Battery Bank Sizing Run Time

Re: Battery Bank Sizing Run Time

Roughly, the current draw would be:

• 2000 watts * 1/23.5 volts ave batt voltage * 1/0.85 eff = 100 amps
• 225 AH / 100 Amp = 2.25 hour rate till dead

Or you can load the spread sheets from Smart Gauge:

• http://www.smartgauge.co.uk/peukert2.html
• http://www.smartgauge.co.uk/peukert_depth.html

Enter in the 100 AH 250 AH and 20 Hour 225 AH rating and get Peukert Factor = 1.07

Then load the next spread sheet, enter in the capacity @ XX hour rating and find out that at that heavy discharge rate the battery will be exhausted in around 1.92 hours until dead.

Normal operation, we would discharge to 50%. Extraordinary operation (where long cycle life is not a requirement (weekend cabin/RV/UPS) you could discharge it down to 20% state of charge (80% capacity usage).

From my standard rule of thumb... I would limit current to ~40% maximum (C/2.5 surge) or ~12.5% maximum (C/8 continuous) based on the 20 Hour capacity... That would give:

• 225 AH * 0.40 * 23.5 volts * 0.85 eff = 1,795 Watts Max Surge
• 225 AH * 0.125 * 23.5 volts * 0.85 eff = 562 Watts Max Continuous

Adding parallel strings will give you slightly longer than simply the extra number of strings--Because each battery is being discharged at a slightly lower rate, so its "apparent" capacity is a bit higher at these lower rates.

Note that while I have accuracy to four decimal places--the actual accuracy is probably closer to two (just showing more numbers so you can reproduce the math--or find my errors ).

Make sense?

-Bill

Re: Battery Bank Sizing Run Time

Batteries have different "apparent storage capacities"--depending on current flow...

The second link shows how to calculate the battery Peukert Factor (exponent) if you cannot get it from the battery company (just need two AH capacity values for two different points (i.e., 225 AH @ 20 hour and 250 AH @ 100 hour) or load an Excel spread sheet and it will do the math.

Once you know the P.F... Then it is a "simple" calculation to figure out how long the battery will supply power at a fixed current (or use the second spread sheet).

Now--you end up with another issue--If you are using an inverter, it is not a constant current device but a constant power device:

• Power = Voltage * Current

So, the operating range of a (for example) 12 volt battery runs from ~15 volts (under charge) to ~10.5 volts (dead)... (or ~30 volts to 21 volts for a 24 volt bank).

So, as the voltage drops, the inverter draws progressively more current to keep up with the constant 120 VAC power...

I have seen a few batteries designed for UPS operation where they list battery capacity in Constant Watt Load instead of Constant Current.

I guess, it is possible to do an estimate based on constant power--your you can pick the low voltage / high current point as worst case--and do the capacity estimate there:

• 2,000 watts * 1/30 volts * 1/0.85 eff = 78 amps
• 2,000 watts * 1/24 volts * 1/0.85 eff = 98 amps
• 2,000 watts * 1/23 volts * 1/0.85 eff = 102 amps
• 2,000 watts * 1/21 volts * 1/0.85 eff = 112 amps

If your intent is not to use more than 50% capacity--perhaps yu can use 1/23 volts as your current estimate point...

Using the spread sheet, we see that ~112 amps would give the 225 AH @ 20 Hr capacity an apparent capacity of 191 Amp*Hours (C@1.7 hour rate).

Interestingly, I used the numbers for the T-105RE in the SmartGuage spread sheet and came up with P.F.=1.07 but the T-105 number from NAWS's battery FAQ is P.F.=1.25 ... If I run the above 112 amps at 225 AH and P.F.=1.25, I get: 123.5 Amp*Hours (C@1.12 hour rate).

So, obviously, there is a disconnect somewhere (T-105 is not the same as T-105RE, or there is an issue in the P.F. calculation spread sheet, or the rating numbers from Trojan are not right, or something else... I don't know). But, again from the NAWS battery FAQ--PF~1.3 for flooded cells and PF~1.10 for AGM's would indicate that the PF=1.25 for the Trojan T-105 is more than likely the correct set of numbers.

In any case--depending on what numbers you use, you can run from 1.1 hours to 1.7 hours a 2kW load on set of T-105 batteries... A fairly large spread...

But--you are probably going to hit another problem... A typical flooded cell battery can manage around C/2.5 or C*40% as a maximum surge capacity... In your case we are estimating around 102 amp load:

• 102 amp / 225 AH (at 20 hour rate) = 0.45 discharge rate

You are pretty much at the point, probably, where the chemistry is unable to keep up with the loading--so you may have the battery voltage collapse as you are exceeding its current capacity (let alone overheating the battery because of the I^2 * R self heating losses from battery resistance).

And the Peukert Factor is not a "pure" mathematically derived equation... It is really a curve fit to an observed set of data:

Peukert's effect is partially a result of slow electrolyte dispersal/diffusion and partially a result of an internal resistance that varies as a function of the discharge rate. This is what makes it so difficult to calculate. If the effect was due purely and simply to one of these effects then it would be much simpler to compute the final results.

Down load the Excel file (it is not password protected--so you can unprotect the cells and study the math) from the Smart Gauge website:

We have written you a simple Peukert calculator in Microsoft Excel format and put it on this website. This will enable you to play to your heart's desire. This calculator uses the exact same equation shown above but rearranged to a more elegant format. You can download it by right clicking here and selecting "save target as" or use the link on the left hand side.

This calculator will allow you to enter the battery capacity, the capacity rating (i.e. 20 hour rating, 100 hour rating etc) and Peukert's exponent for the battery type. It will then calculate a range of discharge currents from very low up to a discharge equivalent to the battery capacity. It then displays what is termed the "peukert corrected amps" (which is the equivalent discharge rate such a load will remove from that particular battery) for each discharge current and the available run time, again for each discharge current (note that the time is shown in hours as a decimal not in hours and minutes). Finally it shows the total amp hours available from the battery at each discharge rate.

So the answer is there is an equation that you can use to estimate the battery's apparent capacity at a set current flow (knowing the Peukert Factor), you can estimate an average current for the Inverter, and you have an estimate of the inverter's efficiency (which varies with load, temperature, battery voltage, etc.)...

And then you have other rules of thumb, which if exceeded, will bring other effects into play (Batteries not at 77F--Hot battery more capacity, cold battery less capacity; too high of current can overheat battery; too high of current can collapse battery voltage because of resistance and lack of electrolyte diffusion at high currents; etc.).

If--you are looking for a battery to supply high currents (as a ratio of battery AH capacity--Basically more than C/8 discharge continuous discharge current or more than C/2.5 surge current)--you should look at AGM type batteries--they are much better suited to that operating regime.

Am I making any sense to you at all? I am not quite sure if I am explaining the multiple issues clearly.

-Bill