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Dare I ask? Batteries question...

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I'm booked in for an 'engine and batteries' course with RCR in December, so I don't want to fully open the can of worms labelled 'batteries' if I can help it. But I think I'm going to have to prise open the lid just a little and peer inside...

So, we bought our boat six or seven weeks ago, complete with this little lot:

FUPpk-03Sq2BJT4NAE4uZQMtPEFyx-dXeGmdgm7RMJXrFxT3YBTG2TgimCxf2O2Le5Iacg1rmw8NLLA3XCFgeC_L2iO7UqBktGBJIKaMoSWJF6-4sN4Exu68GPkn-H4ldCTF50SN16od9v_sfFepyXwJCT4mc8J9CSMghMiHisnB_jVh483TsIsbGhtMrF_ng8tRQscab_BoEAD4j8BMvSIwBsaFtW_qMdP5xjtFvaqwgOpttAY5lEqqocmV_Q2SCTWncrPrRFrsaH3ol9Eu3P99oYinm7RJ_lOEAHMyGFam2JrfN1mNL8k17zr1rPV-RWek_ZAOvyLAUhGdnrwOMGZCeghg3B3LCLSE4guDM35Cyp4s9qzBcgusyMvCMoQUCHXD0fHJ-ofUl683qnbFiJ0odtSthjeb7pOJjOJ3JUcoqxrCAOASPE5SIGaJWwmWYPXB8EDIR0pv5UNQ3EGvWpSotLE37fWWpITYBYmRy4FsxBLlZzfWQmzdyky6EB8RGd3dQkoWuVDkduQrlft1y84Ogv9E9-4v25ajCHdHwk2URc431EYa1lf0IVpHcb6fXA6qRFqYPj7DgMSimYH2m0Xg23WegJmlLVKx0ucfQE4zZcXSHhQ-BRUpEXY9FyUyRhU4TFDWP050eDKt6CXAuIHyHB_I4L2oXg=w358-h637-no

I've been studiously ignoring them up till now, but realistically I'm overdue to at least check whether they need topping up with water. As I understand it this is simple enough in theory - unscrew each of the three white cap thingies on each battery and top up with distilled water to just below the level of the neck where the cap screws on - yes? But I believe you're supposed to make sure the batteries are fully charged before doing this, and I haven't got the foggiest how to check. Obviously I've tried searching but everywhere I look I seem to be assailed with dire warnings about the difficulty of the task and/or blinded with science ('specific gravity' etc.).

My hunch is that if we do it at the end of the 3-4 hour cruise we're planning for tomorrow, we should be fine; we did a similar cruise a few days ago and then left the boat without using any electricity, and we've been doing about 12-16 hours cruising every weekend since we bought the boat with fairly light electricity usage while on board. I'd say the engine has been run for at least 6 hours of cruising after every evening on board, and there's a Sterling alternator to battery charger fitted, so I can't see how we can have failed to keep the batteries charged. But if there's something simple I could/should be doing to make sure the batteries don't end up under- or over-filled with water...

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I check and top up my batteries every 205 engine run hours when I change the oil and filters. I find that is about the right time for me. You have good batteries there so look after them.

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I think I can see a red wire going to a negative terminal :D

This "only top up when fully charged" thing is way over played. Just get them reasonably charged, that will be fine. Topping up whilst discharged would most likely be ok.

Don't fill them to the top, that will give trouble, about 1/4 inch below the bit of plastic sticking down from the filing holes, but again this is not that critical, just judge it by eye

Can you send another picture with that cover right out of way so that we can see the other terminal? then we can discuss (at length) optimisation of battery bank wiring.

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

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4 minutes ago, dmr said:

Can you send another picture with that cover right out of way so that we can see the other terminal? then we can discuss (at length) optimisation of battery bank wiring.

 

Not that it matters much in real life! 

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8 minutes ago, Mike the Boilerman said:

 

Not that it matters much in real life! 

Trojans have fewer but thicker plates so are less able to deliver high currents. This infers a higher (effective) internal resistance. Therefore are Trojans more or less susceptible to non optimised battery bank configurations......discuss?

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

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Those are Trojan batteries, topping up instructions here:

http://www.trojanbattery.com/tech-support/battery-maintenance/

Some safe handling information here:

http://jgdarden.com/batteryfaq/carfaq.htm

If you're planning to live off grid 24/7 then there's probably a bit of a learning curve but the RCR course should cover much of it.

