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#### Gaber Mohamed Boraey

##### Full Member level 4
Hello
Here is a lead acid battery used for UPS , please look at the specifications here

Here is the datasheet for similar battery

In common, I see these batteries with ampere hour capacity shown on the battery , except this one as you see, only 34w/cell/15min/1.67v and that it’s 12v battery
I understand from this that the cell voltage is 1.67 v, and the power is 34w for 15min from one cell

I’ve tried to calculate but not get right estimations
The power = voltage * current, 34w*6 = 204w is the power taken from all battery for 15 mins?
If so, then power taken for 1 hour = the power for 15 mins multiplied by 4 = 816 watt /hr
The battery voltage is 12v
So the capacity ampere .hour = 816/12= 68 AH and not 9Ah

Can you tell how the 9AH capacity calculated ?

I believe the manufacturer specifies that current draw stops when the 12V battery reaches 10.5 V. Then it has delivered 9 A-Hrs (?). All advice says to recharge it at that point.

However if we drain the battery all the way to 0V then it might deliver the higher figure eventually. Depleting it entirely (even one time) reduces the L-A battery's useful life.

Then it has delivered 9 A-Hrs (?). All advice says to recharge it at that point.
No calculations for the 9 AH capacity?

Hi,

the Ah value is litterally given in the datasheet. Indeed multiple Ah values for different discharge levels.

The only thing you need to know is that 1h = 60 min.

Then look at the table "Constant Current Discharge Characteristics Unit:A"
Column "60 min" (= 1h)

Example: If you discharge it down to 1.70V cell voltage ( = 6 cells x 1.70V = 10.2V battery voltage), then you can draw a constant current of 5.57A.
--> 5.57A for 1h = 5.57Ah

Klaus

34W @ 15min-rate to 1.67V per cell @ 25'C spec 204 W in 1/6h = 34 Wh

17.4 A * 1/6h = 2.9 Ah when used at a rate of 120C = 15 minutes
When used at a rate of 20 h = 1C the capacity is given = 9Ah
Therefore 15 MIN efficiency is 2.9/9 = 32% which is good for Absorbent Glass Mat (AGM) type batteries.

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If so, then power taken for 1 hour = the power for 15 mins multiplied by 4 = 816 watt /hr
Ah = Ampere x hours.
Wh = Watt x hours.
15 min = 0.25h

so if you have 204W for 15min then you get
204W x 0.25h = 51Wh

if nominal battery voltage is 12V, then
Ah = Wh / V
Ah = 51W / 12V = 4.25Ah

What the datasheet clearly shows is that different discharge conditions like
* discharge level
* discharge current
* and so on
result in different usable Ah values.
--- Updated ---

34W @ 15min-rate to 1.67V per cell @ 25'C spec 204 W in 1/6h = 34 Wh
15 min = 1/4h = 0.25h
so 204W x 0.25h = 51Wh

Klaus

Must be my cognitive dissonance to make the same mistake as they for 34Wh

Hi,

Must be my cognitive dissonance to make the same mistake as they for 34Wh
The datasheet does not talk about Ah at all ... so there is nothing about 9Ah.
I´d rely on the datasheet value and forget about 9Ah for the time being.

W = V x A. So far so good.
But during discharging and charging the voltage is not constant. It even is not linear.
Thus when we talk about Energy in the unit of Wh we can not directly calculate the voltage nor the current.

Klaus

Must be a magician's deception or my senior's moment to make the same mistake to get 34Wh using 1/6th hour.
Take2:
For 15 minutes = 17.4 Ah / 4 = 4.35 Ah vs. spec. 9 Ah = 48 % surge efficiency ( even better now )
This can only be used for constant current loads/

Yet it would degrade from start to end-of-life (EOL) after 200 recharge cycles @ 100% DOD using plotted min. value.

For the LED industry, they use LM70 or 70% of original Lumen Maintained to determine MTBF, assuming no defects.

What would be the Wh rating in 15 min be for the last cycle? 70% of initial? lower ??

They say

• which reads as 10% to 25% of the initial Pd rating at 1.70V /cell cutoff.
• meaning from "pretty to really dead" at EOL

Only 3rd party reported 9Ah

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