# Battery power

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How can you expect anyone to give a useful answer to such an incomplete question?
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How much power does a 140 AH battery consume (in watts) when it is charging?
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Power is defined as the first derivative of energy w.r.t time...i.e how much energy is dissipated, generated or transmitted per unit time. The rate of the energy provided by the supply to your charging battery depends on the rate of the charging current. The total energy being supplied minus certain losses is going to be the total energy storage of the battery divided by 2 (if it is half charged/dicharged when recharging commences). The total energy can be calculated by multiplying the terminal voltage by the amp hours value not forgetting to convert hours into seconds to obtain the energy in joules. Power is in joules/sec....or watts
• posted
The battery is a 12 V lead acid battery. It is being charged by a UPS (750 Watts). The UPS is getting power from mains (220 V AC). The battery is half discharged.
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A watt describes only an instantaneous condition of power dissipation.
You (or your battery) don't consume watts of energy, but joules of it (watt-seconds, or watt-hours, or even calories).
You can measure approximate charging loss, in watts, by measuring the surface area of the battery, measuring the temperature of the surface in degC, then calculate
P = (Tm-Ta) x As / 1000 (=/- 10%)
P= power in watts Tm= measured surface temperature (degreesC) - averaged over the entire surface would be more accurate. Ta= ambient temperature (degreesC) As= surface area cm^2
Measuring Watt-seconds requires the integration of this value over a continuously monitored period, as the losses will vary with rate of charging and state of charge over the life of the cell, for any chemistry.
Some battery chemistries are more exothermic on charge than others, and all may exhibit abnormalities outside of regular recommended charging conditions.
RL
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It's easier to just take a real world measurement. These work surprisingly well for how inexpensive they are

There's a European 240V version of it out there as well. Plug the UPS you wish to measure into it with no load and look at the wattage drawn, it will be higher during charging than it will once the battery is fully charged.
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legg wrote in news: snipped-for-privacy@4ax.com:
There's another way, likely easier. You can monitor watt-hours coming in, and those coming out, and eventually you'll get a ratio that tell you your average losses, and the longer you monitor, the better the accuracy.
As both methods depend on some device's determination of state or completion of charge, that's the point where most loss occurs, because once you reach upper limit, input energy must either be stopped or diverted. I guess with a mains device that is switched off bar a tiny monitor PSU those losses don't matter much but it is interesting when you use a solar panel array, you become quickly aware of how much useful power is shunted to oblivion, no matter how sophisticated the charger, once the storage is full on a sunny day. Some chargers make it quite clear what you're missing...
As a crude guess, I'd say my 240 AH of 12V storage gives out about 90% of what it received, not counting shunting loss when fully charged. If I want a better reckoning I'll have to use a better meter than what I'm using now. I don't think cheap regulators are very accurate, and they don't come with much documentation either so it's hard to see exactly what they're doing.
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James Sweet wrote in news:gs97p3\$htt\$ snipped-for-privacy@news.motzarella.org:
Neat, but I think it would tell you more about your charger's efficiency than your batteries losses.
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Depends on the battery voltage, chemistry, and charge time largely.
Graham
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You still can't derive power without stating a charge time and even then, you don't normally charge lead-acid types at constant current, so the power input will decrease as the battery gets fully charged.
You can estimate the ENERGY required though.
Graham
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I suppose it depends on what information he's really looking for.
A simple voltage and current test with a multimeter at the battery itself will tell you how much energy is being poured into the battery at that moment.
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On Fri, 17 Apr 2009 10:14:57 +0500, "Papita" put finger to keyboard and composed:
An *ideal* 12V, 140AH battery will deliver 140 amps for 1 hour, or 1A for 140 hours, or 10A for 14 hours, at 12V, before it is completely discharged.
The supplied energy would be ...
E = V x I x t = 12 x 140 x 1 /1000 kWh = 1.68 kWh
If you could fully charge such a battery in 1 hour, then you would require an average power input of 1.68kW. In 10 hours, the power would be 168W.
