# Solar Panel Power Information and Calculator (an error?)



## 110428 (Mar 7, 2008)

I have been playing with the Motorhome Solar Spreadsheet and I think I may have found a number of minor anomalies particularly when used for a 24 volt vehicle.

1) You have calculated the load of each device buy assuming a current consumption (say a fridge running as 3.62 amps) You have then (correctly) calculated a wattage by reference to the specified system voltage (12/24 volts). However whilst mathematically correct I think this is logically wrong. Devices like fridges have a power demand and that translates into a current at a given voltage. A 100 litre compressor fridge normally uses about half the current at 24 volt than the same model at 12 volts. I think it would be better to assume a wattage and (if required) calculate a current. As a result the watt hours per day is dependant on the system voltage.

2) In converting watt hours/day into AH per day you have used 12 volts rather than the system voltage. This figure should therefore be labelled as "at 12 volts" otherwise I think it is reasonable to assume it is at the system voltage.

3) In calculating Equivalent de-rated Wh of Battery you multiplied the Battery Fitted AH by the system voltage (rather than by 12). So we must assume that this AH figure is at system voltage rather than "at 12 volts" This is confusing.) You have also corrected for Round Trip Efficiency. I think this is incorrect. This effects the amount of current required to charge the battery not its capacity.

4) In calculating the % surplus you have used the battery capacity in such a way that this figure changes with system voltage at constant load. This is illogical. I am also concerned about the meaning of this figure. It is presumably a "good thing" and yet it is calculated in way that it gets smaller with a larger battery. Should it not be calculated as a % of daily load?

5) I think it would be better to display reserve days as positive rather than negative, that way it gets "better" as we approach self-sufficiency.

6) I am not sure what "Allowance for retaining charge" is. From the way it is used it seems to relate to either or both of the inability to fully charge a battery, and the undesirability of fully discharging it. If so then I suggest it is too big. Very few solar regulators succeed in charging to more than 90% and to ensure a battery life of many years it would be best not to discharge to less than 50%. So a figure of 40% might be better than 90%.


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## sallytrafic (Jan 17, 2006)

Your point 2. You are quite correct it should be system voltage and I don't know how that error crept in as the version I have does divide by the system voltage not 12.

I'll have to take a look at the others later as I am busy babysitting at the mo


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## DABurleigh (May 9, 2005)

I suggest you get nuke to V&V it, Frank.


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## sallytrafic (Jan 17, 2006)

Point 1

I have only included values for illustration, the accompanying texts make it clear that its your figures that matter.

I would expect people to either enter the actual amperage measured in the green column or overwrite the formula in the yellow column with measured or rated wattages for their actual equipment. That instruction also appears in the notes on the SS

In either of the cases it would work correctly independently of system voltage.

Point 3 is more difficult comprising two separate points so I'll do some more baby sitting and reply later.


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## sallytrafic (Jan 17, 2006)

Zenslo said:


> 3) In calculating Equivalent de-rated Wh of Battery you multiplied the Battery Fitted AH by the system voltage (rather than by 12). So we must assume that this AH figure is at system voltage rather than "at 12 volts" This is confusing.) You have also corrected for Round Trip Efficiency. I think this is incorrect. This effects the amount of current required to charge the battery not its capacity.


The de-rated watt hour of the battery is used to see how your selected battery will perform in the solar pv system.

Taking the first of your points I could have converted the outputs of the solar modules and worked in Ah for the battery but I think things are clearer in Wh. As I am sure you realise (but in part I am explaining this to anybody who is interested) if you get two 70Ah batteries and put them in series you still only have a 70Ah battery. However the Wh has now doubled you can do twice as much work with a 24V battery as a 12V battery of the same Ah rating. So I contend that I am right but will put a note on the spreadsheet so that anyone who is interested in this figure can see why and why the solar panels are treated differently.

Round trip efficiency. It is up to the user to input his own figure 80% is a conservative estimate erring on the side of caution. Yes it effects the charging and discharging rather than the capacity but if I may use an analogy. Say you fill and empty a beaker of water in a very slapdash manner and as a result end up with only 80% of the water that you required one way around it would be to accept your slapdash method and make the container bigger is the first place. Assuming an adequate supply of water it now doesn't matter if you spill the water on the way in to the beaker or on the way out. Or if you don't change the size of the beaker you have to accept that you are just going to get 80% of the water that you started with. So it is with the battery system in this model, but what we do is say effectively that the battery is the battery you have fitted but its 80% smaller than we thought.

