consuming less watts per lumen.

Domestic deep-cycle batteries are available in capacities up to and beyond 80Ah if needed over-the-counter at places such as marine equipment chandlers, etc..

As another option, try connecting more smaller batteries in parallel with the 8.2Ah one, perhaps?

How much battery do you need to run your device? Here is how you estimate it.

Step 0. A little tutorial on measurements of charge. After all, it is electrons that are stored in the battery. In phreshman fisicks we all learned that the measure of charge is the coulomb and that a single electron has 1.602e-19 coulombs of charge. One amp flowing in a wire for one second will use one coulomb of charge.

Q = I*t

where Q is the charge in coulombs, I is the current in amps and t is the time in seconds.

The amount of charge passing through that wire (conducting 1.0 amps) in 60 seconds is 60 coulombs, and in one hour you would have had said "hello" and "goodbye" to 3600 coulombs of charge.

Batteries were evidently developed by engineers who subscribed to the "whatever's easiest" system of measurement. They got tired of pulling out their slide rules to divide by 3600 every time they wanted to know how long 24000 coulombs would last them and came up with the unauthorized unit of amp-hours. Later, when smaller batteries were used they came up with milliamp-hours.

Don't be confused by the hyphen. Amp-hours means amps times hours. Divide by amps and you get hours, divide by hours and you get amps. So it isn't amps, and it isn't amps per hour, it is amp-hours. And, by the way, I have even used the term amp-seconds because when you say "coulombs" everybody goes glassy-eyed on you.

Don't get me wrong, I love amp-hours for units, it is a handy rule of thumb. Amp hours is how much charge is stored in the battery. Since a battery changes voltage during the discharge, it isn't a perfect measure of how much energy is stored, for this you would need watt-hours. Multiplying the average or nominal battery voltage times the battery capacity in amp-hours gives you an estimate of how many watt-hours the battery contains.

E = C*Vavg


Where E is the energy stored in watt-hours, C is the capacity in amp-hours, and Vavg is the average voltage during discharge. Yes, watt-hours is a measure of energy, just like kilowatt-hours. Multiply by 3600 and you get watt-seconds, which is also known as Joules.

As long as we are in the prelude, I might also mention that since the charge in a capacitor is Q=CV that a battery can be rated in farads as well. A 1.5 volt AA alkaline battery that stores 2 amp hours of charge (that's 7200 coulombs) has the equivalent capacitance of 4800 Farads. Of course a battery makes an awfully weird capacitor because the voltage doesn't drop proportionally to the stored charge, it has a high equivalent resistance, and etc.

The following method assumes that you know how many amps you need from the battery. If you know the watts go to Step A below.

Step 1. Back of the envelope

If the current drawn is x amps, the time is T hours then the capacity C in amp-hours is

C = xT

For example, if your pump is drawing 120 mA and you want it to run for 24 hours

C = 0.12 Amps * 24 hours = 2.88 amp hours

Step 2. Cycle life considerations

It isn't good to run a battery all the way down to zero during each charge cycle. For example, if you want to use a lead acid battery for many cycles you shouldn't run it past 80% of its charge, leaving 20% left in the battery. This not only extends the number of cycles you get, but lets the battery degrade by 20% before you start getting less run time than the design calls for

C' = C/0.8

For the example above

C' = 2.88 AH / 0.8 = 3.6 AH

Step 3: Rate of discharge considerations

Some battery chemistries give much fewer amp hours if you discharge them fast. This is a big effect in alkaline, carbon zinc, zinc-air and lead acid batteries. It is a small effect in NiCad, Lithium Ion, Lithium Polymer, and NiMH batteries.

For lead acid batteries the rated capacity (i.e. the number of AH stamped on the side of the battery) is typically given for a 20 hour discharge rate. If you are discharging at a slow rate you will get the rated number of amp-hours out of them. However, at high discharge rates the capacity falls steeply. A rule of thumb is that for a 1 hour discharge rate (i.e. drawing 10 amps from a 10 amp hour battery, or 1C) you will only get half of the rated capacity (or 5 amp-hours from a 10 amp-hour battery). Charts that detail this effect for different discharge rate can be used for greater accuracy. For example the data sheets listed in http://www.powerstream.com/BB.htm

For example, if your portable guitar amplifier is drawing a steady 20 amps and you want it to last 1 hour you would start out with Step 1:

C=20 amps * 1 hour = 20 AH

Then proceed to Step 2

C' = 20 AH / 0.8 = 25 AH

Then take the high rate into account

C''=25 /.5 = 50 AH

Thus you would need a 50 amp hour sealed lead acid battery to run the amplifier for 1 hour at 20 amps average draw.

