SmartGauge Electronics - The importance of small loads

The importance of reducing small loads

The small loads in many installations tend to be ignored on the basis that they don't really make much difference when compared to the effect of the larger loads.

This can sometimes be foolhardy. Once again, we have Peukert's effect to thank for the fact that the small loads can actually make an enormous difference to the available run time.

Let's look at a typical installation. It is very difficult to say what a "typical installation" is, but most that we come across (that have been specified and installed correctly) tend to give about 24 hours run time from a full recharge down to 50% state of charge.

The discharge on this battery bank will be a mixture of very heavy draw for short periods of time, medium draw for longer periods of time, and very small loads on for very long periods, or even on permanently.

Let's compare the effects of reducing a heavy current draw item to that of reducing a small current draw item. What is important here is that we do not consider reducing loads by a certain fraction or percentage. Reducing 2 different sized loads by, say, 25% will increase the available run time by the same amount in each case. What we have to consider is reducing a load by a fixed amount, say 1 amp, or 5 amps, or whatever.

If it wasn't for Peukert's effect then reducing the current draw of any item by a fixed amount (say 2 amps) would make no difference to whether the load was a heavy draw item or a small draw item. The effect on the available run time would be the same. However Peukert's effect is real, and the results of it are real. They are also extremely illuminating.

Due to Peukert's effect being exponential, the smaller the current draw, the more difference it makes to the available run time when this current draw is increased or decreased. Simply have a look at the graph on the Peukert calculator and you will see that small changes in current draw make a much greater difference to the available battery capacity at the left hand side of the graph.

Use the Peukert calculator and enter a battery capacity of 400 amp hours at the 20 hour discharge rate. Set Peukert's exponent to 1.3

In the user input box at the bottom enter a discharge current of 50 amps. You will see it calculate a total run time of 6.08 hours to 0% state of charge. Now change the discharge current to 49 amps (a reduction of 1 amp). You will see the run time has increased to 6.24 hours (remember these are decimal hours - not hours and minutes). This is an increase in run time of 2.6% for a reduction in discharge current of 2% (this can also be calculated by mutiplying the reduction in discharge current as a percentage by peukert's exponent i.e. 2% reduction * peukert's exponent of 1.3 = 2.6% increase in run time).

Now enter a discharge current of 5 amps. The available run time will be 121 hours. Reduce this to 4 amps and the available run time increases to 162 hours, an increase of roughly 34%

So at heavy discharges, a change in discharge current of 1 amp hardly makes any difference. But at light discharge currents, 1 amp makes a huge difference. This is partly due to Peukert's effect and partly due to the decrease in current being a higher fraction of the original current.

If you consider that much of the time a typical installation will have little or no load on it you will understand that the small loads can make a huge difference to the available battery run time.

Let's try an example.

We have a 400 amp hour battery bank with permanent loads on the batteries totalling 5 amps. This is made up of a carbon monoxide monitor, a gas detector, a burgler alarm, a battery monitor, a water tank gauge, a toilet tank gauge, the parasitic draw of the alternator controller (none of them switch off fully when not being used) and the inverter in idle mode.

On top of this we have all the other usual equipment such as lighting, microwave ovens, toasters, TVs, Hi Fis etc.

For ease of calculations let's say all this other equipment and the permanent loads use 200 amp hours in a 24 hour period thus giving 24 hours total run time from fully charged to 50% state of charge. And again for ease of calculation let's say that the other loads are on for 8 hours a day.

So in a 24 hour period the permanent loads will use 24 * 5 amps = 120 amps.

The other loads are on for 8 hours using 80 amp hours in that time.

If the other loads are somehow reduced by 20% (thus reducing the draw to an average of 8 amps instead of 10 amps) then this draw is reduced from a total of 80 amp hours to 64 amp hours. A saving of 16 amp hours.

However, on the permanent loads, if the inverter is switched OFF instead of being left in idle mode this will save 1 amp for 16 hours (it will be switched on for the other 8 hours) = 16 amp hours.

If the water tank gauge, the alternator controller and the burglar alarm are switched off fully, and only switched on when they are actually needed then we can probably save another 1 amp for 24 hours = 24 amp hours.

So just by sorting out the permanent loads properly we can save 30 amp hours per day.

And this is without even considering the effects of Peukert.

Go back to the Peukert calculator and enter a discharge of 5 amps, you will see from the calculator that we get a Peukert corrected amps figure of 3.3 amps. Reduce the discharge to 3 amps and you will see this reduced to 1.7 amps.

If the other non permanent loads remain the same at an average of 10 amps for 8 hours, we get a peukert corrected amps figure of 8.12 amps.

Adding these up we get 3.3 amps (Peukert corrected amps) for 24 hours = 79.2 amp hours PLUS 8.12 amps for 8 hours = 65 amp hours giving a total amp hours consumed of 144 amp hours.

With the permanent loads reduced to 3 amps we get 1.7 amp (Peukert corrected amps) * 24 hours = 41 amp hours PLUS 65 amp hours (same as above) giving a total amp hours consumed of 106 amp hours.

This is a saving of 38 amp hours. A reduction in total power consumption of around 25%

This is just by sorting out the permanent loads. That is a substantial power saving without actually affecting the use of the installation and without making any compromises.

Obviously certain equipment needs to keep it's permanent supply. Gas detectors and battery monitors for example. But many do not. An electronic water gauge is one example, it sits there, powered up permanently, 24 hours a day, yet is actually used for maybe 5 to 10 seconds each day. Why not simply fit a push button switch, and switch it on only when you want to check the water level?

The same goes for a burgler alarm. Powered up permanently, when you're actually on the boat, RV or whatever. Why not remove it's power feed?

I hope you can see just how important all those little items of equipment can be that "hardly use any power".

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