PDA

View Full Version : load test for deep cycle battery

darryl
11-18-2014, 01:47 AM
I have a few deep cycle and marine batteries kicking around, and I'd like to be able to apply a load to gauge their usefulness. I have some stainless band kicking around, and I've done a resistance check to see what length would be needed to apply a known load to a battery. Turns out that 53 inches in length would suck 100 amps at 12 volts. Using the same formula, 106 inches would draw 50 amps from a 12v source, and of course the math works for other lengths and voltages. I want to make up a load resistor of suitable value to test these batteries from time to time.

Just wondering now what would be a good draw to take from a battery as a test? Obviously, if it's a 3 amp hour battery and 12v, it's not going to like this load, but a 220 amp hour battery should find it easy. Is there a current draw rate vs amp hour capacity that would give a reasonable test figure? I know that some battery guys apply a load and can say it's drawing 200, 300, 950 amps, etc. One battery I had checked was a 90 a/h and it apparently was capable of outputting 1000 amps through the guys tester. Don't know how real this result was, but there was heat- that's proof enough for me that the battery was 'good'.

But my real question is whether a 100 amp load on a 220 ah battery is a significant load or not much more than a tickle? I'm thinking that I'll get some voltage drop with that load, but it shouldn't be much, and the drop would be higher the worse shape the battery is in. I wonder if there's a voltage drop vs current draw vs battery capacity formula that would be useful.

My idea for the load resistor is pretty simple- a piece of 2x4 with an aluminum section screwed to each end. The stainless band would be screwed to the aluminum and a lead with a battery clamp attached. The stainless band might have a number of back and forth bends in it so it becomes fairly compact. It might have a central support point as well, also being attached to a piece of aluminum screwed to the 2x4. The aluminum pieces would prevent localized heating of the 2x4 for the short term, and keep the lead wire junctions at a reasonable temperature also. I could build this in little more time than it took to type this post.

I realize that as the band heats up the resistance is going to rise. The current draw will drop but will still be somewhat 'up there'- acceptable to me anyway.

Thoughts?

PStechPaul
11-18-2014, 02:28 AM
This might be a good question to post on the DIY Electric Car (http://www.diyelectriccar.com/forums/) forum, and similar tests are often done for EV batteries. They mostly use LiFePO4 cells which often have charge rates of at least 5C and discharge rates 10C or higher. Some Li-Ion cells are rated at 50C discharge. The "C" figure is nominal amperes based on the A-H rating, so a 220 A-H battery at 1C would be 220 amps. Lithium cells have very good efficiency so you could apply a 220A load for 45 minutes and have 25% charge remaining. You don't want to go much below 80% discharge on any battery (except in emergencies) because it will drastically reduce cycle life. For Lead-Acid batteries, there is the Peukert factor (http://en.wikipedia.org/wiki/Peukert%27s_law), which explains why a 220 A-H battery will only be able to meet that specification at 1/20 C or a 20 hour discharge at 11 amperes. At 1C (220A), the Peukert effect says you will get only 30 minutes, or 50% of rated capacity. For starting batteries, a CCA (Cold Cranking Amps) capacity is often given, and a typical 100 A-H car battery may have 500 CCA, at which point its terminal voltage could drop as low as 1/2 nominal, or 6 volts. This is also the point at which maximum power is obtained. If you use a heavier load you get more amps but the voltage drops and less power is delivered. This is also related to the principles of impedance matching.

I have an on-line calculator that provides battery capacity using the Peukert computation:
http://enginuitysystems.com/EVCalculator.htm

You might want to make a load that is submersed in a large container of water. It will keep the temperature of the element below 212F and the resistance will stay more constant. Plus, when you're done, you'll have a big pot of hot water to make lots of tea! :rolleyes:

macona
11-18-2014, 03:15 AM
Usually the datasheets for the batteries will have a graph of draw vs time and AH. With lead acid your capacity goes to hell pretty quickly when you start drawing at a good portion of the rated current. I have some 12v 150AH SLAs that I am planning on using to run my telescope, anything past about 20 amp draw starts killing your capacity with those. So I am running 4 in series for 48v into a 1kw inverter. I should never pull anywhere near the load of the inverter, I hope.

I am going to guess 100A on a 220AH battery is going to drop capacity pretty severely.

batt-man
11-18-2014, 03:33 AM
Darrell,

To a certain extent it depends on what your planning to do with the batteries.

With a bit of googling it should be possible to find the manufacturer's data sheet on those batteries and that should tell you the expected discharge rates vs capacity. If you can't find the data sheet for whatever reason a good plan is to plot voltage at c/4, c/2, c, and 2c. Obviously you need to charge to a consistent state between each cycle. With this info you'd be able to get a good indication of what the cell(s) are capable of. So saying that 220Ah @ 2c = 440 amps (>5kw at 12v) so probably not practical in a home shop environment.

However - if you have a specific purpose in mind then test to that application (it would still be good to run the 4 tests above though for reference if you don't have that data sheet).

