Hi All,
here is a formular to calculate the energy needed to break the overunity barrier
in HHO gas production.
As the Faraday law states, that 100 % efficient electrolysis required about 2.4 Watthours
of energy for 1 Liter of HHO gas production you can now yourself calculate, what you are using
for your energy input.
Energy input in Watthours= used (Volts x amps) x used seconds / 3600 seconds= used Watts x used seconds / 3600 seconds
So let your HHO gas generator produce 1 Liter of HHO gas and measure the time
it needs to produce this amount of gas.
Then note the Voltage and amps down and multiply it to get the Watts.
Then multiply with the seconds and divide by 3600.
Now if you get less than 2.4 Watthours you have overunity.
If you have more than 2.4 Watthours you have less than 100 % efficiency and are underunity.
Regards, Stefan.
Understanding and applying these calculations above to my Hy-Dro-Gen, with my last video, I have not reached overunity. Sorry bout that guys...
But with my most resent experiments "Not yet video taped, or documented, I did reach overunity with the final calculation at 1.3333.... To my understanding this is something not done before? If this is the case, I'm flattered!
To achieve this I ran my Hy-Dro-Gen at a very low 2 volts.
I had to add a great deal of electrolyte to reach Times 20 Amps. = 40 watts
It took 120 seconds to make 1 liter of HHO gas times 120 seconds = 4800
Then dividing by 3600 = 1.3333333333333333333333333333333333333333 ;)
Is this overunity? And what does this really mean?
Observations:
a: I have found that the temperature increase is directly related to the voltage. The lower the voltage the less heat developed in the electrolysis's process. I'm nearing a point of no need of a cooling system. It seems that the system is balancing it's self as I push closer to efficiency.
b: Also the water "tap" is staying allot cleaner then with the higher voltages.
If this is of interest, please feel free to chime in! If this is nothing new, I'm sorry for waisting everyones time. You may delete this post if it is of no use. Here is the link to my last experiment, "NOT OVERUNITY" I will try to get you proof of my overunity test as soon as possible.
http://www.youtube.com/watch?v=2nAROmhiZsY
Thanks in advance for any help!
Sanction
PS: Stefan thanks for asking me to come here with my story!
Good work Stefan,
That's a great bit of information. I just wanted to remark that it is interesting how we don't count the value of the water consumed. Is that just because water is so plentiful compared to volatile hydrocarbon fuels do you think?
I mean, if we were figuring for overunity in a gasoline or diesel engine application, we would only count the energy stored in the fuel and would never think to include the energy it required to obtain, extract, transport and refine it.
Yet if we look at hydrogen as the fuel, we only count the energy required to extract it and we disregard the value of the source water itself. Just because it is hugely abundant right now. Is my thinking screwed up here? Am I making any sense? Not exactly, I know.
Sometimes, like with the whole ethanol fiasco and the hit put on corn and other feed stocks because of it, I think we get too focused on the immediate small cycle and forget about the wide-frame picture. Oh, well...just another random thought...no particular point.
Here in USA, we gripe about gasoline costs but seem happy to pay a buck or more for a liter of bottled tap water and create a trillion empty plastic bottles in the process! Crazy!
Reminds me of the book "Gaviotas", though. Ever read that one? True story about a team of science and engineering experts trying to bring the gift of technology to a primitive Columbian Indian mountain village...the end result is quite revealing of our collective techno-folly and tunnel vision at times.
Humbugger
Yes, water is almost abundant and when we can run a generator just
on water we will have enough energy to make desalinated water from the
ocean waters and pour it into the deserts to grow food for everyone.
We can clean up then the whole pollution too.
Plus when combining Hydrogen and Oxygen back to convert energy, the byproduct is also water, so for all practical purposes the system should be all recycling.
Good info. I think we had these discussions before too.
Also, combining the stoichiometric ratio of only the two dissociated gases (Hydrogen and Oxygen) it produces pure water in an endothermic implosive reaction.
Quote from: Sanction on August 23, 2007, 11:54:23 AM
Understanding and applying these calculations above to my Hy-Dro-Gen, with my last video, I have not reached overunity. Sorry bout that guys...
But with my most resent experiments "Not yet video taped, or documented, I did reach overunity with the final calculation at 1.3333.... To my understanding this is something not done before? If this is the case, I'm flattered!
To achieve this I ran my Hy-Dro-Gen at a very low 2 volts.
I had to add a great deal of electrolyte to reach Times 20 Amps. = 40 watts
It took 120 seconds to make 1 liter of HHO gas times 120 seconds = 4800
Then dividing by 3600 = 1.3333333333333333333333333333333333333333 ;)
Is this overunity? And what does this really mean?
Observations:
a: I have found that the temperature increase is directly related to the voltage. The lower the voltage the less heat developed in the electrolysis's process. I'm nearing a point of no need of a cooling system. It seems that the system is balancing it's self as I push closer to efficiency.
b: Also the water "tap" is staying allot cleaner then with the higher voltages.
If this is of interest, please feel free to chime in! If this is nothing new, I'm sorry for waisting everyones time. You may delete this post if it is of no use. Here is the link to my last experiment, "NOT OVERUNITY" I will try to get you proof of my overunity test as soon as possible.
http://www.youtube.com/watch?v=2nAROmhiZsY
Thanks in advance for any help!
Sanction
PS: Stefan thanks for asking me to come here with my story!
@Sanction
well done.
YOu seem to be the second after user ravzz who did it.
Please report more about your low voltage cell.
Did you use the former cell construction ?
How did you design the 2 Volts power supply ?
Many thanks.
Regards, Stefan.
@sanc I would love to see a detailed report on that cell.
Hi everyone!
thanks to all of you and especially Stefan to keep this site up and running!
I think H2O electrolysis has great potential towards achieving over 1 output.
I am working on this and theoretically i am on right track! but before i make any claim(!!!!!) i want to be 100% sure and for this i need some info with proper evidence.
Howmuch electricity required in watt-hour (or kWh) to generate 1 lt of H2 from 10 lt of sea-water (or water with electrolyte)?
1. theortically ?
2. practically ?
please provide evidence to back your claim.
Howmuch electricity can be generated in watt-hour (or kWh) from 1 lt of H2 and 0.5 lt of O2?
1. theoretically ?
2. practically ?
please provide evidence to back your claim.
thanks in advance!
I've been playing around with water electrolysis for quite a while now.
A question: does your formula also apply for distilled water splitted in H2 and O2? Because HHO is a little different from H2O.
