Running a set of 12V Trojans down

pvb said:
Using Trojan's own figures.
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Not meaning to say your figures weren't correct, I didn't have Trojan's figures to hand. Just saying that I'd made an estimate based on guesswork and therefore played safe. Even on that basis I had say the extra battery seemed like a good idea. State of Charge still didn't look great with an optimistic view of likely capacity and increasing it a bit to allow for low current draw.
 
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K, figures are All under load, measured at the battery terminals:
3A discharge with the D2 and other cabin stuff running, sadly not running a brothel :(
12.47V pm
12.37V am 10 hours later

That does not look too bad too me, comments ?

No i am not switching it all off and sit in the dark / cold just to check the resting V ;)
 
I disagree with some of the points made about temperature and current effects on useable capacity and hence on cycle life. (I think I've been through this before with mistroma!)
The defining factor for the nominal capacity of a flooded battery (VRLA is different) is the amount of lead oxide converted to lead sulphate in the positive plates during discharge. This will not be 100%. The duration, temperature and end voltage used to define the nominal capacity are simply convention.
Measuring at different durations (rates), temperatures and end voltages does not change the nominal capacity, only the amount of energy you can extract under those conditions.
The amount of lead oxide converted per Ah withdrawn does not change. The stoichiometry is the same, only the reaction dynamics change.
During the conversion, the volume of material changes. This results in gradual cracking leading to shedding of material, poorer conductivity and possibly dead spots. That is the nature of cyclic damage.
If you use a low discharge rate to take out more than the nominal capacity, you aren't getting it for nothing. You are converting more active material, over discharging the battery and doing more damage per cycle.
That's why when you look at the rate related discharge curves for a battery, the low rate curves will end at higher voltages than the higher rate curves, to avoid over discharging. All the curves will in principle end with the same amount of active material converted (this falls down at high rates because of diffusion effects).
Conversely if the useable capacity is reduced by low temperature, it doesn't increase the effective % cycle depth of a given discharge.
If the battery has a nominal capacity of 100Ah, a 50% discharge - for cycle life purposes - means 50Ah regardless of rate or temperature or end voltage.
You will find that most published cycle life vs DoD curves are pretty linear over quite a wide range. This means to a large extent choosing whether to have 2 or 3 batteries is a matter of desired life and reliability rather than economics. 3 will cost 50% more than 2 and last 50% longer. The lifetime cost per Ah extracted will be the same.
There are bits of over simplification in the above.
 
Tnx,

‘This means to a large extent choosing whether to have 2 or 3 batteries is a matter of desired life and reliability rather than economics. 3 will cost 50% more than 2 and last 50% longer. The lifetime cost per Ah extracted will be the same’

Checking the overnight V drop, i’ll Leave it at 2 Trojan house batteries.
 
simonfraser said:
Tnx,

‘This means to a large extent choosing whether to have 2 or 3 batteries is a matter of desired life and reliability rather than economics. 3 will cost 50% more than 2 and last 50% longer. The lifetime cost per Ah extracted will be the same’

Checking the overnight V drop, i’ll Leave it at 2 Trojan house batteries.
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It's your choice. That comment was general, not related to your specific case.
Looking back at your original post, you appear to be taking 90Ah out of 200Ah nominal capacity i.e. 45%.
As a rule of thumb, that is as high as one would wish to go.
What life it should give in terms of number of cycles, I cannot say. I have seen a Trojan graph in the past but can't find it now (probably Mistroma can). I do recall that these 12V types only have about half the cycle life of the normally recommended Trojan product the T-105.
"Some would say" that they do not merit their premium cost.
If you are cycling them regularly, it is critical that you recharge them correctly and promptly. Overcharging is much less damaging than undercharging (unless taken to extremes).
 
Plevier said:
It's your choice. That comment was general, not related to your specific case.
Looking back at your original post, you appear to be taking 90Ah out of 200Ah nominal capacity i.e. 45%.
As a rule of thumb, that is as high as one would wish to go.
What life it should give in terms of number of cycles, I cannot say. I have seen a Trojan graph in the past but can't find it now (probably Mistroma can). I do recall that these 12V types only have about half the cycle life of the normally recommended Trojan product the T-105.
"Some would say" that they do not merit their premium cost.
If you are cycling them regularly, it is critical that you recharge them correctly and promptly. Overcharging is much less damaging than undercharging (unless taken to extremes).
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In view of some of the comments I was wondering if they had any merits when compared with a decent leisure battery!
 
