BATTERY BANK SIZE

[Gibbo]

Peukerts law is really explaining the fact that at higher discharge rates diffusion of battery plates is progressing too slowly, so that voltage becomes to low. To quote Reinout Vader in his book Energy Unlimited "In fact a battery which has been discharged at a very high rate will recover over time and the remaining capacity can be retrieved after the battery has been left at rest for several hours or a day"

 

Except that's partly (not completely) wrong.

 

Peukert's effect results from about 50% diffusion delay and 50% actually wasted power (as in gone forever) that electrolyses the water (yes even under discharge) and also warms the battery up. When a battery is discharged heavily it gets warm. That energy has to come from somewhere. Obviously it comes from the battery in the form of power dissipated in the internal resistance, but the internal resistance is not contant and increases as a square of the load. So the amount of energy wasted under increasing load is disproportionate to the increasing load. That energy cannot be "got back" by waiting for the battery to recover.

 

Peukert's equation was indeed formulated on the basis of a single continuous discharge run until the terminal voltage fell too low for it to be of any use. But that, as you say, does indeed allow some (not all, not even close) of the energy to be recovered by waiting a while. That's why battery monitors have to use a "fudged" Peukert effect formula to try to account for the energy that can be "got back" and that energy that has gone forever.

 

Properly separating Peukert's effect due to genuinely "gone" energy and battery recovery (surface discharge) allows better calculations to be made.

 

Gibbo

[nb Innisfree]

Except that's partly (not completely) wrong.

 

Peukert's effect results from about 50% diffusion delay and 50% actually wasted power (as in gone forever) that electrolyses the water (yes even under discharge) and also warms the battery up. When a battery is discharged heavily it gets warm. That energy has to come from somewhere. Obviously it comes from the battery in the form of power dissipated in the internal resistance, but the internal resistance is not contant and increases as a square of the load. So the amount of energy wasted under increasing load is disproportionate to the increasing load. That energy cannot be "got back" by waiting for the battery to recover.

 

Peukert's equation was indeed formulated on the basis of a single continuous discharge run until the terminal voltage fell too low for it to be of any use. But that, as you say, does indeed allow some (not all, not even close) of the energy to be recovered by waiting a while. That's why battery monitors have to use a "fudged" Peukert effect formula to try to account for the energy that can be "got back" and that energy that has gone forever.

 

Properly separating Peukert's effect due to genuinely "gone" energy and battery recovery (surface discharge) allows better calculations to be made.

 

Gibbo

 

Nice one.Very informative. I'll have to think about that one!

[Martin Nicholas]

The answer to the question: "How do I charge my lead-acid batteries?" is: "You charge them for as long as possible!". Industrial chargers have a 10-12 hour charge time which gives you an idea.

[chris w]

Industrial chargers have a 10-12 hour charge time which gives you an idea................

........................and proper multistage boat chargers too

[Colin Smith]

The capacity discharge of a battery vs current is far from being linear, it is a curve. So manufacturers quote a single point on that curve (normally, but not always) the 20 hr rate in order to be able to state some figure of merit for capacity.

 

An analogy I have used on here before is that of a car's fuel consumption. Let's say it's 25mpg. But we all know that it's not 25mpg at all speeds and we know also that the fuel consumption is not linear. If it's quoted as 25mpg at a steady 40mph, we would not expect it to be 50mpg @ 20mph or 12.5mpg @ 80mph. It will be a curve.

 

So too with a battery, the quoted say 110AH at the 20hr rate is just the same idea as 25mpg @ 40mph for example. It means something but it certainly doesn't mean everything. It means that at a discharge current of 110/20 amps (5.5A) the battery will indeed discharge fully in 20 hours. But it doesn't mean that at 100A it will discharge in 110/100 = 1 hour and 6 minutes. This would be the answer if battery discharge curves were actually straight lines. But, because they're not, we have to employ slightly more complicated maths to work out the answer. For small deviations from the 5.5A above (110/20), the difference in discharge time is insignificant. But for larger currents, it becomes very significant.

 

So, for a 100A discharge the real number is (110/20)^1.3 x 20 / 100^1.3 = 0.46 hours or 28 minutes

 

This is only 42% of the time predicted by the simple 110/100 calculation above - a huge difference. The "1.3" figure in the calculation is known as the "Peukert factor" and will vary slightly from 1.3 depending on the exact composition of the battery but 1.3 is a good figure to use to represent a wet lead-acid battery. Plotting results using the Peukert factor would reproduce the true discharge curve for the battery rather than the straight line implied by the 20 hour rate figure.

 

Using the true maths shows that a 100AH battery is NOT 33% larger in capacity than a 75AH battery of the same hour-rate (ie: 100/75) but actually:

 

(100/75)^1.3 = 45.4% larger in capacity. Ergo, that's an additional 9% gain in capacity (1.454/1.33) by using the larger battery that didn't come from the simple 100/75 calculation.