Step 4. What if you don't have a constant load? The obvious thing to do is the thing to do. Figure out an average power drawn. Consider a repetitive cycle where each cycle is 1 hour. It consists of 20 amps for 1 second followed by 0.1 amps for the rest of the hour. The average current would be calculated as follows.

20*1/3600 + 0.1(3559)/3600 = 0.1044 amps average current.

(3600 is the number of seconds in an hour).

In other words, figure out how many amps is drawn on average and use steps 1 and 2. Step 3 is very difficult to predict in the case where you have small periods of high current. The news is good, a steady draw of 1C will lower the capacity much more than short 1C pulses followed by a rest period. So if the average current drawn is about a 20 hour rate, then you will get closer to the capacity predicted by a 20 hour rate, even though you are drawing it in high current pulses. Actual test data is hard to come by without doing the test yourself.

If you know the watts instead of amps, follow the following procedure

Step A: Convert watts to amps

Actually, watts is the fundamental unit of power and watt-hours is the energy stored. The key is to use the watts you know to calculate the amps at the battery voltage.

For example, say you want to run a 250 watt 110VAC light bulb from an inverter for 5 hours.
Watt-hours = watts * hours = 250 watts * 5 hours = 1250 watt hours

Account for the efficiency of the inverter, say 85%

Watt-hours = watts * hours / efficiency = 1250 / 0.85 = 1470 watt-hours

Since watts = amps * volts divide the watt hours by the voltage of the battery to get amp-hours of battery storage

Amp-hours (at 12 volts) = watt-hours / 12 volts = 1470 / 12 = 122.5 amp-hours.
If you are using a different voltage battery the amp-hours will change by dividing it by the battery voltage you are using.

Now go back to Steps 2-4 above to refine your calculation.

Have a nice day!

Here is the answer to your questions:

Q: I m having battery of 12v 8.2AH/20hr. what it's mean?

The battery voltage is 12 V and ampere-hour (AH) capacity is 8.2 AH. If the battery is discharged at constant rate of (8.2 AH / 20 H) 0.41 A, it will last for 20 hours. This means, you can get 12 V x 0.41 A = 4.92 watts from the battery for 20 hours. The time will proportionally decrease (or increase) if the discharged current is increased )or decreased).

The batteries are generally not allowed to discharge up to full capacity (if done so, the battery life is decreased). Generally, 80% DOD (Depth of discharge) is allowed. With this consideration, you can use only 80% of the battery AH, i,e 8.2 AH x 0.8 = 6.56 AH.

Q: If i connect this battery to my 800watt UPS & i connect 3*(5 WATT) cfls (voltage across each cfl is 230 V)to the ups then in how much time battery will discharge?

Your load is 3 x 5 = 15 watts, 230 Volts. The UPS size 800 watts is enough to deliver the load. However, the UPS output should be 230 Volts (your DC system is only 12V).

Let's assume the efficiency of the UPS inverter (including the output transformer) is 80%. The DC load is then 15 W / 0.8 = 18.75 watts. If the battery supplies the load, the average discharge current is 18.75 W / 12 V = 1.56 A. You can use only 6.56 AH from the battery with 80% DOD level. So, the battery can supply only 6.56 AH / 1.56 A = 4.2 Hours.

Q: How does the battery life & efficiency will effected?

Battery life and efficiency depends on:

- The DOD level

- The number of discharge/re-charge cycles

- The temperature of the battery room

- The type of battery

- Proper maintenance and regular check up of the battery

Q: Can any buddy give me site which will help about battery?

The following sites may help:

http://www.windsun.com/Batteries/Battery_FAQ.htm

http://www.vonwentzel.net/Battery/00.Glossary