Electrolyte specific gravity anyone?

[dmr]

Based on aiming to stay above about 50% charge I would aim for about 12.0 volts,

 

If I was intending to go down to 20 or 30% soc, I would not rely on a loaded voltage measurement so I have not paid much attention in that region.

 

.............Dave

[cuthound]

Specific gravity measurements are accurate at 60 degrees f (15 degrees c).

 

Add 1 point for every 3 degrees f or 1.5 degrees c above this temperature or subtract if below.

 

As others have said, equalise at 15.5 volts and then retake the SG using a refracting hydrometer.

[Loafer]

Ok I'll try just once more.

 

I am not trying to establish a means of determining SOC from on-load voltages!

 

I would just like to know WHAT loaded voltage you had, AT the same time you were at LOW SOC and decided to run your genny to commence recharge. I have my own voltages for higher SOCs, but I have never seen my batteries down as low as 30-50% SOC.

 

I just want to know what order of voltage one would expect whilst still running an average load when it becomes time to recharge.

 

Pretty Please?

[dmr]

Ok I'll try just once more.

 

I am not trying to establish a means of determining SOC from on-load voltages!

 

I would just like to know WHAT loaded voltage you had, AT the same time you were at LOW SOC and decided to run your genny to commence recharge. I have my own voltages for higher SOCs, but I have never seen my batteries down as low as 30-50% SOC.

 

I just want to know what order of voltage one would expect whilst still running an average load when it becomes time to recharge.

 

Pretty Please?

 

My figure above was a "guess" based on what I remember from looking at voltages a while ago.

I have a smartgage and current measurement, and a set of 3 year old Trojans in good condition.

I will try to measure a few numbers for you tomorrow, I am charging right now.

 

...........Dave

 

a looking

[Loafer]

My figure above was a "guess" based on what I remember from looking at voltages a while ago.

I have a smartgage and current measurement, and a set of 3 year old Trojans in good condition.

I will try to measure a few numbers for you tomorrow, I am charging right now.

 

...........Dave

 

a looking

Phew. Thanks Dave. I'm exhausted now, so there's no rush. The nurse is going to take my computer away shortly. Reminder - I'm only interested in LOW SoC with corresponding 'normal load (for you)' voltage.

[bottle]

but that will be Dave's reading with his 'normal' load, with his type of battery, with his age of battery, with his ...

 

How will that help you? (rhetorical)

[Radiomariner]

I think loafer just wants a rough guideline. If two or three people supply the same information as DMR intends, he can work out an average which is what he needs.

[dmr]

but that will be Dave's reading with his 'normal' load, with his type of battery, with his age of battery, with his ...

 

How will that help you? (rhetorical)

 

Yes, but the OP has 4 Trojans and I have 4 Trojans so a measurement from me is probably better than something found on the www,.... as long as the op takes it for what it is, rather than a fundamental truth,

Then again its somewhat cold here today and that may well be a bigger influence on battery voltage than the load effect.

And there might be a dribble of solar going in and that's an unknown effect.

 

Anyway, so far we are still in the 80% charge region and with a draw of between 0 and 5 amps I am seeing battery (terminal) voltages only about 0.1 volt below the published Trojan voltage curve. I am probably not going to get below about 60% today as I will do an engine run this evening to warm the back cabin and comfort takes priority over scientific experiments

 

.............Dave

[smileypete]

As long as volts bounce back above 12.2V, the batts have at least 50% charge, should do for the OP.

 

For 20-30% charge you'd be looking for at least 12V I guess.

 

The volts will wiggle a bit as the fridge cycles on and off...

 

What might be interesting, does it wiggle more as the batts approach a low SoC?

 

cheers, Pete.

~smpt~

[nicknorman]

 

What might be interesting, does it wiggle more as the batts approach a low SoC?

~smpt~

Well it should do, a battery with low SoC has a significantly higher internal resistance than a well charged one, according to the bottom diagram on this link:

 

http://batteryuniversity.com/learn/article/how_does_internal_resistance_affect_performance

[dmr]

Well it should do, a battery with low SoC has a significantly higher internal resistance than a well charged one, according to the bottom diagram on this link:

 

http://batteryuniversity.com/learn/article/how_does_internal_resistance_affect_performance

 

Interesting Link, I had not read that section before. I would hardly call that curve Linear!!

 

Do you know if the off load terminal voltage of a battery will vary with temperature? or just the voltage required for charging????

 

..............Dave

[cuthound]

Well it should do, a battery with low SoC has a significantly higher internal resistance than a well charged one, according to the bottom diagram on this link:http://batteryuniversity.com/learn/article/how_does_internal_resistance_affect_performance

How can that be?

