Easiest way to get a C.O.P of 1.8

[mbrownn]

Farmhand,

What is happening to the 24v voltage on the two batteries in series while the other two batteries are charging up?

If the cop is even equal to 1 then none of the batteries would ever go down in voltage, is this your experience?

Do the battiers have to be manually swapped around over time and when?

Hopes and dreams......

Todd

The tests I did gave a charging efficiency of around 70% so when we flatten two source batteries in series we get two parallel batteries 70% charged

[Farmhand]

Your calculations are incorrect Farmhand.

It is good to see that people are finally catching on to the fact that the circuits can be made to do much more work rather than hooking a battery to it and thus 'killing' it and then charging it from a separate source...HOWEVER...Farmhand, I have run these experiments a long time ago (and much more complex also with radiant methods of pulse charging Parallel bats from a series battery connection.

THERE ARE STILL LOSSES.

1) The inverter is nowhere near 100% efficient.
2) Losses at the load (your light bulb/motor whatever), in heat and conversion to other types of energy.
3) LOSSES in CHARGING the two batteries. When you charge ANY battery, this is a imperfect chemical process and not like taking two coins from one bucket and putting them in another bucket...the 'transfer' is horriblly low depending on your charge method.

I GUARANTEE that your method is not COP 1.8 and not even COP 0.9.

Its efficiency is better YES...but you need to top up the batteries with solar or something.

TP

hi all, At this point I can't argue with that. And I do agree that there will be losses there must. I think my mistake is to use the term C.O.P. .

The theory i'm going by is that the conserved or saved charge once dissapated this way will get me 1.8 times the amount of work for the same amount of charge actually drained from the batteries overall however long it takes both ways.
What would that be a measure of ?

After running the setup with the series batteries starting at say 12.6v and the parallel one's at 12.1v when the parallel one's get to say 13.4v I will stop and connect the parallel ones in series and the series one's in parallel then run it back the other way and keep doing it untill all the batteries are at 12.1v. Then I can compare that to how much work I can get from the same load by just draining 2 batteries in parallel from 12.6 to 12.1 volts where there is no saving and see what i get, doing that would be using the same amount of battery charge for I predict more work, which should in my opinion increase the C.O.P. but if it doesn't then C.O.P is a useless measurement to me.

I will call it the E.W.P. the "effective work performed". And I will just use it as a comparison of conventional use to this way.

If there is an improvement of efficiency this way compared to the other way, then what is saved continue's the use for longer. Giving a longer effective work period for the same load.

I predict I will get 1.8 time's the work this way to conventional. And if it does give me 1.8 times more work but does not improve the C.O.P. then I don't care for C.O.P. and it has no use for me here.

I'm testing now, but it'll take a while.

Load is a 25 watt bulb on the small inverter. I am getting a control by using the load from two batteries in parallel from 12.6 volts to 12.1 volts to get a time. Then I will start the run test with the two series batteries at 12.6v and the two parallel batteries at 12.1v and keep running the load while manually changing the side's from series to parallel and so forth untill all batteries are at 12.1v which will be effectively using the same amount of battery capacity,
but it should last 1.8 time's as long hopefully.

Well thats the plan. My understanding of what C.O.P actually is was wrong. And it doesn't make much sense to me as a measurement to tell the truth.

I am more interested in the work i can get done from the energy availiable. It is impossible to make energy more than it is. But it can be used more efficiently.

Cheers

[Farmhand]

Farmhand,

What is happening to the 24v voltage on the two batteries in series while the other two batteries are charging up?

If the cop is even equal to 1 then none of the batteries would ever go down in voltage, is this your experience?

Do the battiers have to be manually swapped around over time and when?

Hopes and dreams......

Todd

Hi Todd, The two series batteries are draining of course. I will change the thread title.

I don't think a C.O.P measurement does justice to this arrangement as I stated in the previous post.

Cheers

P.S I changed the thread tiltle, I wouldn't want to be intentionally missleading. However the title didn't change. Yet.

[Farmhand]

The other thing I would like opinions on is if people think the power should be calculated using 12 volts or 24 volts when calculating the power through the inverter or the power used.

