Amp hours pulled from a 12V battery?

[tina-duncan]

Hi friends,

I am having a little difficulty understanding how amp hours being pulled from a battery are calculated, any help would be much appreciated!

I have a 12v 125Ah deep cycle battery. Research has told me that for DC appliances you divide their wattage by 12V to give the amps, so for example a 24W DC item would be at 2A and thus with the 125Ah battery would last 62.5 hours?

My question, however is how does this work when an inverter is used? for example I have a 24W AC appliance that I wish to know how long will operate with the battery, is it still a case of dividing the wattage by the battery's voltage or is there more to it than this?

Thanks,
Tina

 

[Sciurus]

My question, however is how does this work when an inverter is used? for example I have a 24W AC appliance that I wish to know how long will operate with the battery, is it still a case of dividing the wattage by the battery's voltage or is there more to it than this?

Thanks,
Tina

Hi, tina-duncan. If your inverter is 100% efficient, then yes. No inverter is, so some of the energy will go into the inverter and be converted to heat, so the time the battery will run an AC device with the same wattage will be less. How much less depends on inverter design.

 

[tina-duncan]

Hi, tina-duncan. If your inverter is 100% efficient, then yes. No inverter is, so some of the energy will go into the inverter and be converted to heat, so the time the battery will run an AC device with the same wattage will be less. How much less depends on inverter design.

Thank you for your swift response!

This indeed confirms my confusion, I have a killawatt meter plugged in between my inverter and appliance. the meter gives a reading of 0.1A. Neglecting the efficiency of the inverter shouldn't my reading be 2A? (24W/12V)

 

[Sciurus]

If the appliance is drawing 24W at 12 Volts Alternating Current, then you should see about 2A AC. Note that if the meter is measuring DC amps, the alternating property of the current will give a small reading. Also if the inverter output AC voltage is higher, say typical line 120 VAC, then the amperage would be lower.

 

[Fish4Fun]

Hey Tina,

Because I think you are looking for a "real-world" answer, there are some things you should consider:

1) Lead-Acid batteries have a much longer life span if they are not depleted more than 50%. This means that your 125Ah battery should not be discharged more than ~60Ah.

2) Inverter efficiencies are typically greatly exaggerated and reflect only the peak efficiency, at some specific load. That is, an inverter that brags "1600W Surge and Up to 90% efficient" may in reality be a 800W inverter that is only 90% efficient at 480W with the efficiency be considerably less with more or less current draw. Frequently Inverters are at their lowest efficiency in low load conditions because the inverter takes some amount of power to operate regardless of load.

For instance, if the inverter draws 2A @ no load, it may well draw 4A @ 24W output making it only 50% efficient. The same inverter with a 480W load may only draw 22A thus giving it the stated efficiency of "Up to 90% efficient".

3) A 24W AC device almost certainly has a power supply in it that converts 120Vac to some nominal voltage, likely between 5Vdc and 12Vdc. It may be considerably more efficient to simply power the device directly from the battery (replace the power supply with a DC/DC converter or regulator), or buy a similar device that accepts a DC input. While purchasing a new device that accepts a DC input may seem "expensive", if you consider the cost of 125Ah batteries and a suitable charging system, the device itself can become a trivial expense.

4) Following are some realistic expectations from a 125Ah, 12Vdc Lead-Acid Battery with a typical low-end, consumer inverter:

120Vac device @ 24W ==> 24W = 120V * I ==> I = .2A
Inverter input = 2A + 24W/12V = 4A
Battery Life: 125Ah/2 = 62.5Ah ==> 62.5Ah/4A = 15.6 hours

So, you might expect to operate your device for ~15 hours between charging cycles w/o dramatically shortening the life of the battery. The device might actually operate for as long as 30 hours or so on a full charge, but this figure will drop with each charging cycle and you will likely not get more than a couple dozen charging cycles before needing to replace the battery.

Typically people ask these kinds of questions in an effort to "go green" and intend to charge the battery with solar or wind power. If this is your intention, you need to understand that charging a battery also involves losses. For instance, a typical lead-acid battery has a charging efficiency of ~85%. This means that if you input 10A for 10 hours you will have "added" 85Ah to the battery's capacity. But be careful, "over charging" a battery can lead to explosion, fire or destruction of the battery!

So, if you were sizing a solar panel to operate your 24W device and you had 5 solar charging hours per day, the math would go something like this:

Device = 24W @ 120Vac ==> battery draw = 4A => 48W @ 12Vdc. Assuming this device is going to be "on" 24/7, you would consume:

48W * 24 hours = 1.152kWh of 12V power.

