Has anyone ever heard of this? Have you implemented this system on your vehicle and if so, has it worked for you? I have been interested in this idea for some time know but I don't know anyone who has acutally done it. Let me know what is working for you or what is not. Thanks.

Just gotten my car to convert to run on water recently. I bought a e-book and DIY the convertion myself and was pretty simple and straightforward.
Cost me like 60 bucks for the materials and my fuel consumption for last month was cut by close to 200 bucks. Got some positive feedbacks from my friend as well from the article and research i did on this as well. Read the article on my signature if you want more information. Share with the rest of your car buddies too man.

Fuel prices are just increasingly tremendously this days.

Has anyone ever heard of this? Have you implemented this system on your vehicle and if so, has it worked for you? I have been interested in this idea for some time know but I don't know anyone who has acutally done it. Let me know what is working for you or what is not. Thanks.
I have installed a device on a 2004 Nissan Sentra a couple weeks ago and have obtained great results--the car averaged 40-50 MPG with 57 MPG as the highest for one trip.

This device does work despite the many individuals on various forums and websites that promote it as a scam. This is understandable for many reasons.

First, many of the websites that are trying to market the product only sell you a pricy e-book to learn about the essential concepts and how to make the device, not a ready-to-install product. The content and design of the websites is usually gaudy and overly-flamboyant in its presentation (wild bright colors, amazing customer testimonials, etc), which should make any potential customer cautious. Not to mention, their attempt to sound scientific in their descriptions--I often see the chemical formula "HHO", which is just a nonstandard way to write the chemical formula of water, or H20.

Second, I believe many people that criticize the device do not understand the basics of how it works. I have spoken with many individuals about the successes that I have had with the device and they state that I am pretty much full of BS. Particularly, among people with science backgrounds (I myself have a science background) their refutation concerning the device's benefits always hovers around the "First Law of Thermodynamics", which basically states that "you can't get more energy out of a system than the amount that you initially put into the system." Yes, this statement is always true; it is a fundamental law of nature that has been verified experimentally countless times. These people that hold dearly to this law as their only source as a refuting argument often don't even care to attempt to use the device because the law says it all, it just can't happen. I believe their thinking comes from their misconception of how the device actually works. From my discussions, it seems that these individuals believe that the water molecules are broken into the products diatomic hydrogen and oxygen (both are gasses) within the device initially via the energy supplied by battery--actually, the alternator contributes and, ultimately, the gasoline is the real source of the energy--and then the products reform back into water which releases energy, which is supposedly better for fuel economy. But this is not how the device works!! If it did work this way, these individuals would be correct--it would be impossible to get back the initial input energy plus the extra energy needed to run the vehicle than what we had to initially put into the system to break the water bonds. It is important to make clear that this device I speak of acts only as a supplementary system to the operation of the gasoline fuel system already installed in your vehicle, not as the vehicle's only source of fuel.

Third, many people cite the episode of Mythbusters that proved the device was a failure. I have only seen a portion of this episode on YouTube, but I believe that these individuals made some serious errors in their design. First, the Mythbusters showed the device in trial operation splitting the bonds of water and only producing a few gas bubbles per second, if that--my device produces a lot of bubbles. Perhaps, and this is purely speculation, they did not use an electrolyte or build the device correctly. Second, when the Mythbusters installed the device on their vehicle and tried to start their vehicle, the car would not start. Personally, this does not make sense to me. With their miniscule production of oxygen and hydrogen gases as seen in the trial, it is clear that not much of these gasses would make it to the engine cylinders for combustion. Therefore, I would expect that their vehicle should start normally via regular fuel injection with practically no added products from the electrolysis device. Another personal point of speculation is that there was no gasoline in the vehicle to begin with and that they were simply trying to run the vehicle only on water--if this is true, this idea will not work as previously mentioned before due to the "First Law of Thermodynamics." I don't know the educational background of the Mythbusters, but from what I saw of this episode they did not do a very good job of giving this "myth" a fair test--that is, if the "myth" was testing the device's application to increasing fuel economy rather than running a vehicle only on gasses obtained from splitting water.

As a last point, I believe many people think that the idea is just too good to be true. This argument makes sense to these people for good reasons. Gas prices are at an all time high, a private company that came out of nowhere is trying to sell the product rather than a reputable car manufacturer, and numerous gadgets in the past boasted of similar claims that proved to be false.

Earlier I mentioned the misconception some people have regarding how the device works, so I will next present how I believe the device works. Note that I have never purchased an e-book, so everything I say is from my own experience. If anyone can shed any further light on the subject I would appreciate a deeper understanding.

The device is connected to the car battery and operates via electrolysis--energy is taken from the battery and passed through two electrodes that sit within a water solution in the device's well to break the bonds of water. This creates a difference in polarity (+ or -) among the two electrodes. Now, the molecules and ions in the solution tend to have a polarity of their own and migrate towards the electrode that is opposite of their polarity. When they make contact, a chemical reaction occurs; they may receive an electron or give up an electron, which depends on the molecule or ion that we are talking about. The net result of some molecules giving up electrons while others gain electrons is a current that flows between the electrodes, and most importantly, that molecules break apart and form new molecules.

The two molecules that we want a lot of are diatomic oxygen and hydrogen in gaseous form. Now, since these molecules are lighter than the solution they rise and are then taken into the air system that will eventually enter into the engine's cylinders. It is likely, that some amount of this oxygen and hydrogen react to form back into water during this trip. However, the gas molecules that do make to the cylinders can be used during the combustion of the gasoline. Personally, I believe that the oxygen is the key input because it is essential to the combustion of gasoline, and since more oxygen is present in the fuel-air mixture the combustion of gasoline is more efficient, creating more energy for work. As a result, the onboard computer of the vehicle, the electronic control module, will modify the amount of gasoline injected into the cylinders, thus conserving gasoline while still maintaining the car's energy needs.

An important and fundamental question to the success of this device is the following: Is the extra energy that is created from the added products for combustion (oxygen and hydrogen) more than what was initially put into breaking the bonds of water. Note that this idea applies the "First Law of Thermodynamics" in a very different manner than the idea of running your car only on water. Here we are talking about the combustion of what otherwise would be "wasted" or unused gasoline molecules that would be changed into new molecules by the catalytic converter or be spit out of the back the tailpipe if they weren't converted, so energy is indeed conserved. My answer is a resounding yes due to actual results.

To break a bond of water we need a certain and precise amount of energy, which is temperature dependent. This energy is supplied by the car battery. However, the car battery is recharged by the alternator, which has a 50-60% efficiency in supplying output energy; the alternator is run by the energy created from the combustion of gasoline. In actuality, we need more energy to break one bond of water due to the inefficiencies of supplying energy to the device than what the minimum energy according to data of experimental chemistry says.

To my knowledge, this device is always installed underneath the hood of the vehicle. This has the benefit of the water solution absorbing the heat given off by the engine to the environment. The molecules in the solution gain more energy through the absorption of this heat, and as a direct result, this lowers the input energy from the battery in general that is needed to break one bond of water, which surely compensates in some manner to our required energy needs as discussed above. The extra energy can and does come from the battery (or more precisely, the combustion of gasoline), but the results more than compensate since the combustion of gasoline releases more energy than the energy needed to split water and that the gasoline is burning more efficiently and releasing a greater total energy.

The following is my current design for the device. I don't recommend buying the e-books, since any ambitious person can research online the essential concepts needed to build the device.

I built the electrolysis device using a one-quart plastic mason jar. For the electrodes, I used two parallel stainless steel plates spaced about three-eighths of an inch apart that have wires attached to both battery terminals--the distance apart between the plates is important (research "capacitors" to find out why). I have used copper in a previous design because it is a better conductor, but I found that the copper reacts funny and gets covered in a build-up rather quickly; the stainless steel, being a poorer conductor, disperses more heat into the water solution, which I think is a good thing, and provided better results. The jar was filled with water and a good amount baking soda--the baking soda acts as an electrolyte, making the solution have a greater tendency to pass current between the electrodes. I used a tube that extends from the top of the device into the air filter for moving the oxygen and hydrogen into the engine cylinders. I wielded a small bracket together that holds the device into place during driving.

As is evident, you don't have to make any modifications to your vehicle although I have saw that some people have done so. I used a clip-on ammeter to measure the current out of curiosity and found it to be about 8 amps, which is deadly. I have also heard that the functioning of the electronic control module (the main computer in an automobile that controls many necessary engine functions, such as the amount fuel injected into the engine cylinders) in some vehicle makes is programmed in such a way that the device isn't very beneficial--the vehicle I that installed this on was a 2004 Nissan Sentra and I had no problems. The only drawback that I can immediately see from using this device is a reduction in engine life (a portion of the oxygen and hydrogen gases is likely to reform back into water before entering the engine cylinders). Another drawback is dependent on the driver; the device requires high-maintenance to be effective.

I hope that this information is beneficial to some of its readers and I would be inclined to discuss any meaningful and constructive thoughts regarding the device's operation and design.

I'm watching this thread to make sure no spam enters... keep it technical and ill leave it open... and no bans will be handed out.

I wonder what is being done to get this news to everybody and I wonder wha the oil companies think of it?

