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Author Topic: TinMan's "Over Faraday HV HHO production"  (Read 39849 times)
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  Brad has started a tutorial On a process to Break The Faraday Limit,
"Gas Production way over what was once thought possible "

Part 1. of the introduction here
https://www.youtube.com/watch?v=p8e7dOVzyTM

also being discussed here
http://overunity.com/16992/tinmans-over-faraday-hv-hho-production/msg496021/#new

All comments welcome.

respectfully

Chet K
   
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Good day all

I just threw together this *rough*  LTspice sim. of what I gathered to be the circuit presented by Brad, (taken from info. off of his Youtube channel)

Interestingly enough it does indeed show a HV reflected pulse on the primary winding of the *crude* MOT mockup.

Of course I don't have access to the *real* component values so I just guess-a-mated.

Judging from the looks of it, will be tough to find a diode with a reverse voltage rating of 9MV.

take care, peace
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Good day all

I just threw together this *rough*  LTspice sim. of what I gathered to be the circuit presented by Brad, (taken from info. off of his Youtube channel)

Interestingly enough it does indeed show a HV reflected pulse on the primary winding of the *crude* MOT mockup.

Of course I don't have access to the *real* component values so I just guess-a-mated.

Judging from the looks of it, will be tough to find a diode with a reverse voltage rating of 9MV.

take care, peace
lost_bro

Thanks for doing the sim lost_bro.
Your scope shot shows very much what rides on the peak of the DC offset AC wave form across the cell.

The schematic is as below.
C1 is a 1uf HV cap from a microwave oven in  my setup.


Brad


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Hi LostBro

I often run into this type of MV HV performance, sometimes by accident other times by design :D

You may be able to tame the circuit a bit by adding some series resistance and some shunt capacitance to the transformer. The cell will be more complex than just a capacitor. It will have series and shunt resistance as well  as a capacitance. Also soften up the supply a bit and reduce the transformer K to .9 to .95. (those MOT's don't have tight coupling) Then maybe a more reasonable diode will work.

Good to see you here. Peace

P.S. if you post your .asc file I can try a hand to tame it.

Regards, ION


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Thanks for doing the sim lost_bro.
Your scope shot shows very much what rides on the peak of the DC offset AC wave form across the cell.

The schematic is as below.
C1 is a 1uf HV cap from a microwave oven in  my setup.


Brad

Good day Brad

I see a have the HHO cell and primary reversed in the Sim from your posted schematic.  I not sure if that makes a difference in the real world, but it does change outcome of the Sim.

When I have time I will play with the Sim. some more.

take care, peace
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Hi LostBro

I often run into this type of MV HV performance, sometimes by accident other times by design :D

You may be able to tame the circuit a bit by adding some series resistance and some shunt capacitance to the transformer. The cell will be more complex than just a capacitor. It will have series and shunt resistance as well  as a capacitance. Also soften up the supply a bit and reduce the transformer K to .9 to .95. (those MOT's don't have tight coupling) Then maybe a more reasonable diode will work.

Good to see you here. Peace

P.S. if you post your .asc file I can try a hand to tame it.

Regards, ION

Good day ION

Yes, your quite right....  any small change will drastically change the outcome of the Sim. 
Basically I just wanted to see *if* the HV spike would show up on the primary as a result of the S.G. (spark gap) shorting of the secondary coil.

This type of circuit reminds me of the many hours I spent investigating Stanley Meyer's  Water Fuel cell design many years ago.......
Still think this type of circuit requires balanced reactance's & impedance matching to get the desired results.
Probably some type of LC , RLC interaction, reflected impedance, between the HHO cell and inductors along with the necessary driving frequency.

I did NOT spend much time or thought on this Sim., just wanted to get an idea of the possible M.O. of the device so as to think about the possibility of a Solid State version of the same. 

I remember that Stanley Meyers also started his investigations using a modified Car Alternator  driven by a (?synchronous) electric motor and then quickly moved on to a Solid State version.

Interesting.......

And, I was just joking about the 9MV diode ;D

Please find attached Sim.

take care, peace
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Thanks for doing the sim lost_bro.
Your scope shot shows very much what rides on the peak of the DC offset AC wave form across the cell.

The schematic is as below.
C1 is a 1uf HV cap from a microwave oven in  my setup.

