Can someone puts this in lameman's terms. It sounds like it should be the opposite.
I read this in my photovoltaic systems book. Thank you for your help.
Maybe in total Amp Hours.
but it is exactly the opposite of what the title of this thread is stating.
If you discharge the battery at higher current, you get more voltage drop across the internal resistance of the battery.
What kind of battery are they refering to? Both say discharged to zero volts yet one says 1.5V/cell and the other 1.75V/cell!! I know that most cells can be discharged to 0V but when the load is removed the voltage will increase.This is what the Power Sonic technical manual states with respect to what you asked:
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Deep Discharge
Power-Sonic defines ?deep discharge? as one that allows the battery voltage under load to go below the cut-off (or
?final?) voltage of a full discharge. The recommended cutoff voltage varies with the discharge rate. Table 1 shows the
final discharge voltages per cell.
It is important to note that deep discharging a battery at high rates for short periods is not nearly as severe as
discharging a battery at low rates for long periods of time. To clarify, let?s analyze two examples:
? Battery A ? Discharged at the 1C rate to zero volts.
?C? for a 4 AH battery, for example, is 4 amps. Full discharge is reached after about 30 minutes when
the battery voltage drops to 1.5V/cell. At this point, only 50% of rated capacity has been discharged (1
C amps x 0.5 hrs = 0.5C Amp. Hrs). Continuing the discharge to zero volts will bring the total amount of
discharged ampere-hours to approximately 75% because the rapidly declining voltage quickly reduces
current flow to a trickle. The battery will recover easily from this type of deep discharge.
? Battery B ? Discharged at the 0.01 C rate to zero volts.
0.0IC for a 4 AH battery is 40mA. Full discharge is reached after 100+ hours when the terminal voltage
drops to 1.75 V/cell. At this point, the battery has already delivered 100% of its rated capacity (0.01 x
100 hrs = 1C Amp. Hrs.). Continuing the discharge to zero volts will keep the battery under load for a
further period of time, squeezing out every bit of stored energy.
This type of ?deep? discharge is severe and is likely to damage the battery. The sooner a severely
discharged battery is recharged, the better its chances to fully recover.
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Isn't it that resistance that cause the temp to rise?
What kind of battery are they refering to? Both say discharged to zero volts yet one says 1.5V/cell and the other 1.75V/cell!! I know that most cells can be discharged to 0V but when the load is removed the voltage will increase.
Yes, I guessed they were speaking of SLAs. The opposite is true when charging, a slow charge is better than a fast charge.This looks like a description of lead acid cells.
Under high discharge rate, you can reach the point where the external cell voltage is 0V, but there is actually considerable chemical energy still stored in the cell; all of cell voltage is being used to overcome internal resistance, with nothing left for the external load. If the load is removed, then the voltage recovers and the battery is not damaged.
If the cell is discharged to 0V at low current, then you actually use up _all_ of the stored chemical energy.
Since discharging the cell actually involves moving chemicals around which act as both energy storage, electrode, and structure, this sort of complete discharge can severely damage some (but not all) cell types.
-Jon
Whoa, not necessarily. I believe you will find that a modern 3 or 4 mode charger needs to provide at least C/10 for the initial (bulk) charge ... the battery engineers I've spoken to say that C/5 is better.Yes, I guessed they were speaking of SLAs. The opposite is true when charging, a slow charge is better than a fast charge.
"... slower discharge rates remove more energy from a battery than faster discharge rates? "
Slower discharge rates may remove more energy from the battery but not as much power ( the time the energy is released). If you want to get your submarine to deep water fast you don't care if you get all the energy out of your battery, you want a lot of it NOW.
Batteries are more complicated than dividing the AmpHour rating of the batter by the hours and figuring that's the amps you get for that duration.
The sizing for large batteries (lead acid) for power plants involved first developing a load pofile ( amps load vs time )and then using the Hoxie method to size the batteries.
Available battery output will vary over time for the reasons stated above and for physical charesterics of the electrolyte in the case. The fluid next the plates changes first and fluid in the bottom or sides of the jar discharge later. Thats a simple answer. I remember reading that some Navy batteries had forced electrolyte circulation. The power that is stored in a battery is in the electroyte (all of it), you get it out from the reaction that occurs at the plates.
There is lots of info on the web about submarine batteries. A lot of the science was developed for Submarines.
http://www.fleetsubmarine.com/battery.html
If get get around a "Squid" who was an electrican or engineering officer and want to know about batteries most of them know the subject very well.
Of course, it's a submarine battery! :roll:I was at Pier 39 in S.F. They have a submarine battery on display, it said the thing weighed 100,000lbs!
It depends what type of battery your refering to. They've had Ni-MH cells charging at .1C for over two years with no damage. With these cells, a slow timed charge is always the best. It's also the best way to "try" to correct a unbalanced pack.Whoa, not necessarily. I believe you will find that a modern 3 or 4 mode charger needs to provide at least C/10 for the initial (bulk) charge ... the battery engineers I've spoken to say that C/5 is better.
Remember that these smart chargers charge at current limit to a voltage, switch to a constant current mode to another voltage, then float at a temperature compensated level ... if they are the good ones.
It is not good to put a C/50 or C/100 on and leave it. In addition to ultimate damage, that often will not fully charge it either!.
Can someone puts this in lameman's terms. It sounds like it should be the opposite.
I read this in my photovoltaic systems book. Thank you for your help.