Is there a patent for the conversion and is there a possiblity of this being offered as conversion by mechanics and car manufacturers? here is more info about the process being a success: http://www.automotiveforums.com/vbulletin/showthread.php?t=910718&highlight=water+gas+fuel

Oil companies obviously dont like this, and car manufacturers wont offer this anytime soon because it costs money for R&D, so until someone MAKES them offer an alternative fuels car, they most likely will drag their feet in coming out with one.

I don't trust that.

Because the converter use the energy of the car (from the engine through the alternator) to produces the hydrogen. So you don't save energy.

And even if you connect it to a spare battery, the pressure of the hydrogen is too low to save a considerable quantity of fuel. Hydrogen need to be compressed at about 10000 psi with a special compressor. In the kit, it's not compressed at all.

It's just a scam.

I wouldnt say its ALL a scam, as me and a friend have sucessfully done this, it wasnt nearly as glamerous as people make it seem and needed a lot of refining to keep the car running, but it is possible

In fact, after installing the kit, the car will run on hydrogen AND gasoline.
99.7% on gasoline and 0.3% on hydrogen.

And the car will use 2% more fuel to supply the kit to cover the energy loss.
In global, fuel consumption will increase about 1.7%.

So your car works, but still with gasoline and a little bit of hydrogen. What is your new and old fuel consumption? I guess it's the almost the same.

Not this run your car on water crap again...:banghead:
:disappoin Read post #9 (http://www.automotiveforums.com/vbulletin/showthread.php?t=896628) (first sentence)


....so if I convert my car to run on water and put pee in it instead, I could recycle AND save money on gas??? :grinno:

:lol:

Buffordboy... a very well-though out post, but the fact remains that you can't get more out of combusting H and O than you spend getting them apart - regardless of the temperature of the water you're splitting. Added heat doesn't loosen the chemical bonds of the H and O. That is a fixed issue.

So... it doesn't matter if you try and run pure H and O, or just use it as a supplement, you're still getting less return than you're spending. To say that splitting H2O then recombining it yields a positive net energy violates the laws of thermodynamics, period - regardless of how much H and O you use to supplement the normal fuel.

Its true; additional oxygen will cause the fuel to burn much more violently and with more heat, but to suggest that you don't use that extra energy (and more) to merely sustain electrolysis is the flaw of your argument.

The bottom line is this. No matter what you do, no matter how many patents are issued, television shows tested, no matter how many catalysts or magical magnets the aftermarket says they have; you CANNOT split water and then recombine it for a net gain of energy. At best using today's technology under the hood of a car, you might see less than 10% of the energy you spend splitting make it back to the electrolysis after you've recaptured the energy from its combustion.

The other flaw I see is that you claim 8A at a supposed 14.4V. That's a mere 115W; not nearly enough to actively produce any appreciable amount of gasses. My guess is that you only have to refill your bottle about every 10,000 miles.

In order to get a good fizz like opening a soda bottle (still not enough to make a difference) you would need somewhere in the vicinity of 40 amps at 14.4V. The other question I have is this... if you put baking soda in that water, and the plates are 3/8" apart, that should be almost a dead short. Not sure why you're only getting 8A from that continuity.

Not this run your car on water crap again...:banghead:
:disappoin Read post #9 (http://www.automotiveforums.com/vbulletin/showthread.php?t=896628) (first sentence)


....so if I convert my car to run on water and put pee in it instead, I could recycle AND save money on gas??? :grinno:

:lol:

You don't understand what I say. I say that it's not working and I explain why. I told to Greenblurr93 that his car run 99.7% on gasoline, so he still pollute.

And like Curtis say, it doesn't produces enough gazes anyway.

You don't understand what I say. I say that it's not working and I explain why..
I wasn't talking about your post, I know it don't work. Thats why I made the joke......

Buffordboy... a very well-though out post, but the fact remains that you can't get more out of combusting H and O than you spend getting them apart - regardless of the temperature of the water you're splitting. Added heat doesn't loosen the chemical bonds of the H and O. That is a fixed issue.

So... it doesn't matter if you try and run pure H and O, or just use it as a supplement, you're still getting less return than you're spending. To say that splitting H2O then recombining it yields a positive net energy violates the laws of thermodynamics, period - regardless of how much H and O you use to supplement the normal fuel.

Its true; additional oxygen will cause the fuel to burn much more violently and with more heat, but to suggest that you don't use that extra energy (and more) to merely sustain electrolysis is the flaw of your argument.

The bottom line is this. No matter what you do, no matter how many patents are issued, television shows tested, no matter how many catalysts or magical magnets the aftermarket says they have; you CANNOT split water and then recombine it for a net gain of energy. At best using today's technology under the hood of a car, you might see less than 10% of the energy you spend splitting make it back to the electrolysis after you've recaptured the energy from its combustion.

The other flaw I see is that you claim 8A at a supposed 14.4V. That's a mere 115W; not nearly enough to actively produce any appreciable amount of gasses. My guess is that you only have to refill your bottle about every 10,000 miles.

In order to get a good fizz like opening a soda bottle (still not enough to make a difference) you would need somewhere in the vicinity of 40 amps at 14.4V. The other question I have is this... if you put baking soda in that water, and the plates are 3/8" apart, that should be almost a dead short. Not sure why you're only getting 8A from that continuity.
This is the one and only refutation that I will offer a rebuttal to. As I said in my original post, my intentions are to discuss the design and operation of the device, which is a constuctive endeavor that I am willing to contribute time towards; trying to make every non-believer believe that the device does work just by arguments is very time-consuming and not very effective due to the amount of garbage that already surrounds this subject.

You are right that adding heat doesn't loosen the bonds. I never mentioned that statement in my post. Instead you probably interpreted something differently than what my intent actually was--I will look at my post and modify it accordingly at some time. Heat does raise the kinetic energy of the molecules, and according to the collision model, faster moving molecules react more frequently.

You are also right the energy is fixed when breaking the bonds of water. The splitting of water into oxygen and hydrogen is an endothermic reaction, which means in order for this reaction to happen we need energy (heat) absorbed from the surroundings. Since the device does absorb heat from the engine, this probably supplies a majority of the heat for the reaction rather than relying solely on the battery as many refuting arguments often cite.

I measured that 8 amp current with the automobile turned off, so this could be why.

To be really clear to all of the readers of this thread, like I said in my orginal post, this device (at least my own personal design) requires high maintenance--during a recent one hour trip I had 1/4 of a quart of water remaining. Originally, I used copper as my electrodes and the one electrode crumbled in a few days and the other had deposits.

Thanks for your support concerning my main argument of how the device works.

Its true; additional oxygen will cause the fuel to burn much more violently and with more heat
I'll crunch some rough numbers for you. A gallon of gas has mass of approximately 2.5 kg. The mass of one-quart of water is about 1 kg, or .9 kg of O2. Suppose that I average 50 miles per gallon on my recent trip, so I used one gallon of gas. According to the amount of water remaining in the device, I put about .7 kg of O2 into the engine with this gallon of gasoline. Now, assuming that the compression ratio is 10:1 (I think this is standard for vehicles), I'll use 25 kg of air to combust with the 2.5 kg of gasoline. Of this 25 kg of air, 20% or 5 kg is O2. So what this means is that I supplied 14% more oxygen to the engine to be used during combustion. Of course, only a portion of this oxygen will be used to increase the efficiency of the gasoline.

If it's all about the oxygen, then why not use a higher compression ratio with more air? Well, for the similar scenario in regards to the addition of the .7 kg of O2, we would be inputting about 2.5 kg of nitrogen, or 13% more air than the standard compression ratio. Higher compression ratios lead to engine knocking, which actually decreases the efficiency of the combustion and may damage your engine. I am not a car expert, but this number seems extraordinary.

I see that you are a respected authority on this forum and I respect your opinion. However, I disagree strongly.

"Extraordinary claims demand extraordinary evidence," as Carl Sagan once said. I have provided directions on how I built my device, so feel free to spend a couple of dollars to try it out. Note that oxygen sensors may provide a problem with some vehicles--sensing more oxygen can cause the ECU to inject more fuel to even out the mixture, which is opposite of the results we seek (thanks to serge_saati for reminding me of this point).

Best Regards,

Buffordboy23

You don't understand what I say. I say that it's not working and I explain why. I told to Greenblurr93 that his car run 99.7% on gasoline, so he still pollute.

And like Curtis say, it doesn't produces enough gazes anyway.

actually, the system i used was 0% gas... we took the gases from the electrolosis reaction injected them using a homemade control box to modify the injector pulse to allow the proper amount of gas into the chamber... (the 'gas tank' had a seperate pump..) But like i said, it was VERY crude, didnt run great and i wouldnt recommend anyone try it on their daily driver... this was done on a scrap motor on a homemade engine stand... im just saying that it can in fact be done.. but as curtis said, is no where near as efficiant as a gasoline motor.

So... it doesn't matter if you try and run pure H and O, or just use it as a supplement, you're still getting less return than you're spending. To say that splitting H2O then recombining it yields a positive net energy violates the laws of thermodynamics, period - regardless of how much H and O you use to supplement the normal fuel.

