Updated 28 May 2016
Peukert’s Law, describing battery capacity, became misunderstood following a flawed US article in the mid 1980s, here’s what Peukert really meant.
Pic: associated with http://www.youtube.com/watch?v=rWr7hdmuLUU
In 1894, Peukert explained a battery’s amp hour capacity can be expressed in terms of its ability to support a constant current load over time. His example was that, if it could sustain a current of 5 amps for 20 hours (at a still usable voltage), that battery could be rated as having a capacity of 100 amp hours.
Around 1984, a US company argued publicly (to the effect) that were that 100 amp hour battery to be subject to a 50 amp load, its stored energy would be depleted by a lot more than 50 amp hours. It referred to this as Peukert loss. This was accepted by many people. Since then the term ‘Peukert loss’ became commonly used. The argument, however, is based on a massively flawed assumption. The author has mistaken energy for power (and vice versa) throughout.
Energy and power defined
Energy is the capacity for doing work. Its base unit is the joule, but can electrically be expressed in amps. A battery’s capacity (amp hours) can be obtained by integrating current flow over time in terms of the hours of its flow.
Power is the rate at which work is done. Its base unit is the watt and is the work done, or expended, at the rate of one joule per second. It is thus totally different from energy: it relates to the rate at which energy is used.
Someone stacking two hundred 1 kg cans on a two metre high shelf (one or two at a time) exerts a certain amount of energy to do so – but not that much power (a child can do it). A weight lifter heaving up 200 kg by two metres in a second or two, however, uses the same amount of energy but needs and uses a great deal more power.
Much of the time using power when meant energy does not matter. But if misused technically, whatever follows may make no sense.
Here is an extract from the flawed US article. (The entire flawed argument is still on Wikipedia. (Google Peukert’s Law)
‘Mr Peukert first devised a formula that showed numerically how discharging at high rates actually removes more power . . . than a simple calculation would show it to do. For instance discharging at 10 amps does not remove twice as much power as discharging at 5 amps. It removes slightly more. . . discharging at higher rates removes more amp hours.’
The writer not only confused power for energy throughout the article, but in reams of flawed maths, and articles thereafter.
What Peukert really meant
When energy is drawn, battery voltage falls. In describing this, Peukert used (as an arbitrary example) that, were a battery able to sustain a current of 5 amps for 20 hours (before falling below a nominal 10.6 volts) that battery could be rated as having a capacity of 100 amp hours.
When a heavier load (such as a motor with, say, a 10.6 volt cut-out) that draws 50 amps is connected across that same 100 amp hour battery, its voltage falls at a greater rate than it would do at 5 amps draw. After (say) 60 minutes of that 50 amp draw, battery voltage is likely to fall below 10.6 volts. The battery can no longer sustain that rate of discharge . The voltage cut-out disconnects that load. That battery’s energy has at that point been depleted by 50 amp hours. No more – nor less.
Lead acid batteries have a very slow internal reaction time. Once that load is removed the battery will slowly recover. A voltage level consistent with a now 50% state of charge is around 12.3 volts. The battery is likely to be able to run that motor again for another (say) 30 minutes before the voltage cut-out operates once again. It has now delivered 75% of its available capacity. The battery’s 25% remaining capacity is still available. It is unlikely to be able to again run that motor. It does however still retain almost all of the remaining (now 25%) capacity were the load current to be reduced to 5 amps.
What Peukert really meant is that rate of discharge does not in any way affect overall capacity. That which it affects is usable or effective capacity if loaded at greater than its rated load. Full rated capacity (energy) is still available but can only be supplied at the intended (rated) current.
(The Peukert ‘loss’ argument also defies a basic law of physics. Energy cannot be ‘lost’ as such: it can only be changed into another form. So where and how, with batteries, does that ‘lost’ energy (that the US company called ‘power’ ) go to? Even heavily discharging batteries do not jump up and down, sing operatic arias or heat up by more than a degree or two.
What Peukert really meant – as an exponent
Peukert suggested that the amount of energy at different constant rates of discharge that is available from a battery can be expressed as an exponent. An (impossible) 1.0 indicates that the rate of discharge makes no difference. Most batteries thus have an exponent of about 1.1 to 1.4. (most LiFePO4’s are likely to be <1.05).
Can what Peukert really meant be proved in practice?
What Peukert really meant can readily be proven. It has been checked practically many times (down to about 20% state of charge) by myself and many others. There is a very minor heat loss (<2%) but too small to measure reliably outside a well equipped temperature controlled testing laboratory.
Checking what Peukert really meant is so easy it is strange that people on Internet forums argue endlessly about it. Take any fully charged 100 Ah lead acid or AGM battery (that is in good condition) and discharge by any load that will draw a constant 50 amps for 60 minutes. It is now discharged by 50 amp hours (to 50%). Then connect a constant load of 5 amps. You will find the battery will withstand that load for a further approximately 10 hours (i.e. a further 50 amp hours.) This video demonstrates this – at some length: http://www.youtube.com/watch?v=rWr7hdmuLUU.
Why what Peukert really meant matters
A 200 Ah battery under a 50 amp load will sustain that load for about 2.3 times longer than will an otherwise identical 100 amp hour battery. In effect it become a 230 amp hour battery (in terms of available power). A 400 amp such battery will perform much as a LiFePO4 in typical RV service.
Big AGM banks are truly effective (for RV use). If their weight is not a critical issue they are still a good, cheaper and simpler alternative if operated from 100% to 60% SoC. The occasional deeper discharge does next to no harm.
The above can be proven mathematically but that the conceptual approach (that I use here and in my books) is easier to understand by most.
Lithium-ion batteries and what Peukert really meant
Lithium-ion batteries are able to sustain their voltage under very high loads (i.e. they have a Peukert exponent of a probable 1.05). It is this that enables an 18 amp hour LiFePO4 battery to jump start a 4WD engine several times. (Engine starting needs a lot of power but surprisingly little energy – a mere 4-5 watt hours). They can likewise be charged at much higher rates (but need expertise to do so).
I acknowledge Julian and IanB (their forum names) for invaluable comments, and Tony Lee for advising of his long term RV experience with a 400 amp hour AGM battery bank.
If you found this article of interest – do please check out my book Caravan & Motorhome Electrics. It covers every aspect. It is even used by auto electricians as their main text. Our books also include the all-new Caravan & Motorhome Book, the Camper Trailer Book, Solar That Really Works (for RVs) and Solar Success (for home and property systems).
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