How long to charge

Joda · Senior Member

How long to charge

Peter_n_Margaret · Guru

If the ammeter is reading 1.5A in sunny conditions, either the battery is virtually fully charged or you have some sort of problem.
What is the regulator and what is the panel output specs?
Cheers,
Peter

Joda · Senior Member

Output of panel is5.65amps & 21V. I,m preparing 3 la batterys for xmas, Charging 1 at a time Brain fade, come home & checked the wrong battery, ie the fully charged one, Just realised should have put red lead of multi meter into 10amp port to read amps. Been using old car analog ammeter which is probably not that accurate as well, will check again tomorrow with multimeter. Thanks for the advice, John

swamp · Senior Member

HI
what are the specs on the solar panel[s] Anything written on them ???
12/24 v ??
Are they 2 x 12v 100w portables ??? =wired in parallel = 10.5 amps roughly on a flat battery
Controller brand and model ??

Wet flooded deep cycle OR AGM batteries
flooded is slightly different voltages to AGM
100% flooded =12.7v
100% AGM = 12.9 v
50% capacity is around 12.00v for a reasonable life span donot discharge further

Warren-Pat_01 · Guru

Hello Joda,
Be wary of letting your battery terminal voltage drop to 11.6v - this is a sure way of reducing its life significantly & hitting your hip pocket. As swamp pointed out, it is generally recommended that you do NOT DISCHARGE a deep cycle battery LOWER THAN 50% of its capacity.
Have a look at this web page (I keep a copy in my van files) - www.energymatters.com.au/components/battery-voltage-discharge/

There is one battery (I've heard) that can be taken down to less than 50% of capacity - the Optima Yellow Top but its cost is much more than a normal AGM.

Another thing to consider is the mounting of your panel. I assume it is mounted the practical way (horizontal) on top of your van - if that is the case, there is a loss of output power (some say 38%, others 25%, from experience, I've lost about a third). This will affect your charge times.

As Peter mentioned above, your 1.5A looks like the float current - ie the battery is fully charged. It doesn't matter whether you use an analog or digital meter, the latter may be more accurate. The voltage on float will read somewhere between 13.2 & 13.7v.
If that is not the case, you need to get someone to check the system out.

Peter_n_Margaret · Guru

Warren-Pat_01 wrote:
Hello Joda,
Be wary of letting your battery terminal voltage drop to 11.6v - this is a sure way of reducing its life significantly & hitting your hip pocket. As swamp pointed out, it is generally recommended that you do NOT DISCHARGE a deep cycle battery LOWER THAN 50% of its capacity.

That is an urban myth.

A deep cycle battery life is determined by total input/output. The depth of the cycles has only a marginal effect.

Cheers,

Peter

Mike Harding · Guru

Peter_n_Margaret wrote:
Warren-Pat_01 wrote:
Be wary of letting your battery terminal voltage drop to 11.6v - this is a sure way of reducing its life significantly & hitting your hip pocket. As swamp pointed out, it is generally recommended that you do NOT DISCHARGE a deep cycle battery LOWER THAN 50% of its capacity.
That is an urban myth.
A deep cycle battery life is determined by total input/output. The depth of the cycles has only a marginal effect.

It is not an urban myth.

Look at the curve for "Cycle Life in Relation to Depth of Discharge"

http://www.ritarpower.com/upimg/201712415147457.pdf

http://www.ritarpower.com/battery/Energy%20Storage%20Battery/DC%20Series/

Joda · Senior Member

Thanks for that link wp 01, have ordered a coloumb meter to put in the set up which will make it easier to keep track of the batterys. This set up is only used at xmas for our 80lt. Waeco fridge, I believe its best to take fridge load off the load terminals of the controller ??(mppt), before we were running fridge direct off a battery while charging (solar) depleted batterys. As i understand it the panel (200w) will run the fridge during the day, depending on fridge draw, (haven,t checked that yet) & run off the the battery at night. Son in law is experienced sparky who admits He does,nt fully understand low voltage stuff, John.

Joda · Senior Member

Just digested Mikes graph He posted, learning curve is starting to flatten out a bit! Have whole set up in backyard & going to monitor it for a while. Taking readings & recording them instead of doing things willy nilly, thanks again guys, much appreciated. John.

Peter_n_Margaret · Guru

Mike Harding wrote:

It is not an urban myth.
Look at the curve for "Cycle Life in Relation to Depth of Discharge"
http://www.ritarpower.com/upimg/201712415147457.pdf
http://www.ritarpower.com/battery/Energy%20Storage%20Battery/DC%20Series/

I suggest that you are miss-reading the graph.