If so, then what about the following figures. To produce 10 cm3 of H2 and app. 5 cm3 O2 I need 2 volts and ,31 A during 300 seconds. And these are probably conservative numbers, because my measuring equipment is not the best there is. So the energy input is:
2V x ,31A x 300 secs/3600 secs
I know this is a slow process, but I got these values over and over again, although I ran into problems using the same water over and over again. It seems to me the characteristics of water are changing.
As power source I use a Philips PE 1512 Lab power supply and a common available electrolyzer.
But I have great difficulty in accepting this to be overunity.
Kind regards
Hi all,
I'm a chemical engineer interested in the Meyer type high voltage "electrolysis", and will when I get the time investige more into this.
BUT; I NEED TO CORRECT THE EFFICIENCY STATEMENT THAT STEFAN PROPOSED;
First, There are two approaches that may be made into calculation efficiency of electrolysis;
1. Using Faraday law..or
2. Using the measured collected H2/O2 gas volume and do a pure mole and energy calculation
Both of these methods should lead to same answer which is;
In a collected volume of O2/H2 gas the amount of hydrogen energy will be 7744 Joules pr liter collected gas.
It is then assumed that the gas holds 20 degC and that the gas is saturated with water vapor.
Electrical energy is as known purely E (Joules) = Voltage * Amperage * electrolysis duration time
So efficiency of any electrolysis of Meyer farcturation of water, where H2 and O2 gas is collected together is;
EFFICIENCY = 7744* V1 / (U*I*t) where
V1= Collected H2/O2 volume measured in liters
U = Electrical input voltage
I = Amperage during electrolysis
t = Time of electrolysis measured in seconds
multiply above by 100 to get % efficiency
So back to points 1 & 2;
I have confirmed that both methods leads to same rsult, and have made a word document that explains the details if any are interested.
I believe Stefan confused Watthrs with Amphrs. According to Faradays law, the electrical consumption is 2,4 amphrs pr liter PURE hydrogen produced.....this at ideal voltage which is 1,47 volts, which leads to 3,52 watt hrs pr liter PURE H2 produced.
Now then; A few on Youtube seems to have reached well over unity with Meyer type system, check out RAVI's results.....impressive allmost 1000% overunity
regards
?ystein
Hi ?ystein,
many thanks for this correction.
Please can you post the WORD document or better in PDF format if you can ?
So you say that 1 liter HHO ( H2 and O2 together) gas
has an energy value of 7744 Joules = 7744 Wattseconds.
So we only have to divide
7744 Wattseconds / 3600 seconds
to get:
2.15 Watthours of energy for 1 Liter of HHO gas.
So it is now the Faraday law barrier of 2.15 Watthours which you must be below to produce
1 Liter of HHO gas to be overunity.
Is this correct now ?
Many thanks.
Regards, Stefan.
IS this over unity or just better than Fariday process. I assume that if you hurn liter of HOH you ate saying it wll release equall amount of energy?
With ICE losses and alternator losses looks OU point and and HOH self powered points are very far apart.
George King
Hi George,
100 % efficiency would only be 2.15 Watthours energy required for
1 Liter HHO gas for going from electric energy to HHO gas production.
Of course with added the low efficiency of the car motor of about 30 percent only
you still need to produce then even more HHO gas for this energy input.
But for the first time, just let us concentrate maybe to even
just half the gasoline consumption, so that would already be quite
an achievement, if you can run your car twice as long for the same
gasoline paid for cash..
I made some calculations too for another purpose but still usable in this situation:
Calculations on electrolysis of water (H2O)
Given
1 gram atom = atomic mass
1 gram mol = molecular mass
Water = H2O, so 2 atoms H and 1 atom O
Question: how is the distribution of mass in 1 molecule of water (H2O) in % H and O.
Calculation
Atomic mass H= 1
Atomic mass O=16
So the Molecular mass H2O=1+1+16=18
Total mass of H in water=2/18x1%=11,11%
Total mass of O in water=16/18x1%=88,89%
Given
1 liter of water has a mass of 1000 gr and equals 1000 cm?
1 liter of molecular H has a mass of 0,09 gr, 1 liter of molecular O has a mass of 1,47 gr
Question: how many liters of H2 and O2 are contained in 1 liter of H2O
Calculation
For H2: Total mass of H in H2O 11,11% (See question above)
so 11,11%x1000 gr=111,1 gr of H
111,1/0,09x1 liter=1.234,44 liter H
For O2: Total mass of O in H2O 88,89% (See question above)
So 88,89%x1000 gr= 888,90 gr of H
888,90/1,47x 1 liter= 604,69 liter O
So in your case from 1 liter of water you get app. 1.800 liters of HHO gas.
Yes, Stefan
of course, you are right, So your original proposal was not far off anyhow.
But since the experimenters collect their gas generally through bubblers and measure volume, I would say my formula is more straight forward to use for calculating efficiency.
Also, this takes care of/assumes any water saturation, and that gas collected holds approximately 20 degC, 1 atm..
One thing that the experimenters should be aware off is that pressure may build up in the bubblers of some set ups, and will distort the mauserment, but not much I think...
Another item; I get warnings from my security software of "spyware" when I open your website, and it states "ads.adbrite.com/adserver/....."
Anything to warry about ?
regards
?ystein
Quote from: oystla on September 02, 2007, 08:37:01 AM
Yes, Stefan
of course, you are right, So your original proposal was not far off anyhow.
But since the experimenters collect their gas generally through bubblers and measure volume, I would say my formula is more straight forward to use for calculating efficiency.
Hi ?ystein,
but your formular for:
3,52 watt hrs pr liter PURE H2
is not so good,
cause all the people working on this mostly directly
produce mixed H2 and O2 gas, so they do not seperate the gases
and to know,
that they just have to break the barrier of 2.15 Watthours of energy is more easy for them,
cause they then can directly measure their HHO gas (mixed H2 and O2 gas) output.
But anyway, many thanks for putting up the right calculations..
Quote
Another item; I get warnings from my security software of "spyware" when I open your website, and it states "ads.adbrite.com/adserver/....."
Anything to warry about ?
regards
?ystein
Well, this is just a different text ad provider, but as it seems, that they
can not provide any ads, I will drop them out again.
Nothing to worry.
Regards, Stefan.
I am having a time trying to convert 2.15 watthours per litre into XXXX (millilitre/minute)/watt. Does anyone want to give it a go for me. I keep getting numbers that are erroneous.
Again:
Energy input in Watthours = used (Volts x amps) x used time in seconds / 3600 seconds =
used Watts x used time in seconds / 3600 seconds
So let your HHO gas generator produce 1 Liter of HHO gas= 1000 milliLiters HHO gas and measure the time
in seconds it needs to produce this amount of gas.
