On our most recent camping trip (
http://campingwithsolar.blogspot.com.au/2015/04/our-first-real-usage-of-solar-at.html) I monitored the number of Amps captured by our solar panels and the power consumed by the fridge, LED lighting and other miscellaneous connections such as charging the mobile devices over a period of 3 days. By the end of the 3rd day we had recorded 48.688A being captured by the sun and 44.038A being consumed by all devices. With all things being equal this would suggest that the state of charge of the battery at the end of the 3 days should be roughly the same as it was when we started, if not a little higher. Looking back, on day 1 we started with 91% available capacity but at the of the third day we only had 80% charge remaining in the battery. Where did I lose 11% of my total capacity?
Just in case you're not a numbers person let's try and put some perspective into why understanding this is so important to me. We have two 100AH batteries in parallel, giving us the equivalent of 200AH. An 11% loss equates to 22AH so if our fridge draws roughly 2A per hour then I have lost roughly 10 hours of running time. That's huge.
Nerd Alert
In my subsequent research I found some very interesting information that I had not known below. I always thought that amp-hours was simply a unit that described a quantity of current over a one hour period. Well, it's not quite as simple as that.
In Physics, the measure of charge is the Coulomb, which is 6.24 x 10
18 electrons. And it's electrons that are stored in a battery (this much I knew).
Now, Q = I * t where Q is the charge in Coulombs, I is the current in amps and t is the time in seconds. This means that if a wire is conducting 1 amp in 60 seconds then this is 60 Coulombs of charge per minute, which equates to 3600 Coulombs per hour.
As it turns out, engineers become frustrated trying to work with the number 3600 when calculating amps or hours for a given number of Coulombs so they invented the unofficial unit now commonly known as
amp-hours. And to make things just a little more confusing the hyphen means times. Yep, amp-hours means
amps times hours.
Someone thank the engineers, seriously. When I'm working with my 100AH batteries I know this means I'll get (theoretically) 100 hours of battery with a constant load of 1 amp. I'd hate to be working with Coulombs as my units.
Coming back that 'theoretical' comment, read on for a little more insight into how battery capacities are defined by manufacturers.
The C-rate
I recently purchased as Sealed Lead Acid battery that was rated as "7.2AH/20". I knew what 7.2AH meant but had no idea what the value of 20 referred to. I've since discovered it is the manufactures way of describing how they arrived at the rating of 7.2AH.
A battery rated at 7.2AH could be interpreted as being able to deliver 1 amp over 7.2 hours, or even 7.2 amps over a 1 hour period. As I'll be describing in more detail later this is not actually achievable. The value of 20 indicates the C-rate, also known as the discharge current rate. For this battery, the manufacturer discharged the battery over a 20 hour period and determined what load was required until it had reached what is known as the "cut off voltage" - the voltage at which the battery is considered "empty". So, in reality, the manufacturer has determined that this battery reached the cut off voltage after 20 hours while drawing a load of 360 mA (0.36A x 20 hours is equivalent to 7.2AH).
As an aside, I don't know what the cut off voltage is for this 7.2AH battery but from what I've read my 12v 100AH lead acid batteries are considered full at 12.7v and empty at 11.9v. This is handy to know because based on this I can determine the capacity of my batteries using some simple math:
Capacity (%) = 125 * (volts - 11.9)
But I digress. All this C-rate stuff is important to know, so read on.
Contributors of lost capacity
Learning about the C-rate didn't help explain my 11% loss but at this point I'd like to mention some factors that contribute to loss of battery performance over time.
- The age of the battery
- The temperature
- Constant charge and discharge
- The rate of charge
- Component corrosion
- Electrical shorts
- Vibrations
- Under and over charging
One of the biggest "rule of thumb" to keep in mind is to always try and prevent the battery from discharging any more than 50%. The more you discharge a battery, the more you degrade the lifetime of the battery.
Recharge rate
Different battery types have different charging requirements so be sure to refer to technical datasheets associated with the battery you're using. My batteries require a charge rate no greater than 10% of the battery capacity. The best recommendation I can make is to purchase a good quality charger suitable for the batteries you're using. I'm using a Ctek 250S-Dual when connected to solar and an ArkPak when sitting at home in the garage. These are 5 and 7 stage chargers, respectively.
At a minimum, you should be looking at a charger that applies at least these 3 stages:
Bulk charge - In this phase a constant current is supplied at a rate determined by the charger (such as the 10% rate I mentioned earlier. This current is applied until the battery reaches approximately 70% capacity.
Absorption charge - In this phase the charger maintains a constant voltage to the battery and the charging current decreases as the battery approaches full charge.
Float charge - In this phase the charger maintains the battery at full charge.
The 5 and 7 stage chargers apply additional techniques, such as desulphation (the removal of sulphate crystals that form as a result of not being fully charged over a period of time), to help extend your battery life. It's worth spending a little extra money on a good charger.
Peukert's Law
After loads of reading I finally stumbled across what is known as Peukert's Law. Finally, some more nerdy stuff.
In 1897 a German scientist by the name of Wilhelm Peukert expressed the capacity of a battery in terms of the rate at which is discharged. Essentially, when thinking about a 100AH battery it is easy to assume this will deliver 1 amp over 100 hours, or 100 amps over 1 hour. Unfortunately, this is not the case. In reality, as the rate of discharge increases, the battery's available capacity decreases.
