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/renewable-energy/3459-proposed-test-find-out-if-energizer-giving-greater-than-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...

https://www.panaceauniversity.org/Bedini%20Monopole%203%20Group%20Experiment.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?

[smw1998a]

@Sucahyo,

Here is for your reference...

https://www.panaceauniversity.org/Bedini%20Monopole%203%20Group%20Experiment.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, [email protected] 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.