To replace that 1.152kWh of power, you would need to input:

1.152kWh/.85 = 1.355kWh and this would imply a solar panel capable of:

1335Wh/5 hours = ~270W solar panel

With a 125Ah capacity battery you would discharge the battery to 40% capacity before the charging cycle began==>

4Ah * (24 - 5) hours = 76Ah ==> 1 - 76Ah/125Ah = 0.392 = 39.2%

Please note this system would fail if there were a single day of no charging, and the battery life would likely be adversely affected by the continual deep discharge cycles.

For comparison, if you were to replace the device with a 12Vdc device (or replace the power supply in the device with a 12Vdc power supply) and we assume that there is no conversion loss or the conversion loss is equal to the existing losses in the internal power supply (the device simply consumes 24W @ 12Vdc) the sizing of the panel would go as follows:

12Vdc @ 2A = 24W
2A * 24 Hours = 48Ah = 576Wh/Day
576Wh/.85 = 677W ==> 677W/5 = 135W Solar Panel

Your battery would have a charge of ~70% at the beginning of the charging cycle:

2Ah * (24 - 5) = 38Ah ==> 1 - 38Ah/125Ah = 0.696 =~70%

I know this seems like a lot to power a "small device", but the reality of "green energy" and batteries is that most of us take power for granted. The "convenience" of plugging in a device and having it "work" masks the tremendous amount of work that is being done to maintain that device's functionality.

As an example,

1) Assuming you simply plugged your 24W device into a standard 120V outlet, it would cost: 24W * 1kWh/1000Wh * 24Hours/Day * 365 Days/Year * $0.10/kWh = $21.02 per year.

2) Assuming you converted the device to a 12V input and used a 125Ah battery + 130W Solar panel:

Charge Controller ~$50
Battery ~$90
Solar Panel ~$800
-----------------------
Up-Front Total: $940

With a Battery Life of 5 years, annual battery cost would be $90/5 = $18. Savings $3/Year, period of time before system "paid" for itself: $940/$3 per year = 313 years.

3) For the case with the inverter and a single 12V, 125Ah battery the system would never "pay for itself" because it is unlikely you would get more than a year or two out of the battery discharging it more than 50% daily. If you used 2 * 12, 125Ah batteries, you might expect to get five years from them, but the annual cost would be $36 ($180/5 years), and again, you would never recoup your initial investment because your annual battery expenditures would be higher than $21/year.

If you were to replace the lead-acid battery with Lithium-Ion battery packs, you might fair better in the long run, but the upfront costs would be considerably higher.

I am certain this was a lot more answer than you intended in your OP, but I got the feeling this is the direction you were heading in, and I thought I would run it through for you rather than you going out and buying and inverter, battery and charger only to find the system did not perform as you had imagined. If you did not intend to charge the battery with solar or wind, then please except my apology for assuming that.

Fish

 

[mheslep]

...Typically people ask these kinds of questions in an effort to "go green" and intend to charge the battery with solar or wind power...

As an example,

1) Assuming you simply plugged your 24W device into a standard 120V outlet, it would cost: 24W * 1kWh/1000Wh * 24Hours/Day * 365 Days/Year * $0.10/kWh = $21.02 per year.

2) Assuming you converted the device to a 12V input and used a 125Ah battery + 130W Solar panel:

Charge Controller ~$50
Battery ~$90
Solar Panel ~$800
-----------------------
Up-Front Total: $940

With a Battery Life of 5 years, annual battery cost would be $90/5 = $18. Savings $3/Year, period of time before system "paid" for itself: $940/$3 per year = 313 years.

...

It may be that these assumptions could use an update.

1. Thin Film 128W PV Panel. $http://www.google.com/products/cata...a=X&ei=HoS4Tc63M5P1gAeqk9hY&ved=0CHYQ8wIwAg#" for quantity 1, not $800. If trends continue, it is likely that panel will cost ~$250 next year and so on.

2. If utility power is available to the OP, and this is not a grid backup application in which case comparisons to grid power are not relevant, then the grid can act as 'the battery.' In larger scale systems a grid tie disconnect would be required, but with this 24 W case a diode OR with the grid or perhaps a simple DPDT on/off timer will do. If utility power is not available to the OP, then a solar comparison should be against a portable combustion generator + noise + fuel availability and transport, and/or a remote utility hookup charge or a long distance buried electrical conduit run. Either way, these approaches do not amortize out to 10 cents / kwh.