I thought of another interesting point for some consideration. The device is under the hood of the vehicle and receives heat that the engine gives off to the environment. Anyone know the approximate temperature underneath the hood during highway driving conditions? I am sure this would vary depending on where we took our measurement.

When water is split into its components, energy (heat) must be taken from the surroundings. Some amount of this heat must come from the engine and the rest from the battery. Of course the battery's energy is supplied by the alternator, whose energy is supplied by the engine, so the input energy will be greater due to inefficiencies. It's hard for me to say to what extent; I see sources cite alternator efficiencies between 50-90%.

Now that the molecules are split, they enter the engine for combustion. Probably a substantial portion of the split water and components reforms back into water. This chemical reaction is the equivalent of the one to split water, but just in reverse; this means that the same fixed value of energy needed to split the water originally (neglecting inefficiencies of the devices) is now released to the surroundings. Note that "whole" fixed value of energy is released within the engine; when we split the water the whole fixed value did not have to come solely from the battery because of the heat given to the surrounding environment by the engine.

I think the description above may illustrate that the totality of this combustion process is not as inefficient as people tend to think. Every argument that I've heard assumes that the battery must supply the "whole" fixed value of energy to split water, while keeping the water solution from cooling and slowing the kinetic energy of the molecules and reaction rate, and neglect the heat given off by the engine. My description does obey the First Law of Thermodynamics too.

I am not saying that "this" process produces more energy than what we put into the system, but I believe the operation of this system is more complex than what many people think. I really think that the extra concentration of oxygen is a key player in improving fuel economy with this device.

I found an interesting article that should help to clear up the debate over whether or not an on board electrolysis device can improve fuel economy.

As the main focus for the paper, the authors developed a model to analyze the effects of hydrogen as an additive to the combustion energy output of methane fuel. In general, the results are positive; an addition of a small amount of hydrogen increases engine output power. There are many studies in the literature that support this idea.

The important part of the paper in regards to this thread was when they modeled the viability of using an on board electrolysis device to produce hydrogen and oxygen. The authors assumed efficiencies of 70% and 30% for the electrolysis device and the overall electric generation and mechanical efficiency, respectively.

Since I can't add attachments to this posting, the graph is located at the following link:

LINK REMOVED BY USER: See post #33 for display of graph.

The line designated "PROD ENERGY" represents the total energy needs to run the electrolysis device. Based on their brief analysis, the authors state that the viable range of beneficial operating conditions is narrow and that the idea is not economically advisable.

So what do these results mean? Can the electrolysis device really work? I guess the answer is that it depends; the main zone of engine operating conditions one consistently drives within plays a large role, which probably lends support to why there is such a divide among people regarding this topic.

I have three important things to point out to the reader concerning this graph. First, the primary fuel used in the analysis was methane rather than standard grade gasoline. Second, their graph is based on a compression ratio of 8.5:1. I believe that most standard vehicles have a compression ratio of 10:1. In the main body of their paper, the authors show that this compression ratio has a power output about .3 kW greater than the 8.5:1 ratio with the addition of hydrogen. Third, the authors assumed that all of the oxygen and hydrogen produced from the electrolysis device would be present at the time of combustion. I believe that actual systems could come very close to this ideal, since the oxygen and hydrogen are produced at different electrodes.


For interested readers, the paper's citation is

S.O Bade Shrestha, G.A. Karim. Hydrogen as an additive to methane for spark ignition engine applications. International Journal of Hydrogen Energy, 24 (1999), 577-586.

We know that the electrolysis of water work and save energy.
But the electric power must be provided by an external source of power. Not from the car itself.
It's like if you're eating yourself for surviving. You will not live very long.
You need to eat something else.

You need a lot of kWh to produce the Hydrogen.

And the hydrogen is not dense like fuel. You need to compress it at very high pressure to store it in a tank of 50L, so that the car can run at least 500km.

So it works, but a stupid kit sold on the Web doesn't save fuel.

We know that the electrolysis of water work and save energy.
But the electric power must be provided by an external source of power. Not from the car itself.
It's like if you're eating yourself for surviving. You will not live very long.
You need to eat something else.

You need a lot of kWh to produce the Hydrogen.

And the hydrogen is not dense like fuel. You need to compress it at very high pressure to store it in a tank of 50L, so that the car can run at least 500km.

So it works, but a stupid kit sold on the Web doesn't save fuel.
I wouldn't necessarily say that a kit sold on the Web doesn't save fuel. It can work, but only under very limited conditions. The graph tends to support this idea although the primary fuel used in the analysis is methane. First, the device must be designed and manufactured so that it is about a 70% efficient. With that said, the quantity of its hydrogen output must be optimized and balanced with the current operating conditions of the engine on a consistent basis and when the engine conditions operate within the narrow and beneficial range, or else the input energy from the car itself will be larger than the energy we get back.

I think we are discussing two different hydrogen technologies. Your post seems to refer to a technology that stores hydrogen on board and accesses it at a later time during driving. The technology I describe and that was analyzed in this article refers to hydrogen that is produced via electrolysis and immediately sent to the combustion chambers. The small amount of hydrogen enhances the flame velocity and thus causes a more complete burn of the gasoline. The addition of oxygen also enhances this combustion.

You are right that adding heat doesn't loosen the bonds. I never mentioned that statement in my post. Instead you probably interpreted something differently than what my intent actually was--I will look at my post and modify it accordingly at some time. Heat does raise the kinetic energy of the molecules, and according to the collision model, faster moving molecules react more frequently.
REACT, yes. Split molecular bonds in the absence of a catalyst? No. You're oversimplifying the heat/energy correlation. Its true that heat is a form of energy, but if you simply add heat to water, you'll reach boiling long before you add any appreciable assistance (if any) to electrolysis. Seriously, I don't have the math here right in front of me, but I would wager that the amount of energy required to take your volume of water from 23C to 99C is about 1/1000th the energy that you use to electrolyze the water. So, unless you have math to support your heat theory helping electrolysis, I think its no longer a possible consideration for your argument. Many of these electrolysis machines specifically have to cool the water to prevent boiling from the electrical resistance inherent in the water. Normally, such high Amperages are required to generate any appreciable gasses, and its only a matter of minutes before the water is violently boiling.

You are also right the energy is fixed when breaking the bonds of water. The splitting of water into oxygen and hydrogen is an endothermic reaction, which means in order for this reaction to happen we need energy (heat) absorbed from the surroundings. Since the device does absorb heat from the engine, this probably supplies a majority of the heat for the reaction rather than relying solely on the battery as many refuting arguments often cite.
Like I said above; a little rise in temperature makes no measurable difference in the rate of electrolysis.

I'll crunch some rough numbers for you. A gallon of gas has mass of approximately 2.5 kg. The mass of one-quart of water is about 1 kg, or .9 kg of O2. Suppose that I average 50 miles per gallon on my recent trip, so I used one gallon of gas. According to the amount of water remaining in the device, I put about .7 kg of O2 into the engine with this gallon of gasoline. Now, assuming that the compression ratio is 10:1 (I think this is standard for vehicles), I'll use 25 kg of air to combust with the 2.5 kg of gasoline. Of this 25 kg of air, 20% or 5 kg is O2. So what this means is that I supplied 14% more oxygen to the engine to be used during combustion. Of course, only a portion of this oxygen will be used to increase the efficiency of the gasoline.
I see a major flaw in your theory here. Compression ratio has nothing to do with how much fuel/air is ingested. The stoichiometric ratio of gasoline to air ais 1:14.7. It doesn't matter if the engine is 6:1 or 12:1 compression, it still ingests 14.7 lbs of air for every 1 lb of gasoline.

Plus, there is nothing to prove that the heat surrounding the engine didn't cause the water to simply evaporate. Underhood temps can reach nearly 200 degrees.

If it's all about the oxygen, then why not use a higher compression ratio with more air? Well, for the similar scenario in regards to the addition of the .7 kg of O2, we would be inputting about 2.5 kg of nitrogen, or 13% more air than the standard compression ratio. Higher compression ratios lead to engine knocking, which actually decreases the efficiency of the combustion and may damage your engine. I am not a car expert, but this number seems extraordinary.
Again... raising the compression doesn't mean more air is being ingested. Raising compression gives more temperature (not heat) to the air/fuel charge, so when its ignited, it is able to convert more of the chemical energy into kinetic energy and therefore greater (and more efficient) power

I think this may be one of the problems, and I mean no offense at all, but you yourself just admitted to not being a car expert. Some of the claims you make about the operation of the device are in contradiction with how an engine works and uses the inputs you give it.

I have provided directions on how I built my device, so feel free to spend a couple of dollars to try it out.
Thank you but I've seen hundreds of them fail before, I don't need to see it on my car again. Just like I don't click on banner ads. You never get the free laptop, but some people are just so fleeced by the prospect of something for nothing that they'll click that ad anyway.

Now that the molecules are split, they enter the engine for combustion. Probably a substantial portion of the split water and components reforms back into water. This chemical reaction is the equivalent of the one to split water, but just in reverse; this means that the same fixed value of energy needed to split the water originally (neglecting inefficiencies of the devices) is now released to the surroundings. Note that "whole" fixed value of energy is released within the engine; when we split the water the whole fixed value did not have to come solely from the battery because of the heat given to the surrounding environment by the engine.

