Battery life

[naahou2000]

Hi everybody,
I have a telecommunication system powered by two 12V batteries in series (24V) and a 220Vac / 24Vdc charger.
I am looking for someone who can help me calculate the autonomy of this system. The two batteries are of the 12V, 115 AH, C20 type. Consumption is around 4,5 A.

Knowing that a discharge measurement with resistance was made according to the attached table
please

[Alain G]

naahou2000

Welcome to Econologie!

It is quite simple to calculate the capacity of the batteries.

you take the voltage x amperages for one hour.

The capacity of your batteries is 24 volts x 115 amps = 2760 watts

So 4.5 amps x 24 volts = 108 watts of consumption.

2760 watts divided by 108 watts = 25.55 hours.

Of course you can exceed this capacity, but at the expense of the long-term life of your batteries and this for new batteries, which will decrease over time.

Note that your batteries seem to have a capacity of 1800 watts which would give 16.66 hours (according to my estimate).

Are they new?

[naahou2000]

Thank you for your answer
my batteries are not new, they are installed in 2002, and for this reason I am trying to estimate their autonomy. I want to know their current state (autonomy) based on the discharge measurements made and reported on the table above.
1) Can the measurement method developed in this table lead to a good estimate, if not what is the right way which allows me to better estimate my batteries.
2) At what battery voltage should the discharge be stopped so as not to damage the battery?
3) To properly measure the capacity of the batteries, should you discharge at C20, so 115/20 = 5,75 A ??
please

[Alain G]

naahou2000

Your 4.5 amp charge (not recharging) is a good indication to check your batteries, recharge the batteries thoroughly, when you have reached with this charge (4.5 amp) 10.5 volts for each battery or 21 volts for both, calculate the time to reach this voltage and you will have a good idea of ​​the capacity of your batteries.

There is no scientific method to calculate the capacity and it is important not to apply too high a load to make the measurement otherwise it will be distorted.

You take the measurement of 21 volts with the load of 4.5 amps and not empty.

the calculation is as follows: hours x volts x amps (about 100 watts on average), so simply multiply the hours to reach 21 volts per 100 watts and you will have a very good idea of ​​the capacity in watts that your batteries can contain .

Example: 100 watts x 22 hours = 2200 watts of capacity.

It all depends on the type of battery you have, forklift batteries for example can take a deeper discharge, almost flat and last very long because the cells are designed for that.
If you have the brand and type of your batteries, I would be able to tell you more about their capabilities.

I don't understand what C20 means, can you explain it to me?

[Christophe]

Ah funny this subject, I did a similar yesterday, you will find some interesting info there I think: https://www.econologie.com/forums/batterie-a ... t7407.html

By cons: be careful when you do your discharge tests not to make a deep discharge, lead batteries do not like at all especially, in your case, that they are already a certain age!

I don't understand what C20 means, can you explain it to me?

If I understood everything on the other subject: it's capacity / 20

[Alain G]

Thank you christophe I just found what it is on the web and sorry for the double subject.

C20 means 20 hours so you're right naahoo2000 it's good 5.75 amps, sorry it's a long time since I did not work in the batteries and I did not know the new standards, but I was not very far with 4.5 amps and I was not mistaken for the 10.5 volt maximum discharge which is exactly according to the standard for this battery.

But you can still calculate the time with 4.5 amps without any problem if I made myself to the specifications.

Your measurement seems to be carried out with the charger in operation, is that the case?

[naahou2000]

Thank you for your answers,
the type of my batteries is: sonnenschein

The measurements made are of course with isolated batteries.
From what I understand I must continue the measurements until I reach the voltage of 21V for the two batteries in series. Then I calculate the power dissipated by the resistance and divide it by the useful power of my equipment.
Thank you