 

That's why it's much better to keep your domestic bank as one large bank rather than splitting it into a couple of smaller banks to run different systems on the boat as some people do. You end up with much more capacity "for free" if you connect them all together.

 

Also beware of specsmanship: a 110AH battery at the 20hr rate could instead be legally quoted at the 40hr rate and the capacity would be 130AH at this rate. You can bet the 130AH figure would be in large bold lettering whilst the 40 hr rate would be in small print!!!

 

Chris

 

Hi Chris,

 

Re: "So, for a 100A discharge the real number is (110/20)^1.3 x 20 / 100^1.3 = 0.46 hours or 28 minutes"

 

For some reason I can't get your calculation to work for me in a spread sheet?? Probably my cack-handed effort with formulas! Do you know of a spreadsheet anywhere that I could download that includes this calculation?

 

Cheers,

 

Colin

[chris w]

Hi Chris,

 

Re: "So, for a 100A discharge the real number is (110/20)^1.3 x 20 / 100^1.3 = 0.46 hours or 28 minutes"

 

For some reason I can't get your calculation to work for me in a spread sheet?? Probably my cack-handed effort with formulas! Do you know of a spreadsheet anywhere that I could download that includes this calculation?

 

Cheers,

 

Colin

Your calculation is spot on Colin. I'll work out the formula syntax for Excel and post it on here but it won't be till Wednesday.

 

Chris

[Colin Smith]

Your calculation is spot on Colin. I'll work out the formula syntax for Excel and post it on here but it won't be till Wednesday.

 

Chris

 

Thanks Chris, very kind of you!

 

Cheers,

 

Colin

[chris w]

Here it is Colin - managed to find some time:

 

+(($D$3/20)^1.3*20/E3^1.3)*60

 

 

AH

110

 

current minutes

100.............28

90...............32

80...............36

70...............44

60...............54

50...............68

40...............90

30...............132

20..............224

10..............552

 

 

 

If you want to keep the battery size constant, eg 110AH and plot the discharge time in minutes against various currents, then you need to keep the AH constant by enclosing it in $ signs.

 

 

In this formula, square D3 contains the AH of the battery, eg: 110. You must write $D$3. The "$" sign tells Excel to always go to the same square, ie: D3 is a constant.

 

Column E contains the current list in the formula with E3 containing 100A in this formula. So 90A will be in E4 etc. Just insert the formula in F3 and then copy and paste to the other rows. Your error may have been in not using the $ signs to assign a constant. If you don't do this, then on the next row after the first, the program will look in D4 instead of D3 for the battery AH figure.

 

Note also (I should have put this in my previous post) that you should only discharge to 50% to avoid damaging the battery. Therefore the formula above should contain a factor of 2 for a real-life situation as follows:

 

+(($D$3/20)^1.3*20/E3^1.3)*60/2

 

so the actual table should be:

 

AH

110

 

current minutes

100.............14

90...............16

80...............18

70...............22

60...............27

50...............34

40...............45

30...............66

20..............112

10..............276

 

 

 

Chris

[Colin Smith]

Here it is Colin - managed to find some time:

 

+(($D$3/20)^1.3*20/E3^1.3)*60

 

 

AH

110

 

current minutes

100.............28

90...............32

80...............36

70...............44

60...............54

50...............68

40...............90

30...............132

20..............224

10..............552

 

 

 

If you want to keep the battery size constant, eg 110AH and plot the discharge time in minutes against various currents, then you need to keep the AH constant by enclosing it in $ signs.

 

 

In this formula, square D3 contains the AH of the battery, eg: 110. You must write $D$3. The "$" sign tells Excel to always go to the same square, ie: D3 is a constant.

 

Column E contains the current list in the formula with E3 containing 100A in this formula. So 90A will be in E4 etc. Just insert the formula in F3 and then copy and paste to the other rows. Your error may have been in not using the $ signs to assign a constant. If you don't do this, then on the next row after the first, the program will look in D4 instead of D3 for the battery AH figure.

 

Note also (I should have put this in my previous post) that you should only discharge to 50% to avoid damaging the battery. Therefore the formula above should contain a factor of 2 for a real-life situation as follows:

 

+(($D$3/20)^1.3*20/E3^1.3)*60/2

 

so the actual table should be:

 

AH

110

 

current minutes

100.............14

90...............16

80...............18

70...............22

60...............27

50...............34

40...............45

30...............66

20..............112

10..............276

 

Chris

 

 

Fantastic! Thanks Chris - owe you a beer for that!

 

I've been playing around with this and got some very interesting results for the bank that I have. Interestingly enough, the figures provided by my battery supplier compared to the calculations using your formula are almost identical.

Next thing Im going to do is look at the charging times required, for a given drain on the battery pack, by charging method - alternators via the Sterling controller or shore-power charger. This should keep my brain busy for a bit!!

 

Thanks again,

 

Colin