I think the description above may illustrate that the totality of this combustion process is not as inefficient as people tend to think. Every argument that I've heard assumes that the battery must supply the "whole" fixed value of energy to split water, while keeping the water solution from cooling and slowing the kinetic energy of the molecules and reaction rate, and neglect the heat given off by the engine. My description does obey the First Law of Thermodynamics too

Assuming that a little engine heat helps electrolysis is like saying you can melt the polar ice caps with a match. I really can't stress this enough; raising the temperature of water by adding a little heat WILL NOT provide a measurable volatility to assist electrolysis. Even if it DID, the law of conservation of energy still exists and would counteract your proposal. Let's say you were able to use ONLY heat and not use electrolysis at all to attain your H2 and O2 from water. Then you burn it in the engine. Of the very inefficient combustion, a percentage goes to kinetic, some to sound, heat, light, friction, etc. The heat that soaks into the engine, gets vented to the atmosphere. Of the heat you put into the water to get it to split, only a teeny fraction makes it back to the jar as heat. If you put 100 units of heat into it to split it, you absolutely have to recapture 100 units of heat from its combustion to sustain the reaction. It doesn't matter where the energy comes from; heat of the engine, electrolysis, windmill power, the fact remains that you are attempting to extract energy from a fixed system. It doesn't matter HOW the energy gets to the water, the fact remains that you are supporting the separation of a molecule and trying to sustain that reaction with heat you get from combusting the same molecule. If you believe that you can extract energy from this system, then you must first deny that chemical equations are true. Let's look at this one:

2H2 + O2 + Ea = 2H2O + energy

Where Ea is activation energy; a spark. So, in longhand, this reads; two diatomic hydrogens and one diatomic oxygen in the presence of a small activation energy combine to form 2 molecules of water and a large exothermy. Now... as soon as you set this chain into motion, what happens? the energy you release is like dropping a box of crickets. You'll never catch them all. They scatter. So, in order for the reaction to reverse the other way, you have to come up with some energy on your own to add to the system. This is the basis of the conservation of energy. You can't use the heat energy from the engine, because its part of the equation. You have to put OUTSIDE input for this to work.

Its not a function of where you can recapture energy, you'll never find it all. You might look behind the fridge and find 30 more crickets, but that's not all of them. You need every single cricket back in the box just to support the reaction, which means even if it were possible to recapture all the energy you released, that means there is zero net energy left for kinetic uses to move the car. Once you release that H and O in the engine, you've dropped the box of crickets. If you successfully find all the crickets, then you've expended a net zero energy.

The problem is not with how or where you find the energy, its the reaction itself. Water IS NOT A FUEL, in the context of a vehicle, water is the exhaust. The reaction itself is taking exhaust and converting it to a fuel. I don't care where you try and recapture some of the energy, its a fixed system. Its closed. You will expend more energy trying to electrolyze the water than you can ever get back when you combust it.

I will say this and I will reference this post anytime this comes up again. Any time you discuss water as an automotive fuel, you are discussing the subject of perpetual motion; not because of inefficiency or energy recapturing, but because it requires the same (or more) energy to split water into H and O than you get back by combusting it. It doesn't matter where you get your energy, the net energy you can achieve by splitting and then recombining water is zero... in fact less than zero. You will waste energy trying to accomplish it, and even in supplementary systems like this one we're discussing, since the net energy is less than zero, you will burn MORE fuel trying to shift around the chemical composition of water for no gains.

I'm all for the experimentation and discussion of this stuff, but in all honesty I'm really tempted to close all threads regarding water fueled cars. It is physically and chemically impossible. Period. You can argue it all you want, but I'm right and I have a few thousand years of proven science behind me.

And for those of you who think I'm just a closed-minded scientist, I will inform you that I'm a Libertarian who holds psychic meeting groups. If anyone should grasp on to this new-age stuff, its me, but I don't because it violates laws of the universe.

My apologies for my confrontational nature in an otherwise very mature thread. I'm not lashing out at you, buffordboy. You have been very adult and open about this discussion and I've enjoyed it. But I am getting so tired of Americans' willingness to click on free laptop banner ads, intake tornadoes, pay for expensive fuel instead of driving less or trading in their SUV, and (twice) electing a lying, murdering coward for a leader. I'm picking my battles carefully, and water-fueled cars are my battle right now. When I start seeing well-known, established scientists start believing in snake oil for their cars, I get involved.

"Extraordinary claims demand extraordinary evidence," as Carl Sagan once said
I wish we could ask this Libertarian Scientist his opinion on water fuel.

Curtis73,

I have no problem with you correcting my errors. I enjoy discussing this topic and through my discussions I try to offer opinions supported by facts, but I am liable to make errors from time to time.

Upon reading your post, I agree with your point about the difference among stoichiometric ratio and compression ratio. My analysis, or "number-crunching" is rather simplistic so it may not be wholly illustrative, and in hindsight, it should be disregarded.



If you believe that you can extract energy from this system, then you must first deny that chemical equations are true. Let's look at this one:

2H2 + O2 + Ea = 2H2O + energy

Where Ea is activation energy; a spark. So, in longhand, this reads; two diatomic hydrogens and one diatomic oxygen in the presence of a small activation energy combine to form 2 molecules of water and a large exothermy.

Usually many people state that the enthalpy (dH) at standard conditions is the energy needed to complete the reaction. However, this value does not vary much with temperature. I see that the entropy (dS) of the system is always absent from refuting arguments. Really then, the best argument should rely on the Gibbs free energy equation:

dG = dH - T*dS

where dG represents the minimum work necessary on the system in order to complete the reaction, and T is the temperature in kelvins. dS and dH tend to be static parameters for a given reaction. The temperature T repesents the temperature of the device, so if the temperature of the solution in the device rises above 25 C, or 298 K, then dG will decrease, meaning that the minimum work (energy) needed for the reaction is lowered. See the link

http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/electrol.html

for more detailed info. With that said, I do agree with you about the temperature of the device rising substantially; if the water gets too hot it will evaporate and that is not a good thing.

Again you refer to hydrogen and oxygen reforming back into water during combustion, implying that we lose energy. This is not totally correct and numerous research studies by scientists with PhDs--their publications appear in scholarly journals, not independent publications more prone to bias--show that hydrogen acts as a combustion ehancement to regular gasoline. Essentially, it increases the flame velocity, which promotes a more complete burn of the gasoline. So our energy back comes from the "potential" energy of what would otherwise be uncombusted gasoline molecules.

See post #18 for a reputable reference that shows that this device can work, but only under limited conditions and thus is not avisable economically.

Ok, I buy that... getting more of the potential energy out of the fuel, but in my little non-PhD brain :) I'm still thinking about the fact that most internal combustion engines are only about 23% efficient. Even if we unlocked every last millijoule from the gasoline, it still would lost 70% or more to other energy losses. I can't see how slightly increasing the efficiency of gasoline (or even massively increasing it) can compensate for the fact that the H and O part of this equation are still confined by thermodynamic law. What you get is what you give, and since you have to give more than you get, it seems like a losing battle.

Exactly, the efficiency of the engine is near 23%. Mean 77% lost.
20% is lost in the incomplete combustion of the gasoline. The rest is lost in heat, mechanical...
80% is burned, so the heat efficiency is 23/80 = 28.75%.

By adding a good quantity of hydrogen, you can accelerate the combustion of the gasoline, because the hydrogen react faster and ignite the gasoline.

You will remove (in best case) all the unburned fuel, the efficiency will be 28.75%.

The efficiency will be increase of 25%.

For that, you still need a considerable quantity of hydrogen. At highway, a normal car use 6.8kg of fuel per hour.
The kit produces approximatively 2g of hydrogen per hour. It's not enough.

The kit need to produces at least 1000g of hydrogen per hour to improve the efficiency.

It is a tough battle to win, but I personally believe it is possible within strict limitations. There are a large number of people on both sides of the debate, but little reputable scientific evidence on the topic exists.

My research showed that practically all of the studies focused on hydrogen addition to methane rather than iso-octane. I don't know exactly why this is. I found a few studies on iso-octane that showed positive results as well, but unfortunately, my university doesn't have access to the journals so I can't comment on what the actual effect may be. Methane and iso-ocatane have similar properties in some regards but also differ, so it's difficult to confidently translate the results of the graph that I posted directly to iso-octane.

I found another study on hydrogen addition to methane that shows the engine thermal efficiency can increase by as much as 10%. For the combustion of iso-octane, the enthalpy (a decent approximation for comparison) of released energy is about 1300 kJ per mole; the enthalpy for the combustion or dissassociation of water is about 285 kJ per mole. If the results of this study carry over to iso-octane to a large degree I would expect that a small window of engine operating conditions exists for the device to be beneficial.

There are numerous patents for these electrolysis devices going back many years, but yet we don't see them on the market except for small private companies whose reputation is indiscernable. I think the material design of the device is a challenge to overcome as well; the process and products of electrolysis seem to quickly deteriorate the device. I believe that platinum electrodes are used in the industrial sector, but can't comment on their overall lifetime.

