Jump to content

Effective battery discharge rate


David HK
 Share

Featured Posts

But you are asking an impossible question. Its not like emptying a bucket, the discharge rate will fall as the capacity diminishes.  You cannot draw 100A all the time, it will fall off quite quickly C5 is the 5 hour discharge figure. Which would suggest seven and a half hours but you will never achieve that!

Link to comment
Share on other sites

The C5 figure for 750Ah means you discharge at 150A and it’s flat after 5 hours. There is no data to show how long it would last at 100A, but more than 7.5 hours (ie more than 750Ah). One could probably do a calculation using an estimated Peukert value, but that is a bit complicated. I’d say around 8 to 8.5 hours.

Link to comment
Share on other sites

12 minutes ago, nicknorman said:

The C5 figure for 750Ah means you discharge at 150A and it’s flat after 5 hours. There is no data to show how long it would last at 100A, but more than 7.5 hours (ie more than 750Ah). One could probably do a calculation using an estimated Peukert value, but that is a bit complicated. I’d say around 8 to 8.5 hours.

 

Thank you for exaclty the form of answer I was looking for.  I was a little confused looking at conversion calculations where it meantioned how dividing by 0.88 gave a C10 rate which then showed 852 amps. However all is clear in that I guess what it was actually describing was the more you suck out, the less it lasts.  Deep cycle LA Forklift 2v cells are normally fine on the 20% to 80% curve and there seems an awful lot of cheap cells about. We were going with LIFEPO4 but I think at the way prices are going we can wait awhile fot that route and save a ton of cash in the budget. Thanks again.

Link to comment
Share on other sites

If the load is purely resistive, you will have to keep increasing it to maintain 100A because as the battery discharges, the voltage drops which causes the current to diminish.

 

If the load comprises of say switch mode power supplies (SMPS), the current will increase as the voltage falls because the SMPS' will adjust to maintain a constant power.

Link to comment
Share on other sites

These figures are sort of conceptual because they are a way to specify a battery rather than something to actually do.

There is a question to which I don't have an answer, or the inclination to experiment.

"Flat" is defined as some voltage, maybe 11 or 11.5, I can't remember.

If we discharge a battery under heavy load its gets to this flat voltage sooner because its terminal voltage is reduced due to the load...

So if we discharge a battery at C5 till it is almost flat, then take the load off the voltage goes up and we can discharge it more at a lower rate.

 

How much of the reduced capacity at high load is due to real reduced capacity, and how much is only due to the lower terminal voltage????

Link to comment
Share on other sites

21 minutes ago, dmr said:

These figures are sort of conceptual because they are a way to specify a battery rather than something to actually do.

There is a question to which I don't have an answer, or the inclination to experiment.

"Flat" is defined as some voltage, maybe 11 or 11.5, I can't remember.

If we discharge a battery under heavy load its gets to this flat voltage sooner because its terminal voltage is reduced due to the load...

So if we discharge a battery at C5 till it is almost flat, then take the load off the voltage goes up and we can discharge it more at a lower rate.

 

How much of the reduced capacity at high load is due to real reduced capacity, and how much is only due to the lower terminal voltage????

10.5v for lead acid, normally. Regarding your last sentence the apparent reduced capacity is due to the battery’s inability to supply the load current at a useful voltage at low SoC. As you suggest, the actual capacity in terms of Ah is not reduced if subsequent discharge is much slower. Of course the capacity in terms of watt-hours is very much reduced due to the lower voltage.

Link to comment
Share on other sites

So are you proposing that amp-hours is a fixed quantity defined by chemistry? If the watt-hours are reduced (which they must be to some extent) this must be manifested as heat?.

I propose tat batteries have a two component internal resistance, a bog standard ohmic resistance (which is very low) due to the metal bits inside the battery and a more complicated and non ohmic "chemical resistance". This is responsible for most of the lost voltage on discharge (and charge) and gives the exponential time response. Even though its non ohmic it must still generate heat.

 

Sitting here on the very upper Thames in sunshine, so a great place to think about batteries.😀

Link to comment
Share on other sites

26 minutes ago, dmr said:

Sitting here on the very upper Thames in sunshine, so a great place to think about batteries.😀

 

Ah so you turned right at Oxford after all! 