It has come to my attention that there are some who say that it should be calculated using 24 volts. I can't say for sure but I would say it should be calculated using 12 volts times the amps measured for a power calculation of the power taken from the series batteries.

I think it should be considered the same as a regulated 12 volts because the inverter is powered by 12 volts. But I might be wrong. It has been implied that the parallel batteries are part of the load and the power should be calculated using 24 volts.

Any opinions on that ?

I dunno. But I would like to hear what others think.

I wonder could Aaron change the thread tiltle for me to. "Easiest way to increase work performed from batteries."

Cheers

[Farmhand]

When using the Tesla type battery circuit on my SSG I get about 70ish percent charging efficiency on the back end batteries, 95% on the Bedini style and about 25ish percent mechanical energy, close to a COP of 2.

The figures were slightly better with my Imhotep/Bedini fan in the same setup but as the numbers were much smaller the margin of error was bigger.

Having not physically tried out my AC supply idea I am a little worried about hitting the batteries with too much power. So does anyone have any advice before I hook this up?

I think that the load will limit the power so if I use a 5W transformer as a load then only 5 watts can be fed into the battery. The voltage drop across the transformer should be enough to bring it down to a safe level to hook up to a 12v battery, am I correct?

Again I believe the size of battery will have to be matched to the load.

I am thinking of having a battery/inverter system for each load, maybe with a little ingenuity it can be made like a UPS.

I don’t think the electric company here would have a problem with people reducing their consumption as the power stations are stretched beyond their limits.

On the efficiency of inverters I assumed around 70% if matched to the load. 70% of 70% is 49%, more or less 50%

PS the voltage supply here is 220 to 240v 60Hz

So will you use a stepdown transformer across the AC supply then a FWBR then the batteries ? I'm not qualified to offer anything about the mains supply. There is a guy here that works with the mains supply all the time and his advice to me is to leave the mains alone. Messing with ignition coils at 40 000 v plus is less dangerous he says. Sounds Ok to me though. Any chance of a quick sketch of the setup ? If it works well.

Unfortunately for me I have been electrocuted by low voltage high current numerous time's having worked as a boilermaker for a few years. During my apprenticeship it was always a good laugh for us to get someone to hold a job together for tacking then shock them under the armpit with the electrode while they are holding the job "ground" Haha. With about 80-100 amps not sure how many volts not many. Gota watch out for the flying arms and parts of the job they were holding though, it's quite dangerous for both.

240 is a different story though.

Cheers

[Farmhand]

The thing that happens when electrocuted by an AC welder back and forth through the arms while actually holding the job in one hand and the electrode in the other is the muscles in the chest arms and hands contract at 60Hz, so it is very difficult to let go of what you are holding. As time seems to slow everthing go's dark and funny noises emanate from the person. Not much fun. And a very scary experience, I did it once but when I fell over the earth clamp was pulled from the job, which open circuited me, and everything became bright again. Then I let go of the electrode, and the pipe I was holding. Lucky escape.

I'm starting to think or I did think it a while ago, that that may have been the cause or a contributing factor to the problem I have with damaged discs in my neck and upper back. Who knows. Anyway I don't recommend it for therapy. It didn't do much for me.


Anyway enough humor.

The control test is nearly done. I started the control test with 2 batteries in parallel at 12.65 volts and run the load from 5.15pm yesterday to 6.15am this morning thats 13 hours work from two batteries taking them from 12.65v resting to 12.10v while still under load.

I will keep discharging the parallel batteries untill they will rest at 12.10 volts then I will start the run test with the two series batteries at 12.65v resting or a fraction under and the other two at 12.10v resting and run back and forth changing the sides with no external input untill all batteries read 12.10 volts under load.

That should be a fair test.

Any other suggestions to get a fair comparison are welcome.

Cheers

[wayne.ct]

The other thing I would like opinions on is if people think the power should be calculated using 12 volts or 24 volts when calculating the power through the inverter or the power used.