One would think that a car manufacturer would exploit the small engine window if there was one. This would require extensive engine studies of the combustion products and energies associated with the equivalence ratios. Then they would have to design an electrolytic device that is durable, not something that lasts one week from my experiences and discussions, and add secondary circuitry that operates the device only within optimum conditions. Perhaps, if this happened the overall result would be a doubling in fuel mileage.

Is it really worth it now for automakers to devote large sums of money for R&D to explore this avenue? Many American companies have cut jobs to save money and the future of automobiles appears that it will be electric- or hydrogen-based. Perhaps it was worth it decades ago when the future of the world gasoline supply didn't appear so bleak--computers were simple devices back then so the necessary engine studies may have been infeasible.

The kit need to produces at least 1000g of hydrogen per hour to improve the efficiency.
Let's assume that the results from the graph directly carries over to iso-octane. Then, according to the graph a concentration of 1% of hydrogen and oxygen relative to isoctane can yield more engine output energy than electrolysis input energy.

Let's assume that with the device installed that we can only obtain 30 mpg, a very low value according to the claims. Now say we are driving at a steady 60 mph.

In one second we would have travelled,
(60 miles) / (3600 seconds) = .017 miles

The total gas used,
(.017 miles) / (30 mpg) = .00057 gallons

Converting this to grams of iso-octane whose density is 688 g/L,
(.00057 gallons) * [(1000 L) / (264 gallons)] * (688 g/L) = 1.5 grams of iso-octane

Since according to the graph, we need 1% concentration of hydrogen and oxygen relative to iso-octane, the total grams of hydrogen and oxygen needed,
(1.5 grams iso-octane) * 1% = .015 grams hydrogen and oxygen

Because the splitting of water creates atoms of hydrogen and oxygen in equation proportions, we have two hydrogen atoms (2 g/mole) for every oxygen atom (16 g/mole), or total combined (18 g/mole). Therefore, the percentage of the hydrogen produced relative oxygen in terms of mass is,

2/18 = 11%

This means that of the .015 g og hydrogen and oxygen %11 is hydrogen by mass,

.015 g * 11% = .0017 grams hydrogen

Since this is the needed hydrogen per second, for one hour we need

(.0017 grams hydrogen) * (3600 seconds) = 6 grams of hydrogen

If our mileage per gallon was greater like the device claims this value would be smaller. I also did the calculation based on mass concentration rather than molar concentration like the graph shows; a molar concentration analysis would yield a smaller value. Keep in mind that this result is only true if we hold the said assumption regarding methane and iso-octane.

Please don't take offense serge_saati. I want to provide a fair analysis with facts regarding the possible viability of this device.

I'll do a second analysis, this time using molar concentration.

From above we use 1.5 g iso-octane per second. Iso-octane C8H18 has 114 g / mol, so the number of moles of iso-octane per second is

1.5 g / (114 g / mol) = .013 moles iso-octane

1% concentration of H2 and 1/2 O2 is

.013 * .01 = .00013 moles H2 and 1/2 O2

The mass of (H2 and 1/2 O2) per mole is (18 g / mole)--two hydrogen atoms, one oxygen. So the total mass of H2 and 1/2 O2 is

.00013 moles * (18 g / mole) = .00234 grams H2 and 1/2 O2

Like above, 11% of this mass is comprised of hydrogen, so the total amount of hydrogen per second is

.00234 grams * .11 = .00025 grams of hydrogen

The amount of hydrogen we need in one hour is

.00025 g * 3600 seconds = .9 grams hydrogen

Now let's take the previous analysis a little farther and determine the energy needs to power the device.

The chemical equation to split one mole of water is

H20 (liquid) ---> H2 (gas) + 1/2 O2 (gas)

In order for this equation to proceed we must input energy to do work; let’s use the enthalpy value (285 kJ / mole for H2O) for the equation to determine the energy needed (this value is higher than what is needed in actuality, see post #22).

Since one mole H2 contains two grams of hydrogen, 3H2 moles contains six grams. Therefore, our new chemical equation that produces six grams of hydrogen in hour is

3H20 (liquid) ---> 3H2 (gas) + 1-1/2 O2 (gas)

and the total energy needed is

(285 kJ / mol) * (3 mol H20) = 885 kJ

Now let’s consider inefficiencies. The author’s assume an overall electric generation and mechanical efficiency of 30% and electrolysis efficiency of 70%. So our total overall efficiency is

.30 * .70 = .21, or 21%

Adding this inefficiency to our input energy needs gives

885 kJ / .21 = 4071 kJ, or 4,071,000 J

Since watts in units of joules per second, the total wattage that we need to produce six grams of hydrogen in one hour is

(4,071,000 J) / (3600 seconds) = 1130 W, or 1.1 kW

Referring to the graph for a 1% concentration, we see that our computed wattage is much higher than the author’s plot “PROD ENERGY” to run the device (if our value was lower then there should be cause for concern). Still, with our computed value, we see that there are three plots for the equivalence ratios that show that our overall net power has increased with the device attached. Note that our computed value relies on one of original assumptions from the previous analysis, 30 mpg; if the mpg value were higher, then our wattage value would be lower if we did similar calculations.

Obviosly, the amperage is large; watts = amps x voltage, so the amperage through the device (for a one-cell) is 1130 W / 12 V = 94 amps. Note that a majority of the assumed values were very different from likely values and that the used values actually side with the arguments provided by the device's antagonists, and that our overall watt value is much larger than the authors (who should be unbiased) but fits the data.

If you carry out the same calculations using the results of the molar concentration analysis, .9 grams hydrogen, you should find that we need about 170 W, or .17 kW to power the device.

Your calculation is good, but where in the graph
http://cid-b8848bf931b28541.skydrive.live.com/self.aspx/Public/Graph.JPG
do you found that the water/fuel mixture is 1%?

I see 20, 40, 60, 80%.

Your calculation is good, but where in the graph
http://cid-b8848bf931b28541.skydrive.live.com/self.aspx/Public/Graph.JPG
do you found that the water/fuel mixture is 1%?

I see 20, 40, 60, 80%.

The x-axis has a scale based on 10% intervals but is labeled at 20% intervals, so the whole graph goes from 0-100% molar concentrations. So 1% can be inferred based on the supplied scale; here the "PROD ENERGY" to run the device is roughly 200 W from what my eyesight can tell and there are four equivalence ratio plots that show the indicated power output would have a net increase in energy with the device attached.

Ok, but 200W is not a lot of power. Usually when we drive, we use at least 20kW of power (for a small car). Mean that we will save 1% of fuel, it's nothing.

Also, I don't understand the 4 "equivalence ratio".

Check this add, very funny.

[link removed by moderators]

Ok, but 200W is not a lot of power. Usually when we drive, we use at least 20kW of power (for a small car). Mean that we will save 1% of fuel, it's nothing.

Also, I don't understand the 4 "equivalence ratio".
The 200W is the total energy, including inefficiencies, that we would need to operate the device and produce 1% molar concentration of hydrogen and oxygen. The "equivalence ratio" is the actual ratio of the fuel/air mixture available for combustion divided by the stoichiometric ratio. When I said four equivalence ratio plots, I am talking about the curved lines. For 1% concentration, four of these lines are higher than the straight line "PROD ENERGY"--this means that we have a surplus of output energy with the device attached. For the best case scenario plotted on the graph (0.75 equivalence ratio), the top-most dotted line at 1% has a value of about 2.5 kW. So our net energy gain at this ratio is

2500 W - 200 W = 2300, or 2.3 kW

I don't know what the energy needs of a car are, but you said 20 kW, so this value gives about a 10% surplus of energy. Again, this result only holds if we assume the graph directly translates from methane to octane with no changes.

I don't understand why when the concentration of HHO is 0%, the power output (at 0.75 ratio) is 2.5kW? It shouldn't be 0? Because you don't have hydrogen.

http://gvlqfa.bay.livefilestore.com/y1p4lQKZQN6QHOc-WfE9ktqJGQlIURyJ9UR1omwFZUtj-PWgF5_AL5AbPjwLY4Ng86vhIWTP89Yx4Q/Graph.JPG (http://gvlqfa.bay.livefilestore.com/y1p4lQKZQN6QHOc-WfE9ktqJIDV9HVwmb1UfCwVLS1mJpfiElnJXlDSqcZgEkPyn-zE-X-x-Nw_vgE/Graph.JPG)

serge_saati,

Excellent point. They should be zero. This definitely isn't obvious by the graph. The authors probably assumed that the reader would determine this. But for some small amount of hydrogen-oxygen concentration they jump enormously in value. So if the authors did start at zero and then drew a line to this large jump in energy, many of these lines would lie on top of each other and would probably detract from the overall quality of the graph. However, they probably should have stated when these jumps occur. If I add a few hydrogen atoms (0.00000000000000001%) to the mixture does such a large jump occur? I would think not.

Another important point is that the 0.52 and 0.55 equivalence ratio lines don't have any kW value until about 15% and 25%, respectively--this could confuse some readers too. This graph shows the "net" energy enhancement with the device, not the "total" output energy of the combustion products. For example, the total combustion energy could be 20 kW for the 0.75 ratio with no hydrogen; with a 10% hydrogen-oxygen concentration at 0.75 ratio, the total output energy would be about 22.5 kW, a net enhancement of 2.5 kW.