Final point - deep discharge batteries tend to be designed for a relatively lower current draw, c/2, c/4, sometimes lower. Not a hard and fast rule but more of a "rule of thumb" when dealing with an unknown battery. So if your sure they are deep discharge then c/2 is probably relatively high; i'd go for about c/4 (unless the data sheet was telling me different). That's still a fair amount of heat your going to have to deal with (~660w) but manageable...

Cheers
Batt...

A.K. Boomer
11-18-2014, 08:14 AM
Don't your autoparts stores up there load test batteries for free? they actually jump at the chance here because they usually get to make a sale if it's bad and you can watch them do it to make sure their not pulling a fast one...

you should be able to find the amp rating on the battery or if not find it on line...

DFMiller
11-18-2014, 08:18 AM
Get some SS tubing. Make a coil and run water thru it. 3/8" is about 1 ohm per 100 feet. Never warms up. :-). 500 foot roll and some booster cables makes a high wattage inexpensive stable load.
Dave

J Tiers
11-18-2014, 08:18 AM
Essentially, your idea is fine.... load the battery and see what the voltage drop is. Measure of internal resistance. Final voltage or specific gravity also measures capacity, if you have before and after measurements, and a known amount of discharge, generally significant in terms of nominal capacity.

The heating WILL affect the current, and WILL make the measurement less accurate. Might be OK for a rough measurement anyhow, or not.

The Peukert effect is a complicated way of saying internal resistance.

You NEVER lose any energy in a battery, you ALWAYS get it ALL out again.

The problem is that you may get more or less of it as heat, which is where the Peukert effect comes into play, it predicts what you get as electric vs heat energy.

Cuttings
11-18-2014, 12:51 PM
As Batt-man mentioned, be careful if they are true deep cycle batteries.
They were not designed to be used for heavier loads like starting.
In the case of deep cycle batteries you want to use a smaller load over a longer period of time.

PStechPaul
11-18-2014, 03:03 PM
It may be interesting to note that lead-acid batteries get much of their voltage from relativistic effects:
http://phys.org/news/2011-01-car-batteries-powered-relativity.html
http://physics.aps.org/story/v27/st2

The Peukert effect is not totally due to internal resistance.
http://www.discover-energy.com/what-peukert-effect
http://www.bdbatteries.com/peukert.php
http://jgdarden.com/batteryfaq/carfaq7.htm (much good information about batteries)
http://batteryuniversity.com/learn/ (everything you could ever want to know about batteries and charging)

There is also a lot of good information on batteries and charging and many other electrical/electronic subjects here:
http://www.mpoweruk.com/

For instance, a 12V 100 Ah battery may have 0.01 ohm internal resistance. At 100 amps, the terminal voltage will drop by 1 volt to 11 volts, which means an internal dissipation of 1V*100A = 100 watts, and load power of 1100 watts. Peukert effect shows that the effective A-h will be about 50, so it can provide this power for 1/2 hour. Thus internal energy loss due to resistance is 50 W-h and delivered energy is 550 W-h. The same battery with a 10 amp load will have a terminal voltage of 11.9 volts and an internal dissipation of 0.1V*10A = 1 watt, and load power of 11.9V*10A = 119 watts. But it will exhibit a capacity of nearly the full rated 100 A-h, so it will have internal energy loss of 1 watt-hour and deliver 1190 W-h over the 1 hour discharge. ;)

darryl
11-18-2014, 10:30 PM
Good info, fellas. But I was asking more about the testing than the usage. I'll give another example- suppose I have a nicad in my hands, rated at 1.3 a/h- a pretty common size. If I set up a load to draw .65 amps from it, then measure voltage drop when the load is applied, will I get a reasonable indication of the state of the battery from that info? A nicad is normally capable of putting out many amps, perhaps up to about 30 or more for a short while. While this may affect its lifetime if used with this high of a discharge rate, it's nonetheless possible for it to do that. My test would only draw about half the amp/hour capacity, so my question is whether that's 'enough' of a test to indicate the condition of it.

With the lead/acid 220 a/h batteries I have, the test would also be at about half the a/h capacity, even though we are now dealing with 100 amps into a test load. Correct enough, that does translate into a lot of heat, and I don't want to burn wiring, overheat connections, etc, so I want to keep the test current down to 100 amps or less. The question again is whether this is 'enough' of a test.

When these batteries are called upon to deliver current, I don't expect to be drawing more than about 50 amps, and probably more like 20 or 30 amps would be an average. If I'm just running some lights as emergency lighting, it would be more like 3 amps. I won't know what the full capacity of the battery is unless I do a near-complete discharge at a known rate which is well below the a/h rating, but probably what I'd be more interested in is whether I could trust it to deliver several tens of amps short term if called for, and what kind of voltage drop to expect.

Going back to the question and phrasing it a little differently- suppose I dropped a 100 amp load on the battery and the voltage went from a near-full charge level of say 13 volts down to 10 volts. What would that mean for a 220 a/h battery? Is it ok, or does it have higher internal resistance than it should have. This example would tell me that I could expect to draw say 50 amps, and the voltage wouldn't drop below about 11.5 or so- and this would give me some assurance that it will function when needed. For how long, well this would be gauged by how quickly the voltage continued to drop with the test load applied. If it couldn't hold voltage for more than five seconds, I'd have to consider it shot. At the same time I don't want to keep the test load applied for minutes at a time, 'cause my cheap and simple load resistor is going to get really hot- which presents a handling problem.

