How to measure the COP of a battery?

[sucahyo]

Ok, this is stupid question, you can laugh but please answer how can I measure the COP of a battery?


Use this for an example.
YouTube - 2 minutes charging for 3 minutes of power
"2 minutes of charging at 0.55A, using circuit that consume 1.2A, give 3 minutes of power for 0.7A load"


There are other thread, but I still not sure if I can apply it to my battery because I am not and will not measuring empty battery or to discharge it until empty.
http://www.energeticforum.com/renewa...han-1-cop.html

[Inquorate]

I'd charge the battery with known volt x amps, from a known voltage to a known voltage, and take note of the time. That will show how much energy it takes, in joules to charge the battery.

I'd then discharge the batteries into supercaps (sent you some of those, 55F) successively, keeping note of the joules expended, until the battery has fallen back to the same voltage it started at.

OR

I'd get the 13v battery (when loaded) to move an electric motor with a large wheel that is marked clearly at the edge, and using a potentiometer, keep the revolutions at a specific RPM, noting the amps. I'd see how long it took for the battery to drop to a specific voltage, about 1v, to 12v.

Then I'd take a 100mA wall outlet power supply at 12.5v (average of the battery voltage) loaded, and use potentiometer to get the same amps and revolutions. I'd then take all the values for the battery and power supply, figure out the joules..

Something like that. Very time consuming.

I'll be comparing 4 batteries in parallel and a motor at a known rpm, seeing time to drop a certain voltage.

Then I'll get my circuit (bedini spike tesla switch hybrid) to do the same, and compare the times.

Don't know if that gives you some ideas good luck

[selamatg]

@Sucahyo,

Here is for your reference...
http://www.panaceauniversity.org/Bed...Experiment.pdf

Hope will help.

Selamatg

[smw1998a]

About Batteries

Ok, this is stupid question, you can laugh but please answer how can I measure the COP of a battery?

Hi Sucaho,
I don't view this is a stupid question at all. Here is a re-post of a post I wrote elsewhere some time ago primarily aimed at the SG/SSG. I hope you find it useful.

Regards Lee..

>>>...
I have been load testing for a long time, mainly small capacity SLA batteries but I do have large 110Ah diesel batteries and several leisure batteries of various sizes. There are two important characteristics you need to know of any new battery, the state of health (SoH) and the state of charge (SoC). The SoH of a battery should be 100% at new, that is, if the manufacturer states a capacity of 55Ah then a SoH of 100% means the battery can store 55Ah worth of charge. SoC, on the other hand, is a measure of how much charge is left in the battery in relation to its SoH. SoC provides important data in which remaining load time can be calculated before the battery will need recharging but as SoC is relative to SoH, We need to find out just how healthy our batteries are.

When I first get a battery, new or otherwise, after checking the physical condition of the battery and topping up, I give it a unique name, 5Ah SLA 1 or 110Ah A1 and this name stays with that battery. The next step is to find as much as I can about the battery from the manufacturer’s data sheets. Charge and discharge curves are particularly useful. Now I can go ahead and calculate the state of health (SoH) of my battery.

Before I continue, a word about depth of discharge (DoD). When you continually cycle batteries and do work, how much energy or capacity you take out of the battery is called depth of discharge. DoD is important because the more energy you take from a battery during a single discharge cycle, greatly reduces the expected working life of that battery. SLA manufacturers suggest a maximum charged cell voltage of 2.4v per cell (14.4v) for a 12v battery and a maximum discharge to 1.75v per cell (10.5v) under a C20 load. So, if 10.5v is considered as 100% DoD at C20 then a battery with an SoH of 100% should take 20 hours to reach 10.5v from fully charged. A 100% DoD per cycle is really not good for any battery and should be avoided.

Depth of discharge in a six cell 12v SLA battery
100%DoD=10.5v 75%DoD=11.47v 50%DoD=11.95v 25%DoD=12.92v

A word of warning, discharge curves are not linear but the percentages above represent a good indicator of what you should expect from your battery during a load test. If you use a load termination point of 12.4v as many of us did in the Monopole group, we were only taking 33% of the rated capacity of the battery. In conventional terms, 33% of a fully charged 1.2Ah battery is 400mA, with a 100% SoH you should see a eight hour discharge at C20 to 12.4 volts. I don’t recall ever achieving this but as a datum it is very useful.