VicS said:
In view of some of the comments I was wondering if they had any merits when compared with a decent leisure battery!
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Trojan a up-front in publishing the performance data for their products. Before Trojans can be compared with what you suggest; you'll have to define "leisure" in this context. :)

John
 
Plevier said:
I disagree with some of the points made about temperature and current effects on useable capacity and hence on cycle life. (I think I've been through this before with mistroma!)
The defining factor for the nominal capacity of a flooded battery (VRLA is different) is the amount of lead oxide converted to lead sulphate in the positive plates during discharge. This will not be 100%. The duration, temperature and end voltage used to define the nominal capacity are simply convention.
Measuring at different durations (rates), temperatures and end voltages does not change the nominal capacity, only the amount of energy you can extract under those conditions.....
Click to expand...

I'm probably just being forgetful but don't remember previously going through a discussion of current effect vs. useable capacity and lifespan. Must have been a while ago and don't remember your explanation of why Trojan's published data was not correct.

I was basing my comments about current on Trojan's own published data (see link below).
http://www.trojanbattery.com/pdf/datasheets/T105Plus_Trojan_Data_Sheets.pdf

This clearly gives different capacity for different discharge rates.
5hour......(37A) .... 185 Ah
10 hour.(20.7A) .. 207 Ah
20 hour.(11.25A) .. 225 Ah
100 hour.(2.5A) .. 250 Ah

Trojan say that the figures relate to:
The amount of amp-hours (AH) a battery can deliver when discharged at a constant rate at 80°F (27°C) and maintain a voltage above 1.75 V/cell. Capacities are based on peak performance.

I didn't question this with their technical people as I could see a good explanation for the 5 hour rate. I doubt you'd get anything like 185Ah if discharging at a very high current (say 100Ah). I had just accepted Trojan's data as something explained by Peukert's law.
 
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I stand by what I said.
Trojan are not following normal practice in giving those ratings to the same end voltage at such widely different rates.
You will note that they say "can deliver". That is not the same as saying the actual capacity is changed.
Using their figures, if you take out 125Ah at 2.5A i.e. 50% at the 100hr rate, you will do more damage than if you take out 92.5Ah at 37A i.e. 50% at the 5hr rate.
Here chosen at random from a quick Google https://nuclearcat.com/images/lead_acid_discharge_curve2.gif is a more typical presentation - note the end voltage varying with discharge rate.
Commonly it will range from 1.6vpc at very high rate to 1.85vpc at low rates. (Historically 1.75vpc for 8hrs at 25 Deg C has been the US standard and 1.8vpc for 10hrs at 20 deg C the UK one. There is nothing absolute about it, what matters is consistency.)
Peukert's expression - which is just empirical, not a law - merely describes the variation in deliverable capacity at different rates.
 
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simonfraser said:
Tnx, 90Ah is during the winter weekends, summer no prob afloat.
I have shore power with a sterling charger, so they are fully charged by Monday till the weekend.
I am assuming fully charged, as i can see the V peak to 15 from the charger and then drop down to float.
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My Sterling charger reverts to float long before the batteries are fully charged, often after only an hour in absorbtion mode. I usually have to reset it 3 or 4 times before the Amps drawn are low enough to consider the batteries fully charged - which is irritating.
 
Plevier said:
I stand by what I said.
Trojan are not following normal practice in giving those ratings to the same end voltage at such widely different rates.
You will note that they say "can deliver". That is not the same as saying the actual capacity is changed.
Using their figures, if you take out 125Ah at 2.5A i.e. 50% at the 100hr rate, you will do more damage than if you take out 92.5Ah at 37A i.e. 50% at the 5hr rate.
Here chosen at random from a quick Google https://nuclearcat.com/images/lead_acid_discharge_curve2.gif is a more typical presentation - note the end voltage varying with discharge rate.
Commonly it will range from 1.6vpc at very high rate to 1.85vpc at low rates. (Historically 1.75vpc for 8hrs at 25 Deg C has been the US standard and 1.8vpc for 10hrs at 20 deg C the UK one. There is nothing absolute about it, what matters is consistency.)
Peukert's expression - which is just empirical, not a law - merely describes the variation in deliverable capacity at different rates.
Click to expand...