1 hour ago, dmr said:

Trojans have fewer but thicker plates so are less able to deliver high currents. This infers a higher (effective) internal resistance. Therefore are Trojans more or less susceptible to non optimised battery bank configurations......discuss?

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

Maybe not the best topic to discuss it in, it will lead to 'expert hell' for the OP :D

Edited by smileypete

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1 hour ago, dmr said:

Trojans have fewer but thicker plates so are less able to deliver high currents. This infers a higher (effective) internal resistance. Therefore are Trojans more or less susceptible to non optimised battery bank configurations......discuss?

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

 

Good question!

Let's imagine cable configurations identical on two imaginary banks of batteries, and that the (tiny) resistances of every single crimped joint and every single clamped battery terminal junction in each set are also identical (vanishingly unlikely).

On the other hand let's not. Let's have a nice cup of tea and a sit down.

http://www.nicecupofteaandasitdown.com/

 

 

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18 minutes ago, Mike the Boilerman said:

 

Good question!

Let's imagine cable configurations identical on two imaginary banks of batteries, and that the (tiny) resistances of every single crimped joint and every single clamped battery terminal junction in each set are also identical (vanishingly unlikely).

On the other hand let's not. Let's have a nice cup of tea and a sit down.

http://www.nicecupofteaandasitdown.com/

 

 

What about the large resistances due to a bloomin' terminal not being tightened up enough? :-/ 

Maybe another reason to run with as few batteries as you can.

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55 minutes ago, Mike the Boilerman said:

 

Good question!

Let's imagine cable configurations identical on two imaginary banks of batteries, and that the (tiny) resistances of every single crimped joint and every single clamped battery terminal junction in each set are also identical (vanishingly unlikely).

On the other hand let's not. Let's have a nice cup of tea and a sit down.

http://www.nicecupofteaandasitdown.com/

Nonononono!

Do not imagine anything, just get a clamp meter out... 2 mins of measurements, then have a well earned cuppa.

An ounce of real world measurements is worth more than a pound of supposed expert theoretical wibble - aka 'expert hell' :)

Edited by smileypete

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13 minutes ago, rowland al said:

What about the large resistances due to a bloomin' terminal not being tightened up enough? :-/ 

Maybe another reason to run with as few batteries as you can.

 

Exactly. I suspect the resistances of the various and occasionally poorly made interconnects, will dwarf the internal resistances of the batteries, however clever the configuration. 

But I don't know as I've never measured any.

 

2 minutes ago, smileypete said:

Nonononono!

Do not imagine anything, just get a clamp meter out... 2 mins of measurements, then have a well earned cuppa.

An ounce of real world measurements is worth more than a pound of supposed theoretical expert wibble - aka 'expert hell' :)

 

Yeahbutnobut.... how will measuring the currents in only one installation help answer Dave's question?!

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9 minutes ago, Mike the Boilerman said:

 

Exactly. I suspect the resistances of the various and occasionally poorly made interconnects, will dwarf the internal resistances of the batteries, however clever the configuration. 

But I don't know as I've never measured any.

 

Well if most electrical issues are down to poor or broken connections, maybe 'Check all your connections are tight' could be in your V1 list? :)

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17 minutes ago, Mike the Boilerman said:

Yeahbutnobut.... how will measuring the currents in only one installation help answer Dave's question?!

If the problem is waaaayy less than expected, it does points to something being amiss with the expert theoretical wibblings...

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7 minutes ago, rowland al said:

Well if most electrical issues are down to poor or broken connections, maybe 'Check all your connections are tight' could be in your V1 list? :)

 

How did you reach THAT conclusion?!!!

I was commenting that poorly made connections have a higher resistance and a higher significance than the exact wiring configuration of Trojans, or batteries of lower internal resistance. A very fine and narrow point, in response to DMR's fine, narrow and very theoretical question. 

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3 hours ago, dmr said:

Trojans have fewer but thicker plates so are less able to deliver high currents. This infers a higher (effective) internal resistance. Therefore are Trojans more or less susceptible to non optimised battery bank configurations......discuss?

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

Less to do with the batteries internal resistance and more to do with the plate surface area. The greater the plate surface area exposed to the electrolyte, the more chemical reaction can take place and the more current can be delivered in a short space of time.

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Just now, cuthound said:

Less to do with the batteries internal resistance and more to do with the plate surface area. The greater the plate surface area exposed to the electrolyte, the more chemical reaction can take place and the more current can be delivered in a short space of time.