 

I'm just off to Reading shortly, to clean out a horribly and completely sooted up floor standing Potterton boiler about 25 years old. I think I might just about be able to see the sun through the haze of soot which might become suspended in the air, unless the cloth filter in my Henry turns out to be really good...

 

 

 

 

Link to comment
Share on other sites

36 minutes ago, dmr said:

So are you proposing that amp-hours is a fixed quantity defined by chemistry? If the watt-hours are reduced (which they must be to some extent) this must be manifested as heat?.

I propose tat batteries have a two component internal resistance, a bog standard ohmic resistance (which is very low) due to the metal bits inside the battery and a more complicated and non ohmic "chemical resistance". This is responsible for most of the lost voltage on discharge (and charge) and gives the exponential time response. Even though its non ohmic it must still generate heat.

 

Sitting here on the very upper Thames in sunshine, so a great place to think about batteries.😀


On the “chemical resistance” this is a bit above my pay grade, since the actual chemical reaction is surprisingly complicated. I think the point is that whilst the chemistry mobilises a fixed number of electrons (the capacity in Ah), the potential at which they are mobilised depends on the reaction rate, the concentration of the chemicals (SoC) and the localised concentration of chemicals at the plates. So it see it more that the reaction mobilises electrons with lower potential and thus heat is not dissipated.

 

However, that of course flies in the face of conservation of energy, so it must be wrong! But is the loss of Wh capacity transferred to heat gain in the battery? I wouldn’t have thought so, even though batteries do have high thermal mass. It’s a conundrum!

Link to comment
Share on other sites

59 minutes ago, dmr said:

So are you proposing that amp-hours is a fixed quantity defined by chemistry? If the watt-hours are reduced (which they must be to some extent) this must be manifested as heat?.

I propose tat batteries have a two component internal resistance, a bog standard ohmic resistance (which is very low) due to the metal bits inside the battery and a more complicated and non ohmic "chemical resistance". This is responsible for most of the lost voltage on discharge (and charge) and gives the exponential time response. Even though its non ohmic it must still generate heat.

 

Sitting here on the very upper Thames in sunshine, so a great place to think about batteries.😀

Correct, any difference between what you put into a battery and what you take out must appear as heat, it can't magically vanish down some chemical wormhole since the reactions are reversible over many charge/discharge cycles and conservation of energy applies.

 

These losses are quite large for lead-acid batteries on both charge and discharge, hence the Peukert effect (and why they get hot at high currents) -- depending on the battery design and charge/discharge rates, you only get back maybe 60% or what you put in, or even less at high rates -- and for lead-acids, "high" means anything above a small fraction of C. For example, looking at the Leoch lead-carbon batteries (which are better at high rates than most lead-acid batteries), you get the following for the 210Ah (10h rating) battery:

 

Leoch.png

 

The power/energy loss fits quits well to a constant internal resistance up to about 1.5C rate and then starts to fall apart -- not surprising since by this time half the energy has been lost. This resistance will be from all causes including electrical resistance but also the energy losses in moving ions around inside the cells, there's no easy way to say which is which.

 

With LiFePO4 the losses are much lower (round-trip efficiency typically >90%) and you get full capacity up to rates of 1C or higher, depending on construction.

Edited by IanD
Link to comment
Share on other sites

So, does the last post actually mean anything to 99.999% of people ? Looks a bit like ego preening. 

For me, Just going to load up a stack of batteries and see if I need more .  The Paracetomol effect ( because electrical always gives me a migrane)  or any other sciencey stuff is irrelevant to what I need. I just wanted to make sure I understood what I was reading on the label. No PhD involved.

  • Greenie 1
Link to comment
Share on other sites

7 minutes ago, David HK said:

So, does the last post actually mean anything to 99.999% of people ? Looks a bit like ego preening. 

For me, Just going to load up a stack of batteries and see if I need more .  The Paracetomol effect ( because electrical always gives me a migrane)  or any other sciencey stuff is irrelevant to what I need. I just wanted to make sure I understood what I was reading on the label. No PhD involved.

I was answering the question from dmr. If you think that's ego preening, that's your I-can't-understand-sciencey-stuff problem not mine 😉

Edited by IanD
  • Greenie 1
Link to comment
Share on other sites

4 minutes ago, David HK said:

I have seen many of your other posts so I think I will go with my original thought on the post.