Sorry, not trying to be awkward but must be having brain fade.

I realise that Peukert's expression is just empirical but is often referred to as a law, so just going with the flow. You say "merely describes the variation in deliverable capacity at different rates". I thought that's what I described.

i.e. The exact capacity depends on the size of the load. A larger load (Amps) results in a lower capacity. Of course it is obvious that a larger load will discharge a battery faster—but the product of amps x hours will be less than with a much smaller current.

I know that high current drain will drag the voltage down, too obvious to state and I was simplifying. I also avoided Wh as well because I assumed a 3A drain wouldn't impact Voltage hugely initially and a lot of kit works just as well with slightly lower voltage. e.g. Lights get a bit dimmer with a small drop but not enough for you to spot so you seem to get extra energy for free. Obviously not true but effect seems to be that user gets more Ah so thinks there's more energy (if he/she thinks about it at all :D:D).

The paragraph above was pretty much what ran through my head the first time I saw Trojan's table and it seemed reasonable to use a slightly higher capacity figure with a low amperage. This was of some interest when I installed a Smartgauge and wanted to check manufacturer's claims were at least vaguely correct. Actually seemed with a couple of tests at 3A and 15A. Not wildly far from claims for T105 capacity (after about 40 odd charge cycles) and Smartgauge accuracy.
 
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Try this analogue.

You have a tall water tank, full to the brim. In the bottom is a porous sponge disk. Below that is a draw off pipe with a tap. You have a pressure transducer in the pipe.
Lets say that the gauge is showing 2.3 units as the static head with the tap shut.
As soon as you open the tap, the pressure drops because of the flow resistance of the sponge. The more you open the tap and the faster the flow, the bigger the drop.
After a while you see the pressure dropping further as the static head reduces too as the water level drops.
You test how much water you can get out from a full tank at different flow rates, always terminating when the pressure drops to 1.75. There is still some water in the tank at that point.
You find that at low flow rates, you can get more water out.
But that doesn't mean you have made extra water or that the tank has become bigger.
The capacity of the tank is a constant.
It means you have increased the proportion of the tanks contents you have taken out, i.e. you have increased the depth of discharge.
You derive a mathematical expression to predict how much you can get out at any flow rate and you call it Mistroma's Expression (or Law if you wish!). It does not in its basic form tell you depth of discharge, but you can easily add that by dividing by tank capacity, which is a constant.

Now imagine the water is dirty and the sponge gradually clogs up, increasing the pressure drop and reducing the amount of water you can get out in each test.
Is the sponge going to be clogged less by taking out x litres at a lower rate than by taking out x litres at a higher rate?
 
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Plevier said:
Try this analogue.

You have a tall water tank, full to the brim. In the bottom is a porous sponge disk. Below that is a draw off pipe with a tap. You have a pressure transducer in the pipe.
Lets say that the gauge is showing 2.3 units as the static head with the tap shut.
As soon as you open the tap, the pressure drops because of the flow resistance of the sponge. The more you open the tap and the faster the flow, the bigger the drop.
After a while you see the pressure dropping further as the static head reduces too as the water level drops.
You test how much water you can get out from a full tank at different flow rates, always terminating when the pressure drops to 1.75. There is still some water in the tank at that point.
You find that at low flow rates, you can get more water out.
But that doesn't mean you have made extra water or that the tank has become bigger.
The capacity of the tank is a constant.
It means you have increased the proportion of the tanks contents you have taken out, i.e. you have increased the depth of discharge.
You derive a mathematical expression to predict how much you can get out at any flow rate and you call it Mistroma's Expression (or Law if you wish!). It does not in its basic form tell you depth of discharge, but you can easily add that by dividing by tank capacity, which is a constant.