Hope this give a better insight into what I'm considering.

PStechPaul
11-19-2014, 12:16 AM
A lead-acid battery's state of charge (SOC) can be estimated pretty well by the terminal voltage (from http://www.mpoweruk.com/soc.htm):

http://www.mpoweruk.com/images/soc_voltage.jpg

The voltage is also related to temperature, but if your batteries are indoors that will not be as much a factor as in a vehicle outside. There is also some stabilization that occurs after a high current charge or discharge. The last time I had a problem with a car battery was about 4 years ago when all seemed fine on a drive to Harrisburg and I even parked and then started again to move to another location with no sign of problem. After spending about four hours at a dog event "Woofstock" I returned to the car and it was totally dead. I found someone to give me a hotshot and it started right up and I let it idle for a while, but when I put my foot on the brake the engine immediately died and even the dome lights barely lit. I had the vehicle towed to a garage (that was fortunately open on a Sunday afternoon), and he confirmed that it was indeed "dead" and would not accept a charge or put out even minimal current. Previously, my batteries had slowly become weaker and seemed to charge up quickly but also discharge quickly, so the capacity had diminished.

So, in your case, I would suggest charging the battery to its maximum float voltage, and then apply a moderate load of about C/10 (22.5 amps). It should supply current easily for 2 hours, at which point it should be at 80% SOC approximately. The terminal voltage might drop a little bit at that load, so perhaps it might be 13.2 volts no load and 13 volts at 22.5 amps, which will give you an approximate internal resistance of 0.2/22.5 = 0.01 ohms. After two hours, determine the SOC from the voltage and the chart above. 12.75V would indicate about 80% SOC, and the battery should be good. If it drops to 12.5 volts that indicates 60% SOC and the battery has lost about half of its capacity. It may still be usable, but probably "on its last legs".

You could get a new battery of similar type and do a comparative test.

J Tiers
11-19-2014, 12:44 AM
The Peukert effect is not totally due to internal resistance.

Most of those really DO comment on what is effectively a battery resistance issue. Maybe not true DC resistance , but resistance type effects, resulting in voltage drop under load. If it walks like a duck......

There are really TWO voltages to consider, but easy access is had to only ONE of them. And it is the WRONG one.

The obvious voltage is the one you measure on the terminals. *After* all the voltage drop effects. That one is made of lies and deceptions. It may tell you what you can do with the battery in existing state and rate of discharge, but it does not give an indication of ampere hours remaining.

The real "voltage" you want for capacity is the "driving" voltage, the one you would measure with no load. The one BEFORE all the various resistance type effects. Unfortunately that one is not very accessible.

Rich Carlstedt
11-19-2014, 08:57 PM
It may be interesting to note that lead-acid batteries get much of their voltage from relativistic effects:
http://phys.org/news/2011-01-car-batteries-powered-relativity.html
http://physics.aps.org/story/v27/st2

The Peukert effect is not totally due to internal resistance.
http://www.discover-energy.com/what-peukert-effect
http://www.bdbatteries.com/peukert.php
http://jgdarden.com/batteryfaq/carfaq7.htm (much good information about batteries)
http://batteryuniversity.com/learn/ (everything you could ever want to know about batteries and charging)

There is also a lot of good information on batteries and charging and many other electrical/electronic subjects here:
http://www.mpoweruk.com/

For instance, a 12V 100 Ah battery may have 0.01 ohm internal resistance. At 100 amps, the terminal voltage will drop by 1 volt to 11 volts, which means an internal dissipation of 1V*100A = 100 watts, and load power of 1100 watts. Peukert effect shows that the effective A-h will be about 50, so it can provide this power for 1/2 hour. Thus internal energy loss due to resistance is 50 W-h and delivered energy is 550 W-h. The same battery with a 10 amp load will have a terminal voltage of 11.9 volts and an internal dissipation of 0.1V*10A = 1 watt, and load power of 11.9V*10A = 119 watts. But it will exhibit a capacity of nearly the full rated 100 A-h, so it will have internal energy loss of 1 watt-hour and deliver 1190 W-h over the 1 hour discharge. ;)

Awesome description Paul !

Thanks !

Rich

darryl
11-19-2014, 10:50 PM
Some pretty good reading there- thanks Paul.

I've decided to go ahead and build the .12 ohm load resistor. Gathering parts, I have a base (piece of maple) the element (piece of stainless banding material) some heat spreader plates (aluminum pieces) and connecting wire with terminal clamps. Looked around and didn't find any reasonably heavy wire- then I remembered I saved a jumper cable from the scrap heap. Great- already has clamps on it, and I can cut the other two off and make my connections with the bare ends. So I go to cut the wire and it's a piece of cake. Hmm- I didn't expect it to be so easy. Turns out it's about 90% insulation surrounding what looks like 10 ga copper wire. I could probably just twist the ends together and use that alone as the load resistor :(