A discharge to 80% (for instance) provides DOUBLE the power per cycle that a discharge to 40% provides, so you only need to have half of the discharges, so you could have 1 battery instead of 2. The choice is an economic Vs weight one. For example, buy 0ne battery every 3 rears or 2 batteries every 6 years. Same cost. Your choice.

The life of a battery is better expressed in total power in and out, rather than in cycles.

Cheers,

Peter

Mike Harding · Guru

They are not linear functions Peter. Neither in area under the curve for discharged energy nor in the ratio of depth of discharge to number of cycles.

Peter_n_Margaret · Guru

The Ritar chart that you linked to clearly shows that at constant 100% discharge that the life is 300+ cycles (equals about 300 x 100Ah = 30,000Ah for a 100Ah battery).
It also shows that at 50% discharge that the life cycle is 600+ cycles (equals about 600 x 50Ah = 30,000Ah for a 100Ah battery).
The life of this 100Ah battery is about 30,000Ah total power delivered.
Yes, there is a small reduction for constant deep discharging, but it is marginal and suggesting that just the cycle count on its own represents the battery life is simply misleading in the extreme. The difference in life between 100% DOD and 50% DOD is likely 10 or 15%, not the half that the cycle count alone suggests.

The conclusion is that the considerations for how deeply to discharge on a regular basis are influenced by --- weight of batteries, $ invested each time they are replaced, how often they are replaced and the reserve storage required, NOT total power delivered life.
Other manufactures show data that is consistent with this.

Cheers,
Peter

Mike Harding · Guru

Fullriver:

http://www.fullriverbattery.com/product/batteries/DC55-12

Look at the shape of the curve Peter.

Peter_n_Margaret · Guru

Sure, but what does it mean?
What is the total life in Wh (or Ah if that is easier) between 50% and 100% DOD?

I suggest that it supports my previous statement.....
The difference in life between 100% DOD and 50% DOD is likely 10 or 15%, not the half that the cycle count alone suggests.

Cheers,

Peter

Warren-Pat_01 · Guru

Joda,
We carry a 60L Trailblazer in the car sometimes. To save the battery at night, we crank up the thermostat (so it's colder) while we have power (car running) later in the afternoon + a couple of hours without & find that frozen food is still frozen in the bottom of the fridge next morning. We separate the frozen food from the unfrozen with a piece of marine carpet as recommended by the manufacturer.

I'm not certain how the two fridges compare.

Joda · Senior Member

Hi WP 01, Our set up is only used at xmas up on the murray to cater for our family camp. Its an older 80LT waeco, We have 3 deep cycle flooded cell lead acid batterys & a 200watt. Solar panel plus another 72 cell panel for charging batterys on their own. Going to pick up the fridge soon & set up everything in the backyard & do some real world tests using the info. Gained from various people on this forum. Did,nt know whether to run the 3 batts. In parallel & connected to the 200 panels thru a m.p.p.t. reg. Or run the fridge off 1 batt. at a time & swap as needed & charge the used batt. off our other panel. Have,nt looked at putting the 2 panels in parallel yet, have no specs. for one of the panels so will have to measure its output to see if the controller can handle the 2 connected panels. Have a better understanding of solar after my forum crash course & will post results, spoken to people in shops about solar bits & pieces but they don,t come close to the info I,ve gleaned from this forum, John.

T1 Terry · Guru

Peter_n_Margaret wrote:
Sure, but what does it mean?
What is the total life in Wh (or Ah if that is easier) between 50% and 100% DOD?
I suggest that it supports my previous statement.....
The difference in life between 100% DOD and 50% DOD is likely 10 or 15%, not the half that the cycle count alone suggests.
Cheers,
Peter

So we are comparing apples with apples I'll use a Full River chart that shows the cycle life for a battery charged and discharged at the optimal manufacturers rate .... they wouldn't publish something that was the optimum would they wink

Fullriver cycle life 2.JPG

Using this chart and a 100Ah battery as a base figure to make the calculations easier, let's look at the 30% depth of discharge (DOD) graph first. The test conditions show the test load was 0.4CA, so 100Ah divided 0.4 = 25amps per hr. The 30% DOD shows the discharge was for 0.9hr x 25 amps = 22.5Ah x 1200 cycles = 27,000Ah

Now let's use the same formula for the 100% DOD graph. 25 amps x 3hrs = 75Ah (not 100Ah because Peukert has a serious effect on higher discharge rates) 75Ah x 200 cycles = 15,000Ah. 15,000/27,000 = 55%, a long way from the 10% to 15% you are suggesting Peter.

If you have factory backed graphs and logical maths that show a different result I'd love to see it Peter.