Then note the DC Voltage and DC amps down and multiply it to get the Watts.
( e.g. 12 Volts and 10 amps )
Then multiply with the seconds it needed to produce 1 Liter and divide this by 3600.
Then you have the Watthours of energy it needed to produce 1 Liter of HHO gas.
Now if you get less than 2.15 Watthours you have overunity.
If you have more than 2.15 Watthours you have less than 100 % efficiency and are underunity.
So the above example would be, if it takes 60 seconds to fill
up the 1 Liter bottle of HHO gas with 12 Volts at 10 amps :
12 Volts x 10 amps x 60 seconds / 3600 = 2 Watthours per Liter
and you would be overunity, cause it is less than 2.15 Watthours
Regards, Stefan.
Alright taking the time to figure it, it now looks like 7.75 (ml/min)/watt is the goal. Sorry stephan I don't like the kilowatt measurement, just a personal method, right now my best data is good but not good enough.
Yes, you are right,
7.75 milliLiters of HHO per Wattminute, that means for
1 Watt of input power 1 minute long applied to your electrolyzer.
Example:
If you have a 12 Volts DC battery and you power your electrolyzer with it,
you draw 0.0833 amps from it ( 1 Watt) into your electrolyzer for 1 minute constantly,
you must be able to produce more than 7.75 milliLiters of HHO gas.
If you draw more amps, so you have an input power of e.g. 100 Watts, you
must produce in that minute more than 100 x 7.75 mL= 775 mL= 0.775 Liters of HHO gas.
So to reach 1 Liter of HHO gas, you must not use more than 129 Watts of power
1 minute long.
So if you use a 12 Volts battery, you must use less than 10.75 amps for 1 minute long
to produce 1 Liter of HHO in 60 seconds to be overunity.
Or in other words:
if you apply 10.75 amps at 12 Volts ( 129 Watts) you must be able
to produce 1 Liter HHO gas in
less than 60 seconds to be overunity !
If you need more than 60 seconds for 1 Liter HHO , then you are underunity....
Hope this helps.
Regards, Stefan.
OK, once more;
Anyone collecting H2 and O2 gas together in a chamber and measuring the total volume in liters produced can use the following formula to calculate the efficiency of the electrolysis cell ;
EFFICIENCY = 7744* V1 / (U*I*t) where
V1= Collected H2/O2 volume measured in liters
U = Electrical input voltage
I = Amperage during electrolysis
t = Time of electrolysis measured in seconds
multiply above by 100 to get % efficiency
regards
?ystein
I had a post (Do not remember what forum) from someone Running (external powered) Electrolysis to Fuel Cell and its water by product back into Electrolyer cell. He reported that after a while FC woould take longer and longer to convert GAS, lower power from FC and at some point water had to be replaced, He felt that he may have been creating heavy water, but did not have analysis equipment.
George King
georgeking@cosmicsalamander.com
Quote from: xingu on September 01, 2007, 06:13:06 AM
I've been playing around with water electrolysis for quite a while now.
A question: does your formula also apply for distilled water splitted in H2 and O2? Because HHO is a little different from H2O.
If so, then what about the following figures. To produce 10 cm3 of H2 and app. 5 cm3 O2 I need 2 volts and ,31 A during 300 seconds. And these are probably conservative numbers, because my measuring equipment is not the best there is. So the energy input is:
2V x ,31A x 300 secs/3600 secs
I know this is a slow process, but I got these values over and over again, although I ran into problems using the same water over and over again. It seems to me the characteristics of water are changing.
As power source I use a Philips PE 1512 Lab power supply and a common available electrolyzer.
But I have great difficulty in accepting this to be overunity.
Kind regards
So the question I have is how many watts is required to run a typical car?
What about an SUV?
If we uses Miles per Gallon to figure for cars, how many Watts per Gallon is Water?
Let's assume that it takes 12 Volts times .5 Amps to produce 1 liter of Hydrogen within 60 seconds...
Is that enough Hydrogen to run a car? What about an SUV?
It sure would be great to have a spreadsheet that did all of this calculations for us.
That way we could figure out that once we reach X number of liters of hydrogen per second, we have a usable system that will produce enough energy to run a car.
Anybody want to tackle this? (it's a little out of reach of my math skills)
FB
Quote from: ForeverBlissed on September 07, 2007, 12:36:35 AM
So the question I have is how many watts is required to run a typical car?
What about an SUV?
If we uses Miles per Gallon to figure for cars, how many Watts per Gallon is Water?
Let's assume that it takes 12 Volts times .5 Amps to produce 1 liter of Hydrogen within 60 seconds...
Is that enough Hydrogen to run a car? What about an SUV?
It sure would be great to have a spreadsheet that did all of this calculations for us.
That way we could figure out that once we reach X number of liters of hydrogen per second, we have a usable system that will produce enough energy to run a car.
Anybody want to tackle this? (it's a little out of reach of my math skills)
FB
One horsepower is 745.7 Watts. I'll contribute that well-known fact.
Also, it sounds like 6W per liter per minute would be a fantastic achievement, about 21 times better than what Stefan is suggesting represents unity.
For me, the formula is rather simple...
mpg(w/cell) > mpg(wo/cell)
anything else is just gravy...
The amount of hho to run a vehicle has not been tested in a fashion the tells anything. No one to my knowledge has ran a car on hho and the published the results. So no know one knows if an engine need to consume 100% hho or if it an be mixed with air and still get enough power out of it to run the vehicle. The sad state of affairs with accurate data on hho is that the best info on burnable ratios comes from youtube and he was useing spice jars, and didn't seal it very well. And the test was not done under pressure as a piston in an engine would be. I am working in that direction, but first thing is first and thats to quantify the basics.
WHEN TESTING THE EFFICIENCY OF AN ELECTROLYZER::::::::
The above posted calculations for calculating the efficiency are designed for electrolysis of Pure Water.
Contaminants (even electrolytes) effect this calculation.
Even if you plan to USE electrolytes in your device, efficiency tests should be done with Pure Water, as it s difficult to know the precise energy values of the electrolyte, and dissolution %, ect.
if you use an electrolyte during the efficiency testing, the above calculations may show over unity, but incorrectly so.
recent electric vehicles were reported to consume 200A @ 36v in order to travel 45 MPH.
so Given 7744 Joules / (200x36) = roughly 1.08 seconds of motor run-time from 1 liter of gas?
that is assuming your fuelcell is converting H2+O into electricity+water at 100% efficiency.
As we all know, calculating/measuring the INPUT power is pretty easy. Just hook an ammeter and voltmeter right after the battery.