Brad

Brad, before you get away for the weekend --- what have you found is the sweet spot (RPM wise) for the smart motor?
mine is the old style 42 coil, 1mm wire

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Brad, before you get away for the weekend --- what have you found is the sweet spot (RPM wise) for the smart motor?
mine is the old style 42 coil, 1mm wire

Thanks, Ron

Ah so you have the original 100 series stator.
They use to run on about 40 volt's,and were the lowest voltage motor they built.
These are rare,as everyone wants them for wind generators.

Anyway,the RPM/frequency will depend on a lot of thing's.
The system needs to be tuned to your cell.
Changing things like spark gap distance,capacitor value,and RPM,changes both frequency and voltage across the cell.

I started by way of just visual gas output.
Straight tap water will vary in conductivity,and only trial and error is really the only way to tune the system.

I would start out getting some gas production number's from your cell,using low voltage ,high current method's-standard method's. Not doing this,means you will have no idea whether or not your are gaining in efficiency,or going backward's--so you need a bench mark from your cell first.

Im guessing you have all the system complete,so as you can get accurate MMW measurements?.--like the displacement measuring device,a dryer and condenser ?.


Brad


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Im guessing you have all the system complete,so as you can get accurate MMW measurements?.--like the displacement measuring device,a dryer and condenser ?.

Brad

Nope, working on the reservoir atm, still missing a flow meter, dryer and condenser.

What I was looking for was a ball park starting RPM number, like what was your pony motor RPM? Here that would be 1750 or 3400 RPM

Thanks, Ron
 
   

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Nope, working on the reservoir atm, still missing a flow meter, dryer and condenser.

What I was looking for was a ball park starting RPM number, like what was your pony motor RPM? Here that would be 1750 or 3400 RPM

Thanks, Ron

As we are on 50Hz here,our standard RPM is 1440 and 2880 for motors.
The reduction is close to 2:1
But that was just a demo setup,as the actual setup has to have a variable speed for tuning.
The demo setup will show you the wave form needed across the cell though,so it's a good simple start to see what you are looking for.


Brad


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Had a nice Chat with Darrell [Brad's cell builder]
will be speaking with him again later today

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Exciting - thanks, Brad for sharing and helping.
Below is a photo of an H2/O2 collector I used with experiments a while back - very straightforward, collect all gases in a graduated cylinder over an interval of time.  The graduated cylinder allows quick measurement of the volume of gases produced.

Using a flow-meter, dryer, condenser etc - add to the complications.  With a factor of 2 or so more volume of gases than expected, a straightforward graduated cylinder system should suffice... then add bells and whistles later as desired!

Let's remember - OK, IMHO - that the FIRST device that clearly demonstrates "overunity" (I would prefer "tapping a previously untapped source of energy") is the one that will get the attention and First-Place status... along with its inventor.

 And there are several inventions looking good right now, serious contenders for First Place Finish - including Mikes's and Floor's/Luc's.  IMHO.
   
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Physics Prof:

Assuming the Faraday limit, can you or anyone refute the calculations in the video?

https://www.youtube.com/watch?v=qn480IwYi2g

Granted if you can exceed Faraday, everything changes, but by how much? Possibly a lot more than we imagine when considering overall system efficiencies.

In the example given, the author does not even consider the cost to the system (IC engine plus alternator) of producing the HHO gas. (system petrol to electrical inefficiency)

When hooked to a feeble car alternator, which are made as cheap as possible, as a generator of electricity, an IC engine plus alternator is notoriously inefficient.

So the cost of the electricity produced is at best only about 20 to 30% efficient when considered as a system.

If the IC engine plus generator system could produce electrical energy at 100% efficiency, then you would only need to exceed the Faraday limit by a small amount to have a gain in the overall IC engine plus generator plus HHO gas generator.

But this is not the case and due to the extreme inefficiency of system electrical production, the Faraday limit must be exceeded many fold to just break even.

This needs serious discussion.