T1 Terry

Peter_n_Margaret · Guru

Sorry Terry, but using 25A for a 100Ah battery in a typical RV is simply BS.
Try the same calculation running (say) a compressor fridge at 3.5A x 25% duty cycle and Mr Peukert won't have much if any effect.
Cheers,
Peter

T1 Terry · Guru

Peter_n_Margaret wrote:
Sorry Terry, but using 25A for a 100Ah battery in a typical RV is simply BS.
Try the same calculation running (say) a compressor fridge at 3.5A x 25% duty cycle and Mr Peukert won't have much if any effect.
Cheers,
Peter

I didn't do the chart Peter, Full River batteries made that chart so it must reflect the best outcome possible, surely, why would a manufacturer publish something that made their batteries look bad?

T1 Terry

T1 Terry · Guru

Sorry, I didn't read Full Rivers info correctly, the charge rate was 0.4CA max. constant current then constant voltage at 14.7v and took 6 hrs to recharge after? Who knows whether it was the 30% DOD or 100% DOD, they seem to have omitted that bit of info.
The discharge rate was 0.25CA but that still equates to a 25 amp per 100Ah discharge rate. The 0.4CA actually equates to 40% or 40 amps per 100Ah capacity.
It's gets confusing when at one point the specify discharge rates as C20 or C10 which is the number of hrs taken to fully discharge the battery, then in the next graph they switch to CA which is the capacity divided by the number before the CA

T1 Terry

Peter_n_Margaret · Guru

Also Terry, you seem to not understand Peukert.
High current discharge rates do NOT cause a loss of energy. Energy can not be created or destroyed.
The apparent loss of energy (seen as voltage drop from the battery) during high current discharge rates is simply the battery chemistry not being able to keep up with the demand.
When the high current discharge rate is terminated, the battery chemistry catches up and the apparent loss of energy disappears completely.

So, even if a substantial proportion of the batteries capacity is used under high current discharge rates, after resting (or a slowing down of the demand), the total nominal capacity of the battery can still be delivered at lower current discharge rates.
Believe that, or not.
I have not seen any graphs showing this effect. They all tend to continue with the high current until the battery voltage reaches "empty". There are some papers on this subject though.

In any case, not much of this applies to the typical RV battery application.
Cheers,
Peter

Joda · Senior Member

Me again, I,ve hooked up My 2 solar panels in parallel, where is the gain in doubling the panel size? Voltage output is the same obviously, is it measured in watts, amps, amp hours or does it havest more efficiently? Forgive the dumb question, 40 years as a toolmaker does,nt help much when it comes to this stuff, John.

Jaahn · Guru

Joda wrote:
Me again, I,ve hooked up My 2 solar panels in parallel, where is the gain in doubling the panel size? Voltage output is the same obviously, is it measured in watts, amps, amp hours or does it havest more efficiently? Forgive the dumb question, 40 years as a toolmaker does,nt help much when it comes to this stuff, John.

Hi Joda smile

Back to the real solar questions !!

Having two panels gives double the power you can get, which in real terms means the current is doubled. That could mean you get the batteries charged in half the time. Speaking generally of course. hmm

Or you could use the extra power to do other things at the same time. Like charge the phones, use the computer, run another fridge, watch the telly or listen to the radio/cd.

Jaahn

PS There seems to be a difference between those panels. The one with the regulator is about 110W and 12V.

The other one with 72 cells would be normally wired as a 24V panel but may be wired as a 12V. What voltage does that panel put out in full sun when not connected to anything. Also what are its measurements. We could estimate something with those.

-- Edited by Jaahn on Tuesday 30th of October 2018 08:57:36 PM

-- Edited by Jaahn on Tuesday 30th of October 2018 09:08:02 PM

Joda · Senior Member

Thanks Jaahn, by current do You mean amps? My m.p.p.t. controller is rated to 20 amps & did,nt want to kill it, I imagine 200watts of panel would not do this? Have a columb meter coming so I can better monitor whats going on. I feel better if I can understand how it all works plus can troubleshoot armed with the correct info. John.

Joda · Senior Member

Both panels open circuit voltage is around 18.3 to 19.4v, My newer panel is rated 5.65 amps. Cheers John.

Older panel has p.w.m. controller, I wired the 2 panels so I can isolate this one via a switch.

When not using it to charge a battery Its switched on to give Me the double output for running Our waeco.

From what You tell Me Jaahn I can just leave the 2 panels connected all the time & just put a discharged battery on to the 2nd. Panel to recharge.

John.