Now imagine the water is dirty and the sponge gradually clogs up, increasing the pressure drop and reducing the amount of water you can get out in each test.
Is the sponge going to be clogged less by taking out x litres at a lower rate than by taking out x litres at a higher rate?
Click to expand...

I fear this analogy is flawed.
Charge going into a lead acid battery is not like water going into a bucket. Not every electron gets to meet the right ions. At best maybe 95% of the charge will be available to discharge again. The % efficiency varies with both charge and discharge currents as ions go astray in both processes.
 
Oh sure analogies are never perfect and certainly my last paragraph in particular is a considerable oversimplification.
However I think the first part gets over the important point that by discharging a battery more slowly to a given end voltage, you are not increasing its real capacity, you are increasing the depth of discharge relative to nominal capacity (i.e. you are converting a higher proportion of the active material) and the cycle life effect is commensurate. That's the point under discussion here, not recharge efficiency.
There's no such thing as a free lunch.
 
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OK thanks, the analogy didn't really help that much as I already understood that part but I think the "i.e. you have increased the depth of discharge" bit helped. My problem was that you seemed to be stating that electrical capacity was fixed and not affected by things like discharge rate. That is obviously incorrect as very rapid discharge results in loss of energy so "useable electrical energy is reduced". e.g. Pull out a few hundred amps. and energy is lost, heat being the most obvious one.

However, we were talking about pretty low discharge rates so relative conversion efficiencies wouldn't differ hugely. I guess that you are saying that taking a fixed number of Wh (not Ah) from a battery gets you to the same SoC whether 3A, 5A or 10A. It obviously breaks down at large currents such as 200A but we can ignore that.

Difficult to comment on that as I haven't carried out many experiments. Everything about batteries is difficult to measure with sufficient accuracy unless you have very expensive kit. You never actually know the capacity figure even with a brand new battery. I did seem to get figures in line with Trojan's at 3A and 15A with similar temp. corrected SG an hour after discharging stopped. I only worked between 80% to 60% because I was more interested in checking Smartguage than anything else.

Trojan say you can get 250Ah with a 3A draw but only 207Ah with 22.5A if fully discharging both batteries.
You are saying that taking 250Ah will result in a more discharged state than taking 207A. Reasonable only if conversion efficiencies are very similar.

Are you saying that taking 113Ah would result in the same stable SG in both cases above rather than higher SG at end of 3A test? My limited experience would have led me to expect 1.17 vs. 1.48 based on a couple of tests (60% & 40% SoC respectively).

I am interested as my Smartgauge is currently back with Merlin's technical department. It started giving readings much lower than expected and Merlin are testing it. They were surprised at the detailed information I was able to supply and it grabbed their interest. It is being tested for free and replaced for a reasonable fee if damaged (or free if it is an obscure memory error). I can see if your thoughts allow them to shed any new light on difference between Smartgauge and SG readings.
 
See comments interposed in your quoted message.

Mistroma said:
OK thanks, the analogy didn't really help that much as I already understood that part but I think the "i.e. you have increased the depth of discharge" bit helped. My problem was that you seemed to be stating that electrical capacity was fixed
* yes the ultimate chemical capacity is - like the volume of the water tank
and not affected by things like discharge rate. That is obviously incorrect as very rapid discharge results in loss of energy so "useable electrical energy is reduced". e.g. Pull out a few hundred amps. and energy is lost, heat being the most obvious one.
* the chemical reaction produces a fixed amount of electricity (measured in Ah or coulombs or whatever) for every increment (gms or oz or whatever) of plate material reacted, whatever the rate of the reaction. It doesn't "care" whether that electricity is used externally or in internal resistive losses.

