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Conventional alternative energy systems => All other conventional alternative energy creation systems => Topic started by: dandman on April 18, 2011, 01:52:04 AM
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I have discharged 2 batteries 3 times for 1amp hour at a time and recharged those batteries from another single battery back to their original charge state for the loss of .01 of a volt per charge hour. IS THIS UNUSUAL
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Maybe Im short on detail so here goes. The discharging batteries started at 12.64v and were discharged for 30 mins at 2.1 amps. Then connected to the feeding battery, which started at 12.83v. The discharged batteries took 2 hours to be returned to 12.64v and feeding battery on completion had dropped in charge to 12.81v. This was done 3 times without interval between discharge and recharge and the subsequent drops in voltage of the feeding battery were .01v per hour. In my untrained ignorance, I am pretty sure that is more than 1 from 1 but would appreciate an educated persons plain language explanation if it is not. Happy to discuss further and thanks in advance.
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You ever heard the term, A bad battery will make a good battery go bad?
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Thanks for the sarcasm. I charged these batteries fully and discharged them at 68.8 amp running a 700watt 240v drill through the 1kw inverter before starting this alternative way of recharging them. Does anyone have any educated input please .
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I don't think are including the amps in your measurements. You may show a certain voltage but you need to know the amps as well. The best way is to load test the battery used to charge both before and after and then load test the ones charged before and after you think they are fully charged. Once you do that, you will then find out exactly what was lost.
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Great, thanks for the input. The amperage reading is .7 of an amp from the feeding battery. When I put a 10w 12v bulb across the feeding battery directly, its voltage decline is hugely different to when it is feeding the other batteries but I am doing this ever so slightly differently to just connecting them in parallel. Any thoughts on why the battery may be acting differently?
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Great, thanks for the input. The amperage reading is .7 of an amp from the feeding battery. When I put a 10w 12v bulb across the feeding battery directly, its voltage decline is hugely different to when it is feeding the other batteries but I am doing this ever so slightly differently to just connecting them in parallel. Any thoughts on why the battery may be acting differently?
What is the " ever so slightly differently to just connecting them in parallel " ? That may be a key factor. Overall judging a battery's power by it's open circuit voltage reading is not a good way to determine it's available power. Rather it's all about watts as in amps times volts. Are these lead acid batteries? I'm not an expert on this but I'm sure some others here may be able to tell you the best way to determine if you have something anomalous here. Getting accurate measurements of actual power used and power added to other batteries can be more difficult than it seems.
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I am introducing a degree of resistance. These are standard lead acid batteries 2 x 600cca and 1 x 500cca. I understand its about watts. If I discharge at 2.1 amps over 30 minutes then recharge for 2 hours to return to the original voltage and repeat the same thing 3 times over the same time periods. Is that not an accurate enough measure of watts used watts reinstated through charge?. The anomally or query is about the apparent lack of discharge rate of the feeding battery in this situation as opposed to under normal circumstances?. I am pretty sure I could recharge a torch battery more than twice the total value of the 2 batteries used.
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I dont think you understand what I said. You have to load test the battery's to know how may amps they contain. I am assuming you are working with 12 volt car battery's, am I right? If so, the .7 amp reading you have must be what the dead batterys are drawing from the charged battery.
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Yes I am using 12v, lead acid car batteries. I dont know how to load charge a battery. So Im relying on deductive reasoning and what what knowledge I do have. Please see my last reply and let me know if that rationale is sufficient?. It will also give an indication of why I believe there is an anomally in readings of amperage versus actual energy used from the feeding battery. Thanks for bearing with me but I am self taught and do appreciate the feedback.
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Hi dandman, thanks for sharing your tests.
Could you explain what the little difference in charging is.
Also, i assume the two 12 volt charging batteries in parallel and the input 12 volt battery is in parallel with the two charging batteries.
Or are you using some kind of boost charger in between them.
Also, have you load tested in anyway the single 12 volt input battery to see if it can still carry a decent load or could it be some surface charge making it appear as if it has not lost much capacity, thanks.
peace love light
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Yo skywatcher. I have only tested the discharge rate of the single battery. it discharges at the same rate under the same load over the same amount of time as it did before the recharging. As mentioned previously, I am introducing resistance while charging. There appears to be no degradation to the single battery's ability to carry the same load as it did previously. I believe, that I may be deliberately
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No for the love of putunia's,
of course transfers are not lossless
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Are you saying that you charge batteries by hooking them in parallel?
If so I have seen this effect before in capacitors. One capacitor has more charge than the other capacitor and hooking them up in parallel, positive to positive and negative to negative the capacitors will charge to a higher voltage than what the original capacitor had.
http://www.youtube.com/watch?v=wbn4vede2us
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Hi, It's not simply hooking them up in parallel. I believe Im forcing energy to back up in the charging batteries. The losses are the value of the resistance I mentioned earlier instead of the value of the energy discharged.
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Hi All, Thanks for the feedback. It's been invaluable in helping me to prove that I am doing what I think I am. What I now am looking for is commercial interest. If you know anyone, feel free to point them in my direction. Please keep the input coming.
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Greed=bad.
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It makes the world go round and everyones gotta eat
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If you load an 12,9volt battery with exactly 12,0 volt ... so this battery take not any load , not any amps , not any power. !
The battery will only load /fully) IF YOU load wit 20% PLUS in Voltages.
so 13,8 to 14,2 volt must be load !!
Also the energy that you UNload before ..
you MUST charge up with 20 % more Energy To the battery again.
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The load of your battery from (example) 10,0 to 12,0 volt , have NO SENSE - NO any Power (only "unergergized voltage!
So i hope, by bad language is technical to understand.
Gustav Pese
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Hi Pese. Thanks, understand that. So we are agreed that any voltage above 12v is power. So, what I am achieving is discharging 2 x 12v batteries from 12.64v to 12v and recharging them to within 98 percent of the supplying battery's value over and over. I am introducing resistance in a unique way such that the supplying battery loses the value of that resistance and not the value of the amperage it is replacing. I do not claim to be able to fully recharge the batteries.
If you attach 2 batteries in parallel the stronger battery will drop and the weaker rise until they appear to hold an equal voltage. This system does not. It allows use and recharge of one battery in excess of the total power available in the 2 batteries used. If it sounds too sensational or upsets the values of the scoffin boffins then they can think of it as more efficient use of existing chemical energy. Cheers