OUTPUT measurement is more tricky and whatever was posted above will only give you theoretical output value. It is not necessary that you'd be able to use all that energy.
The only way to get its value is to BURN the gas (thats collected over a specified time (say 1 min)) and let it heat a known amount of water in a calorimeter. The rise in temperature of water will give you the useful energy produced. Of course it means that you need access to a precise calorimeter or a lab.
Another method is to drive a tabletop IC engine and connect a known load on its shaft. Use up the collected gas. The work done on the load is your actual output.
Another method is to use the heat generated by the gas to heat a thermoelectric junction and discharge it through a known resistor and measure the current as usual.
HHO just sitting there in a jar is totally useless. You must get the work out of it, that work is your actual output. Now you can happily divide it by the input and check whether its gives an USEFUL overunity.
I've seen many projects on HHO, that yell OU, but I've never seen anyone do a serious measurement. Whats with that?
Hello Stefan,
I'd like to make some clarification regarding your calculations of overunity for electrolysis.
1. Because a liter is a unit of volume, the mass of HHO gas in a liter will vary with ambient pressure and temperature according to the Ideal Gas Law, i.e. PV = nRT. Since energy release from burning HHO depends on mass rather than volume of the gas, it's better to assess overunity based on HHO mass produced by electrolysis.
2. Taking into account the rate of energy release from burning hydrogen, which is 142.35 kJ/g H, one can derive the following formula for estimating the expected energy release (Eh, kJ) from burning HHO gas produced via electrolysis of W grams of water:
Eh = 15.817*W (in kJ)
This formula assumes that the HHO gas is pure and contains no other gases in its volume such as water vapor or air. So, for example, the electrolysis of 1 kg of water (i.e. 1000 g H2O) will produce HHO gas, which after burning will release 15.817*1000 = 15817 kJ of energy.
3. To assess the efficiency of electrolysis and whether it's overunity, one needs to calculate the electric energy (Ee, kJ) consumed to split W mass of water. This can be done using the formula:
Ee = 0.001*V*A*T (in kJ)
where V is average voltage used (Volts), A is the average current consumed (Amps), and T is the time (in seconds) needed to split W mass of water. The constant 0.001 serves to convert Joules into kilojoules, so that Ee can be compared to Eh.
The efficiency of electrolysis (Eff, %) can then be computed as:
Eff = 100*(Eh/Ee) (%)
Efficiency over 100% will indicate overunity.
Thank you,
- Ned Nikolov
Quote from: nedh20power on September 24, 2007, 01:20:41 PM
Hello Stefan,
I'd like to make some clarification regarding your calculations of overunity for electrolysis.
1. Because a liter is a unit of volume, the mass of HHO gas in a liter will vary with ambient pressure and temperature according to the Ideal Gas Law, i.e. PV = nRT. Since energy release from burning HHO depends on mass rather than volume of the gas, it's better to assess overunity based on HHO mass produced by electrolysis.
OK;
1. Most experimenters I've seen measure the actual gas volume production.
So therefore we should of course propose a simple formula they may use to calculate the efficiency based on actual gas volume produced. Gas volume will here mean the O2 / H2 mixture together with some water vapor.
2. The formula I propose and which is accurate to wathever degree is required here is:
EFFICIENCY (%) = 100 * 7744 * V /(U*I*t) where
V= Volume H2/O2 gas produced in liters
U = Voltage
I= Amperage
t= Test period in seconds
The above formula is calculated from ideal gas law and Daltons law.
Assumptions included in the formula
- Gas collected holds 20 degC. However, If it is instead 10 degC or 30 degC, it will only have marginal consequences.
- Gas collected holds 1 atmospheric pressure, which is very close to the truth in most cases.
- The collected gas is saturated with water vapor, since most experimenters use a bubbler to collect the gas.
And yes, Higher heating value of Hydrogen is 141,9 KJ/g.
So forget about mass basis calculations. My formula is fit for purpose and is more than accurate enough for any experimenters needs.
regards
?ystein
there might be an overunity the other way.
lol
for example you make hydrogen and oxygen from water then on the higher place you combine air oxygen with extracted hydrogen.
the extracted oxygen go to the floor and the oxygen from the ceiling is burned with the extracted hydrogen.
it is like moving oxygen up without moving it.
i guess such process should take heat from surrounding
In that case... Does this look right?
In a collected volume of O2/H2 gas the amount of hydrogen energy will be 7744 Joules pr liter collected gas.
1L = 7744J
1000cm3 = 7744J
1cm3 = 7.744J
1J = 1 / 7.744 = 0.1291cm3
H2 + O2 gas at normal atmospheric pressure has 7.744J/cm3 or 0.1291cm3/J
The Average small home uses 1000kWh per month
1kWh = 3,600,000J
1000kWh * 12 = 12,000kWh per year
12,000kWh / 52 weeks = 230.7692kWh per week
230.7692kWh per week / 7 days = 32.9670kWh per day
32.9670kWh / 24 = 1.3736kWh per hour
1.3736kWh per hour = 4,944,960J per hour
4,944,960J per hour * 0.1291cm3/J = 638,394.3360cm3 of gas per hour or 638.3943l per hour
638.3943l per hour / 60 = 10.6399l per minute to supply a 1000kWh per month house?
(This is assuming 100% efficiency converting it from H2 and O2 to Electricity)
If a fuel cell was 60% efficient then you'd need another 30% of gas to make up the thermal losses.
Did any of you run the calculations to run a ICE on hydrogen? You guy's are just to funny.
The amount of oxyhydrogen needed to run an internal combustion engine is spectacular. Idling a small engine (e.g. 5hp) would require 500-1000 LPH (liters per hour), while idling a car engine would probably consume about 3000LPH of oxyhydrogen. Driving down the highway would probably consume 20000-30000 LPH of oxyhydrogen. Have fun.
well I guess Meyer Boyce Dingle etc etc etc must be comedians or magicians
Well I am sorry to have be the one to throw a wrench in but it's true. In all of meyers videos he lucky to be producing 1LPM. I have been expiermenting with a mock up cell so I have pretty good idea visually.
Quite frankly I think they are all full of shit and good at magic. I mean what the hell? David blain can walk on water and shit. I will have you know I work with the best and I have 200 HP car dyno at my disposal as well a one million dollars worth of research equipment so if anyone really wants to show me they can. There is just one catch. you won't be fooling me becuase I will hook up every piece of equipment including the gas samplers. In all the videos on youtube there is one flow meter and it's telling the truth. 40 amps for 2 LPM.
No one has proven anything and those that claim crazy numbers can't be contacted. Hmmmm, what would that be? This is bull shit but have fun.