Also, their research results were based only on data from combustion in a "single" cylinder CFR (Cooperative Fuel Research) engine. This is important in quantifying the results more properly.

I think that the graph show the total output power of the engine. Not the surplus. It make more sense. It's why at 0% the power is 2.5kW at 0.75 ratio.

The engine is probably a small 50cc 2.5kW engine used in small scooter.

So at 1% molar, the increase of power is 0.025kW/2.5kW = 1%.

The power is increased by 1%, it's nothing.

If you want a 10% increasement, you need 30% of concentration.

For the 0.6 ratio, you need 5% concentrate for 500% increasement. It's more interesting. At this ratio, the air fuel mixture is too rich, it's why the addition of oxygen help.

In the car, the air fuel mixture is correct, it will be similar to the 0.75 ratio curve. Mean that we need at least 30%/1%*0.9g = 27g of H2 per hour.

I think that the graph show the total output power of the engine. Not the surplus. It make more sense. It's why at 0% the power is 2.5kW at 0.75 ratio.

The engine is probably a small 50cc 2.5kW engine used in small scooter.

So at 1% molar, the increase of power is 0.025kW/2.5kW = 1%.

The power is increased by 1%, it's nothing.

If you want a 10% increasement, you need 30% of concentration.

For the 0.6 ratio, you need 5% concentrate for 500% increasement. It's more interesting. At this ratio, the air fuel mixture is too rich, it's why the addition of oxygen help.

In the car, the air fuel mixture is correct, it will be similar to the 0.75 ratio curve. Mean that we need at least 30%/1%*0.9g = 27g of H2 per hour.
On page 582, it states verbatim "In Fig. 10, the net energy enhancement due to the addition of hydrogen and oxygen to the main fuel methane is compared with the energy required to produce the same amount of hydrogen and oxygen in situ by electrolysis of water." This production energy includes the inefficiences I stated earlier.

Your welcome to whatever interpretation that fits your judgement. Since the primary fuel is methane, the results may not directly carry over to iso-octane. I would like to see a source URL link if you have one that supports your statements regarding the concentration of hydrogen needed for gasoline fuel ineconomy to increase.

Ok, but I still not understand why at 0.75 ratio, the jump is too high from 0% to 0.0001%? Usually the slope of a log graph doesn't change that much.

Also, what is the original power of this engine (w/o hydrogen)? I want to know the increase ratio of power.

I want to know also what is the air/fuel mixture at 0.75 ratio?
You said that equivalence ratio = mixture/stoichiometric ratio
so 0.75= mixture/2 --> mixture = 1.5???
The optimal mixture for methane is 20.

I think that the mixture is wrong for the 0.75 ratio, it's why the HHO resolve this problem. The mixture at output of the electrolyser is perfect. It add a lot of oxygen to lean the mixture.

At 0.75 ratio, the original power is probably lower than the one at 0.52 ratio.

Ok, but I still not understand why at 0.75 ratio, the jump is too high from 0% to 0.0001%? Usually the slope of a log graph doesn't change that much.
Check out this link (particulary, the top-right graph showing three different colored lines):

http://en.wikipedia.org/wiki/Logarithm

This could help explain why the graph appears the way that it does--I agree with you that the plots closely approximate a logarithmic behavior.

I want to know also what is the air/fuel mixture at 0.75 ratio?
You said that equivalence ratio = mixture/stoichiometric ratio
so 0.75= mixture/2 --> mixture = 1.5???
The optimal mixture for methane is 20.
The equivalence ratio (ER) is given by,

(ER) = [ (f/a) / (fs/as) ]

where f = the actual mass of the fuel, a = the actual mass of the air, fs = stoichiometric mass of fuel, and as = stoichiometric mass of air.

Multiplying both sides by (fs/as) gives,
(ER)*(fs/as) = (f/a)

According to this link:

http://www.process-heating.com/CDA/Articles/Energy_Notes/abdb99d56e268010VgnVCM100000f932a8c0____

Pure methane has a stoichiometric air-gas ratio of 9.53 to 1 on a volume basis (the weight ratio is 17.2 to 1).
So with ER = .75

(.75) * (1/17.2) = .044 = f/a

Inverting the quantity gives,

a/f = 1/.044 = 22.7 (a lean mixture)


For your other question, I will look through the document to see if I can find anything. I believe CFR engines are standard for combustion research. You may be able to find the specs online.

Here's some links of studies done on hydrogen addition to gasoline that showed positive results, in case anyone was wondering what the effects were:

http://sciencelinks.jp/j-east/article/200519/000020051905A0769657.php

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V3F-498MBG5-103&_user=10&_coverDate=07%2F31%2F1994&_rdoc=1&_fmt=high&_orig=article&_cdi=5729&_sort=v&_docanchor=&view=c&_ct=1073&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=958425c4c864d4c849aafbb0fe46c1c5

http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B7GWS-4M2WTP3-2&_user=10&_rdoc=1&_fmt=&_orig=search&_sort=d&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=508b068a066e13adffc8be5cd0c8bb58

Unfortunately, I can't access them on-line.

EDIT: In regards to the two offered interpretations of the graph, mine and that of serge_saati, his interpretation is correct. The article stated that the test engine was a one-cylinder CFR engine, but never gave the specifications of the engine. Waukesha has a few different CFR engines used for engine research that are associated with a certain fuel type used. It makes sense that for use of methane as in the article that the CFR F-1 engine would be used. Serge_saati showed me the following link that has specs on the engine, or at least probably close to the one used in the article (?):
http://www.geocities.com/runyardj/octane1.html
According to serge_saati's calculations, the power of the engine is 2.5 kW. Therefore, his interpretaiton makes most sense, so I apologize for this misinformation. My limited knowledge of automobiles affected my judgement; I thought 2.5 kW would be rather a small output power for a single cylinder, since four-cylinders are the standard minimum on automobiles.

As an example, the correct way to the read the graph is the following: Say you have 5% concentration relative methane. Draw an imaginary vertical line from 5% to the top of the graph box and draw a dot where this line intersects the lines already on the plot; this is your output power at 5% for a particular equivalence ratio. To find the increase in power for a particular equivalence ratio with the added concentration of hydrogen and oxygen, subtract the value at 5% from the value at 0% concentration. To find the net increase/decrease in output power with the electrolysis device attached, subtract the power for the electrolysis device needed to produce the 5% concentration from your value that was obtained from the previous statement.

Ultimately, to get a large increase in power you must run a very lean engine while using methane. This has the drawback of a lower total output power.

Since I have contributed much to this thread, I will leave you all with this one last post for final consideration. There are some assumptions that needed to made about midway through but look at the underlying logic when making your determination. You can even manipulate your own values from my analysis if you choose. A graph (http://cid-810ca62460f8699f.skydrive.live.com/self.aspx/Public/Graph2.JPG) from the following journal article is presented as well and also the experimental setup and test engine specs (http://cid-810ca62460f8699f.skydrive.live.com/self.aspx/Public/page|_2.jpg (http://cid-810ca62460f8699f.skydrive.live.com/self.aspx/Public/page%7C_2.jpg)):

Maher Abdul-Resul Sadiq Al-Baghdadi. Performance study of a four-stroke spark ignition engine working with both of hydrogen and ethyl alcohol as supplementary fuel. International Journal of Hydrogen Energy 25 (2000) 1005-1009.


The following was one of my posts on another forum.

Originally Posted by atom888 http://www.physicsforums.com/Prime/buttons/viewpost.gif (http://www.physicsforums.com/showthread.php?p=1805914#post1805914)I just did a rough calculation of air ratio by your value given.

Alright, giving 14.7 g of air going into combustion. The oxygen is 3 gram and 11.7 gram of mostly nitrogen. The energy loss to nitrogen is the different in enthalpy between nitrogen at ambient temperature and nitrogen enthalpy of exhaust (I lost my thermal book to look up table value). Let say we introduce 2.5 gram of electrolysis gas. The ratio is .5g of hydrogen and 2g of oxygen. The gas intake now is offset by 14.7 - 2.5 = 12.2g .

Continue with his later I gtg...This is way too much gas from electrolysis in 1 second though. I'll give a description of some important ideas for determining how much additional gas products that we can put in using electrolysis.

First, let's calculate the theoretical work (energy) needed to produce 1 gram of hydrogen. Our chemical equation is the following:

2(H20) (l) ---> 2(H2) (g) + (O2) (g) dH = 483.6kJ

Since this equation produces four grams of hydrogen, multiply it by 1/4.

1/2(H20) (l) ---> 1/2(H2) (g) + 1/4(O2) (g) dH = 120.9kJ

Now the amount of work we need to input into the system to split water is given by the Gibbs Free Energy equation:

dG = dH - TdS

where dG is the change is free energy, dH is the change in enthalpy, T is the temperature in kelvins, and dS is the change in entropy.