 

If you connect a battery with a low SoC to a charger it will draw a high current.

 

If you connect a battery with a high SoC to a charger it will draw a low current.

 

Seems to defy logic and the laws of physics.

[dmr]

How can that be?

 

If you connect a battery with a low SoC to a charger it will draw a high current.

 

If you connect a battery with a high SoC to a charger it will draw a low current.

 

Seems to defy logic and the laws of physics.

 

I imagine you have to hypothesise a "chemical battery" inside the actual battery and this has some voltage. The charge current is then the difference between the applied charge voltage and the "chemical" voltage, divided by the total internal resistance.

A flat battery has a low voltage so will take a big charge current despite an extra bit of internal resistance.

A charged battery has a high internal/chemical voltage so takes very little current despite a lower internal resistance.. This makes sense to me because when we first turn the charger off we actually see this higher voltage.

 

...............Dave

[nicknorman]

Yes it is not just as simple as a model comprising a resistor. As Dave says, you have to add an voltage source as well, and it is a lot more complicated even than that. But I think we should confine our theorising just to the discharge phase. I think it is intuitive to think that a freshly charged battery is capable of putting out loads of current with relatively little reduction in terminal voltage, whereas a well-discharged one is not, and is likely to exhibit a greater voltage drop when the same current load is put on.

 

Interesting Link, I had not read that section before. I would hardly call that curve Linear!!

 

Do you know if the off load terminal voltage of a battery will vary with temperature? or just the voltage required for charging????

 

..............Dave

No I don't know, and it seems hard to find definitive answer on t'internet. However it does make sense that it should, although I think with a lower co-efficient than the one we apply for charging voltage. Certainly though, we can say that the on-load terminal voltage will vary with temperature due to the differing rates of chemical reaction.

[Loafer]

The last time I took my Trojans down a bit, I noted that after a total of 187Ah drawn, from a 450 Ah bank, the loaded voltage was 11.67V. The load was generally our evening telly average, between 4 and 9A, with fridge and lights intermittent.

 

This is the sort of data that I'm looking for, purely for interest. Remember chaps, I'm only INTERESTED - I'm not trying to determine SOC from loaded voltage. I already know that from both 'SOC' and 'Consumed Energy' Ah readings on the Victron BMV600S.

 

Please take a look when you're at your normal lowest SOC, as SmileyPete and dmr kindly did for me, and let me know. Many thanks in advance.

 

but that will be Dave's reading with his 'normal' load, with his type of battery, with his age of battery, with his ...

 

How will that help you? (rhetorical)

It will help me! That is what I'm looking for. In Dave's case, it will be useful info unless he's running an unusual load like a power tool. He knows better than that though, and would remark on that.

 

Anyway, many thanks for ALL or your replies, I know what you were trying to get me to do. XXX

[cuthound]

I imagine you have to hypothesise a "chemical battery" inside the actual battery and this has some voltage. The charge current is then the difference between the applied charge voltage and the "chemical" voltage, divided by the total internal resistance.

A flat battery has a low voltage so will take a big charge current despite an extra bit of internal resistance.

A charged battery has a high internal/chemical voltage so takes very little current despite a lower internal resistance.. This makes sense to me because when we first turn the charger off we actually see this higher voltage.

 

...............Dave

Yes it is not just as simple as a model comprising a resistor. As Dave says, you have to add an voltage source as well, and it is a lot more complicated even than that. But I think we should confine our theorising just to the discharge phase. I think it is intuitive to think that a freshly charged battery is capable of putting out loads of current with relatively little reduction in terminal voltage, whereas a well-discharged one is not, and is likely to exhibit a greater voltage drop when the same current load is put on.

 

No I don't know, and it seems hard to find definitive answer on t'internet. However it does make sense that it should, although I think with a lower co-efficient than the one we apply for charging voltage. Certainly though, we can say that the on-load terminal voltage will vary with temperature due to the differing rates of chemical reaction.

Thanks guys,

 

That has cleared up my misunderstanding.

[dmr]

I will hopefully get a graph here later, it involves a bit of format conversion and messing about.

 

My measurements today suggest that with a light load (5-7amps) the battery terminal voltage is at most 0.1 volts lower than the Trojan Voltage Curve

With an 11 amp load its closer to 0.15 volts. The big unknown here is the temperature effect. Does the battery voltage do the same thing as the recommended charging voltage by unceasing at 30mV/degree? The batteries were only about 5 degreesC

 

So if you really started out at 100% and took out 187Ah that would be 58% state of charge, giving about 12.23 volts "rested".