Quote from: magnetrecharge on April 23, 2008, 08:25:10 PM
Did any of you run the calculations to run a ICE on hydrogen? You guy's are just to funny.
The amount of oxyhydrogen needed to run an internal combustion engine is spectacular. Idling a small engine (e.g. 5hp) would require 500-1000 LPH (liters per hour), while idling a car engine would probably consume about 3000LPH of oxyhydrogen. Driving down the highway would probably consume 20000-30000 LPH of oxyhydrogen. Have fun.
actually i did calculations right here and i agree with u that it needs a huge amounts of HHo to run a car:
http://www.overunity.com/index.php/topic,4424.0.html
I gave also good arguments on HHO addition in other theads.So when somone says he is running his car on HHO only i just smile.Anyhow HHO addition can shift the lean burning limit of existing fuel mixture,so it makes some space for engine tuning to boost mpg,BUT u actually have to adjust your engine ,otherwise i see it difficult to work as a pure add-on system.
MAGNA nice to have you here your obviosly interested in the truth so am I and others here at the top of this Forum on left side is a board click on Zero point energy a fellow wrote an article yesterday on this and has links to the new devices BIG production people that should talk to you and want to[ I hope] Chet PS heres the link also another article on GEET above this one jibbguy/ http://www.opednews.com
Well let's just say I have an open mind. I am not a nay sayer but no one has provided me with anything but claims.
I will say this. I try and stick to the law's. That said I have observed strange things. I have a very large iron core choke
that came out of a frequency drive system. It's weighs about 15 pounds!I am using a 150 MM h2 flow meter. Without the choke the cell is producing about 40mm. With the choke installed it climbs to 70. In either case there was a definite increase with pulsed frequency. I can say without a doubt that there is something going on with PWM. I am only using a simple circuit with a signal generator. I am current limited to 2 amps. This is what is puzzling. On the scope I'm reading upwards of 80 volt PP spikes on the scope. Note that I wm working in the 20 volt range. It seems to produce the best in the 15 to 50 KH range. Note that I am not using the fancy Lawton circuit yet.
So here is my report and itst 100% unbiased accurate.
I will try and retest today.
No pulse at 2 amps current limited. 20 MM HHO gas.
Pulse " " 30 MM " "
Pulse with large choke in series
with the positive side. 70 MM " "
Magna your the guy thats been missing here someone to verify the results LanMasterd Feynman Z monkey and others are simplifying the PWM to a computer device you should look in alot of hard work already done by very skilled and determined men Chet
http://www1.eere.energy.gov/hydrogenandfuelcells/tech_validation/pdfs/fcm03r0.pdf
ICE's are one of the least efficient methods for using Hydrogen imho... It would make much more sense to use a Hydrogen fuel cell, batteries and an electric motor.
If platinum was cheaper or an alternate was found... we'd all be driving electric cars.
Quote from: ramset on April 24, 2008, 08:47:28 AM
Magna your the guy thats been missing here someone to verify the results LanMasterd Feynman Z monkey and others are simplifying the PWM to a computer device you should look in alot of hard work already done by very skilled and determined men Chet
I can't verify without info or unit to test. That said I have some new numbers. I am not a numbers guy but
I have done many controlled tests and I am very good at it. Here is the latest updated numbers.
Straight juice no pulse with a 6 cell Myer style. 316 stainless tubes.
Current draw 4.3 amps. Supply over current limit so voltage is low. Maybe .03.
Produces 60 CC of Gas Steady.
Pulsed circut with no choke in line.
Volatge is 30 across cell at 2.85 amps.
40 CC of gas steady. 40 volts PP on scope.
Pulsed circut with choke. 2.5 amps 30 volts across cell.
68 CC of gas. 75 volts PP on scope
Note that the increase in gas corresponds with the scope reading.
This suggest that Back EMF is indeed playing a role here.
If anything I am very intrigued by what I see.
Because I can monitor gas flow real time it's easy to experiment and I will be doing just that
with different circuits and chokes.
Quote from: Tacmatricx on April 24, 2008, 10:43:59 AM
http://www1.eere.energy.gov/hydrogenandfuelcells/tech_validation/pdfs/fcm03r0.pdf
ICE's are one of the least efficient methods for using Hydrogen imho... It would make much more sense to use a Hydrogen fuel cell, batteries and an electric motor.
If platinum was cheaper or an alternate was found... we'd all be driving electric cars.
Here is the problem in a nut shell and don't count HHO out yet,
People want and need a cheap way to boost millage. My agenda is simple and
maybe practical. Most people can't afford electric cars and want or need the power of an ICE.
Modifying a car to run on gas and electric is expensive and most people don't have the skills
to make one them selves.
It may be possible with a 70% effeicent cell to produce enough HHO to get a 80% increase in fuel economy
using battery's charged up at home. I have plenty of room in the back of my car for battery's.
The concept is cost and return based to me not efficiency as it would be with most consumers.
If one could get a cell to produce 50 LPM for even 40 minutes on the battery's that would get me to work and back
and could provide a huge boost in millage. The idea is to charge the battery bank at home on the grid and would come at a cost of about 1 dollar for say, 4 battery's. 50 LPM or 3000 LPH is 80% sufficient to for cruising speeds and would require little or no fuel from the engine discounting acceleration's hills and wind. This is based on simple dyne that I do all the time. In other words you are using 80% HHO and 20% fuel. The typical Road load horse power is about 10 to 15 at 60 MPH depending on the weight of the vehicle.
I don't and can't subscribe to claims of HHO boosters getting 60MPG. It does not add up. Those cells are using brute force at 50 to 60 amps. That's a cost of 1 horse power from the cars alternator. So one can see that at cruising speeds that's a 10% millage penalty. Consider that with the fact that you are producing less then 1/2 horse power worth of HHO. People are overlooking one simple fact.
It's dirt cheap to charge battery's. The standard electric car cost's about 2 cents per mile to charge. Yep, that's right. 100 miles
would cost you 2 bucks at today's electrical rates. I think my idea warrants investigation. Especially when you consider you could use solar or wind power to charge your battery's. I believe that we overlook very possible things that would help us all out a great deal while trying to do the impossible. It could very well be that Stan Meyer made a very efficient cell perhaps upwards of 90%. I don't believe he had over unity because I just don't think that's possible but that doesn't mean I won't keep an open mind. I really respect you guy's but I do think a lot of you take large steps without taking baby ones first. In other words how about we all focus our efforts on unity and then we can look more into over unity? The nay sayers are your friends to but if they see that we can accomplish what I said above that will get some attention to the cause. If Meyer had a cell that was upwards of 90% that is not something to sneeze at! I propose that some of you work with me in creating 70% efficient electrolysis because I believe it is attainable in the near future. If anyone thinks they have already done that I would like to work with you and incorporate my idea in a test car. In the quest for the impossible we often over look the possible. Of course the opposite is true. Sometimes in the quest for the impossible we discover the possible! If we are to have mainstream
science take us seriously we must strive to use good science and real numbers.