A new battery is quite straight forward if you have the data sheet, simply select a suitable C20 resistive load and begin a timed discharge to 12v, 50% of DoD, a data logger and a voltage controlled switch are crucial for this. Charge your battery with a suitable conventional charger and repeat this exercise. If your battery is rated at 10Ah your C20 load of 500mA should run for 10 hours from fully charged to 12.00v (50%DoD=5Ah @ 500mA/10), it rarely does, especially with old batteries. Say your load test only managed 6 hours, that’s fine, you now have your SoH in conventional terms, it is 60% of rated capacity of the battery. You now know the SoH of your 10Ah battery and that it only has 6Ah of useable capacity, as your DoD is only 50%(LTP 12.00v) of the SoH, your working capacity is 3Ah from fully charged. Your state of charge (SoC) can now be calculated, observing voltage between fully charged and 12v (50%DoD), discharge being 6 hours@ C20. Again, it is not linear but experience will improve the accuracy of your SoC calculations. It takes a while to get the hang of this but it is worth the effort. Keep a battery diary, your initial charge/ discharge cycles should be done conventionally to obtain the SoH of an individual battery and the data from the tests should be the first entry in your diary before they go near the SG/SSG.

I had toyed with the idea of basing the C20 load on the SoH of the battery and not the batteries rated capacity, this was because several of my 110Ah diesel batteries had a SoH of less than 6% and the C20 load at the rated capacity made discharges very short and the battery would recover almost all its voltage over the following few hours after the initial load test. Like the 10Ah battery of the example, if the C20 rate were based on the SoH of 6Ah (60%) instead of the rated 10Ah, would this be beneficial to the reconditioning of the battery once introduced to the SSG?

Once your batteries are charged on the SG/SSG, watch your load times, I found that, initially, I got far shorter load times than the conventional charger and SoH suggested. As you know the SoH of your battery it should not be too many cycles before your energiser matches the conventional SoH of your battery under load and then should improve on it. Sadly, my work on this battery monitoring regime ended last year due to pressures from work etc. As a basis for performance measurement of any energiser it should work well. The old energy in, energy out arguments are moot because I have deliberately ignored energy consumed by the energiser, the method focuses solely on the performance of the battery being charged.

>>>...

[smw1998a]

@Sucahyo,

Here is for your reference...
http://www.panaceauniversity.org/Bed...Experiment.pdf

Hope will help.

Selamatg

While I commend the BM3 group for encouraging this type of load testing, sadly, this method of of calculating the energizers C.O.P as a independent charging system is deeply flawed and misleading.

Regards Lee...

[sucahyo]

I'd charge the battery with known volt x amps, from a known voltage to a known voltage, and take note of the time. That will show how much energy it takes, in joules to charge the battery.

I'd then discharge the batteries into supercaps (sent you some of those, 55F) successively, keeping note of the joules expended, until the battery has fallen back to the same voltage it started at.

Thanks, I rather surprise to see a capacitor with such capacity .

Do you assume that what measured by amp meter is what actually transfered to the battery, especially the charging current?

I'd get the 13v battery (when loaded) to move an electric motor with a large wheel that is marked clearly at the edge, and using a potentiometer, keep the revolutions at a specific RPM, noting the amps. I'd see how long it took for the battery to drop to a specific voltage, about 1v, to 12v.

Then I'd take a 100mA wall outlet power supply at 12.5v (average of the battery voltage) loaded, and use potentiometer to get the same amps and revolutions. I'd then take all the values for the battery and power supply, figure out the joules..

It compare battery output and power supply output? You don't compare battery charger input?

I'll be comparing 4 batteries in parallel and a motor at a known rpm, seeing time to drop a certain voltage.

Sorry, I have no idea what you are trying to conclude from the experiment.

I use simple battery swap to measure COP, different experiment, but I still have no clue what COP means in battery swapping experiment. If I loose 0.01V on each 2 minutes, what is my COP for two 7Ah battery?
YouTube - Swap Charging front and back battery

Here is how I wire the relay (thanks for them too ), I am using different power source for the relay:



@selamatg, thanks, but unfortunately I also agree with Lee because using that method my circuit COP is either too high or too low.