If we take 0.1 to 0.15 volts off this for your load then that's 12.08-12.13 which is a fair bit higher than your figure of 11.67..

 

Do you measure your voltage at the battery or after an isolator/fuse etc? How cold were your batteries?.

 

.............Dave

[dmr]

Yes it is not just as simple as a model comprising a resistor. As Dave says, you have to add an voltage source as well, and it is a lot more complicated even than that. But I think we should confine our theorising just to the discharge phase. I think it is intuitive to think that a freshly charged battery is capable of putting out loads of current with relatively little reduction in terminal voltage, whereas a well-discharged one is not, and is likely to exhibit a greater voltage drop when the same current load is put on.

No I don't know, and it seems hard to find definitive answer on t'internet. However it does make sense that it should, although I think with a lower co-efficient than the one we apply for charging voltage. Certainly though, we can say that the on-load terminal voltage will vary with temperature due to the differing rates of chemical reaction.

 

I also have failed with the www.

In my model of the universe most chemical reactions get more lively with temperature. So I can see that we might need to charge a cold battery harder to get the reaction going, but conversely I would expect a hot battery to make more voltage on discharge. But maybe the reaction has to be endothermic one way and exothermic the other ????? (always preferred physics to chemistry at skool)

 

...........Dave

[Loafer]

I will hopefully get a graph here later, it involves a bit of format conversion and messing about.

 

My measurements today suggest that with a light load (5-7amps) the battery terminal voltage is at most 0.1 volts lower than the Trojan Voltage Curve

With an 11 amp load its closer to 0.15 volts. The big unknown here is the temperature effect. Does the battery voltage do the same thing as the recommended charging voltage by unceasing at 30mV/degree? The batteries were only about 5 degreesC

 

So if you really started out at 100% and took out 187Ah that would be 58% state of charge, giving about 12.23 volts "rested".

If we take 0.1 to 0.15 volts off this for your load then that's 12.08-12.13 which is a fair bit higher than your figure of 11.67..

 

Do you measure your voltage at the battery or after an isolator/fuse etc? How cold were your batteries?.

 

.............Dave

The voltage was measured by my BMV600S, which equates quite well to the several other voltmeters I have tried, applied directly to the battery terminals.

 

The reason for my interest stems from Trojan's own specifications, namely that their definition of 'capacity' is that the batteries will deliver 450 Ah by the time they reach 10.5V.

 

Therefore, I'm expecting that the loaded voltage vs Ah used should still to be within their spec, i.e. at 450 Ah used, I should see a minimum of 10.5V loaded voltage.

 

I'm wondering if my 11.67V after my 180-odd Ah 'used' is reasonable, or possibly normal.

 

Hence my search for other users able to provide similar statistics.

 

Please, also, if you have the means, can you refer to actual voltages seen, rather than saying they are 0.15V below Trojan's curves? That means I have to keep looking up the curves! Don't forget I'm a lazy c

 

Edited to remove part of a word

Edited February 2, 2015 by Loafer

[dmr]

This is the results of todays experimentation.

 

 

 

(sorry its not bigger, will try harder next time)

 

The solid line is the published Trojan Voltage vs Charge.

The dots are my measured battery voltage
I took state of charge from the Smartgage.
I also took battery voltage from the Smartgage.
Smartgage is not the worlds best voltmeter,and only displays to 0.05 resolution, and jumps about a fair bit.
When it was obviously jumping between two values I estimated an in-between value.
The discharge current also jumps about a bit so I made a visual average.
Current was mostly 4.5 to 7 amps, though one point was 7.8.
The very first point was 0.18 amp discharge before I turned the computer etc on.
The last point was 6 amps, but I then increased this to 11 amps for 5 minutes and took another reading.
The odd value at 67% is a mystery, I repeated it and got the same reading.
Battery temperature was low at 4 to 6 degrees C

So I conclude that a light load reduces the terminal voltage by a bit less than 0.1 volts.
BUT I have not done any temperature correction.
I was not able to go below 63% charge but there is no sign of any divergence over the measured range so I would be surprised if anything dramatic happens at lower charge.

 

...............Dave

[Loafer]

PS to dmr

 

Please don't go to any great lengths on my behalf! Just a couple of examples will be just great.

 

Has anyone else seen 11.6V, with an alleged 'plenty' of charge left in their batteries, at a 'normal boaty' 4-8A draw?

 

I'm sort of wondering if my Trojans are as able as I thought they would be. I cannot discharge to 30% because the solar panels interfere and keep them way above 30% (even in February!).