We could have taken the analogy further and gone back into the original battery sizing and made the required battery bigger. This option exists when after seeing how well your system will perform in the model you decide to buy bigger batteries.


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## davesport (Nov 12, 2006)

Coulpe of points about this post.

Eh....you've got a Unimog......well, where the pictures ? I'm sitting here salivating. Please post some pictures  

Could someone prde me with a linky to the Solar calculator please. 

many thanks, Dave.


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## sallytrafic (Jan 17, 2006)

Zenslo said:


> 4) In calculating the % surplus you have used the battery capacity in such a way that this figure changes with system voltage at constant load. This is illogical. I am also concerned about the meaning of this figure. It is presumably a "good thing" and yet it is calculated in way that it gets smaller with a larger battery. Should it not be calculated as a % of daily load?
> 
> 5) I think it would be better to display reserve days as positive rather than negative, that way it gets "better" as we approach self-sufficiency.


Can I say at the outset that this table (Reserve days and % surplus) is hopelessly wrong (although which months are in surplus and which in deficit is correct)

The point that I had also missed apart from cocking up the formulas somewhat was that if your battery is fully charged at the start of any period then by definition it will last the period of autonomy that you have selected (in the example two days) even if there is no solar energy from the pv modules at all, that after all was what we set out to do. If you manage to put back all the power that you have used the reserve days will be infinite within the life of the battery and you will have reached solar nirvana.  in the example of two days autonomy this will occur as you approach 3 days.

I will change the column to show actual reserve days but of course these are only average conditions. The formula will be =IF(K33<0,($J$46+$I33)/$J33,"") etc

and I think a note on the spreadsheet would also help.

You are quite correct the surplus is independent of the battery and the formula should have been =IF($K36>0,$K36/$J36,"") etc


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## sallytrafic (Jan 17, 2006)

Zenslo said:


> 6) I am not sure what "Allowance for retaining charge" is. From the way it is used it seems to relate to either or both of the inability to fully charge a battery, and the undesirability of fully discharging it. If so then I suggest it is too big. Very few solar regulators succeed in charging to more than 90% and to ensure a battery life of many years it would be best not to discharge to less than 50%. So a figure of 40% might be better than 90%.


You know on first looking at neither was I  So I went back to the full spreadsheet that I was responsible for and yes its as you suspect a figure of the allowable depth of discharge.

Our home built regulators were quite capable of fully charging a battery as indeed any that allow high voltage excursions which are I agree the better ones. So its a question of whether you put 90% or 50% to protect the battery.

I think it should be left as it is with a note to explain the user input choices available. My reasoning is as follows. You may be willing to buy new batteries every three years or so and are willing to accept this cost compromise versus the cost, weight and size compromise of only allowing the batteries to drop to 50%. Then there is the practical issue of judging when the battery has fallen to a charged state of 50% of capacity or 10%.

I would always recommend never more than half discharging a battery but if the choice came down to freezing or using my blown air heating system I know what I would choose.


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## sallytrafic (Jan 17, 2006)

Ok I'd like to thank Zenslo for bringing it to my attention now just to pm Nuke about the new version.

Incidently I've just found a problem in my version of Microsoft Office when I save a current version of the spreadsheet it doesn't overwrite the old one.

That could explain a lot  Something else to sort out don't I just hate MS 

Anyway Its also attached here


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## sallytrafic (Jan 17, 2006)

Anyone who has watched the proceeding post in the last few minutes will have seen three versions of the spreadsheet  . I have had assistance from a 14month old, now I just need to wipe the marmite off this keyboard :lol:


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## sallytrafic (Jan 17, 2006)

For the unimog click on Zenslo's profile then on his website.

I hope he come sback on here again obviously knows his stuff.

BTW still finding out why microsoft office is still treating my file as if it were read only but not telling me so when saving. I've opened the file in another spreadsheet application (not MS Excel) and the file can be overwritten as indeed can a renamed version...bizarre.

Still we've nearly weaned ourselves off of Office now but there is one file that I will have to keep as Excel for reference purposes and that is the original solar model which I was responsible for and helped develop. At about 1MB and including many macros (vba) lots and lots of conditional formatting asking to convert it to another spread sheet is probably a step to far even though in the dim and distant I converted it from Lotus and wrote all the vba etc. This was much more than a solar model you could input energy from generators wind etc as well as try unusual configurations like having some of your pv panels pointed North


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## 110428 (Mar 7, 2008)

Many thanks for the multiple replies. I certainly was not expecting such a detailed response. I accept nearly all the points you make.