Let's now compute dS. dS is given by

dS = SUM[nS(products)] - SUM[mS(reactants)]

where n and m are the coefficients of the substances in the chemical equation. My chemistry textbook gives values the following values in units of J/(mol*K):
H2 (g) -- 130.7
O2 (g) -- 205.2
H20 (l) -- 69.9

If you solve for dS, you should find dS = 0.0817 kJ/(mol*K)

Substituting our computed dH and dS values into the Gibbs Free Energy equation, we have

dG = 120.9 - T*(0.0817)

in units of kJ. It is evident that as temperature increases, the input energy (work) is reduced--the temperature of the water will rise because of heat dissipation due to various car components and inherent resistances.

Assuming the standard condition temperature of T = 298 K, you will find that the theoretical work needed to produce 1 gram of hydrogen is dG = 96.55 kJ.

Now let's assume some efficiencies. Assuming that the electrolysis device ranges from 60-80% efficiency and that the electrical energy generation and mechanical efficiency of the automobile is 30%, our total range of efficiency is 18-24% (I quoted these values on a previous post from a journal article). Taking these efficiencies into account gives the following values at 298 K:

18% efficiency -- dG = 536.39 kJ
24% efficiency -- dG = 402.29 kJ

Now divide these values by 3600 seconds to find the joules per second or watt-hours needed to produce the gram of hydrogen:

18% efficiency -- 149.00 W*h
24% efficiency -- 111.75 W*h

So how much hydrogen then can we produce in hour. This is where my results become speculative. An alternator can typically supply a maximum current of 100 amps and has voltage at about 14.4 volts. If the head lights and all accessories are off, then the car typically needs less than 10 amps to operate the necessary electrical devices (ignition coil, etc.). So let's assume that 90 amps are available for electrical work, then by the relation P=I*V,

P=1,296W*h.

This is the upper ceiling for the power that we can use to produce hydrogen while keeping the battery charged. However, I can't say for certain if the battery can output this amount power any given moment--the cold cranking amps commonly have values above 100 amps--because it may be constrained by the internal chemical reactions. Let's assume that it does, so that we can calculate the maximum amount of hydrogen that can be produced via electrolysis. Before we do, note the following: since we are looking at the power from the alternator being transferred into work at the electrolysis device, we don't have to take the 30% efficiency value from earlier (it is still important though) but only the efficiency of electrolysis. If you take our theoretical work energy needed 94.55 kJ from earlier and apply the electrolysis efficiency range using similar manipulations from above, you find the following work requirements.

60% electrolysis efficiency -- 44.70 W*h
80% electrolysis efficiency -- 33.52 W*h

If we divide our maximum output power from the alternator/battery by these values we find the maximum number of grams of hydrogen that can be produced in one hour:

60% electrolysis efficiency -- 28.99 g/hr
80% electrolysis efficiency -- 38.66 g/hr

Multiply these values by a factor of 8 to find the number of grams of oxygen produced.

Now we are ready to look at the maximum amount of hydrogen that we can put into the engine for combustion each second. Divide by 3600 seconds:

60% electrolysis efficiency -- 8.1 x 10^-3 g/s
80% electrolysis efficiency -- 1.1 x 10^-2 g/s

A good approximation for the amount of gasoline used each second by a car is 1.0-1.5 grams. The best case scenario for this analysis is 1.1% mass fraction of hydrogen relative to gasoline. You can determine the rest of your values from here.

As a last analysis, let's determine how many grams of gasoline we need to produce 1 gram of hydrogen. The combustion energy per kilogram of stoichiometric mixture is 2.79 MJ. The stoichiometric ratio of air to gasoline is 14.7:1, so the amount of gasoline needed to produce 2.79 MJ of energy can be found from the relation

x + 14.7x = 1000 grams

which shows that x = 63.69 grams of gasoline

We can now determine our objective from the following relation:

grams of gas to produce 1 gram of hydrogen = (Work to produce 1 gram H2) * (gasoline mass per combustion energy ratio)

Note that we do not need to consider the efficiencies if we use one of dG values that already includes them. For example, we previously determined that we need dG = 536.39 kJ based on an 18% total efficiency:

grams of gas to produce 1 gram of hydrogen = [(536.39 kJ) / (1 gram of hydrogen)]*[(63.69 g gasoline)/(2.79MJ)] = 12.23 grams of gasoline

This amount will obviously be consumed over whatever time interval is necessary to produce 1 gram of hydrogen. It also doesn't imply that the idea works, only that the gas consumption to produce one gram of hydrogen is not very much (about .5% of a gallon of gasoline, or a reduction of .15 miles per gallon if you get 30 miles per gallon).

i can imagine Brita coming out with some sort of water filter to help water mileage if it ever becomes a nationwide fad

pretty much any thing help these days even if it is a small increase

that just my:2cents:

and why do so many people precive that it just runs on the water itself because that is entirely not true

Everybody know that it doesn't run only on water.
And it doesn't make any improvement on fuel consumption, so it doesn't help at all.

well obviously not or the title of this thread would not of been "run your car on water"

and are you for sure that this idea would never work effcintly?

btw 1% of anything is still something

I don't start this thread myself. I just defend my opinion on the subject.
The improvement is not 1%, it's -0.5%.

I don't say that the idea is impossible, I say that it's impossible the way they make it.

To make it work, it's more complicated than that.
First, we should take a pickup, like the F-350 and load it completely with large capacity 12V rechargeable battery. For exemple rechargeable lithium battery. I would say about 100 batteries.
Second, connect 6 electrolysis devices, that's sold on the Web (12V*10A), per batteries. At total 6*100batts=600 devices. It will cost about 600*100$=60000$ just for the devices. More than the price of an electrical vehicle.
Third, fill the electrolysis with water. It will uses 32ml per devices, so at total 600*32ml=19L of water (5 gallon).
Fourth, connect 600 hoses from output gase of the electrolysis (1 per device) to the hose between air filter and throttle body.
Finaly, it will work!

It will save fuel, but it will last 1h if you run at 100mk/h (62mph) and it will only save 30% of fuel. If you want more, you need more batteries and devices! If you run less than 100km/h, it will still last 1h but it will save more than 30%. For exemple, if you run 30 km/h (19mph) it will save 100% of fuel in 30km. So the total energy saving is the same per km.

After 1h, you need to charge all the batteries! It will take 3 day to charges all the batteries. So for the persons that run more than one hour per day, it's totaly useless.

so what are you saying you would need more efficient charging system or more efficient electrolosys?

btw i missread thought it said 1% efficency not power


and do you have any ideas that might make it actually work

It's possible to have a faster charge, but we need to plug the charger directly to the transformer of the house without going through the fuse.

It will take 5x less time to charge, mean only 7 hour which is a good improvement.

But we must shut down the electrical power in the house during the charge.

And yes it will work.

@wade623

Personally, I have obtained 30% increase in fuel economy using one of these homemade devices. Of course, this claim can be argued as occurring due to the placebo effect, i.e. changing my driving habits, since I can't experimentally prove my results.

I put forth a lot of wrong ideas, but also some enlightening ones in past posts. The problem I see is that the scientific research regarding hydrogen addition only looks at power output from the engine in isolation from the electronic components. In real-time the sensors and ECU should modify the pulse-width for fuel injection based on the combustion reactants and products. Therefore, according to the research, the engine must run lean to achieve an improvement in fuel economy. Of course, any lean engine will improve fuel economy, but at the risk of potential engine-damaging effects. The role of hydrogen addition permits the engine to run under lean conditions without these negative effects. Therefore, the energy content from hydrogen combustion is not what's important, which is the discussion in many posts, but rather the properties of hydrogen and its effect on the combustion.

So, the question is then, how would hooking one of these devices to my car that runs at stoichiometric improve fuel economy? Well, here is my new theory, so I appreciate any criticism/feedback. I think that addition of hydrogen modifies the combustion products so that less free oxygen is available after combustion; or put another way, more pollutants such as nitric oxides are created as a result. This would be a result of a better gasoline burn because hydrogen is very diffusive and has a high flame velocity relative to gasoline vapor. So when the exhaust gases are read by the O2 sensor, the engine appears to be running rich because of the reduction in free oxygen. As a result, the ECU shortens the pulse width for fuel injection--now the car is running lean and the reading by the O2 sensor is now what the sensor expects at stoichiometric conditions, although the car is actually running lean. This idea cotradicts the research evidence that hydrogen addition lowers emissions. However, the research evidence is based on copious amounts of hydrogen (50-250 grams per hour), while these homemade kits would generally produce 5 grams or less.