@Lee, thanks. Basically I need to know SOC and SOH first? What is the relation of SOC and charging and discharging joule?

The problem is to know when to stop charging. When my battery full the voltage gone very high even when the charging amp also rise, give false impression of more efficient charging. My battery is a broken one, max voltage is 10.4V. So I avoid charging it to full too.

Also my stingo charger has erratic behaviour. On one battery connecting booster circuit to source negative increase charging amp up to 1V or 3V on good battery, on other it decrease but raise charging voltage when connected to coil CT. On other it activate higher charge when connected to source positive. Weird circuit.

That is why I need a way to measure the charger output vs battery charge. But still not sure how to measure battery charge.

As of measurement:
Input : 12V x 1.2A x 2 minutes = 1728 joules
Charger: 10.5 x 0.55A x 2 minutes = 693 joules
Discharge: 9.5 x 0.75A x 3 minutes = 1282.5 joules

Method used for COP calculation do not seems to have connection with how I loose 0.01V per 2 minutes if I use my circuit in battery swapping test. I expect at least the COP of the charger energy divided by discharge energy to be lower than 1. Otherwise there is other thing flow to the battery according to BM3 manual...

[juju]

dead battery

Hello guys, maybe this is off topic but... someone advice me to put destilled water in my dead battery, on other topic that i cant find now...

i finnaly get it and fill the battery (12V 7ah), i found strange that it fill up with just a little bit of water, compared to the size of the battery.. i fill all the 6 valves.

the battery shows 13, 14 volts when its being charged.. but cannot hold more than 7 Volts before 2 days of charge.. someone can guess why?

[smw1998a]

@Lee, thanks. Basically I need to know SOC and SOH first? What is the relation of SOC and charging and discharging joule?

The problem is to know when to stop charging. When my battery full the voltage gone very high even when the charging amp also rise, give false impression of more efficient charging. My battery is a broken one, max voltage is 10.4V. So I avoid charging it to full too.

Hi sucahyo,
State of Health and State of Charge are relative measurements of a particular battery only. You could have two physically identical batteries of the same age that have lead very different existences, as a result, the state of heath of these two identical batteries will be very different. As I mentioned you must asses the condition of your battery in conventional terms. I do this by charging the battery with a conventional charger suited to the battery in question. I then discharge the battery through a known constant resistive load at the C20 Rate. If the battery is a 14Ah SLA then the C20 discharge is 700mA. I would use a load termination point of 12v under load. This represents 50% of the batteries capacity and on paper this should give a load time of 10 hours@700mA assuming State of Health is 100%. However, If you only get 7.5 hours to the LTP of 12v then your batteries state of health is only 75%. You know you can only draw 700mA from this battery for 7.5 hours before it reaches 50% depth of discharge (or 12v under load). You could chose to consider this battery as a 10.5Ah battery and adjust your C20 load load to suit 525mA for 10 hours as we only discharge 50% of the batteries capacity.

Once you understand how much work your battery can do, over 10 hours at it's c20 rate of discharge you can calculate it's state of charge at any time during that discharge simply by reading the voltage under load. You can still work your Joules out as you have been doing but everything is relative.

Experience has taught me that you can not judge the remaining capacity of a battery by standing voltage alone, particularly when charged from pulsed capacitors or inductors. You must first asses what the battery can deliver after conventional charging in order to see benefits of alternative charging methods, this can only be done via load tests in the order of hours, ten or twelve, never in minutes, Lead acid batteries are slow and cumbersome creatures and do not like to be rushed.

As for measuring input. This is even more controversial. IMHO you need a scope and you can measure avarage current during the pulse and deviding it by duty cycle etc. As an example a rotor bassed SSG with a frequency of 170Hz and a average current of 500mA and a duty of 33% will draw a current of 1.5Amps while the transistor conducts so all is not as it may first seem.

Regards Lee...

[sucahyo]

the battery shows 13, 14 volts when its being charged.. but cannot hold more than 7 Volts before 2 days of charge.. someone can guess why?