However I still think the calculation of "Equivalent de-rated Wh of Battery" should *not* be corrected for "Round trip efficiency" particularly in view of the fact that it is later used with "Daily Balance" to calculate both "Reserve Days" and "% Surplus" when "Daily Balance" has *already* been corrected for "Round trip efficiency" when "Wh per Day" was calculated.

I think you may be double counting?


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## sallytrafic (Jan 17, 2006)

Zenslo said:


> Many thanks for the multiple replies. I certainly was not expecting such a detailed response. I accept nearly all the points you make.
> 
> However I still think the calculation of "Equivalent de-rated Wh of Battery" should *not* be corrected for "Round trip efficiency" particularly in view of the fact that it is later used with "Daily Balance" to calculate both "Reserve Days" and "% Surplus" when "Daily Balance" has *already* been corrected for "Round trip efficiency" when "Wh per Day" was calculated.
> 
> I think you may be double counting?


Yes its a moot point but the argument in favour of using it in both de-rating the battery storage part of the system and the solar power generating side runs along these lines, and if I may use the beaker of water analogy again.

If you accept that due to slapdash handling you can waste water both filling and emptying the beaker you will also need more water to completely fill the beaker as well. Of course this depends on whether having derated the beaker you buy a bigger one. 

I have tested this model against the much more sophisticated model that is owned by my previous employers and it compares well. Each year we would hold a meeting to discuss amongst other things that solar model new battery figures and testing feedback. Originally we only used RTE on the storage side of the model. One year an engineer who had been working on diesel cyclic charging systems as well as solar, proposed and we were persuaded that we had to model it into the supply side as well mainly because of the RTE he had to run the engine for longer. Our R&D tested the model against a real systems by measuring instantaneous current in and out of the system. At the moment it may err on the side of caution however, as in practice using 5,600Ah traction batteries and autonomies in the region of *30 days* the RTE wasn't as low as 80%


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## 110428 (Mar 7, 2008)

Thank you for that. Now having found somebody who clearly knows what he is talking about I have a question.

I have been involved in a discussion about microwave ovens, inverters, too small batteries and fuel cells for some time and I wont bore you with the details, but the gist of the debate end up as:

Assume a 70% charged deep cycle 100AH (at the 20 hour rate) 12 volt battery.

Take 100 amps from it for 5 minutes each day (to run a microwave for example) and nothing for the rest of the day.

At what continuous current (24 hours a day) do you need to re-charge it to maintain 70% charge.

One school of thought says 100 amps "converts" to 245 amps (Peukert etc) so to restore the 20.4 AH used you need about 0.85 amps for 24 hours. Call that the 0% Recovery scenario.

The other school of thought says Peukert only relates to continuous discharge and with 5 mins on and 23 hours an 55 minutes off you should do the calculation with 100 amps (not 245) so you only need 0.35 amps. Call that the 100% recovery scenario. 

Others have suggested 50% recovery.

Can you point me to a good reference for this sort of calculation?


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## sallytrafic (Jan 17, 2006)

Zenslo said:


> Thank you for that. Now having found somebody who clearly knows what he is talking about I have a question.
> 
> I have been involved in a discussion about microwave ovens, inverters, too small batteries and fuel cells for some time and I wont bore you with the details, but the gist of the debate end up as:
> 
> ...


Not sure I can my experience lies in the opposite direction with trying to find the C720 rate (including self discharge) for batteries designed for a daily deep discharge cycle. The best reference I can find on the net is here >smartgauge<  and I wonder if approaching them with your question will get a result. Certainly I have found their technical section very helpful in the past.

There are also two or three on here with both academic and practical knowledge but as a non-subscriber you can't pm them. I'll see if they are able to help. You might consider subscribing


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## DABurleigh (May 9, 2005)

Frank,

Thanks for pointing this one out to me. Been more out than in recently. Visited Tutankamun exhibition in the Dome yesterday, boat trip to Tower Bridge, late lunch at Prohibition bar/restaurant at St Katherine's Dock, then walked back to Waterloo over Millenium Bridge. Nice day.

I have mulled the issue, but have nothing to add beyond the exchange Stephen has already had with MMM's 12V technical consultant on the topic.

