slower discharge rates remove more energy from a battery than faster discharge rates?

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zappy

Senior Member
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.
 

iwire

Moderator
Staff member
Maybe in total Amp Hours.
Bingo.

If you draw 100 amps for an hour you will drain the battery more than if you draw 10 amps for 10 hours.

The higher the draw the more the battery temp raises, the energy to raise the temperature is coming from the battery so it is discharged quicker.
 

charlie b

Moderator
Staff member
Bob, that makes sense, but it is exactly the opposite of what the title of this thread is stating. I think that that is why Zappy is asking the question. :confused:
 

winnie

Senior Member
My understanding is slightly different from Bob's, but the conclusion is the same.

If you discharge the battery at higher current, you get more voltage drop across the internal resistance of the battery.

For the same amp hours used, a greater % of the watt hours gets lost in the internal resistance rather than being delivered to the load.

On top of this, the internal voltage drop means that you reach your 'end of discharge' voltage sooner, meaning that you will use fewer amp hours available from the battery.

As I understand it, if you draw 100 amp hours from a battery, you will use up essentially the same amount of chemicals in the battery, no matter if it is a fast or a slow discharge. So 100 amps for 1 hour will 'drain' the battery essentially the same amount as 10 amps for 10 hours, however as noted above less energy is delivered to the load.

-Jon
 

ELA

Senior Member
This is what the Power Sonic technical manual states with respect to what you asked:
*********************
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.

***************************
 

wptski

Senior Member
This is what the Power Sonic technical manual states with respect to what you asked:
*********************
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.

***************************
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.
 

winnie

Senior Member
Isn't it that resistance that cause the temp to rise?
Exactly. The energy is being dissipated in the resistance, causing the temperature to rise, rather than being delivered to the load.

Like I said, we have a slightly different understanding of the situation, but reach similar conclusions.

As I understood what you said: If you discharge the battery at higher rates, 100 amp-hours delivered to the load will use up more of the charge stored in the battery, with the difference being wasted as heat in the battery.

What I said: If you discharge the battery at higher rates, 100 amp hours delivered to the load will use up the same amount of stored charge, but deliver a smaller fraction of the energy to the load, with the difference being wasted as heat in the battery.

-Jon
 

winnie

Senior Member
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 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
 

ELA

Senior Member
wptski,
winnie does a great job of explaining this all in laymens terms and yes they are Sealed Lead Acid batteries.
I was just providing a link to a more technical description for those interested. I should have provided the entire link as it is a good read here:


http://www.power-sonic.com/index.php?id=42

Under SLA there is a technical manual that provides some good info.
 

wptski

Senior Member
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
Yes, I guessed they were speaking of SLAs. The opposite is true when charging, a slow charge is better than a fast charge.
 

swei

Member
All battery manufacturers will quote capacity (in AH) at various discharge rates. Only a few will actually supply Peukert's Exponent, which will allow you to estimate capacity at any rate of discharge.

Slower charging alone is definitely not better for deep cycle lead-acid batteries, as it will not sufficiently agitate the plates to break up the lead sulfate. You really need at least C/10 to get things moving inside. SLAs differ by construction in their requirements (and often have specific setpoint requirements - best to check with the manufacturer before you buy a charger.
 

GeorgeB

ElectroHydraulics engineer (retired)
Yes, I guessed they were speaking of SLAs. The opposite is true when charging, a slow charge is better than a fast charge.
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!.
 

BJ Conner

Senior Member
"... 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.
 

zappy

Senior Member
"... 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.
I was at Pier 39 in S.F. They have a submarine battery on display, it said the thing weighed 100,000lbs!
 

LarryFine

Master Electrician Electric Contractor Richmond VA
I was at Pier 39 in S.F. They have a submarine battery on display, it said the thing weighed 100,000lbs!
Of course, it's a submarine battery! :roll:





(You don't expect a battery that heavy to float, do you? :D)
 

wptski

Senior Member
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!.
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.
 

Electric-Light

Senior Member
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.
Battery capacity in Ah is rated in "20 hour rate", or C/20.

So a capacity of 10Ah is achieved by discharging it at 10/20 or 0.5A for 20 hrs.

At a discharge rate of C, or 10A, I believe the actual usable capacity is something like 5Ah.

In theory, paralleling two 10Ah batteries doubles the usable capacity, but when the discharge rate is high, it will more than double it, because 10A load = 1C on a 10Ah battery, but on a 20Ah battery, it is 0.5C.

The usable relative capacity at 0.5C is greater than at 1.0C rate.
 
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