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Thread: Is it realy possible to make enough hho to run 100%hho

  1. #21

    Default is it real?

    guys i've recently seen(on u tube...other internet places) a couple of guys running their pick up on 100% hoo gas. they say - and show its hydrogen on demand, anyone seen it/verify it. in us somewhere...guess i shoulda got a link now i come to think of it...but it sure looks impressive and sorta gives me a kuta( kick up the ass) if its straight up

  2. Default

    I'm surprised at the lack of chemical knowledge around here.

    The answer is easy to find and I'll show you all the math behind it:

    Water has a density of 1g/mL, so 1L = 1000g.

    Water has a molar mass of 18.01g. Therefore, the number of moles of water in one liter is 1000/18.01 = 55.52 moles.

    One mole of water decomposes into one oxygen and two hydrogen atoms. As such, the decomposition of one mole of water yields one mole of hydrogen gas and one-half mole of oxygen gas through the following equation:

    (H2O) --> (H2) + 1/2(O2) (remember that the gases are diatomic)

    Subsequently, 55.52 moles of water decompose into 55.52 moles of hydrogen gas and 27.76 moles of oxygen gas.

    According to the Ideal Gas Law, at standard temperature and pressure (1 atmosphere, 293.15 Kelvin), one mole of any gas will occupy 22.4 liters of volume.

    One liter of water has produced 55.52 moles of H2 and 27.76 moles of O2 for a combined total of 83.28 moles of gas. Each mole of gas occupies 22.4 liters, so 22.4*83.28 = 1865.5 liters of gas.

    And thus we arrive at the answer: One liter of water makes 1865.5 liters of gas.

    __________________________________________________ ______________

    As far as running an engine on nothing but the gas mix, generation problems aside:

    The lower flammable limit of hydrogen in air is 4% by volume. Therefore, it would be theoretically possible to run an engine on a blend of oxyhydrogen and air which would amount to 4% hydrogen delivered at the manifold.

    Air is a mixture of many gases, mainly nitrogen and oxygen. The average molar mass of air is 28.97. This is the average weight of 22.4 liters of air at STP.

    To achieve 4% hydrogen by volume in air by adding a mix of 2:1 H2:O2 takes a bit of simple math. The equation goes like this:

    (% gas A)(molar mass of gas A) + (% gas B)(molar mass of gas B) = (100%)(molar mass of mixed gases)

    Therefore, we can solve for %HHO needed as long as we know the molar mass of the target gas. The target gas in this case is 4% by volume hydrogen and the rest air. Let us determine the molar mass:

    1 mole of air is 28.97g and occupies 22.4 liters. 4% by volume of 22.4 = 0.896. 22.4-0.896 = 21.504. Therefore, the target gas has 21.504 liters, or (.96 moles) of air and .896 (or 0.04 moles) of hydrogen. Cumulatively, the molar mass is (.96*28.97) + (.04*2) = 27.89 grams per mole.

    This makes sense because we know that adding hydrogen to air will decrease its molar mass because it is much less dense than air.

    Anyway, now to find the amount of HHO needed to make air 4% hydrogen. We come back to the equation:

    (%HHO)(molar mass of HHO) + (100-%HHO)(molar mass of air) = (100%)(molar mass of mixed gases)

    And insert the values

    (%HHO)(18.01) + (100-%HHO)(28.97) = (100%)(27.89)

    and then solve for %HHO, which comes to 9.9% after a bit of algebra.

    Therefore, you need to add at least 9.9% HHO to air to make it combustible.

    __________________________________________________ ______________

    So what does this mean? How about an example:

    In a 2-liter engine at 1000RPM, you need a fuel/air intake of 1000 liters per minute. (500 2-liter intakes in 1000 rotations in one minute) 9.9%vol of the gas has to be HHO in order for combustion to take place, so 9.9% of 1000L = 99 liters.