However, we were talking about pretty low discharge rates so relative conversion efficiencies
* chemical conversion efficiency doesn't vary it is always 100%
wouldn't differ hugely. I guess that you are saying that taking a fixed number of Wh (not Ah) from a battery gets you to the same SoC whether 3A, 5A or 10A.
* Ah not Wh - as voltage declines during discharge (mainly due to acid s.g reducing) an Ah at end of discharge is fewer Wh, even though it has taken the same amount of active material to produce it
It obviously breaks down at large currents such as 200A but we can ignore that.
* no it doesn't

Difficult to comment on that as I haven't carried out many experiments. Everything about batteries is difficult to measure with sufficient accuracy unless you have very expensive kit. You never actually know the capacity figure even with a brand new battery.
* Agreed!
I did seem to get figures in line with Trojan's at 3A and 15A with similar temp. corrected SG an hour after discharging stopped. I only worked between 80% to 60% because I was more interested in checking Smartguage than anything else.
* I don't disbelieve their figures. We are arguing about what they mean in terms of depletion of the battery.

Trojan say you can get 250Ah with a 3A draw but only 207Ah with 22.5A if fully discharging both batteries.
* no you are not FULLY discharging the battery in a chemical sense, if you do it will be difficult or impossible to recharge, everything is related to an arbitrarily defined NOMINAL capacity which is substantially less than the theoretical chemical capacity. There is always capacity left that you choose not to extract - like the water left in the tank in my analogy
You are saying that taking 250Ah will result in a more discharged state than taking 207A.
* yes, whether it's at high rate or low rate
Reasonable only if conversion efficiencies are very similar.
* conversion efficiency of active material to electricity is 100% regardless of rate

Are you saying that taking 113Ah would result in the same stable SG in both cases above rather than higher SG at end of 3A test?
* yes
My limited experience would have led me to expect 1.17 vs. 1.48 based on a couple of tests (60% & 40% SoC respectively).

I am interested as my Smartgauge is currently back with Merlin's technical department. It started giving readings much lower than expected and Merlin are testing it. They were surprised at the detailed information I was able to supply and it grabbed their interest. It is being tested for free and replaced for a reasonable fee if damaged (or free if it is an obscure memory error). I can see if your thoughts allow them to shed any new light on difference between Smartgauge and SG readings.
Click to expand...

I suspect you are still thinking of the battery as a device that stores electricity, akin to a capacitor.
It isn't.
It's a chemical reactor.
During discharge you consume chemicals to produce electrical energy. Producing 1Ah of electrical energy will always consume the same amounts of the chemicals - lead oxide at +ve plate converting to lead sulphate, metallic lead at -ve plate also converting to lead sulphate, and sulphuric acid becoming more dilute - regardless of reaction rate. The "efficiency" does not vary. That electrical energy is then dissipated partly in the internal resistance and partly in the external load. The voltage reduces also as the acid specific gravity reduces during the discharge so a constant current is not a constant power.

(During recharge, the process efficiency to regenerate the chemicals does vary a lot according to charging conditions and state of charge, that's an entirely different case.)

Courtesy of Wikipedia (to save me scratching my head to remember) -

EDIT - dammit all the formatting has gone, look at it here https://en.wikipedia.org/wiki/Lead–acid_battery in the section headed "discharge"

Note the last sentence that says "The sum of the molecular masses of the reactants is 642.6 g/mol, so theoretically a cell can produce two faradays of charge (192,971 coulombs) from 642.6 g of reactants, or 83.4 ampere-hours per kilogram (or 13.9 ampere-hours per kilogram for a 12-volt battery)." It doesn't say anything about that being discharge rate - i.e. chemical reaction rate - dependent.
AFAIR doesn't your battery weigh about 25kg? (You can't calculate accurately from that because of the weight of the case, terminals etc, and the non-stoichiometric balance of reactants in a practical battery which will be +ve plate limited.)


Negative plate reactionPb(s) + HSO−4(aq) → PbSO4(s) + H+
(aq) + 2e− Release of two conducting electrons gives lead electrode a net negative chargeAs electrons accumulate they create an electric field which attracts hydrogen ions and repels sulfate ions, leading to a double-layer near the surface. The hydrogen ions screen the charged electrode from the solution which limits further reactions unless charge is allowed to flow out of electrode.
Positive plate reactionPbO
2(s) + HSO−
4(aq) + 3H+
(aq) + 2e− → PbSO
4(s) + 2H
2O(l)The total reaction can be written as
Pb(s) + PbO
2(s) + 2H
2SO
4(aq) → 2PbSO
4(s) + 2H
2O(l)
 
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