There is dead cell or unresponded cell. It is inevitable that cell may not get full at the same speed. And get worse over time. The one get fulls receive most the current and the voltage from the charger, the one half empty get less. If we can charge the cell individually it would be better.

Experience has taught me that you can not judge the remaining capacity of a battery by standing voltage alone, particularly when charged from pulsed capacitors or inductors. You must first asses what the battery can deliver after conventional charging in order to see benefits of alternative charging methods, this can only be done via load tests in the order of hours, ten or twelve, never in minutes, Lead acid batteries are slow and cumbersome creatures and do not like to be rushed.

Thank you .

Do stopping at 12V on normal battery always correspond to 50% capacity? How about battery that have one cell broken or half broken? How many percent the amp output of a battery with 50% capacity?

Just thinking that maybe the increased voltage by radiant charge is permanent.

I usually measure with 1 Amp load.

As for charging, I use FWBR from two radiant charger that being tapped is positive and negative. I think it do not came in pulse. But I agree that scope may be required for single diode charging part.

[smw1998a]

There is dead cell or unresponded cell. It is inevitable that cell may not get full at the same speed. And get worse over time. The one get fulls receive most the current and the voltage from the charger, the one half empty get less. If we can charge the cell individually it would be better.

Thank you .

Do stopping at 12V on normal battery always correspond to 50% capacity? How about battery that have one cell broken or half broken? How many percent the amp output of a battery with 50% capacity?

Just thinking that maybe the increased voltage by radiant charge is permanent.

I usually measure with 1 Amp load.

As for charging, I use FWBR from two radiant charger that being tapped is positive and negative. I think it do not came in pulse. But I agree that scope may be required for single diode charging part.

Hi sucahyo,
A load termination point of 11.95v under a C20 load seems arbitrarily considered 50% of battery capacity in conventional terms. Personally I would not continue with a battery that may have a dead or shorted cell as this would constitute physical damage and the bad cell will have a negative impact on the performance of the other five cells, no amount of radiant charging will improve this. Your load should be relative to your batteries 'rated' capacity. In your case 16% less than the rated capacity due to your dead cell. I would consider your battery 100% discharged at 8.75v (1.75v per cell x 5) and 100% charged at 12v (2.4v per cell X 5) 50% depth of discharge would be around 9.9v under a C20 load. Your C20 load must match the modified rated capacity of your battery. eg. A 60Ah battery minus one cell could be considered as a 50Ah battery, the C20 load is calculated as 50/20=2.5amps, 10v@2.5A would require a resistance of 4 ohms with a minimum power rating of 25Watts, in this case. The ideal load time for this battery is 10 hours to 9.95v. Unfortunately, these figures are based on conventional charging and as your battery is clearly damaged you cannot charge the battery conventionally, therefor you have no benchmark to judge charging performance. As I said, I would strongly urge you not to continue experimenting with a damaged battery.

Regards Lee..

[sucahyo]

Thank you for the explanation .

I am not into experiment that observe battery charge to see how can I improve my circuit. I am into experiment to see how my circuit behave on any kind of battery, and to see if I can calculate efficiency of my circuit on such battery.

I am into practical implementation. I improve my circuit with theory. The experiment is to confirm battery reviving and cold charging effect. One of them failed means my circuit has too much conventional current.

I rarely see C20 load on real life. The closer implementation of my circuit are portable 6watt CFL lighter. Where the circuit can become lighter when carried around, and become a charger at home. The input power requirement is about 600mA at 10V. Charging output measured at 200 to 400mA, sometimes 700mA depend on the battery.

I intend to use load like what real life requires it. I try to see if my circuit can make the battery live longer with that condition.

People around my place is hard to convince. If I can just show the benefits of the circuit, at least they learn not to throw broken CFL as long as the bulb still intact. And also reuse broken motorcycle battery, which usually have one or more cell broken.

Measuring how the circuit perform on broken battery is important. Considering broken SLA is sold around half dollar (more for lead acid) there is potential saving for anyone just need power from battery. They may not need to buy normal battery if the broken one serve what they need. If one battery is not strong enough, they can choose to use two or three instead since the circuit do not have problem being powered from 3V or 24V.