Quote
Gasoline (petrol) engines
Modern gasoline engines have a maximum thermal efficiency of about 25% to 30% when used to power a car.[citation needed] In other words, even when the engine is operating at its point of maximum thermal efficiency, of the total heat energy released by the gasoline consumed, about 70-75% is rejected as heat without being turned into useful work, i.e. turning the crankshaft.[1] Approximately half of this rejected heat is carried away by the exhaust gases, and half passes through the cylinder walls or cylinder head into the engine cooling system, and is passed to the atmosphere via the cooling system radiator.[2] Some of the work generated is also lost as friction, noise, air turbulence, and work used to turn engine equipment and appliances such as water and oil pumps and the electrical generator, leaving only about 25-30% of the energy released by the fuel consumed available to move the vehicle.
« Last Edit: 2016-11-28, 15:28:48 by ION »


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Physics Prof:

Assuming the Faraday limit, can you or anyone refute the calculations in the video?

https://www.youtube.com/watch?v=qn480IwYi2g

Granted if you can exceed Faraday, everything changes, but by how much? Possibly a lot more than we imagine when considering overall system efficiencies.

In the example given, the author does not even consider the cost to the system (IC engine plus alternator) of producing the HHO gas. (system petrol to electrical inefficiency)

When hooked to a feeble car alternator, which are made as cheap as possible, as a generator of electricity, an IC engine plus alternator is notoriously inefficient.

So the cost of the electricity produced is at best only about 20 to 30% efficient when considered as a system.

If the IC engine plus generator system could produce electrical energy at 100% efficiency, then you would only need to exceed the Faraday limit by a small amount to have a gain in the overall IC engine plus generator plus HHO gas generator.

But this is not the case and due to the extreme inefficiency of system electrical production, the Faraday limit must be exceeded many fold to just break even.

This needs serious discussion.

His calculations may be right,but his assumption that using a HHO system will not improve your car's MPG is incorrect.

I have seen a number of  wanabe debunkers run tests with and without HHO system's,and they keep finding that the HHO system actually decreases MPG  C.C
The reason this keeps happening,is because they keep using bloody late modle car's with ECUs,and--wait for it--O2 sensors  :D

So,when the HHO is added to the fuel/air mix,the O2 sensor tells the ECU that the car is running to lean,because the HHO allows a much cleaner and complete burn of the fuel.
So now the ECU thinks the car is running to lean,so it starts feeding the motor more fuel  C.C, .

And so they come up with this rubbish that HHO systems actually cause your car to use more fuel.


Here is the truth ION,and the system as a whole dose not get much more inefficient than this--but still an increase in run time when the HHO system is put into use.

First up,we have a cheap carbureted chinese honda imitation 6.5HP engine--not so efficient.
Then we have a gen-e-motor from WW2 to power the cell--also very inefficient.
To top that off,we have my very first attempt at a dry cell,made from sheets cut from a S/S water tank,that were all dented--very inefficient,with an MMW of just over 3  :'(

So,in the video,we use the same metered amount of fuel in both runs.
The generator(gen-e-motor) is unloaded on the first run,and HHO cell is not powered-engine running on gasoline only.
In the second run,the generator is now loaded,as it is powering the HHO cell.
You will see a current draw of about 25 amp's,and a voltage across the cell of around 11.2v
So in the second run,we have a load of around 280watts on the generator/engine.

First run using the metered amount of fuel,no load on the engine,and no HHO=1 minute 54 seconds
Second run with same metered amount of fuel,280 watt load on the engine to run the HHO cell,and HHO included=2 minutes 11 seconds
The battery is only for starting the engine,and is disconnected once the engine is started,and never reconnected.

So,taking into account all the vast inefficiencies in the system as a whole--where did the extra 17 seconds run time come from?.

We just increased our efficiency by 14%--and that was using the worst of the worst equipment.
So,using that setup-as bad as it was,if we could run our engine for an hour on a LTR of gasoline,then it would run for 68.4  minutes,instead of 60 minutes.

Darrell !that is about to join us here soon!,run's a HHO cell in his vehicle--i'll let him tell you how much more efficient his vehicle is now--people just wouldnt leave HHO systems in there vehicles,if they did not work.
A 30% increase in efficiency is not hard to get,using a good HHO system-providing you remap the ECU if your vehicle has one.


https://www.youtube.com/watch?v=O-YWHj31Y20


Brad



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Dear All.

" Faraday's limit is an MMW of 9.28 "

Is this definitive? What does it actually mean? Could someone redefine it to say, Watts per Litre please?

Once a suitable benchmark is had, going " over Faraday " will be very easy to establish.

Kind regards, Graham.



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This is taken from wiki:

The electrolysis of water in standard conditions requires a theoretical minimum of 237 kJ of electrical energy input to dissociate each mole of water, which is the standard Gibbs free energy of formation of water. It also requires energy to overcome the change in entropy of the reaction. Therefore, the process cannot proceed below 286 kJ per mol if no external heat/energy is added.