It has come to my attention that there are some who say that it should be calculated using 24 volts. I can't say for sure but I would say it should be calculated using 12 volts times the amps measured for a power calculation of the power taken from the series batteries.

I think it should be considered the same as a regulated 12 volts because the inverter is powered by 12 volts. But I might be wrong. It has been implied that the parallel batteries are part of the load and the power should be calculated using 24 volts.

Any opinions on that ?

I dunno. But I would like to hear what others think.

IMO, you will have a confusing and probably inaccurate picture if you use a round figure like 12 or 24. The actual voltage will change over time, as I am sure you know. But, you can't be sure what exactly you are measuring when you have AC or intermittent or spikes on a battery terminal. Your results would be more convincing to me if you had a high frequency capacitor across the battery terminals of each battery to buffer the battery voltage. This would reduce the measuring of spikes at the battery terminals. You would not lose any power in a capacitor but it would help assure that you are actually measuring the EMF of the battery and not some spikes. Use a scope, if possible to see the nature of the potential on the battery terminals.

When computing power follow the basic rule of dividing your circuit into SOURCE and LOAD. Assume, for example, that the series batteries make up the source and you have two loads in series, one of which is the two parallel batteries. Do you have one current or two? If I understand correctly what you are doing, you have only ONE current that you need to measure. Measure all the voltages and the current or currents at the same time or at least as quickly as you can. Measure across each component of the load, the parallel batteries and the inverter. You will end up with some algebra problems to calculate the power dissipated in each part of the circuit. For good measure, calculate the power going into the inverter and the power (current and volts) out of the inverter.

If head and light are the work-goal you want then you should be able to create a lot of heat and light either way. I will be interested to hear whether you get more with your approach than if you go a more conventional route.

[Farmhand]

IMO, you will have a confusing and probably inaccurate picture if you use a round figure like 12 or 24. The actual voltage will change over time, as I am sure you know. But, you can't be sure what exactly you are measuring when you have AC or intermittent or spikes on a battery terminal. Your results would be more convincing to me if you had a high frequency capacitor across the battery terminals of each battery to buffer the battery voltage. This would reduce the measuring of spikes at the battery terminals. You would not lose any power in a capacitor but it would help assure that you are actually measuring the EMF of the battery and not some spikes. Use a scope, if possible to see the nature of the potential on the battery terminals.

When computing power follow the basic rule of dividing your circuit into SOURCE and LOAD. Assume, for example, that the series batteries make up the source and you have two loads in series, one of which is the two parallel batteries. Do you have one current or two? If I understand correctly what you are doing, you have only ONE current that you need to measure. Measure all the voltages and the current or currents at the same time or at least as quickly as you can. Measure across each component of the load, the parallel batteries and the inverter. You will end up with some algebra problems to calculate the power dissipated in each part of the circuit. For good measure, calculate the power going into the inverter and the power (current and volts) out of the inverter.

If head and light are the work-goal you want then you should be able to create a lot of heat and light either way. I will be interested to hear whether you get more with your approach than if you go a more conventional route.

Hi Wayne, That is a very good explanation for measuring the power I think I can follow that but I need more accurate equipment. I can borrow some. And I will more than likely get the calculations wrong a few time's. But your explanation make's sense to me. Thank's.

The amount of total work is the task at hand for me at the moment, I would like to get a real life look at all of the work done in comparison in a fair way.
And to my dismay it took a whole hour of draining at 2.6 amps to move the parallel battery's from 12.10v to 12.09v it was painfull to watch. So I must give another hour to the the control test. 14 hours conventional from 12.65 to 12.10v with the control load.

I'll keep track of and deduct time for swaps during the run test.

Cheers

[Farmhand]

Directory:Pete's "Zero Amp" Electronics - PESWiki

Boguslaw that is an interesting link. You don't miss anything . Cool.

I think there might be a couple of reasons he can actually do that.

Doesn't the battery charger he is using have a battery in it ? One of those jump starter charger type things. I can't tell so I can't say either way.

Interesting to think about and I don't think it is as hard as it would seem.

Very interesting. Thanks for posting that.