3. While the US average electric rate is indeed a bit over 10 cents / kwh, it varies substantially by location, which the OP did not specify. In Puerto Rico for instance, the rate is 65 cents / kwh. So in Puerto Rico w/ a Thin Film panel and perhaps an inverter, the pay back period drops to maybe three years. Finally, and this just wild speculation here, the average US electric rate may not remain at 10 cents / kwh for the next 313 years.

 

[RV-Man]

Wow, this is an old question, but I want to point out something that no one else dared mention on here, and without it, a lot of everything stated is null-and-voided on the spot.

First, let me offer a bit of my experience with this. Before I retired, I was a Director of Engineering for a major RV company. In our numerous designs, a couple of things were constant - customers demanding absolutely as much electrical power as they could get, and our constant fight to economically provide them with not only 'shore power' (connecting to the standard AC power grid) but 'boondocking' power, as well (where there's just you, bears, and nice lakes). As a result, we fought constantly with power, DC systems, charging, charging speeds, and DC loading issues.

Along the way, we kept stumbling over one thing - our deep cycle batteries weren't lasting like our calculations said they should (this was some time back). Customers complained they couldn't get 30, 45, etc. minutes of operations time in 'boondocking mode' without having to start their generators to charge their deep cycles. This is a big no-no late at night in state parks, and when your batteries drive things like your refrigeration compression cycle, your heater igniter, and more, well, you get the picture - it's important.

Inadvertently, I happened to stumble across a man who had worked for years in the Ambulance-design-business. His electrical designs were quite impressive, and over lunch, he explained this little thing called 'Peukert's curve' to me. In essence, it is the function of a battery under load, and how discharge is not a linear event, but a curved parabolic function, which means that the more load you place on a battery, the less 'run time' you get before you reach your desired discharge level (where you want to effect recharging without doing damage to it). So, I took my fancy Quattro Pro (think 'Microsoft Excel Competitor', before Excel became popular), plugged in a few formulas to ease calculation, and voila, projected data! The end result now actually matched what our customers were seeing in the field, quite accurately. It also gave us a sick feeling that we thought we should get a lot more power than we did, out of those batteries. In a nutshell, THIS is what drove RV manufacturers to start adding multiple battery configurations, larger chargers, etc. The customers loved us, until they saw how much all that extra equipment was now going to cost them - but at least we could now offer them actually-accurate data on what each system design could achieve for them when 'boondocking' it.

Take a look at Peukert's Curve for what you are looking for.
Peukert's Curve = T = (C/((I/(C/R))^n) x (R/C))
C=Battery Rating (amp-Hours, average type 72 deep cycle is around 100 amp hours)
T=Time-rating @ load
n=battery efficiency (common deep cycle wet-cells use 1.2 as start value)
I=Current-amps
R=Rating Method (20-Amp Hour, 100-Amp Hour, etc. (this is the testing process, not the 'C' rating of the battery itself); value is input in aH's)

When calculated, a 10-amp load, on a 100-amp-hour rated deep cycle which was rated in the 20-amp-hour rating-method with an assumed efficiency of 1.2 will reach discharge condition in 8.71 hours.
T = (C/((I/(C/R))^n) x (R/C))
T = (100/((10/(100/20)^1.2) x (20/100))
T = (100/(2^1.2)x0.2
T=8.71 hours

Note, current is a factor in the calculation, but voltage is not a value in the calculation, as the equation assumes your current is correct for the given voltage system being calculated. You can also use the equation with parallel circuits, by simply multiplying the T value against the number of parallel batteries in the circuit, such that two of the batteries described above, in a parallel circuit, with that same 10-amp load, would then yield 17.42 hours of capacity. This is the ONLY way to actually design DC electrical systems. Any other way is really just a shot in the dark. It also allows you to 'spec in' a given battery type for warranty and performance-sake.

 

[RV-Man]

By the way, Peukert's Curve is 'theoretical', not applied, so it has no value for 'fudge'. In design, you must agree upon some 'slop', or 'leeway' (in mechanical engineering, you would call this stacked tolerances), to make up for ever-changing electrolyte density, plate conditions, temperature variations, surges, etc. With RV's for general application, we allowed a 20-percent 'safety' factor (i.e.: we reduced the projected time this much, to have a safe-zone of operability that we could still be comfortable advertising). The 'n-value' really is the only 'slider' in the equation, and you would only need to 'calculate' it once, usually at the start of using the battery when new. 'n' is battery efficiency, which should NEVER be confused with your inverter's efficiency. They are separate values for separate components.

