[modest]

Originally Posted by DanGray

OK, tell me how many BTU's in a 2 liters of hydrogen/oxygen in either kind hydrogen that is produced from the generator.

1 BTU = 1054 Joules

I gave the figure "13,536 joules per hour" above for 2.184 liters per hour. 13536/1054 = 12.84 BTU/hour is what you're looking for. In North America BTU/hour is usually equivalent to "BTU" as a rating for an appliance.

Originally Posted by DanGray

I know of the loss in conversion, like say in , running a 5 HP ele motor set up to run a 120 generator feeding the motor of course it can't be done..

An electric motor turns electricity into mechanical power. When well-manufactured, they're about 80-90 percent efficient at doing so. A hydrogen generator is no different in concept from an electric motor. It is truing electricity into mechanical power, but it is generally agreed to be less efficient at doing so. The figures I gave above are a best case scenario and show 74% efficiency. Add to that the real world problems of excess heat in the alternator and electrolyte and so on and you're easily down under 50 percent in a very well made system.

It is simply not advantageous to use a hydrogen generator over an electric motor and it is utterly ridiculous (conceptually) to use a hydrogen generator over readily combustible gasoline. Think about it...

You've got this really well burning fuel - gasoline. You could put it in your engine and make your car go, but instead you'd prefer to put it in your engine, make your alternator turn, make some electricity, use the electricity to make some hydrogen, then use the hydrogen to make the car go. There can be no sensible explanation for why a reasonable person would do that. Every one of those steps looses you useful energy that would have gotten you down the road.

Originally Posted by DanGray

what's happing here is more than that , producing it may not be = to the power consumed to make it but in the process some how it's causing the Gasoline to be more efficient thus a mileage increase,

No. The hydrogen is generated with the gasoline. The gasoline literally has to turn the motor to make the hydrogen. So the amount of energy used getting the hydrogen must be equal or greater than the energy ability of the hydrogen as a point of fact.

So the only question left is the design of the engine. In theory you could design an engine capable of burning gasoline/hydrogen efficiently. In a perfect world that vehicle (with a hydrogen generator) wold work as well as a vehicle with no generator that is designed to burn only gasoline. But, in our world we loose useful energy through the conversion of fuel to electricity to hydrogen.

Originally Posted by DanGray

I don't know just what's going on but it is working, and not by braking your laws, so if you can get your minds way from , it can't work because of the laws, try and figure out why it is working..

Reputable institutions (such as popular mechanics) have tested these setups and came to the very understandable conclusion that they don't work.

~modest

[modest]

Originally Posted by DFINITLYDISTRUBD

BTW. diamond wave, square wave etc. really ain't going to make much of a difference in H and O2 output...straight DC provides the highest gas yield per minute.

I agree with DD on this - it's a very good point. DC is most efficient at electrolyzing water.

DanGray, whatever your intentions with the square wave, it sounds wrong. Your car is DC and your generator would be most efficient as DC... no need to add another energy-loosing step.

~modest

[DFINITLYDISTRUBD]

will give you 1.0068 g/hour of hydrogen/oxygen. 1.0068 g is 2.184 liters via the ideal gas law. So - the first thing to recognize is that you're only getting about 2 liters of gas per hour. Burning 1.0068 grams will give you 13,536 joules per hour or 3.760 joules per second.

First a lil question here is this H alone or H and O2 total weight? The difference would be quite significant with regard to how much power you would get out of burning it (to the detriment of Dangray's arguments I believe.)

Secondly feeding a 4-stroke internal combustion engine water/water vapor does work to boost power and efficiency (in proper measure)) BUT it's prone to causing detonation the wrong proportion. Blowing a nice hole in the cylinder head or the piston, and or bending or breaking your con rods, etc... AND it (the water) has the nasty tendency of blowing past the rings and mixing with the oil in the crankcase...resulting in lack of lubricity and sooner rather than later the need for a new engine.

Lastly as to the so called benefits the hydrogen generator...What becomes of the O2 generated is it burned as well or vented? If it's burned this would be most likely of more benefit to the engine than the H. (More O2 =hotter more violent fire...That said though the quantities available of either would result in a negligible increase in power or efficiency even without the drag of the generator....Think about it Dan a two liter
pop bottles worth of gas at atmospheric pressure generated per Hour...Your average car burns over 100CFM of air every minute with gasoline vapor consumption being about (1/15) to (1/20th) the volume of the air ("15-1" was considered average when I took engine repair in highschool.) I don't know how many liters to the cubic foot but I'm pretty sure it works out to requiring a heck of a lot more than two liters per hour.

[modest]

Originally Posted by DFINITLYDISTRUBD

First a lil question here is this H alone or H and O2 total weight? The difference would be quite significant with regard to how much power you would get out of burning it (to the detriment of Dangray's arguments I believe.)

It is both H and O. I believe this is the proper way to consider it because electrical energy is going to be used for both reducing and oxidizing regardless - so we might as well use both the hydrogen and oxygen. The calculations I show assume that both oxygen and hydrogen are produced and both are put in the engine and all heat energy achieved combusting them is successfully delivered to the engine as power.

~modest

[DFINITLYDISTRUBD]

Originally Posted by modest

It is both H and O. I believe this is the proper way to consider it because electrical energy is going to be used for both reducing and oxidizing regardless - so we might as well use both the hydrogen and oxygen. The calculations I show assume that both oxygen and hydrogen are produced and both are put in the engine and all heat energy achieved combusting them is successfully delivered to the engine as power.