MAG sounds good Have you looked in on the cold electricity thread ? DR Stiffler has apparently done a lot of research on hydrogen production and says he is going to share this knowledge here at overunity he is bringing a group of people up to speed on a circuit right know and then will go to hydrogen production however I wood gladly assist you in any way possible because the ability you have to test in real time is just what is needed here Chet
Quote from: ramset on April 24, 2008, 06:52:53 PM
MAG sounds good Have you looked in on the cold electricity thread ? DR Stiffler has apparently done a lot of research on hydrogen production and says he is going to share this knowledge here at overunity he is bringing a group of people up to speed on a circuit right know and then will go to hydrogen production however I wood gladly assist you in any way possible because the ability you have to test in real time is just what is needed here Chet
That would be great. Please keep me informed! I will check out the other threads! I am a labview programmer by the way.
I have hardware to do the testing as well.
Quote from: oystla on September 01, 2007, 10:12:56 AM
I'm a chemical engineer interested in the Meyer type high voltage "electrolysis", and will when I get the time investige more into this.
...
1. Using Faraday law..or
2. Using the measured collected H2/O2 gas volume and do a pure mole and energy calculation
Both of these methods should lead to same answer which is;
In a collected volume of O2/H2 gas the amount of hydrogen energy will be 7744 Joules pr liter collected gas.
It was my understanding that the faraday law stated something along the lines of 250KJ per mole of water. This is roughly 250KJ per 1.5 liters of collected gas - i.e. 1 mole of H2 and .5 moles of O2 from one mole of water, or 18 grams. This neglects any H2O vapor in the collected sample, but this is certainly not going to double the weight (and hence the volume at 1atm) of the sample?
I have used a two-plate device to generate hydrogen with virtually zero ohmic heating (sub-microsecond pulses @ 200KW) and came up with a result of approximately 240KJ per liter of hydrogen. This also agrees with Faradays law.
Just so you know, about this point in the post, my forward slash and quotation mark keys stopped working. My laptop is also running at about ten billion degrees celsius. Not sure what is going on all of a sudden.
annnnyhow.... Faraday gives the result that at 14.5VDC, 30A, (keep in mind that voltage does not matter with Faraday)
I should get .0045 moles of hydrogen in 14.5 seconds.
In the same period of time, I have seen a half liter of HHO produced. By my back of the envelope calculations, this runs to about .33 moles of H2.
The cell in question is running at 435Watts - DC, so P=VI works here without RMS considerations. multiplying by the time, to get joules, we see 6307.5J for .33 moles. Multiply by 3, and thats about 18.9KJ per mole of hydrogen. OR, 18.9KJ per 1.5 liters of HHO. Even if we guess that 100% of the gas is monoatomic, we thus double the joules per mole H2, and wind up with just heavy of 37KJ per mole of hydrogen, or per 1.5 liters of HHO, monoatomic.
This is my question - why does Nernst and Faraday indicate 250KJ per mole, but everyone is quoting 7KJ per mole or so?
Where does this ridiculoulsy low number come from? If I recall correctly, the heat of formation of water is the same 250KJ per mole... Just look it up in a CRC manual.
Here is a nice link that illustrates why I think this cell is only using 7% of the energy it should...
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html (http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html)
ok now i will reboot and attempt to edit this post with forward slashes. GRRR!
* forward slash problem fixed*
other than that, heres my opening shot on the forum so to speak. Take care, and happy hunting for overunity!
Someone plug this in
130v @ 10 amps /12 L/M ...
http://www.youtube.com/watch?v=I90IcfqS9MU&feature=related
if this is true then lots of peolpe are getting overunity. The above post would be 180% efficient, and other's I no of would be around 200% efficient.
Hallo,
is there a "step by step guide" to build a cell like that in the video of the first thread?
I am also building HHO cells, but my cells are not very effective, some help and information would be great.
felix
Quote from: hartiberlin on August 23, 2007, 05:06:45 PM
Yes, water is almost abundant and when we can run a generator just
on water we will have enough energy to make desalinated water from the
ocean waters and pour it into the deserts to grow food for everyone.
We can clean up then the whole pollution too.
Where are you going to put the waste salt by product :)
test
Quote from: ashtweth_nihilisti on July 23, 2008, 09:55:56 AM
Where are you going to put the waste salt by product :)
On chips (french fries)
There are around 42 trace elements in "salt" the table salt you eat has 2 of them(bad) , so is not therapeutic. Also if you study ecology you know that the earth cannot balance salt elsewhere.
it is at best a Small solution,(not viable long term) salt destroys land and this salt cannot be eaten with proper health effects, French fries are bad for you too ;)
Ash
I am new to this HHO generation and have just started some testing on a small scale. I wanted to post some of my findings, and since the posts on this forum seem to include test results, I decided to post here. Hopefully, my findings will spark some new ideas.
I am using a 12V battery from my Makita drill. My ?cells? are constructed using a nylon bolt, washers, and nuts with 2, 1 ?? SS washers as the plates, and a 14g copper wire (against each plate) for battery connection. I am (for now) putting these cells in a small plastic bowl of water and cannot measure HHO production, so I decided to measure the circuit characteristics. The circuit is defined as the battery and one or more cells connected together. For HHO generation, I am simply observing how much bubble production there is, assuming there will be some equivalent amount of HHO to bubble ratio. Like everyone else?s testing, I have noticed bubble production is based on the separation of the cell plates, how many cells there are, and how the cells are connected.
My battery started at ~13.7V and has drained down (with testing) to ~12.5V. First, I tested 1 cell with 3 spacers between plates (~2mm) and got a current of ~.38A and decent bubbles. Next, I took one spacer out of the cell (~1.5mm) and the current went up to ~.48A with much better bubble production. I settled on this cell configuration and built a second cell. I connected the two cells in series in the circuit and measured ~.18A with a total bubble production (between both cells) of maybe the same as just one cell. Next, I connected the two cells in parallel and measured ~.68A to ~.88A with 2 ? 3 times the bubble production as one cell! (NOTE: current started at ~.68A and went up to .88A in about 30 sec.)