-- Edited by Joda on Tuesday 30th of October 2018 09:57:43 PM

T1 Terry · Guru

Peter_n_Margaret wrote:
Also Terry, you seem to not understand Peukert.
High current discharge rates do NOT cause a loss of energy. Energy can not be created or destroyed.
The apparent loss of energy (seen as voltage drop from the battery) during high current discharge rates is simply the battery chemistry not being able to keep up with the demand.
When the high current discharge rate is terminated, the battery chemistry catches up and the apparent loss of energy disappears completely.

So, even if a substantial proportion of the batteries capacity is used under high current discharge rates, after resting (or a slowing down of the demand), the total nominal capacity of the battery can still be delivered at lower current discharge rates.
Believe that, or not.
I have not seen any graphs showing this effect. They all tend to continue with the high current until the battery voltage reaches "empty". There are some papers on this subject though.

In any case, not much of this applies to the typical RV battery application.
Cheers,
Peter

Last bit on this topic Peter, but worth answering none the less. The statement about energy not being created or destroyed but rather converted to another form is quite right, so you should consider the other forms of energy used during a high rate discharge or recharge in a lead acid battery.

The first one is heat, that heat energy has to be converted from some other form of energy and the only one available is electrical energy. The internal resistance is a lead acid battery is well documented, the effect of resistance in an electrical circuit is also well documented, it converts electrical energy to heat energy, so how can you just dismiss this loss of electrical energy to heat energy as not a part of what makes up the losses referred to as the Peukert factor? You can't get that heat energy back into electrical energy just by letting it stand and cool, it is lost from the total store of electrical energy in the battery.

The second electrical energy conversion is in the form of electrolysis. when a lead acid battery is rapidly discharged or recharged there is some electrolytic action in the water that makes up part of the electrolyte. It is well documented that lead acid batteries produce hydrogen and oxygen gas when charging and discharging and the higher the current the greater to gas production. It is also well documented that the electrolytic action that causes the breaking down of water to hydrogen and oxygen is very energy inefficient so quite a bit of electrical energy is used so that is also lost from the total electrical energy store within the battery.

Apply this information to rational thought and these two by-products of high current discharge and recharge logically make up a part of what is bundled together in the term Peukert factor, it is lost energy that can't be regained by letting the battery stand.

T1 Terry

-- Edited by T1 Terry on Wednesday 31st of October 2018 04:50:43 PM

T1 Terry · Guru

In case anyone is still awake and is wondering what the hell that last post has to do with the original question ...... charging a lead acid battery faster than it can accept the current and perform the chemical reaction required to hold that energy in storage, creates a Peukert factor, so simply adding the current (amps) together and multiplying it by the number of hrs does not add up to the number of Ah returned to the battery, the losses need to be added in as well. The closer to fully charged or plate saturation, the less efficient the chemical reaction and the greater the losses to both heat and electrolysis. This explains why a faster than normal charge rate makes the electrolyte bubble (electrolysis) and the case to get hot (current through a resistor) That electrical energy might have been fed into the battery but it may have been converted to heat and gas generation and not stored in the battery as electrical energy that can be retrieved.

T1 Terry

Peter_n_Margaret · Guru

According to Peukert then, drawing 600A (or so) from a battery for 30 seconds (at 25C) until it can not deliver it any more, the battery is FLAT.
This is called CA (cranking amps). Even Fullriver provide that data for their AGM batteries. I use one in the OKA for cranking.
We all know that the battery is NOT in fact flat and that it will recover quite quickly and do it all over again in a few minutes.
Nor does this battery get hot.
And a few minutes on the alternator and it is fully charged again. The actual power used is quite minimal.
As I said earlier, "Believe it or not", I don't care, but do some research if you are a sceptic and are genuinely interested.
Cheers,
Peter

Joda · Senior Member

Groan n n n , side cutters at 20 paces please gentlemen! I asked the time & got told how to build a clock, , anyway, My problem is resolved using the info. provided. Joda.

T1 Terry · Guru

Joda wrote:
Groan n n n , side cutters at 20 paces please gentlemen! I asked the time & got told how to build a clock, , anyway, My problem is resolved using the info. provided. Joda.

:lol: The problem is you asked for a simply answer to an extremely complex problem that requires a lot more than simply looking up a chart. If you had asked the same question about a lithium chemistry battery the answer would have been simple. Amps x time, or the required number of Ah divided by the charging current (amps) available would give the time need to reach very close to 100% state of charge.

Unfortunately lead acid batteries don't work like that because there is a complex chemical change involved with either storing or retrieving the stored electrical current because it isn't actually stored as electricity. This chemical process is very inefficient and the faster you try to make that process occur the more inefficient it becomes.

Basically, no simple answer to the question you asked so you can't expect to get a truthful simple reply because it doesn't exist, too many factors need to be included.

T1 Terry