[mbrownn]

So will you use a stepdown transformer across the AC supply then a FWBR then the batteries ? I'm not qualified to offer anything about the mains supply. There is a guy here that works with the mains supply all the time and his advice to me is to leave the mains alone. Messing with ignition coils at 40 000 v plus is less dangerous he says. Sounds Ok to me though. Any chance of a quick sketch of the setup ? If it works well.

Unfortunately for me I have been electrocuted by low voltage high current numerous time's having worked as a boilermaker for a few years. During my apprenticeship it was always a good laugh for us to get someone to hold a job together for tacking then shock them under the armpit with the electrode while they are holding the job "ground" Haha. With about 80-100 amps not sure how many volts not many. Gota watch out for the flying arms and parts of the job they were holding though, it's quite dangerous for both.

240 is a different story though.

Cheers

No it is a series arrangement, The AC load can be any appliance. In most cases the load is a transformer or fluorescent tube so I used a transformer as an example.

The bridge rectifier is in series on the neutral side, of course there is no circuit unless a battery is connected to the bridge rectifier.

I have been running computer simulations and everything seems to be fine other than the voltage seen by the appliance is lowered by the battery voltage, this shouldn’t be much of a problem for me with a 220 to 240v supply but may be more significant in the US with an 110v supply.

The simulator does throw up an error when the load is inductive. This may be because of the switching of the diodes in the rectifier causing a spike in the inductor (possibly a source of radiant)

Just so long as the load does not exceed the charging rate of the battery, I think it will be ok. Any comments from the electricians out there?

As usual with any battery system the cost of the batteries required to do this for a whole house would be huge.

On the safety aspect, yes this is dangerous as we are working with lethal power, all components have to be treated like it is mains power at all times.

See the circuit below

The graphs on the image represent from left to right

The bridge rectifier switching

The AC supply

The load

The charging battery

It is interesting that the simulator shares the power between the load and the battery, but as we know the power that passes through the load must pass through the battery too so actual charge rates will be interesting.

On the arc welder shocks, as apprentices we used to do that all the time but I'm sure it can’t have been good for us, nowadays you would be dismissed and probably sued for it.

[Farmhand]

Thanks for the drawing. Simulator result is interesting. Hmm battery voltage drop from the supply, I suppose it would matter more with some appliances than others.

Thanks again

[Aaron]

thread title

Hi Farmhand,

We're unable to change the thread titles. I'm sure there is some
plugin that lets us do that but we don't have it.

You could edit the first post with a comment.

[Farmhand]

Ok no probs, Thanks Aaron. Did that.

[Farmhand]

Hi all, Progress report #1 - I've done the first side change at 2:40 am the test started at 5:15pm that is 9 hrs 25 minutes for one way. The swap took 7 minutes, during the swap I measured the two batteries that were in parallel Batt #1 and #2 were reading 12.79 volts and the two that were in series Batt #3 and #4 measured 12.2 volts if they had sat resting for longer the higher one's would have relaxed down and the lower one's would have relaxed up.

Looks to me like reletively the battery voltages remained fairly equal but opposite to how they started.

This could take a very long time. At this point I'm going to go out on a limb and say that I think I can easily get more than double the effective work performed and still have way more charge in the batteries than if they were discharged conventionally.

Once I get past double the work I am going to stop because my batteries are still new and I want to continue to bring them up to speed with spike's. I will have proved my point at least to myself, and that is all that really matters to me.

Double the work will be 28 hours compared to 14 hours maybe I'll go to 30 hours for a round figure. So it will need to continue untill 11:15 pm tonight to acheive that. That will be another two swaps. Should be no problem.

Looks like it would take far too long to get all the batteries to 12.10 volts this way. So far that is how it looks to me.


Cheers

[Farmhand]

Umm if I was to stop the test after 15 hours with no external input and the same batteries all have a higher voltage than they started with, what would prove to myself ? I'm mad ? Or my batteries are getting better as I'm using them ?

I would have to wait for some voltage dropping after resting to get the correct voltage's . So I can't be sure how it would end up. I suppose I have to continue. Don't seem to have lost much charge yet.
13 hours gone.