Since each mole of water requires two moles of electrons, and given that the Faraday constant F represents the charge of a mole of electrons (96485 C/mol), it follows that the minimum voltage necessary for electrolysis is about 1.23 V.[29] If electrolysis is carried out at high temperature, this voltage reduces. This effectively allows the electrolyser to operate at more than 100% electrical efficiency. In electrochemical systems this means that heat must be supplied to the reactor to sustain the reaction. In this way thermal energy can be used for part of the electrolysis energy requirement.[30] In a similar way the required voltage can be reduced (below 1 V) if fuels (such as carbon, alcohol, biomass) are reacted with water (PEM based electrolyzer in low temperature) or oxygen ions (solid oxide electrolyte based electrolyzer in high temperature). This results in some of the fuel's energy being used to "assist" the electrolysis process and can reduce the overall cost of hydrogen produced.[31]

However, observing the entropy component (and other losses), voltages over 1.48 V are required for the reaction to proceed at practical current densities (the thermoneutral voltage).

In the case of water electrolysis, Gibbs free energy represents the minimum work necessary for the reaction to proceed, and the reaction enthalpy is the amount of energy (both work and heat) that has to be provided so the reaction products are at the same temperature as the reactant (i.e. standard temperature for the values given above). Potentially, an electrolyser operating at 1.48 V would be 100% efficient. (end quote)

Now the efficiency to produce 2H2+O2 (4moles H+ 2moles of O) cannot go above 100% Faraday, but, if you use the O2 to produce energy through chemical change, 33% of the volume of HHO, then the same volume of H2 would be 133% of Faraday, and you take the O2 from the air for free when you burn it.

Now I'm not trying to flog my SMD (electrolysis), but you have to look at the same system of breaking the water molecule, electrolysis with electrolysis. Unless there is something special about HHO and not H2+O2, O can't live on it's own without combining with another atom, as in CO or O2 or NO etc, "HHO" will live an exceedingly short life. So out of H2+O2, only the H2 has the energy and the O2 is the oxidiser.

Now breaking the water molecule through non electrolysis is another thing, a different ball game as it is not eggs for eggs as in Faraday electrolysis.

The best electrolysis is around 80-85% but has to run at around 90c to reduce the resistance heating of the power input, so extra energy is put in in the form of heat.

Hope that helps

Regards

Mike 8)



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Hi Mike.

This " simpleton " needs simpler!! This chemistry malarkey is over my head..... :)

How many Watts to make a Litre of gas? Using normal Faraday equation.

Then with the most basic of Wattmeters we can establish whether we can actually achieve " over Faraday " production.

Kind regards, Graham.


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Hi Mike.

This " simpleton " needs simpler!! This chemistry malarkey is over my head..... :)

How many Watts to make a Litre of gas? Using normal Faraday equation.

Then with the most basic of Wattmeters we can establish whether we can actually achieve " over Faraday " production.

Kind regards, Graham.

Hi Graham

MMW of 9.28 should read 9.28mltrs per 1mwatt of power consummed, 0.00928ltrs per miliwatt

Regards

Mike 8)


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Hi Mike.

So.... 0.928 Litres per Watt ? That doesn't seem correct to me. I wasn't that good at mathematics either!! ;D

Kind regards, Graham.


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Hi Mike.

So.... 0.928 Litres per Watt ? That doesn't seem correct to me. I wasn't that good at mathematics either!! ;D

Kind regards, Graham.

yes does not seem right to me either ;D

Regards

Mike 8)


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The watts is watts/minute and not hour and mils/minute

Regards

Mike 8)


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It's turtles all the way down
Not just Watts, but Watt hours per Liter

or watt-minutes per milli-liter

maybe even Watt-seconds per micro-liter

The original calculator link Brad posted early on has a time in "seconds" field to be filled in:

http://www.hho-generator.de/en/hho-mmw-calculator.htm

But calls the calculator Milliliters pro Minute pro Watt

MMW = Milliliter pro Minute pro Watt

Part of the confusion is that the abbreviation MMW is in minutes while the calculator is in seconds

Maybe I would state the data input to the calculator as:

Milliliter per Milli-Watt-Seconds or MMWS

but that is just my preference since a Watt-Second= 1 Joule

There is a lot of information on the web if you type in "watt hours per liter for electrolysis"

Stefan Hartmann says this:

http://overunity.com/3107/formular-to-calculate-energy-per-liter-of-hho-gas/msg47169/#msg47169

The rest of that page is a good read, but I have not checked it for correctness.