    Your gas generator must produce, at minimum, 99 liters of HHO per minute. This guarantees, however, that the fuel/air entering the engine is combustible. This makes no statement as to whether the combusting mixture will release enough energy to keep the engine spinning, much less move a vehicle. That calls for a whole other set of equations.


    Last edited by ElectroNut; 01-25-2011 at 12:08 AM.

  3. Default

    Assuming you people will want "That next set of equations," I've compiled them here:

    We'll have to work with a practical example in this case. Let us examine again the 2-liter engine spinning at 1000 RPM. How much energy can it output running on only 4% hydrogen? This is an easy calorific equation which can be readily calculated.

    We know that when hydrogen reacts to form water, a certain amount of energy is released. We also know that of the 1000 liters of air entering the engine each minute, 40L, or 4% of it, is hydrogen gas. 40L of hydrogen gas, according to the Ideal Gas Law, is equivelant to 1.786 moles.

    Two moles of hydrogen react with one mole of oxygen to produce one mole of water and some heat. The heat of combustion of hydrogen and subsequent formation of water is 285.83 kilojoules per mole of hydrogen reacted. Therefore, 1.786 moles of hydrogen per minute will generate 510.49 kilojoules of energy per minute. This can be converted to horsepower since one horsepower is equivalent to 44.74 kJ/min.

    As it turns out, 510.49 kilojoules per minute equates to 11.04 horsepower. This is, of course, assuming that all of the heat energy released is converted to mechanical energy. We know this to be very untrue - the efficiency of a modern internal combustion engine is around 19%. This means that just about 2.09 horsepower would be mechanical - not even enough to keep the engine spinning - while the rest would be lost as heat.

    So how much HHO would it take to produce something like 50 horsepower with a 19% efficient engine? We simply run the equations backward: 50hp = 2237 kJ/min divided by 19% = 11774 kJ/min. One mole hydrogen is equivalent to 285.83kJ, so that makes 41.19 moles of hydrogen needed. In liters, 41.19 moles is 922.7 liters of hydrogen per minute.

    Assuming you'd be feeding the engine pure HHO at this point, we know that half a mole of oxygen is needed for every mole of hydrogen, so 922.7 + 461.4 = 1384.1 liters of HHO per minute, or 23 liters per second. As a fun fact, if the engine displacement is 2L, it would have to spin at 1384.1 RPM while producing this 50hp.

    On a last note, let's compare this to gasoline. Gasoline has a listed average energy density of 34,200,000 joules/liter when reacted with air. One liter of hydrogen (or 1/22.4th of a mole) contains (1/22.4)*285,830 = 12,760 joules per liter when reacted with air. Take 2/3 of the 12,760 to account for the oxygen in the mix and you end up with 8,507 joules per liter of HHO.

    This shows that every liter of HHO is equivalent to just 0.25mL or 0.00845 fl. oz. of gasoline with respect to energy content. Conversely, one gallon of gasoline is worth 15,140 liters of HHO.

    Last edited by ElectroNut; 01-25-2011 at 12:58 AM.

  4. #24
    Join Date
    Feb 2010
    los angeles , california


    don't mean to disrespect your intelligent but you are comparing apple to oranges. The gasoline engine is not made to run on hho, it is design to run on gasoline. Maybe you should do some more research on the car that run on pure water. I believe one cup is equal to around 100 miles I might be a little off. I 've already explain many time how to run on pure hho. and I'm not gonna explain again go do your research and or check out some of my post and read them you will understand what I'm talking about. Go to youtube and check out this car made in japan and tested there.