Yes, but if the battery can't deliver high currents (for whatever physical reason) then the voltage Must fall when a high current is demanded, and R=V/I   but this is why I said "Effective" resistance. I am pretty sure the volt drop is non ohmic so maybe should not be called resistance but I'm saving that one for another day.

However the main theme of the post was pure mischief, to create further hot air over the battery bank configuration debate :D.

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

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3 hours ago, dmr said:

Trojans have fewer but thicker plates so are less able to deliver high currents. This infers a higher (effective) internal resistance. Therefore are Trojans more or less susceptible to non optimised battery bank configurations......discuss?

The question has a simple answer. Less. 

 

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10 minutes ago, cuthound said:

Less to do with the batteries internal resistance and more to do with the plate surface area. The greater the plate surface area exposed to the electrolyte, the more chemical reaction can take place and the more current can be delivered in a short space of time.

I was also just thinking this morning that the smaller plate surface area on 'domestic' batteries also means they won't take as much current when charging. If you never connect to a shore line and rely on the engine and alternator to charge them up, this could be a problem unless you run the engine for long periods or have a very low power usage. 

This is why I wonder whether starter batteries are more suitable in that situation?

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8 minutes ago, dmr said:

Yes, but if the battery can't deliver high currents (for whatever physical reason) then the voltage Must fall when a high current is demanded, and R=V/I   but this is why I said "Effective" resistance. I am pretty sure the volt drop is non ohmic so maybe should not be called resistance but I'm saving that one for another day.

And you’re 100% correct. There’s no such thing as the battery’s internal resistance but we refer to the ‘effective’ internal resistance to make the maths nice and easy. 

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2 minutes ago, rowland al said:

I was also just thinking this morning that the smaller plate surface area on 'domestic' batteries also means they won't take as much current when charging.

But that ain’t the case :)

Lead Acid Batteries will generally take as much current in Amps as the ‘missing’ charge in Ah. So a 50% charged 200Ah battery will accept somewhere around 100A. When it’s 75% charged it will take around 50A and so on. 

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2 minutes ago, WotEver said:

And you’re 100% correct. There’s no such thing as the battery’s internal resistance but we refer to the ‘effective’ internal resistance to make the maths nice and easy. 

Success! I have totally derailed an honest battery question thread and turned it into a lovely CWDF battery discussion :D

The internal "wiring" and conductive parts of the plates will constitute a true internal resistance but the chemical side of things is at best an effective resistance, it might even be an impedance with a long time constant, but I suspect that the "equivalent circuit" of a lead acid battery would be a non linear nightmare. Where's that Gibbo when you need him?

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

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3 minutes ago, dmr said:

Success! I have totally derailed an honest battery question thread and turned it into a lovely CWDF battery discussion :D

The internal "wiring" and conductive parts of the plates will constitute a true internal resistance but the chemical side of things is at best an effective resistance, it might even be an impedance with a long time constant, but I suspect that the "equivalent circuit" of a lead acid battery would be a non linear nightmare. Where's that Gibbo when you need him?

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

Nahh... the equivalent circuit is a voltage source with a resistor and capacitor in series ;)

Edited by WotEver
Fixed a typo

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3 minutes ago, WotEver said:

But that ain’t the case :)

Lead Acid Batteries will generally take as much current in Amps as the ‘missing’ charge in Ah. So a 50% charged 200Ah battery will accept somewhere around 100A. When it’s 75% charged it will take around 50A and so on. 

I would have thought that the smaller the total plate surface the less charge it will take due to there being less chemical effect. 

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4 minutes ago, WotEver said:

Nahh... the equivalent circuit is a voltage source with a resistor and capacitor in series ;)

But if you really want to go all esoteric...

https://www.mathworks.com/tagteam/40542_SAE-2007-01-0778-Battery-Modeling-Process.pdf

2 minutes ago, rowland al said:

I would have thought that the smaller the total plate surface the less charge it will take due to there being less chemical effect. 

Yup, you’d think so, but it isn’t the case. 

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22 minutes ago, WotEver said:

The question has a simple answer. Less. 

 

Correct! The battery’s “internal resistance” (aka chemical reaction speed) is a greater proportion of the overall resistance and thus minor variations in interconnect resistance become less significant. Consider for example our battery bank, with its 70mm^2 interconnects, at 50% SoC, Rested voltage ~12.2v. Add 2 volts extra from charging device, current of say 150 A flows. Voltage drop in interconnects less than 0.005V. Voltage drop across battery around 1.995v. The bigger the ratio between battery voltage drop and interconnect voltage drop, the less the already insignificant effect of differing interconnect lengths is.

Edited by nicknorman

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