Here is something else quite interesting. The National Hydrogen Association has as one of its members a Canadian company that markets and installs these water electrolyzers. There are many notables among the Board of Directors. Here is a description of the company article:
http://www.hydrogenassociation.org/newsletter/ad101_canada.asp
NHA members which includes the Canadian business:
http://www.hydrogenassociation.org/about/members.asp?sort=2
membership application which states that applicants must be approved by the board:
http://www.hydrogenassociation.org/join/regApplication.pdf
NHA Board of Directors:
http://www.hydrogenassociation.org/about/board.asp

to serge satti
i meant for normal car use i wast talking about your idea regarding charging sys. and devices
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ok so say how many grams of hydrogen would it take to run say a midsize car and increase fuel econemy by say 20%

and the reduction in free oxygen would be caused by a more efficent burn in the cylider am i right

how would you make electrolosys have more output without increasing energy use?

how much water would be used @ 50 gram per hour

ok so say how many grams of hydrogen would it take to run say a midsize car and increase fuel econemy by say 20%
I cannot answer this question, because I really don't know. My car is 2004 Nissan Sentra and I probably produce 3-4 grams per hour to obtain the results that I claimed. Many people say that they attach the device and it doesn't work. There are many explanations which I won't comment upon.

and the reduction in free oxygen would be caused by a more efficent burn in the cylider am i right
This is how I am interpreting it. Under normal combustion most of the gasoline is combusted anyways. Hydrogen addition will likely contribute to combustion of the remaining gasoline vapor. Hydrogen acts a "connector" between the localized fuel charges, especially at lean conditions, which promotes this more complete combustion. See this link for more info:

http://en.wikipedia.org/wiki/Hydrogen_Fuel_Injection

how would you make electrolosys have more output without increasing energy use?
how much water would be used @ 50 gram per hour
The most efficient setup is probably a series of parallel plates, where the outermost plates are positive and negative and where the internal plates are neutral plates--when the electric potential is applied, a charge will be induced on the internal neutral plates. The minimum voltage to split water is 1.23 volts. I use only two plates (electrodes) to obtain my results. I can't comment if high hydrogen production would lead to a greater improvement in fuel economy. However, according to the scientific research (see the previous link) if sufficient hydrogen is introduced, then it appears possible that you can reprogram your ECU to run leaner without large risks of engine-damaging effects--the extra mileage would come from the leaness of the mixture, not the energy content of the hydrogen.

I don't think you can produce 50 grams per hour unless you make some radical and probably impractical modifications to your car. A typical alternator has a max output of 80 amps. Assuming the alternator outputs this amperage constantly, the battery output has a maximum of 12V*80A = about 1000 watts-hours of power for the battery to maintain its internal charge. Scientific literature says 33 watt-hours is needed to split water to produce 1 gram of hydrogen per hour; however, this is the theoretical value (100% efficiency), so including efficiencies your now talking probably around 50 watt-hours per gram, which limits you to 1000/50 = 20 grams per hour, but this is probably unrealistic due to the other electrical needs of the vehicle.

Many scientific articles explored devices called hydrogen reformers, which essentially modify a portion of the gasoline on-board into copious amounts of hydrogen.

ok so say how many grams of hydrogen would it take to run say a midsize car and increase fuel econemy by say 20%

and the reduction in free oxygen would be caused by a more efficent burn in the cylider am i right

how would you make electrolosys have more output without increasing energy use?

how much water would be used @ 50 gram per hour

For a midsize car, it will take 4.75kg of electrolysis gases (H2+O2) per 100km for an increase of 20%.

Yes, the reduction in free oxygen will be caused by a more efficent burn in the cylider, because hydrogen burn completely, it's a gases, not a liquid.

If you want to increase the efficiency of the electrolysis, you must built the electrolysis device by yourself.

You should increase voltage and decrease current to reduce the electric loses and increase efficiency.
Also increase the conductivity of water by adding a lot of salt for exemple, to reduce electric lost.
And finaly, increase the volume room for the gases to decrease the pressure and reduces energy lost.50g/h is nothing. It will just uses 71g of water.

For a midsize car, it will take 4.75kg of electrolysis gases (H2+O2) per 100km for an increase of 20%.

...
Also increase the conductivity of water by adding a lot of salt for exemple, to reduce electric lost.
Your 4.75 kg figure implies that hydrogen addition and its subsequent combustion is to derive energy to assist in powering the car. Many people think that this is the role of hydrogen, but it is not. Since you have to produce the energy to split water to produce the hydrogen, you will always lose--it's thermodynamic law. The mileage increase likely comes from the car leaning out the mixture in reponse to the modification of combustion products, and therefore, not much hydrogen may be needed.

I wouldn't use salt as an electrolyte. It's composed of calcuim and chlorine ions. Therefore, your likely produce chlorine gas, Cl2. There are better electrolytes that don't produce such dangerous gases. Many people use baking soda.

Your 4.75 kg figure implies that hydrogen addition and its subsequent combustion is to derive energy to assist in powering the car. Many people think that this is the role of hydrogen, but it is not. Since you have to produce the energy to split water to produce the hydrogen, you will always lose--it's thermodynamic law. The mileage increase likely comes from the car leaning out the mixture in reponse to the modification of combustion products, and therefore, not much hydrogen may be needed.


No, I will not loose energy, because the electrical energy come from the 100 extra 12V batteries at the rear of the pick-up, not from the alternator. This is in my idea version, not in the kit version. And I will able to recharge the 100 batteries from the electrical power of my house.

And it will not make the mixture leaner because all additional atoms of oxygen created by electrolysis will react completely with the hydrogen.

i understand that as of now hydrogen only assits gasoline in burning so adding hydrogen prettymuch just leans out the engine and thats what helps it ge better milage

i understand that as of now hydrogen only assits gasoline in burning so adding hydrogen prettymuch just leans out the engine and thats what helps it get better milage

i understand that as of now hydrogen only assits gasoline in burning so adding hydrogen prettymuch just leans out the engine and thats what helps it get better milage

No, you don't understand. The device doesn't produces just Hydrogen. It converts water into Oxygen AND Hydrogen in good stoechiometric ratio. The both exit at the output of the device by the same hose. And they doesn't react with each other until they get sparked by the ignition of the engine.

They doesn't affect the richness of the engine.

A rich mixture is: too much fuel and not enough oxygen.

A lean mixture is: too much oxygen and not enough fuel.

Hydrogen is used as fuel substitute from gasoline. It burns with oxygen and produces heat and pressure. So the engine use less gasoline.

And it will not make the mixture leaner because all additional atoms of oxygen created by electrolysis will react completely with the hydrogen.
I think this grossly oversimplifies the combustion process. If there were true, then why would a car put out a wide variety of pollutants? Check out the link for the products of gasoline combustion:

http://www.newton.dep.anl.gov/askasci/chem99/chem99583.htm

According to the chemical equation, gasoline and oxygen produce carbon dioxide and water only. Where do all of the other negative emissions, such as nitric oxides, come from then?

Originally Posted by Buffordboy23
According to the chemical equation, gasoline and oxygen produce carbon dioxide and water only. Where do all of the other negative emissions, such as nitric oxides, come from then?

Because in air we don't have just oxygen. We have a lot of azote. And azote react with the excess of oxygen to form NO2 and NO3. Because in the engine, air/fuel mixture is a little bit leaner to have a better combustion, so it have an surplus of oxygen.

Also, in gasoline, we don't have just octane. We have a little bit of sulphur. This one react with the excess of oxygen to form SO2.

And because gasoline is a liquid and not a gases, it makes time to burn and not burn completely. So engine emit also fuel vapor.

But with the eletrolysis, it's different. Pure oxygen and pure hydrogen exit at the output of the device. No azote, no fuel vapor, so sulphur... And because they are both gases, they react instantly and completely.

And because they are in perfect stoechiometric ratio, they burn completely. So they don't affect the actual air/fuel mixture of the engine.

But with the eletrolysis, it's different. Pure oxygen and pure hydrogen exit at the output of the device. No azote, no fuel vapor, so sulphur... And because they are both gases, they react instantly and completely.
Thanks for your reply on the combustion process and pollutant formation. However, this quoted statement is wrong. Your right that the oxygen and hydrogen are likely formed in equal amounts, but they don't react instantly. If they did react instantly, you would have water; but you "can" ignite the mixture from the electrolysis outflow, so this means that the products have not reacted and will mix with the ambient air that is fed into the intake.

Thanks for your reply on the combustion process and pollutant formation. However, this quoted statement is wrong. Your right that the oxygen and hydrogen are likely formed in equal amounts, but they don't react instantly. If they did react instantly, you would have water; but you "can" ignite the mixture from the electrolysis outflow, so this means that the products have not reacted and will mix with the ambient air that is fed into the intake.

Right, the H2/O2 mixture will mix with the ambiant air. But my point is still the same, which is: electrolysis doesn't make the mixture leaner, because it adds oxygen AND hydrogen. The oxygen leans and hydrogen enriches the mixture. So in global, the mixture is the same.

All the hydrogen will burn, so all extra atom of oxygen (from air or electrolysis) will react.

the hydrogen burns when ignited but the extra oxygen helps it burn like adding oxygen to acetaline it makes it burn more effcintly

If we add too much hydrogen without adding oxygen, the mixture will become too rich and a lot of fuel (more likely gasoline) will stay unburned.

If then, we add the extra oxygen, the mixture will become normal and almost all the fuel will burn.

The extra power will come from the combustion of the hydrogen.

The extra power will come from the combustion of the hydrogen.
Serge_saati,

This could be true, but I still think combustion modification plays a role--scientific research looks at hydrogen addition at ultra-lean conditions.

I could test this idea by gathering data from the O2 sensors, emissions, or fuel-injection pulsewidth for with and without hydrogen supplementation. Ideally, I would like to do this while driving. Does anyone know of a way to accomplish this in a cheap manner? Obviously, I would need a labtop computer, and some type of probes(?). Thanks.

dont they have programs that measure that stuff for the computer and you just plug it in under the dash


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