Brad:
Regarding your reply #13, I would not extrapolate those numbers just yet, not until you could completely describe the test setup. e.g. were the tests run consecutively if so was the engine and oil allowed to cool to the same value before the HHO test was performed? A two minute run barely gets the engine to it's peak operating efficiency with regards to temperature. A much longer test is required before you can extrapolate as the margin of error also gets blown up.

Was the air density the same in each test? What were the total elapsed killowatt hours delivered to the load each time? What was the total count of elapsed revolutions of the shaft? RPM variances?

Total elapsed kilowatts-hours delivered to the load would take into account many variables and is a good benchmark. The longer the run, the more accurate will be the result.

I think I would have more faith in a much longer run time, perhaps 15 to 30 minutes and tighter data gathering, but I make no claims for HHO so don't have the burden of proof.

ION



« Last Edit: 2016-11-28, 21:22:08 by ION »


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 author=ION link=topic=3371.msg58411#msg58411 date=1480361573]



Quote
Brad:
Regarding your reply #13, I would not extrapolate those numbers just yet, not until you could completely describe the test setup. e.g. were the tests run consecutively if so was the engine and oil allowed to cool to the same value before the HHO test was performed?

Ok,maybe you did not watch the video?,but the engine was never switched off between the two tests(with and without HHO in play).
I had been running the same test  a number of times before the video,and so the engine would have been at running temperature.

The generator is what supplies the power to the HHO cell,and as you can hear in the video,the engine becomes loaded when i switch the cell on.
The engine RPM is governed by the governor,so the same RPM exists in each run.
In the first timed run,the only load on the engine is the four brushes on the armature-which is present in both run's.
In the second timed run,the load is the 280 odd watts being drawn from the generator to power the HHO cell-plus the losses due to inefficiencies. This load result's in producing HHO gas that is returned to the motor,and also the heating of the water in the HHO cell system,and the cell it self--so we have a heat output.
The motor is also now working harder,and so one would expect that it too would be producing more heat--or would it?. ;)

Quote
Was the air density the same in each test?

As i stated above,the engine was never switched off between tests,and so i doubt the air density would have changed much in 5 minutes.

Quote
A two minute run barely gets the engine to it's peak operating efficiency with regards to temperature.

Also as stated above,the engine had been running on and off for an hour prior to the video test.

Quote
A much longer test is required before you can extrapolate as the margin of error also gets blown up.

I agree--a longer test would be better.

Quote
What were the total elapsed killowatt hours delivered to the load each time?

First run-no load,and so no killowatts hours were delivered to a load.
Second run-power to the load is 280 watts for 2 minutes 11 second's.

Quote
I think I would have more faith in a much longer run time, perhaps 15 to 30 minutes and tighter data gathering, but I make no claims for HHO so don't have the burden of proof.

Now this we can do,and with a much more efficient cell and alternator.  O0
Will be done by sunday.


Brad


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Dear Brad

Thanks for addressing my concerns. I may have seen the video a while back and forgotten some of the details.

So first run is with no load on the generator?

And second run is with the generator loaded with 280 Watts or thereabouts.

If that is the case, something special must be happening, because I would have thought the unloaded generator would have used  far less gas than when it is loaded.

Granted 0.37 HP equivalent load is not much for that size engine, and the reflected load on the engine considering the inefficiency of the generator might be closer to 0.5 HP.

Still it is amazing that you got a better run time with load.

Regards, ION


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Dear Brad

Thanks for addressing my concerns. I may have seen the video a while back and forgotten some of the details.

So first run is with no load on the generator?

And second run is with the generator loaded with 280 Watts or thereabouts.

If that is the case, something special must be happening, because I would have thought the unloaded generator would have used  far less gas than when it is loaded.

Granted 0.37 HP equivalent load is not much for that size engine, and the reflected load on the engine considering the inefficiency of the generator might be closer to 0.5 HP.

Still it is amazing that you got a better run time with load.

Regards, ION

ION

Take the time to watch the video again,and see if you can pick out any mistakes i may have made.


Brad


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