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Go Back   SailNet Community > On Board > Gear & Maintenance
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  #41  
Old 09-08-2006
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cardiacpaul is a jewel in the rough cardiacpaul is a jewel in the rough cardiacpaul is a jewel in the rough
You said...

"Cardiac? "gimme".
Sorry, Santa says you'll have to get them yourself. I'm not going to find your local yellow pages and chase down your local industrial suppliers.
What I'm sailing on right now requires a lot less power, I go A/B and alternate two Group31's and that's enough for my modest needs. No reefer, no stereo, and no big budget to be invested. Now, seize on that and proclaim that means I'm unable to do the math and planning for a larger installation, by all means. That would be like saying the design team for the International Space Station are unqualified, because they've never been in orbit before."

I have a double major, in CompSci and Math, so I'm more than qualified to separate theory from practice, and practice from BS.
You provided no sources (not even from your own town) no real world, just theory. your "cred" just tanked. Sorry.

BTW, no offense to anyone, but do you know the difference between a Theory and a hunch? ... an MBA. I'm going sailing.
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  #42  
Old 09-09-2006
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Dave-
"BTW, that current represents energy going from one battery to the other. "
Not really. It is modelled on a resistive load and resistive loss, the weaker battery might as well be a light bulb as I understand it.

But I'm surprised no one put two and two together. Run the model, and now run it with the case of one bad cell in one battery. What's a bad cell? Well, if it fails open...that battery is out of the circuit. If it fails by shorting (an internal short, which appears to be the most common reason cells fail)...Ah, Dave, I would have thought you'd see that one by now.
Battery with all good cells: 12.6V. Battery with one bad (shorted) cell, 10.4V. I figure that one shorted cell, which normally would "just" weaken one battery, would now cause a 275 Amp current to flow down from the good battery.
That's gonna hurt, no matter where you think the current goes. Especially if the batteries don't each have a primary fuse right on them. And in two paralleled batteries...that's what will happen if one cell in either of them shorts out. A potential problem--and fire--that the series cells wouldn't have.
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  #43  
Old 09-10-2006
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Hellosailor,

“"BTW, that current represents energy going from one battery to the other. "
Not really. It is modelled on a resistive load and resistive loss, the weaker battery might as well be a light bulb as I understand it.”

Then why does a battery charge? If it only represented a resistive load (as your comment indicates) then there would be no such thing as charging.

Your model is incomplete. Yes there is resistive component, but much more then that is the capacitive element of the model. A battery is roughly modeled by a capacitor of 1 farad per amp hour in series with the resistive component. The capacitor prevents the battery from behaving in the way you describe. The resistive element dissipates power while the current is flowing, but if you calculate the wattage dissipated using W=(I*I)*R, you will discover that the wattage lost is small, very small at the currents between batteries when paralleled. The capacitive element stores the charge transfer from one battery to the other and prevents the battery for behaving like you describe.

Just as a preliminary report, the batteries under test drew 300 mA when first connected and now after 24 hours is drawing 5 mA. There has been no current reversal, just a steady geometric regression to zero. I will continue the experiment for the next half week.

I now understand why you don't think paralleling batteries work. Reexamine your objections based on this new model. Yes this model is correct.
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  #44  
Old 09-10-2006
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Dave,

Good job! And very useful info. Thanks for your efforts on this question.

Bill
S/V Born Free
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  #45  
Old 09-10-2006
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Right on Dave

I have been waiting for you to fire the capacitor effect of batteries. In fact they make great filters.

Seems you are holding your own so I'll just lurk and enjoy the show.
This is almost like writing my paper for my Masters in electronics.

Good job and fair winds

Cap'n Dave
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  #46  
Old 09-10-2006
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umm...

Cap'n,

Just trying to understand his objections, I made the mistake of not defining basic models and getting everyone on the same page, and I know better.

This has been enjoyable and allowed me to reorganize concepts that I have been using for years and haven't had to formally present for a decade. I hope everyones getting the understanding they need for their own system. I have tried to keep it from getting too technical so everyone can follow. Let me know if I stray.
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  #47  
Old 09-11-2006
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I paralleled 3 of 100 Ahr Interstate "Deep Cycle" batteries in Houston in 1992, and they lasted me 10 years. That wasn't too bad, really, and they were not expensive. I have a dedicated starting battery also.

I bought a great big single unit 275 Ahr beast in 2002, and it cost me a whopping Ł325.... about $550. It had kevlar construction, or something marvellous like that.

I have had it 4 years and it refuses to lose charge (noticeably) if you leave it for months.

Long may it continue.

On the earlier threads.... I never would parallel a good battery with a known under-performer.

The chap was right about paralleling not making a difference to discharge.... it really does not. If you have 300 Ahr available and draw 1/3 of it, that is eaxctly the same as drawing the parallel bank down by the same amount, no matter how many batteries you have in parallel.....assuming both scenarios have the same capacity.

I only run the fridge when the engine is running, so tha battery is rarely discharged more than about 24 Ahr. Most of the sailing we do no is just in a Scottish canal. It must be very different for the long haul merchants in the tropics though.
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Old 09-15-2006
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Experiment Results

I have completed the experiment as far as I am inclined to at the moment. I thought that you would like the results. I will be unavailable for the next week and a half, not sailing, but my daughters wedding. I will respond to any questions on my return, or just let this thread die it there is nothing further. I hope this has been useful to some.

Both batteries were fully charged over several days (< 0.1 amp draw @float voltage) and left dormant, open circuit without charge or load, for 48 hours. The open cell voltage was then recorded. The negative terminals connected together using an 8 AWG cable. The positive terminals where connected together using a Fluke 8060A 5 ˝ digit (three decimal places, at the 12 V range. It will measure 1/1000 of a volt) DMM in current mode measuring in the 2A to 1 milliamp range between both positive terminals. I recorded the current every 15 seconds for the first hour and a half and then at increasing intervals. The temperature of the batteries was also recorded.

Battery #1 had an open circuit voltage of 12.628V and battery #2 an open circuit voltage of 12.681V before the start of the test. At time zero, at initial connection the current between the batteries was 30.35 mA (I had stated before that the current was 300 mA and I apologize, I was reporting from memory, never a good idea) [1 mA = 0.001 Amps] The scale on the DMM was reduced to 200 mA.

The results are shown in the attached graph. Additional observations are that there was a current reversal where the temperature and other fluctuations within the batteries caused changes in the open cell voltage. The current measured at this point was less then 1 mA and stayed less then 1 mA. (Just a quick calculation, IF the 1mA drain was into a resistive load and NOT into the capacitive load of the battery, i.e. the 1 mA was lost and not recovered, that current drain would represent a battery life in excess of 15 years with no recharging [discounting self discharge]. What this means is that this drain is less then the self discharge rate). The battery temperature remained stable at between 77 deg F and 80 deg F, fluctuating with the ambient temperature. The final voltage of the batteries at the end of the test were battery #1 = 12.629V and battery #2 = 12.681V. Battery #1 is the one at 90% capacity and battery #2 the one at 55% capacity. Yes, the open cell voltage for the bad battery was higher then the good battery.
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  #49  
Old 09-15-2006
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dave-

Just curious, why the huge gap in data points?
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Old 09-15-2006
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