- Franc Zabkar
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Thanks. So if everything is ideal and your calculation is correct then the UPS will draw power at an average rate of 168W or 168 Joules/sec right? What do you mean by 'In 10 hours, the power would be 168W' ? I just want a figure in Watts to try to compare it with other electric appliances e.g 100W bulb etc. Also I want to know if my 2 KV generator would manage to charge it alongside running 3 ceiling fans (of about 150W each)
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On Sun, 19 Apr 2009 06:51:09 +0500, "Papita" put finger to keyboard and composed:
I was economical with my language. What I meant was, if a completely discharged battery were to be fully charged in 10 hours, then the required power input would be 168 watts. Of course I'm talking about a *perfect* battery and a perfect charging system. The figures are really only a theoretical minimum.
You really need to know what sort of charging current your battery could tolerate. Furthermore, a 12V lead-acid battery would normally be charged at a constant voltage of between 13.8V and 14.1V, so the current draw would be dependent on its internal resistance. A partially discharged battery would need to be current limited so as not to overheat it.
- Franc Zabkar
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I don't think the OP mentioned anything about a UPS. Just wanted to point out that a thermometer, ruler and timepiece can all be useful in electronics, as can pencil and paper.
Unless the wattmeter can be stuck in the actual charging circuit, your measurement is only going to be vaguely related to what you're trying to measure.
I'd be interested to know how these useful little meters react to low voltage DC......
RL
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He did, in more than one post he said this is a battery in a UPS.
Those watt meters only work with AC, but as I said, it depends on what he really wants to measure. If his goal is to know at what rate energy is being put into the battery, then a multimeter to measure DC amps while charging is the appropriate tool for the job. If what he wants to know is how much energy it draws from the wall to charge the battery, taking into account energy lost as wasted heat in the charger and battery, then the watt meter is an easier way to get all the data and measures true power, taking power factor into account, just as the kilowatt meters used for billing do.
In a nutshell, he hasn't given enough information regarding what data he wishes to measure.
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"Papita" wrote in news:gse02f\$5nh\$ snipped-for-privacy@news.motzarella.org:
Your generator will definitely handle it, it's just a question of how fast you can go with the available power. If you want a quick guide go back and (re)read my posts about a solar power charger into 210 AH (said 240 AH earlier, that was an error).
Battery charging is a complex study despite the steady and well known qualities of lead acid batteries. You've been given good guidance for calculating, based on ideal conditions, but by the time you add in corrections for conditions only you can assess because you're there and we aren't (like temperature, charge/discharge rates, age of batteries, etc) your best bet is to get detailed anecdotal evidence based on long term observations. That's why I tried to give you some.
If you really want to measure it, you need two cheap multimeters, one reading the voltage across the battery, the other reading millivolts across a small length of heavy wire whose resistance you know as exactly as possible. This second gives you the current, and it's better than switching the meter to ammeter mode and placing it directly in circuit because it's safer, and you get to use that meter on other stuff easily if you want. Take readings frequently. And observe the polarity of the reading on that second meter, that one tells you whenther the flow is going in, or coming out. If you end up wanting a greater guide, consider some cheap analog to digital system so you can log the readings automatically. For a UPS though, you'll likely only need to do frequent readings during each charge or discharge run.
If you want a fast guide, and to avoid a lot of work monitoring them over time, you really need to read accounts of average yield in real situations. Look for web sites and forums about DIY solar power for such accounts, look closely at any that seem similar to yours, in capacity, charge times, etc.
And if all you want is a guide to what is drawn out of a mains connection, get a Kill-A-Watt meter like James Sweet pointed out. Those do all your monitoring and averaging for you because they're designed to watch long term use of intermittent loads, like fridge compressors, and unsteady ones like TV's and hi-fi's.
But in the end if you really want to know the ratio of input to output yield, you're going to have to look more closely at the batteries themselves. Starting with existing observations in other systems is the easiest and fastest way to good a clear idea.
Also, get the maker's data sheet for whatever batteries you have. If the Powersafe data I have is anything to go by, you'll likely get all the info you need, except for the charger efficiency.