So, what kind of electrical component are these cells? I started out assuming they were acting like resistors. When I tried to measure the resistance across a cell (in dirty water), the measurement started at ~400 Ohms and climbed to well over 3000 Ohms over ~30 sec. When I put the cell in clean water, could not get readings. Then, I decided to measure voltage. Well, in (dirty) water the cell had ~.44V across the terminals! I took the cell out of the water, but is was still wet, and it still measured ~.2V. After about 1.5 hrs of drying out, it finally measured 0V. So, what kind of electrical component are these cells? Resistors? Capacitors? Semiconductors? Or some other (new) kind of electrical component? Will understanding this help in determining WFC design and what kind of circuit with how much total power would be the most efficient?
hi first time posting here after doing research...
I have not seen here or anyone else use gram measurements of cell before and after electrolysis always measurement of liters or ml of gas produced. Seeing as volume fluctuates due to temp...
p.s. I have not done this yet
Andy
lol... you know i hope none of you here with these calculations of thousands of liters per minute actually get a cell or setup that can actually create this... Your gonna blow your entire engine apart... Your calculations are based purly on the petrol numbers... i think you should consider the explosive force and rate at which HHO burns before you start throwing numbers around on what you need to run a car. Lets consider some points first. You will need something in your cylinder besides HHO and atmosphere in order to slow the reaction down (and to cool your exhaust valves) alot of people run steam to do both in high end performance cars. There is now a 3rd item to put into your ratio equation. HHO : Atmosphere : Steam ratio's.
Also being the budding scientists that you are (yes all sarcasm intended) what is the explosive force of HHO compared to petrol??? I hear numbers of around 2.5 times being thrown around (and that is calculated on explosive force) so now to get the same bang per cycle you should be dividing your required amount by 2.5 minimum. Next off the bat is to those of you wanting to run a closed system with HHO... It has been reported that pure HHO when ignited IMPLODES at 4x the power of normal reaction. THATS 4 TIMES THE POWER OF YOUR 2.5 TIMES THE POWER OF PETROL!!! AND IMPLODES i can't see anyone flooding their engine with excessive amounts off straight HHO having it implode mid cycle causing any problems when their cyclinders try to eat their block from the amounts of force they will have to withstand (be aware that engines are designed to run on EXplosive reactions not INplosive) BUT!!! i may be wrong and if you guy do run engines on a closed cycle HHO can you film it and also if/when your engine eats its own cyclinders can you put it on youtube...
Ok so now the sarcasm is out the way some serious thoughts... Off a guestimate from previous experiances i would say you would need a way of distributing your HHO better. From the look of things most peoples cells "might" be able to run an engine (don't get excited) but your delivery system is now where you need to look at things. i would say its now your "air to fuel ratio" that is screwing this up. There is no way you will be able to create enough HHO to feed the vacuum of your air intake to run your car BUT if your not going to use petrol there is now a "free" port that could be used as a controlled delivery system...
Hi Inventor81,
Quote
It was my understanding that the faraday law stated something along the lines of 250KJ per mole of water. This is roughly 250KJ per 1.5 liters of collected gas - i.e. 1 mole of H2 and .5 moles of O2 from one mole of water, or 18 grams
Now if you go back to the hyperphysic-link you can read that
237 kJ are necessaray to create 1 Mol of H2 and 1/2 Mole of Oxygen. All numbers there are based on the SATP-Condition ( Standard Ambient temperature ) that is 298 K which is about 25 Celsius.
1 Mol of hydrogen has a volume of 22,4 Ltr and 1 Mol of Oxygen 11,2 Ltr
but these numbers are related to 273 Kelvin or 0 celsius !So numbers are given in different books based on different conditions which is very confusing.
Now, I am not well versed in thermodynamics and have to research the literature for the correct calculation of the Gas-Volumina of 1 Mol Hydrogen at 25 Celsius.
Based on this : 1 Molvolumen of H - lets make
a guess of 28 Ltr â€" will reduce the figures of necessary production-energy for 1 Ltr H-Gas to :
237 Kj / 28 Ltr = 8.46 KJoule per 1 Ltr H-Gas whis is the same as 8460 Watt-sec. 8460 Watt-sec / 3600 sec = 2.35 Watt-Hours.
Now compare this with what Stefan has stated in the first post
in this thread.
Best thing to do a good measurement is to produce a bigger amount of gas ( l would say 150 Ltr HHO ) because the influence of the measurement-failure-rate is bigger with a small volumina produced.
I would produce 150 Ltr of HHO then divide 100 Ltr/ 28 ltr ( 1Mol) = 3.57 Mole and the multiply 237 KJ x 3.57 = 846 KJ.
846 Kj equal 846 000 Wattsec/ 3600 sec = 235 Watt Hours.
With a power-supply of 12 Volt and 10 Ampere this 150 Ltr HHO should be produced in a little less than 2 hours.
You also have to take into account the rising temperature during the process and this is another reason that you should first let the cell run until temperature has leveled and then start Volumen-measurement.
By the way 250 Kj for 1.5 HHO-Gas- as you have stated - is totally wrong.
Regrads
Kator01
Yes the only real way to get a real reading instead of an approximation is to run a real time digitally monitoring program and a lot of sensors.
Any formula that disregards temperature and pressure will be inaccurate.
In electrolysis, the change in Gibbs Free Energy (Î"G) is 237.18 kJ. This energy represents the total electrical input energy required to electrolyze 1 mole of H2O with 100 percent energy efficiency (at 298K with one atmosphere pressure (or 101.325 kPa))...
1 mole of H2O electrolyzed will produce one mole of H2 (gas), and a half mole of O2 (gas).
We can figure out the actual volume with the following:
Using the ideal gas law: pV = nRT
http://en.wikipedia.org/wiki/Ideal_gas_law
or: V = (nRT / p)
We know 1 mole of H2 gas has a volume of 24.453 L (at 298K and 101.325 kPa) and .5 moles of O2 will have a volume of half of this: 12.226L
The total H2 O2 gas generated would be 36.679 L
So with 237.18 kJ input energy... or 65.883 Watt-Hours
you should generate (36.68L / 60 Minutes) .6116 Liters per Minute (LPM) at 65.883 W
And generate:
1 LPM for every 107.71 Watts (at 298k / 101.325 kPa)
The MMW would be 9.28
If the temperature of your gas is higher, or the pressure is lower it will appear that you're generating more H2/O2 than you actually are...
Also, the electrolyte temperature will change Î"G as well.
Hello... Newbie to the forum... I think ??? I'm sure I was a member a few years ago but I can't remember my password or user name... ;D Had a bad accident and it's taken a few years to get back on my feet... fine now though.
I've been using Oystla's formula to calculate for years and it works fine for my general experiments.