Goody for you. Some people on the forum actually enjoy discussing stuff like this even if you don't...

Link to comment
Share on other sites

36 minutes ago, IanD said:

Goody for you. Some people on the forum actually enjoy discussing stuff like this even if you don't...

Most of this stuff about batteries is highly testicle greek to me at first reading.

However,on re-reading several times slowly,some chinks of light appear.

Keep it up ianD, I for one enjoy being educated.  😄

Link to comment
Share on other sites

1 hour ago, IanD said:

I was answering the question from dmr. If you think that's ego preening, that's your I-can't-understand-sciencey-stuff problem not mine 😉

 

I find your posts well reasoned, but can be a bit difficult to grasp at first reading. Keep it up, especially in the face of commercial interests.

Link to comment
Share on other sites

4 minutes ago, Tony Brooks said:

 

I find your posts well reasoned, but can be a bit difficult to grasp at first reading. Keep it up, especially in the face of commercial interests.

Thanks. I know my posts are often long and detailed, but a lot of these issues are complicated and a simple answer can just be misleading and raise a lot of questions...

 

"For every complex problem there is an answer that is clear, simple, and wrong" -- H. L. Mencken

Edited by IanD
Link to comment
Share on other sites

7 hours ago, IanD said:

Correct, any difference between what you put into a battery and what you take out must appear as heat, it can't magically vanish down some chemical wormhole since the reactions are reversible over many charge/discharge cycles and conservation of energy applies.

 

These losses are quite large for lead-acid batteries on both charge and discharge, hence the Peukert effect (and why they get hot at high currents) -- depending on the battery design and charge/discharge rates, you only get back maybe 60% or what you put in, or even less at high rates -- and for lead-acids, "high" means anything above a small fraction of C. For example, looking at the Leoch lead-carbon batteries (which are better at high rates than most lead-acid batteries), you get the following for the 210Ah (10h rating) battery:

 

Leoch.png

 

The power/energy loss fits quits well to a constant internal resistance up to about 1.5C rate and then starts to fall apart -- not surprising since by this time half the energy has been lost. This resistance will be from all causes including electrical resistance but also the energy losses in moving ions around inside the cells, there's no easy way to say which is which.

 

With LiFePO4 the losses are much lower (round-trip efficiency typically >90%) and you get full capacity up to rates of 1C or higher, depending on construction.

If I'm reading this table correctly it seems the chemistry for PbC cells is slightly better than for Lead-acid: you get back 100/106% of the energy put in.

Somehow I doubt it's that simple in reality.

Link to comment
Share on other sites

29 minutes ago, George and Dragon said:

If I'm reading this table correctly it seems the chemistry for PbC cells is slightly better than for Lead-acid: you get back 100/106% of the energy put in.

Somehow I doubt it's that simple in reality.

No, this table is just looking at effective capacity (starting from fully charged) at different discharge rates, to see how much capacity drops at higher currents.

 

These 210Ah cells are specified at C10 (210Ah=100%) by which time there's already some capacity drop, at C100 they'd be about 106% (223Ah). I used these PbC cells as an example because detailed numbers are available, it's harder to find them for traction cells but these have worse losses at higher discharge rates than PbC.

 

On top of this there's another set of losses when charging the cells, both due to charging voltage being higher than discharge voltage and the gap between the two increasing with current -- and again at higher currents you have to put more energy in, the loss numbers are probably similar to the discharging ones.

 

Plus lead-acid cells have to be equalised to get them fully charged and prevent sulphation, especially flooded traction cells which are designed to have this regularly, and this wastes more energy (and generator running time) -- PbC cells are supposed to be better with this (partial SoC operation) and have lower losses at higher currents but still need occasional equalisation charging, every week or two IIRC.

 

Put all these losses together and you'll be doing well to get 70% of the energy that you put in back out with lead-acid batteries, in fact 60% is a figure often used for traction cells -- PbC will be a bit better, they might get to 70%. Lithium cells not only have much better performance at high currents and smaller charge-discharge voltage difference but also don't need long equalisation charges, 90% round-trip efficiency is commonly used.

Edited by IanD
  • Greenie 1
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.