Dave


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## Boff (May 10, 2005)

Hi Zenslo!



Zenslo said:


> Assume a 70% charged deep cycle 100AH (at the 20 hour rate) 12 volt battery.
> 
> Take 100 amps from it for 5 minutes each day (to run a microwave for example) and nothing for the rest of the day.


I calculate 8.33Ah if you take 100 amps for 5 minutes.



Zenslo said:


> At what continuous current (24 hours a day) do you need to re-charge it to maintain 70% charge.


Fact is that the microwave has consumed 8.33Ah, which corresponds to 8.33% of nominal charge.

In an _ideal scenario_ (Peukert constant eqals 1), the battery would have lost exactly these 8.33Ah, and this would be exactly the load to charge back over 24 hours. So in this ideal scenario a constant current of 0.35A would be sufficient to bring it back to 70% charge within 24 hours.

Now let's assume a _"semi-realistic"_ scenario: The battery has a Peukert constant of 1.2, but still behaves "ideal" when being charged. In this case, while the microwave still took 8.33Ah, the battery would have lost 20.9Ah instead of 8.33Ah. So you would need a charging current of 0.87A over 24 hours.

In a _fully-realistic_ scenario, we would not only have to consider the Peukert effect while discharging, but also the fact that the battery has an internal resistance, so a part of the charging current will always be just converted into heat and therefore not contribute to the charge. This is typically about 20 to 30% loss. If we assume 30% to be on the safe side, you would need a charging current of *1.13A over 24 hours.*

Best Regards,
Gerhard


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## DABurleigh (May 9, 2005)

Two important points being missed so far:

1) The current generating ohmic losses still plays a FULL part in the electrochemical processes in the battery.

2) Peukert is an empirical formula that represents steady-state surface chemistry behaviour, or put another way, reasonably steady-state currents.

Dave


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## sallytrafic (Jan 17, 2006)

Which is my underlying reason for derating the solar pv modules by the RTE of the battery to cater for ohmic losses during charge. Or were you referring to something else Dave.


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## sallytrafic (Jan 17, 2006)

If anyone has nuke's ear he has an unanswered pm about a new spreadsheet


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## DABurleigh (May 9, 2005)

sallytrafic said:


> Which is my underlying reason for derating the solar pv modules by the RTE of the battery to cater for ohmic losses during charge. Or were you referring to something else Dave.


http://www.motorhomefacts.com/ftopicp-389663-.html#389663

Dave


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## 110428 (Mar 7, 2008)

Once again thanks for all the input.

To avoid "mission creep" with my question I would like to re-state it slightly more precisely and try and summarise the answers offered so far.

Assume a 70% charged deep cycle 100AH (at the 20 hour rate) 12 volt battery *with a Peukert's Exponent of 1.3 and a round trip efficiency of 80%.*

Take 100 amps from it for 5 minutes each day (to run a microwave for example) and nothing for the rest of the day.

At what continuous current (24 hours a day) do you need to re-charge it to maintain 70% charge.

One school of thought says 100 amps "converts" to 245 amps (Peukert) so to restore the 20.4 AH "used" with an RTE of 80% you need about 1.06 amps for 24 hours. Call that the 0% Recovery scenario.

The other school of thought says Peukert only relates to continuous discharge and with 5 mins on and 23 hours and 55 minutes off you should do the calculation with 100 amps (not 245) so you only need 0.434 amps. Call that the 100% recovery scenario.

As far as I can tell *Boff* seems to be opting for *0% recovery* (he has used slightly different values but I think I am reading his response correctly).

*DABurleigh* whilst not committing himself to numbers, seems to be favouring both *a non-0% and a non-100% recovery.*

*CliveMott* in a related post opts for "recovers quite well". I will tentatively translate this into *80%*.


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## DABurleigh (May 9, 2005)

Yup, you have my position characterised correctly - it is somewhere in between. I don't think there is much documented evidence of Peukert in the non-steady state situation you describe, and as even the Peukert exponent is empirical for steady state discharges for a given battery design, you can bet it is even more so for devil-in-the-detail "duty cycle" discharges such as yours.

In such situations, I would steer away from documented empirical "theory" and go with analogous experience any day of the week. I don't have any, beyond when Alison uses the hairdryer, it doesn't seem to stick my Victron Battery Monitor out of kilter/ synchronisation too much 

Dave


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