~modest

Sorry to be such a pain but in your math the output was for the H and O2 in the proportions yielded by the electrolysis (compensated)? How did you sort out the volume of each gas? I don't mean to insinuate that your math is flawed I'm sure it isn't....It's just my curiosity getting the better of me.

[DanGray]

Originally Posted by modest

Reputable institutions (such as popular mechanics) have tested these setups and came to the very understandable conclusion that they don't work.

~modest

the trouble with that is, did they test one that doesn't not work and could not work anyway ?
that could be the biggest problem here, if they tested a ( and I've seen ones that I know can not work).. scammed unit for the money then of course it will fail their tests.. was it a duel unit?
then you all read what they have to say after testing , now it's in stone and a fact to try and reason with but still only read from others work and test .. do you see what I'm saying here..

but regardless, still they are being used with results and I'm sure it's not the one popular mechanics used obviously.. if I could type the 100+ pages in my paper work in less time than building this, I would post it here.. but that isn't going to happen too soon I'm sure .. I'm a two finger pecker on the key board.. and being on Dial up JPEG,ing it is out of the question.

we'll see what happens with my unit ,,,

[DFINITLYDISTRUBD]

Originally Posted by DanGray

the trouble with that is, did they test one that doesn't not work and could not work anyway ?
that could be the biggest problem here, if they tested a ( and I've seen ones that I know can not work).. scammed unit for the money then of course it will fail their tests.. was it a duel unit?
then you all read what they have to say after testing , now it's in stone and a fact to try and reason with but still only read from others work and test .. do you see what I'm saying here..

but regardless, still they are being used with results and I'm sure it's not the one popular mechanics used obviously.. if I could type the 100+ pages in my paper work in less time than building this, I would post it here.. but that isn't going to happen too soon I'm sure .. I'm a two finger pecker on the key board.. and being on Dial up JPEG,ing it is out of the question.

we'll see what happens with my unit ,,,

You do realize that with jpegs you have the ability to seriously reduce file size right?

I will assume that you mean paper paperwork and as such would be scanning the documents (another opportunity to reduce file size)?

[modest]

Originally Posted by DFINITLYDISTRUBD

Sorry to be such a pain but...

On the contrary, my intention was not to be too didactic in post 35 - thinking that somebody would ask if they were at all interested and you did, so... it's all good

Originally Posted by DFINITLYDISTRUBD

in your math the output was for the H and O2 in the proportions yielded by the electrolysis (compensated)?

Yes. Every instance of “grams” or “liters” in my post (both when referring to water and gas) considers both hydrogen and oxygen together in a 2:1 mixture. There are some qualifications to this involving HHO which I'll explain...

Originally Posted by DFINITLYDISTRUBD

How did you sort out the volume of each gas?

I never did. The reason is simply that I had a source which allowed me to convert directly from grams to liters of "Brown's gas" (or HHO gas).

Thus the output of a Brown's Gas generator operating at 5 amps would be 1.678 grams of gas per hour. At 500 amps the output would be 167.8 grams per hour and so on.
A cell running at 500 amps would produce 364 liters of Brown's Gas at atmospheric pressure every hour.

-source

According to this source 167.8 grams of water will electrolyze into 364 liters of HHO. All I did was solve this:

so there was no need to think in terms of each component of the gas. But, there are some issues that come along with this quick (and to be honest - unverified by me) source.

“Brown's gas” (HHO) which the source uses above will not have the same volume as . In either case, a gram of water will make a gram of gas, so the calculation I did holds true either way. But, with HHO, one molecule of water will make 3 molecules of gas. With 2H2:O2, it takes two molecules of water to make three molecules of gas. I cannot stress enough that this is completely trivial for the calculation I did in post 35. The end result of that calculation never relied on liters of gas - only grams - so the volume was not instrumental. That's actually the reason I didn't verify the source. But, that said...

If I were to do this myself rather than using the source above I would completely ignore the claims of “brown’s gas” and just assume we’re making diatomic H2 and O2. I'll do that calculation now:

• We are electrolyzing 1.0068 grams of water per hour.
• Water is 18.015 g/mol. (a mol is a number of atoms or molecules, like a "dozen" only much larger)
• Dividing 1.0068 / 18.015 tells us that we are electrolyzing 0.0559 mols of water per hour.
• Two mols of water will make 3 mols of gas. This can be seen by looking at the equation: where two molecules on the left side make three total gas molecules on the right side.
• Multiplying (3/2)0.559 we get 0.08385 mols of gas per hour.
• The ideal gas law says that one mol of gas is 22.4 liters of gas (at standard atmospheric temperature and pressure)
• Multiplying 22.4 L X 0.084 mols gives us 1.88 liters per hour of total gas produced

This is smaller (as expected) than if we’re considering (or assuming we're making) HHO which the source above considers. But, either way we’re right around two liters per hour and in neither case does our calculation depend on volume. I just thought it was interesting that such little volume is being made.

To answer your question more succinctly - In both cases (HHO and 2H2:O2) the final gas mix will be 2 parts hydrogen to one part oxygen. So, three liters would be one liter of oxygen and two liters of hydrogen.

~modest

[modest]

Originally Posted by DanGray

the trouble with that is, did they test one that doesn't not work and could not work anyway ?

I honestly wonder how you thought I'd answer this, but, Yes... They did test one that "did not work" and "could not work anyway".

~modest

[DFINITLYDISTRUBD]

Thanks Modest I think what has been tripping me up understanding-wise is the difference in the density(?) of the two gases...I know hydrogen is lighter by volume (I was going to guess by about half... and I would have been wrong I see as hydrogen is 0.084G/L and oxygen 1.33G/L according to Kalzium.)