[nicknorman]

I noted that after a total of 187Ah drawn, from a 450 Ah bank, the loaded voltage was 11.67V. I'm not trying to determine SOC from loaded voltage. I already know that from both 'SOC' and 'Consumed Energy' Ah readings on the Victron BMV600S.

MAJOR MISUNDERSTANDING ALERT!!,

 

You most certainly do not know your SoC from the BMV600, you only know the consumed AH. This is because the BMV presumes that the battery has its original capacity. However if you had only 11.67v after taking out 187AH, unless that was with a very large load, you batteries don't have their rated capacity or were not fully charged (don't forget that the BMV shows 100% at some point depending on the settings, but certainly before it is fully charged). This is corroborated by the low specific gravity readings. I recommend you give them a good charge at 14.8v (adjusted for temperature) and then do an equalisation charge at at least 15.5v for a few hours, until the specific gravity stops rising.

[Loafer]

This is the results of todays experimentation.

 

So I conclude that a light load reduces the terminal voltage by a bit less than 0.1 volts.

BUT I have not done any temperature correction.

I was not able to go below 63% charge but there is no sign of any divergence over the measured range so I would be surprised if anything dramatic happens at lower charge.

 

...............Dave

That is brilliant work Dave, thank you very much. Virtual pint for you, but an ACTUAL greenie for your time!

 

It would appear from that graph then, that my Trojans are a bit under-performing.

 

Should I expect that, by virtue of them being almost new?

 

I'll continue to be very interested in others' examples. Thank you!

[Loafer]

MAJOR MISUNDERSTANDING ALERT!!,

 

You most certainly do not know your SoC from the BMV600, you only know the consumed AH. This is because the BMV presumes that the battery has its original capacity. However if you had only 11.67v after taking out 187AH, unless that was with a very large load, you batteries don't have their rated capacity or were not fully charged (don't forget that the BMV shows 100% at some point depending on the settings, but certainly before it is fully charged). This is corroborated by the low specific gravity readings. I recommend you give them a good charge at 14.8v (adjusted for temperature) and then do an equalisation charge at at least 15.5v for a few hours, until the specific gravity stops rising.

I agree completely Nick, I don't rely on the BMV's SOC readings. I only mentioned it above because the reading was available.

 

I know I started from 100%SOC because the previous charging tail current was less than 2% of the c20 capacity (450 Ah).

 

I also know that I saw -187Ah drawn, which must equate to 58% charge in a 450Ah bank innit?

 

(I think the BMV 'SOC' was higher than that, but I didn't record it, and neither does the BMV's history log, because I know how useless the figure really is.)

 

It's all a bit of an anomaly, one which I would like to get to the bottom of!

Nick and Dave, I have already done 2 Victron-controlled equalisation charges, once when new, and again yesterday (30 days later, as recommended by Trojan)

 

 

I have yet to discharge 'proper' since that last Eq charge, so the next few days might provide some more examples.

 

I'll be back with observations if any of you are still interested.

 

Thank you for that interest again, by the way.

[dmr]

The voltage was measured by my BMV600S, which equates quite well to the several other voltmeters I have tried, applied directly to the battery terminals.

 

The reason for my interest stems from Trojan's own specifications, namely that their definition of 'capacity' is that the batteries will deliver 450 Ah by the time they reach 10.5V.

 

Therefore, I'm expecting that the loaded voltage vs Ah used should still to be within their spec, i.e. at 450 Ah used, I should see a minimum of 10.5V loaded voltage.

 

I'm wondering if my 11.67V after my 180-odd Ah 'used' is reasonable, or possibly normal.

 

Hence my search for other users able to provide similar statistics.

 

Please, also, if you have the means, can you refer to actual voltages seen, rather than saying they are 0.15V below Trojan's curves? That means I have to keep looking up the curves! Don't forget I'm a lazy c

 

Edited to remove part of a word

 

>>> If you start at 100% and take out 187aH, with a light load I would expect about 12.1 volts, so your 11.67 is too low <<<

 

Big problem is that none of us knows how the cold is affecting our results (especially mine)

I might try to do something tomorrow. It means turning everything off but maybe I can do that when I walk the dog.

As Nick says an amp-hour counter can get things badly wrong and this may well have resulted in some sulphation.

(Equalisation will fix this).

 

Lets have a shot at working this out...

11.67 volts. Add 0.1 for your load effect gives 11.77. Thats about 28% charge (ouch) Add your 187aH (58%) and this suggests your batteries started out at 86% rather than 100%.

Did you say your specific gravity was just into the green? ( = 1.250? = 87% charge !!!!!!!!)

 

This is all a very rough estimate because you won't get the full 450aH with a cold battery.

 

.............Dave