  5. #25

    Default understanding ...

    maybe i worded my question wrong, sorry...i like yes or no answers..and i know thats not always possible, but i would have to go back to school to work out the numbers, but 1 thing i am fairly sure of is the mixture of petrol/air dos'nt equate to those numbers perhaps i'm missing something but anyway i'd still like to know if anyone has actually seen a 100%hho on demand car

  6. #26

    Default sense

    starting to get the hang of your post electronut, but now my heads sore, i'll keep at it anyway cheers

  7. Default

    Quote Originally Posted by waterbugs View Post
    don't mean to disrespect your intelligent but you are comparing apple to oranges. The gasoline engine is not made to run on hho, it is design to run on gasoline. Maybe you should do some more research on the car that run on pure water. I believe one cup is equal to around 100 miles I might be a little off. I 've already explain many time how to run on pure hho. and I'm not gonna explain again go do your research and or check out some of my post and read them you will understand what I'm talking about. Go to youtube and check out this car made in japan and tested there.
    Hear me out:

    You "believe" one cup is equal to 100 miles, but based on what? Something you saw on YouTube? No disrespect intended, but how is that even an equality? It's like saying a half cup of gas gets a car one mile... very ambiguous. Which car? How old is it? What is the altitude/barometric pressure? What is the temperature? What kind of road are you driving on? How full are the tires? Are you going uphill or downhill? Is the road wet? How much does the car weigh? Is there wind?

    But that's enough questions! I understand that these are averages. You told me you could be a little off, so I'm about to find out.

    Pretend you have a car that gets 25mpg on gasoline. You want to go 100 miles. Thus, you'll need 4 gallons of gas to do it. A gallon of gasoline contains around 129,276,000 joules of energy, so the whole trip used 517,104,000 joules.

    Your car's engine is about 19% efficient. Let us see how much of that gas went to actually getting you there and not out the exhaust as heat. 0.19*517,104,000 = 98,249,760 J. So, if your engine were 100% chemical energy-to-mechanical energy efficient, you would use 98,249,760/129,276,000 = .76 gallons of gasoline. This is 131.5mpg - as you can clearly see, 100% efficiency is absurd. Hang in there; there is a point to all this jabber.

    Let us now examine how much energy is in a cup of water. First of all, the water has to be split into HHO. I will ignore this step because physics says it takes exactly the amount of energy to split the water as will be released when it burns, completely ruining the need to generate HHO at all. Therefore, let us pretend that we can split the water using no energy whatsoever.

    One cup of water is 237mL and contains 13.2 moles of water since water is 18g/mole. Each mole of water decomposes to a half mole of oxygen and one mole of hydrogen, leaving you with 13.2 moles of hydrogen and 6.6 moles of oxygen. When one mole of hydrogen reacts with one half mole of oxygen to form water, 285.83kJ of energy is released. Therefore, 13.2 moles would produce 3,772,956 joules of energy. We now know how much energy is made when re-forming one cup of water from HHO.

    Now plug this into the 100% efficient car model. We eliminated engine efficiency since, as you thoughtfully mentioned, "gas engines are not designed to run on HHO." Therefore we will pretend we have an HHO engine which is 100% efficient - the best-case scenario. It takes 98,249,760 J of energy to get the car 100 miles when engine efficiency is not taken into account. How much water is this? The cup of water had 3,772,956 J, so it would get you 100*(3,772,956/98,249,760) = 3.84 miles.

    Yes. If your engine was 100% efficient AND it took no power to make your HHO, a cup of water would get you 3.84 miles in the typical 25mpg-sized car. (I use this 25mpg model since Stan's dune buggy would probably get about 25mpg if it ran on gasoline)

    Even the most efficient heat engines found in power plants in the form of rankine-cycle turbines only barely approach 43% efficiency. Now we're down to 1.54 miles, assuming you've also come up with some magic HHO engine which is top-of-modern-technology efficient. Is your engine about as efficient as a petrol engine, 19%? Try 0.73 miles, or a bit short of 4000 feet.

    Don't forget to account for the energy it would take to make the HHO in the first place, assuming you could break the laws of physics and make it with less energy that you'd eventually get out of it.

    See my point?

    Anyway, as for the "yes or no" answer:

    Yes if you believe you can make HHO with less energy than you get out of it. Much less.
    No if the laws of physics mean something to you.

    Sorry if I stepped on any toes. Math gets rather impersonal sometimes. As a side note, I really hope I can open some eyes here and perhaps direct amateur research toward less-futile ends.