I've been using rain water successfully for years without additives... acidic rain, I know...
I've been concentrating on heat reduction... but with reasonable hydrogen production and my latest set-up just might be the ticket... LOL... anyway, it looks good at the minute... I'll be testing it over the next few days to make sure that I haven't messed up.
Quote from: CrazyEwok on January 05, 2009, 08:48:51 PM
lol... you know i hope none of you here with these calculations of thousands of liters per minute actually get a cell or setup that can actually create this... Your gonna blow your entire engine apart... Your calculations are based purly on the petrol numbers... i think you should consider the explosive force and rate at which HHO burns before you start throwing numbers around on what you need to run a car. Lets consider some points first. You will need something in your cylinder besides HHO and atmosphere in order to slow the reaction down (and to cool your exhaust valves) alot of people run steam to do both in high end performance cars. There is now a 3rd item to put into your ratio equation. HHO : Atmosphere : Steam ratio's.
Also being the budding scientists that you are (yes all sarcasm intended) what is the explosive force of HHO compared to petrol??? I hear numbers of around 2.5 times being thrown around (and that is calculated on explosive force) so now to get the same bang per cycle you should be dividing your required amount by 2.5 minimum. Next off the bat is to those of you wanting to run a closed system with HHO... It has been reported that pure HHO when ignited IMPLODES at 4x the power of normal reaction. THATS 4 TIMES THE POWER OF YOUR 2.5 TIMES THE POWER OF PETROL!!! AND IMPLODES i can't see anyone flooding their engine with excessive amounts off straight HHO having it implode mid cycle causing any problems when their cyclinders try to eat their block from the amounts of force they will have to withstand (be aware that engines are designed to run on EXplosive reactions not INplosive) BUT!!! i may be wrong and if you guy do run engines on a closed cycle HHO can you film it and also if/when your engine eats its own cyclinders can you put it on youtube...
Ok so now the sarcasm is out the way some serious thoughts... Off a guestimate from previous experiances i would say you would need a way of distributing your HHO better. From the look of things most peoples cells "might" be able to run an engine (don't get excited) but your delivery system is now where you need to look at things. i would say its now your "air to fuel ratio" that is screwing this up. There is no way you will be able to create enough HHO to feed the vacuum of your air intake to run your car BUT if your not going to use petrol there is now a "free" port that could be used as a controlled delivery system...
Stanley mayers talked about recirculating Co back into gas mixture to slow burn rate...
Hello everyone... I need a little help please.
I worked with piezo electrics many moons ago and have finally managed to built an hho cell incorperating peizo electrics... the results have me baffled... can you help by taking a look at my video
http://www.youtube.com/watch?v=C_4sCArs1-0
Am I really achieving extra/free Volts and amps?
I checked with both volt meters and have been very conservative with all values... but 70 to 80% efficient... surely not! ???
Hello,there is about 49 mls of heavy water in every liter of regular water that comes out of the tap.I see no one ever talk about "heavy water" in electroylsis of water.Basically it takes more energy to break it apart than normal water and gets left behind in a water bath.In time all that is left is heavy water.Triffid
I checked with both volt meters and have been very conservative with all values... but 70 to 80% efficient... surely not! ???
[/quote]
Hi all, I found that my measuring tube was only 0.1L and not 0.2L as I had remembered... please disregard my efficiency test. :-[
I've built an online efficiency calculator for HHO. It's available here:
http://nicksrealm.com/HHO/Calculator.html
If you have any problems, instructions are available in the drop down menu or if you have specific questions, I will be happy to try and answer them in the forums.
Nick
Quote from: IM2L844 on September 16, 2009, 03:33:06 PM
I've built an online efficiency calculator for HHO. It's available here:
http://nicksrealm.com/HHO/Calculator.html
If you have any problems, instructions are available in the drop down menu or if you have specific questions, I will be happy to try and answer them in the forums.
Nick
Hey! Very handy Nick! ;D
http://www.disclose.tv/viewvideo/29990/Dr__Steven_Greer_The_Promise_of_New_Energy/
One the gibbs energy comparison
Ok now take that info and set it asside, and use the energy content of H2 to find the mmw needed to make the hho. in other words, at 100% eff. the energy content should equal the energy input.
then cross refernece that answer.
when your done with that compare the actual energy output of a generator running on hho, vs the production cost, and the energy eff. of the engine, to see what mmw would be needed to self run.
calc it from many different angles and see if the numbers match.
Short answer, they don't
Quote from: IM2L844 on September 16, 2009, 03:33:06 PM
I've built an online efficiency calculator for HHO. It's available here:
http://nicksrealm.com/HHO/Calculator.html
If you have any problems, instructions are available in the drop down menu or if you have specific questions, I will be happy to try and answer them in the forums.
Nick
Excellent! This should be the ultimate reference for all HHO makers, presuming your master data is sound.
I would be extremely, EXTREMELY interested to know
which technology is "beating" this calculator by the greatest factor. I have reason to believe, and this is not my idea originally, that with an efficient combustion engine, a bubbler doesn't need many magnitudes greater HHO production to be able to make the energy loop, and have a car running on water almost as if it were low-temperature combustible like petrol. Fill up with water, drive 1000 miles or so with your family car. Bring a jerrycan of more water, for practical purposes and when just missing a fill-up station.
Thanks!
[Off topic, but I just figured tat gas stations might make more
profit selling tap water at 1/5th or 1/10th the price of current petrol, than petrol at current prices. There will be plenty of people selling you the convenience of water supply. Even if it's near free from the tap or river, convenience will always be worth it. I, too, would pay the neighbour's kid a couple of Euro's to go through the effort of filling up my car with my own tap water. A full tank of petrol costs almost 100 Euro's now.]
Has anyone, looked into the metal equation here?
electrochemical potential of metals?
half reactions?
Certain electrodes give off a higher quality of gas than others?
Wouldn't this throw off the calculations, and degrade or improve efficiency?
What is the hp to amp conversion for the alternator?
These formula's are of great interest to me. To get the proper answers/results.
Has anyone analyzed the gas produced from the cell with varying metals and configurations, voltages/amps etc to find the best "sweet spot"? I contacted a conventional producer of hydrogen gas a couple of years ago, and they stated they use iron nickel electrodes, and no more than 8 VDC / AMPs. After that the production quality suffers.
I would as well as many of us would I'm sure, like to get a set of irrefutable, factual calculations, that can describe in crayon to the naysayers what these units do, how, and why. Then be able to lay out a standard answer for the calculation.
Is there a (in a nut shell) formulaic version of what to do, how to do it, and why, to get X results that cant be argued to death?