David Bowling's Continuous Charging Device

[tinman]

In the 3 battery system Work out the power used by the inverter by measuring the potential across its input terminals  and the amps! I don't see this voltage in your 3 battery test.

I posted that in my second last post,where the volt meter is placed across the inverter,and a voltage of 12 volts is maintained. As i said,those are the measurements from the next video,which i will post here as soon as it is downloaded--still 60% to go,as internet is slow tonight for some reason.
Power used by inverter in 3 battery test is 12 volts @ 2.986 amps=35.83 watts.
The power output to the 240v bulb was 24.6 watts
In the single battery test,the power used by the inverter was 34.73 watts.
The power output to the 240v bulb was 24.2 watts.
So the inverter efficiency is nearly exactly the same,and this increase in efficiency is something to do with the battery system/inverter combination.
With the 3 battery system,the charge battery and load will see those high voltage spikes,but in the single battery system,the battery will not see those high voltage spikes.
I believe that is why the charge battery is charging so fast,as it is receiving high voltage spikes,but also a deep full current flow as well.

But a long way to go on this system yet,so lets see where we end up.


Brad

[pomodoro]

Looks like a problem with the calcs to me. Have a look at the picture , which has some easy numbers. There is 1000W flowing and the top device is 60% efficient, the bottom 100%. The top uses 10Vx10A = 100W but being only 60% efficient we get 60W out. This symbolizes your inverter.

The battery being charged and the 12v bulb are 100% efficient, well of course they are not but you have no efficiency value for these so lets assume 100%.  so 90Vx10A =900W used and also somehow turned into a different energy 100% efficiently.

All of a sudden the calcs show 96% efficiency.   The average is not 96% either, looks like it has to be a weighted average instead, since the input power is not equally shared, but in your case it will be when the battery is fully charged, 12V across battery and inverter.
The volts across the battery being charged times the current give you the power going into it, but how much is actually charging the battery, and how much is heating it, how much is electrolyzing the acid into hydrogen and oxygen? The efficiency seems to be taken as 100% in your calcs, hence the overall efficiency of the whole system goes up, like in the picture.

[tinman]

Looks like a problem with the calcs to me. Have a look at the picture , which has some easy numbers. There is 1000W flowing and the top device is 60% efficient, the bottom 100%. The top uses 10Vx10A = 100W but being only 60% efficient we get 60W out. This symbolizes your inverter.

The battery being charged and the 12v bulb are 100% efficient, well of course they are not but you have no efficiency value for these so lets assume 100%.  so 90Vx10A =900W used and also somehow turned into a different energy 100% efficiently.

All of a sudden the calcs show 96% efficiency.   The average is not 96% either, looks like it has to be a weighted average instead, since the input power is not equally shared, but in your case it will be when the battery is fully charged, 12V across battery and inverter.
The volts across the battery being charged times the current give you the power going into it, but how much is actually charging the battery, and how much is heating it, how much is electrolyzing the acid into hydrogen and oxygen? The efficiency seems to be taken as 100% in your calcs, hence the overall efficiency of the whole system goes up, like in the picture.

I guess you will have to wait for the video to be posted to understand what im saying here.

With the new improved circuit,the voltages across the inverter and charge battery are fixed,due to the simple resistive regulation circuit i have included.
The voltage across the inverter remains at 12 volt's,and the voltage across the charge battery remains at 11.9 volt's,due to the resistive load placed across that charge battery(3rd battery),which can be adjusted by the reostat.

The volts across the battery being charged times the current give you the power going into it, but how much is actually charging the battery, and how much is heating it, how much is electrolyzing the acid into hydrogen and oxygen?

That is all correct,and we would think that by adding another 2 batteries to the system,that we would decrease the efficiency of that system,but the overall efficiency go's up,not down.
As i stated above,the 3rd battery(the charging battery)now has a resistive load across it,and most of the current is flowing through the load to complete the circuit--not the battery,as the charge batteries voltage remains a constant,indicating that very little of the power is going into charging the battery,and that also means that losses associated with charging a lead acid battery are omitted.

In my next test,i am going to try and remove the 3rd battery altogether,and just run the resistive load in it's place,and then once again,check the overall efficiency of the system.

Lots more to do yet,and no conclusive answer has been reached,but only preliminary results presented so far.


Brad

[tinman]

Here is the third video and test done on the 3 battery system.
This time we have made the voltage more stable,to allow for more accurate measurements to be taken.

Lots more to do yet,before any conclusions are made,and the results so far are just what has so far been found.

https://www.youtube.com/watch?v=i4URpy_aQA8


Brad

[minoly]

Here is the third video and test done on the 3 battery system.
This time we have made the voltage more stable,to allow for more accurate measurements to be taken.

Lots more to do yet,before any conclusions are made,and the results so far are just what has so far been found.

https://www.youtube.com/watch?v=i4URpy_aQA8


Brad

If this is too basic a question or if I'm missing something elementary feel free to ignore...


In the 3 bat efficiency data, why are you using 24 volts for P/IN to the inverter? shouldn't the P/IN watts to the inverter be calculated using the voltage the inverter is actually receiving 12?


I've not tried to calculate Power in/out on the fly like this before, I usually just keep rotating the batteries to prove to myself that it works. It sure would be nice to be able to use on-the-fly measurements with this...




+++++++++++++++++++++++++++
nevermind, i c you are calculating the 24 volts as total going to the inverter as well as the 3rd battery....