    Last edited by ElectroNut; 01-26-2011 at 08:17 PM.

  8. #28


    Electronut>> Really, really good posts! You've explained things very well in easy terminology.
    There are so many people that think that running an ICE on HHO is economically feasible, it's crazy.
    I believe that 99.999% of us that understand this technology at all realize that the induction of oxyhydrogen is merely a catalyst used to promote a more complete burn in the combustion chamber.
    Kudo's to you!
    1998 Explorer 4x4, 4.0
    14 cell / 2 stack 6x9" drycell reactor 28%KOH dual EFIE, MAF enhancer, IAT & ECT controllers, 2.4 LPM @ 30 amps. 6.35 MMW http://reduceyourfuelbill.com.au/forum/index.php

  9. Default

    Thank you. It's great that you mention the part about the catalysis.

    I have been working on an expression to determine the feasibility of that very process. Unfortunately, it's very involved and I still need to do a whole lot of research.

    It basically comes down to an emissions problem. In the perfect combustion of a hydrocarbon, the only products are water and carbon dioxide. In the real world, we know that many things like carbon monoxide, methanol, formaldehyde, nitrogen oxides, etc. are produced during normal combustion of a hydrocarbon in air, especially under compression. What I need to find is a reliable assay of the content of exhaust gas for several engines under different load conditions. From there, the chemistry is easy. It takes energy to break chemical bonds and energy is usually released when they re-form. If I can re-arrange all the unburned and partially reacted stuff from the exhaust into CO2, H2O, and N2, I can find out how much extra energy those re-formations will produce.

    If the HHO can catalyze these reactions to completion in the cylinder, the engine's performance will benefit corresponding to that amount of energy. I really have no clue as to how much energy this really is. However, I know that catalytic converters are designed to re-arrange around 90% of unburned hydrocarbons and CO, according to the wikipedia article (not reliable, but good enough for preliminary research). This means that a lot of this unused energy is expended at the catalytic converter instead of the engine. This would raise the temperature of the catalyst as well as post-converter exhaust significantly. In fact, wikipedia states that "Any condition that causes abnormally high levels of unburned hydrocarbons — raw or partially burnt fuel — to reach the converter will tend to significantly elevate its temperature, bringing the risk of a meltdown of the substrate and resultant catalytic deactivation and severe exhaust restriction."

    Catalytic converter elements don't usually get significantly heated during operation, so I'm already a bit skeptical as to how much extra energy you could really extract from a more complete burn. I cannot make a prediction at this point; the math will have to do the talking.

    The other difficult part is finding the load of the cell on the engine and whether the catalysis extracts enough energy from the gasoline to be beneficial. This will require an efficiency value for the alternator, the power draw of the cell, and the gas output. The gas is burned and therefore adds its energy back to the system less the efficiencies of generation, another bonus. All of this will correspond to an engine load value in "raw crank horsepower" which is mechanical energy and can be compared to the gain from the more completely-burned fuel.

    Of course, none of all that takes into account the possible effect on, for example, the oxygen sensor. A study will have to be performed to determine whether the more completely burned fuel eats up extra oxygen (probable) and how less O2 in the exhaust would correspond to the ECM possibly leaning out the injectors.

    Food for thought. If anyone can find pre-catalytic converter exhaust emission profiles for engines under load, I'd really appreciate them!

    Last edited by ElectroNut; 01-27-2011 at 12:19 AM.

  10. #30
    Join Date
    Feb 2010
    los angeles , california


    okey Im gonna explain this how you run on pure water. On vacuum stroke about 1/4 way down water, yes water mist is spray into cylinder a split second right after hho is ignited, this will turn water in stream and pushes the cylinder. this is how the pure hho motor work, check it out on youtube how a pure hho engine work. Remember about just less than a quarter way on vacuum stroke it get ignited not like the old four strokes where vacuum than compress than ignite without water mist. You know we all gonna be running on pure water and generating electricity on pure water soon, thanks to youtube.

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