The worst thing you are going to run into, is the stacked losses in a complex system. If you get into panels, voltage regulators, batteries and inverters, it is quite easy to stack up a 40-45 percent system loss when all is said and done. Again, this is where you get into multiple (parallel, or serial-parallel battery systems) to provide the power you need. When you go putting this into an 'on-road rolling vehicle' where mass really does matter on that holy-grail bottom line called 'GVWR' (gross vehicle weight rating), and you tell a customer that the 2-hour battery system they want to support a 30-amp system is going to cost them $1600 and 600-pounds of weight, you quickly see why they back down to a 1-hour, 10-amp system, in a hurry.

Oh, and depending on your application, what we personally found (when fuel was a $1 a gallon) was that it was far more efficient to put a 2nd alternator on the vehicle and let it charge your batteries, than to alternatively rely on generator charging. One other point we learned in all of our battery testing was 'easy does it on the charging', with a wet-lead-cell deep cycle battery, the best long-term-care charging you can do, is 10-percent charging. By that, what I am saying is, never charge your battery more than 10-percent of it's 'top-off' amperage per given time, or you will damage it and reduce it's life-span. For example, a 100-amp hour deep cycle, discharged to 50-percent, would require 50-amp hours to be recharged (in a perfect world), so ideally you would want to introduce a 5-amp charge for a 10-hour period, to bring it back up. In reality, 6-amp chargers are more common, and would be ideal to offset losses in the charge circuit. Faster charging than this will bring it up quicker, but heat is your enemy, and over-charging by too fast of rate will cause plate material loss inside the battery that will greatly shorten it's overall lifespan. With advanced modern batteries, like AGM (absorbed glass mat) and gel batteries, this is far less of a concern, or a non-issue at all - then you can throw 20-amp hours of charge into it, and bring it back much faster with far less physical damage to the battery itself. You get what you pay for, in the end. AGM's are pricey, but if you want fast-charging you don't have to wait for, they are great. Old 'wet-cells' are still great with most solar panel charging systems with their lower charge wattage.

Hope all this helps the 'random browser' that stumbles on this post - it was a lot of hunting, testing and research that led to a lot of this information, and I've only seen it in a few places where I wasn't the one posting it.

 

[Fiona F]

RV-Man ... simply awesome. I'm one such 'random browser' and quite the beneficiary of all your 'hunting, testing and research'. Thank you. ~Fiona

 

[anorlunda]

Great info from several of the posters on this thread. The topic is complex. Here are a few things not mentioned by others.

1. So-called "smart chargers" with a 3-stage process, and with a battery temperature sensor that moderates rate of charge. These do a much better job of charging wile avoiding life shortening overcharging.
2. Temperature of the batteries is a major factor. If your battery is close to a hot engine, it may run at 130 F much of the time. That is instant death for gel cell batteries, and it dramatically reduces the allowable charging rate for flooded batteries.
3. Sulphation is a major problem in lead acid batteries. Overcharging helps prevent sulphation but reduces battery life. The correct solution is a "equalizing" (also called de-sulphating) charge cycle. My charger's equalize cycle is automatic. It charges to 15.2 volts for two hours once every thirty days. With equalizing, 5 year lifetimes are possible. Without equalizing, one year life is common.
4. A http://www.xantrex.com/documents/Discontinued-Products/Link10(445-0195-01-01_rev-A).pdf is very helpful because people's use patterns and tolerance for low voltage varies. For example, freinds on neighboring boats allow their batteries to discharge 50% of rated amp-hours, wheras I almost never allow more than 20% of rated amp-hours discharge. I also have devices that shut themselves off when voltage drops below 11.8 volts. You can let the monitor adaptively learn the CEF (charging efficiency factor) which is roughly (energy delivered during discnarge)/(energy consumed during charging). CEF of 0.8 is very good. You can also set the .Peukert's exponent to tune it. Most important, you learn your own charge/discharge typical use pattern. The monitor also measures and displays battery temperature.
5. If you are using solar panels in a mobile application, remember weather and latitude. Cloudy days may cut your energy prouction by 50%, stormy days can cut it 80%. Size your batteries for 3 or more consecutive cloudy days. Panels at 25 degrees latitude produce 50% more per day than the same panels at latitude 45.

For example, my consumption is 45 amp-hours per day. I have 205 watts of solar. At 25 degrees latitude and a sunny day, the batteries are fully charged by 11 AM. At 45 degrees latitude and a cloudy day, it takes every minute of the day to recover my consumption.

Technical sources tell you to ignore amp-hours and to use energy in watt-hours instead. That doesn't work well. Amp hours are more practical.

 

[berkeman]

This thread is truly timeless...