Earlier I spoke of the C-rate. As it turns out, this information is relevant to Peukert's Law. For example, a 100AH battery with a C-rate of 20 will fully discharge over 20 hours when the discharge current is 1 amp,
Peukert's Law describes the relationship between discharge current (normalized to a base rated current, such as the above-mentioned 1 amp) and delivered capacity (normalized to the rated capacity) over some range of discharge currents. In simple terms, Peukert's Law describes how a battery discharged over a shorter time with a higher current results in reduced capacity.
Mathematically, Peukert's Law is stated as the following:
Cp = Ikt,
where:
Cp is the capacity at a 1 amp discharge rate (expressed in amp-hours)
I is the actual discharge current (load) (expressed in amps)
t is the actual time to discharge the battery (expressed in hours)
k is the Peukert constant
From what I can gather, this equation will only work on batteries that are specified at the "Peukert Capacity" - the 1 amp discharge rate. Manufacturer's specify the capacity at a given hour rate, such as 100AH at 20 hours. For this reason, the formula has been modified to take this into consideration. The revised formula, solving for T, is:
T=C(C / R)k-1 / Ik
or
T=R(C / R)k / Ik
where:
C,
I,
t and
k are as defined above and
R is the hour rating (such as 20 hours, 10 hours, 5 hours, etc)
The closer the Peukert constant is to 1.0, the better the battery's ability to deliver a capacity that is independent of the current being drawn. The constant does not take into account the effect of temperature or age of the battery; this needs to be adjusted by either re-calculating the value (by experimenting with various discharge rates) or apply an additional fudge factor (say between 0.1 and 0.5) for each contributing variable.
Naturally, I have no interest in trying to determine what the Peukert constant is for my battery so I went hunting for its technical datasheet. The datasheet did not provide such a value but it did contain a chart showing discharge rates for the battery when rated at C-5, C-10 and C-20. Under the chart was a table that provided the AH rating for these C-rates. Finally, some information that I understand, and can use.
The table indicated the following:
- A C-20 rating was equivalent to 100AH
- A C-10 rating was equivalent to 90AH
- A C-5 rating was equivalent to 80AH
Just as a refresher:
- The C-20 rating indicates that the battery was fully discharged (reached the cut off voltage) after 20 hours. This implies a discharge rate of 5 amps (100 / 20).
- The C-10 rating indicates that the battery was fully discharged after 10 hours. This implies a discharge rate of 9 amps (90 / 10).
- The C-5 rating indicates that the battery was fully discharged after 5 hours. This implies a discharge rate of 16 amps (80 / 5).
Now I just needed to find the formula that would algebraically determine the Peukert's constant based on two ratings (I was too lazy to sit down and work it out). Well, here it is:
For the following:
- C1 = Capacity rating #1
- R1 = Hour rating #1
- C2 = Capacity rating #2
- R2 = Hour rating #2
Then Peukert's constant is calculated as:
k = ln(R2 / R1) / (ln(C1 / R1) - ln(C2 / R2))
(Note, ln is shorthand for loge)
As an example, using the following information from the datasheet:
- A C-20 rating was equivalent to 100AH
- A C-10 rating was equivalent to 90AH
- A C-5 rating was equivalent to 80AH
I tried the following combinations:
With C1 = 100, R1 = 20, C2 = 90, R2 = 10, Peukert's constant evaluates to 1.18
With C1 = 100, R1 = 20, C2 = 80, R2 = 5, Peukert's constant evaluates to 1.19
With C1 = 90, R1 = 10, C2 = 80, R2 = 5, Peukert's constant evaluates to 1.20
I don't know for sure which value would be the best to use but I imagine you'd want to choose a discharge rate that closely matched your usage. For me, I'd be drawing less than 5 amps per hour (on average) so I'd probably use a value of 1.18 as a starting point. If you'd prefer to be more conservative then go with a higher value.
Calculating available runtime
Now that we know how to calculate Peukert's constant we should be able to determine the battery capacity required to support a given discharge rate. The effective capacity (
It) at a given discharge rate,
I, is calculated like so:
It = C(C / (I * H))(k-1)
As an example, with a 100AH battery (at a rating of C-20) and a constant load of 2,5A we would have:
C = 100, I = 2.5, H = 20 and k = 1.18
Giving an effective capacity of 113AH. So at 2,5 amps per hour, this would last 45.29 hours. But, we know we should not let the battery discharge more than 50% so the effective runtime is half this - 22.65 hours.
Conclusion
I set out to determine where my 11% was lost. My discharge rate would not have even come close to 5 amps per hour (the C-20 rating) but I guess some loss can be contributed to the intermittent load of around 3 - 3.5A when the fridge compressor kicked in. I don't have enough data to calculate this (it would require constant data logging to know when the compressor kicked in, and for how long it continued drawing the load).
Additionally, I have to assume energy (in the form of heat) was also lost during the charging process and there simply wasn't enough solar energy available to compensate for this.
I don't know how much of this would account for the 11% loss. I don't really need to know the exact breakdown - I just need to know how to take it into consideration on our next camping trip.
Can't let the beer get warm !
Disclaimer
I am no authority on this subject. The information I have provided is based on my understanding of the information I have read. My examples are based on values I plugged into a spreadsheet I created using the mentioned formulas so I'm reasonably confident of the content.
Useful Links
Various links relating to Peukert's Law:
