Anyone tried HHO on a diesel?

[oldbeaver]

Originally Posted by j.kansas

Looking to buy a diesel car and add some type of hho producer to increase milage. Talk to me.

Well, I will try to answer your question: I did installed a HHO generator in my diesel car. Read on before form an opinion...

It is a 6-in-line diesel, normally aspirated.

It is my third version. It produces so much HHO that I have to install a pump to blow it out of the producing "engine", otherwise, it is filled with the HHO gas and the producing surface diminish too much.

I estimate the production to be about 10 times as much as my first version, using only 70% of the total current.

The system has a lot of problems to be solved, though.

First of all, before continuing to develop it, I need a hard figures numbers test to see if it is worth the try.

Yes, I have read lots of literature and opinions about it. From people saying it is a failure to people saying they got 50% improvement. Passing for people saying it cannot work as it would violate the second law of Thermodynamics or such.

My approach is different: I will measure it with a reliable mean. What mean?
Something that can give me the car performance MPG in real time. Something with which I can turn HHO on and off and watch the difference.

"Fundamentalists" say I am wasting my time, because, as it is a closed system, then it will spend more energy than energy produced. It would be true, if it was a closed system. But it isn´t: water is coming in as a source of energy, as well as petrol or diesel is coming in.

So, in my opinion, there is only one way to demostrate, for good or for bad: testing it in a correct way.

Besides, you may get a system that is 20% efficient (it produces 20% of energy spent), and turn it into a 50% efficient system, or 80% efficient, etc. The idea of engineering is precisely that: to make a system more and more efficient.

I have some ideas to improve my system further. But even if I construct a 120% efficient system, the HHO generation has some extra problems to be solved, before become a trustable system.

For instance, make the electrolite solution last long and not become degraded. Avoid the production of foam (my pump allows lot of HHO to come out, but produces foam that interfere with the necessary contact between steel and the solution).

I do not trust on my prototype as it was the last time I tested it. So far.

Other inneficiences it had: it was always producing, even going down a slope. Going down the gas was wasted, but the battery continued to make electrolisis. I need to make production controled with throttle position, for instance. This will improve efficiency further.

All that potential improvements are waiting my MPGuino (or such) to work in my car. I am developing a version of it suitable for diesel cars.

That is it.

Hope I answered your question.

OldBeaver

[chuckm]

"Fundamentalists" say I am wasting my time, because, as it is a closed system, then it will spend more energy than energy produced. It would be true, if it was a closed system. But it isn´t: water is coming in as a source of energy, as well as petrol or diesel is coming in.

I'm going to be a thorough as possible here.
Water, from a thermochemical standpoint, is not a energy source. Hydrogen gas (in combination with oxygen, of course) is an energy source. The process of splitting the chemical bonds between the hydrogen and the oxygen is an energy intensive process, as evidenced by the electrical input requirements. The oxidation of hydrogen gas is an energy releasing process, as evidenced by the heat produced. From a physical chemistry standpoint, that means the potential thermochemical energy of water is lower than that of oxygen and hydrogen, at the same pressure and temperature. The energy balance is one to one; that is, the energy expended dissociating the hydrogen and oxygen would be exactly equal to the energy produced by hydrogen oxidation, but only in a 100% efficient system. Inefficiencies, however, are unavoidable. If any component of your HHO generator gets the slightest bit warm, then your system is consuming more energy than it is releasing in useful (ie motive) energy. Unless the energy to dissociate the water was brought in from outside your car (externally charged battery, solar cell, etc), it is absolutely impossible that there can be a net energy gain. Like I said earlier, TANSTAAFL. The laws of thermodynamics will not be mocked.

[chuckm]

But even if I construct a 120% efficient system...

Sigh. Nevermind the above post. Have fun. Let me know when you start working on anti-gravity, zero point energy generators, or transporter beams. May the Force be with you.

[oldbeaver]

Ok, ok.

Please, explain me clear: is water a source or energy or not?

You wrote "water is not a source of energy" ... etc.

But lower, you also wrote "the thermochemical energy of water .... "

We are talking of energy. It doesn´t mater how God put energy inside water, wether by chemistry, thermochemitry, physic-chemestry or whatever means, it is energy.

You also wrote that hydrogen combinated with oxygen IS an energy source.
And what is water? From the best of my knoledge, it is a combination of two atoms of hydrogen with one atom of oxygen... or not?

What we are trying to measure is how much energy-intensive is the process of breaking water into oxygen and nitrogen. If it overcomes, or not, the combustion of them.

You also wrote: "From a physical chemistry standpoint, that means the potential thermochemical energy of water is lower than that of oxygen and hydrogen, at the same pressure and temperature".

I don´t see how it is lower. When I use compression to burn diesel I don´t see why that compression energy will be higher than the energy produced by diesel combusion.

Figures are missing. And "movement is proved by moving", as the Greek said.

If you are right, figures will show it. Don't dispair.

Second, the major gain on using HHO together with diesel (or gas) is that it actuates as a combustion fastening and helper, of diesel itself. They say. I want to see myself.

So we need to measure the whole result of it. In practice, not in paper.

OldBeaver.

Originally Posted by chuckm

I'm going to be a thorough as possible here.
Water, from a thermochemical standpoint, is not a energy source. Hydrogen gas (in combination with oxygen, of course) is an energy source. The process of splitting the chemical bonds between the hydrogen and the oxygen is an energy intensive process, as evidenced by the electrical input requirements. The oxidation of hydrogen gas is an energy releasing process, as evidenced by the heat produced. From a physical chemistry standpoint, that means the potential thermochemical energy of water is lower than that of oxygen and hydrogen, at the same pressure and temperature. The energy balance is one to one; that is, the energy expended dissociating the hydrogen and oxygen would be exactly equal to the energy produced by hydrogen oxidation, but only in a 100% efficient system. Inefficiencies, however, are unavoidable. If any component of your HHO generator gets the slightest bit warm, then your system is consuming more energy than it is releasing in useful (ie motive) energy. Unless the energy to dissociate the water was brought in from outside your car (externally charged battery, solar cell, etc), it is absolutely impossible that there can be a net energy gain. Like I said earlier, TANSTAAFL. The laws of thermodynamics will not be mocked.

[AmpEater]

Water is not an energy source.

Despite that, HHO injection can improve fuel economy. Any combustion process is not 100% efficient, there is always unburned fuel that ultimately gets expelled with the exhaust gasses.

Don't look at HHO as a fuel, look at it as a catalyst.

I think the claims on petrol engines are hard to believe, but it is well known that the addition of propane to diesel combustion increases power and significantly reduces the amount of unburned fuel in the exhaust by promoting a more complete burn.

[oldbeaver]

AmpEater,

Your name sounds as you use HHO !

Have you ever measure the result of HHO as a catalyst?

Anyway, after I install my MPGuino kind of instrument in my car, I will be able to measure the use of Propane Gas together with diesel.

They sell several systems in Argentina, however, the liquid gas tank is the most expensive, I think. And heavier.

Have you ever test that system with a diesel?

I doubt of all that performance figures that use one fuel for going and another for coming back, refueling tank in each case. They are not very precise, as tank do not full equally and road is not the same conditions: slope, wind, traffic, temp, etc.

Do you know any such system being sold in the US?

Thank you for your comments.

OldBeaver

[oldbeaver]

For Chuckm and the Forum,

I gave the definition for that. Maybe you didn´t read it. I repeat it here:

20% efficient: it gives 20% of output energy compared to input electricity spend in electrolysis (taken from the engine via alternator).
50% efficient: it gives 50% output energy compared to electricity spent.
80% efficient: it gives 80% of energy taken and so on...

More precisely:

As an overall, the sum of burning HHO and its catalytic effect, generates more energy than the extra energy taken from the engine via alternator-battery ?

My third system produces a lot more HHO with 70% of energy spent, compared to the first one. Will that big amount of HHO obtained improve the overall engine performance? I don´t know. It has to be measured.

Anyway, under my definition, gas and diesel produce more energy than the energy used to burn them, so they are more than 100% efficient. A lot more. Otherwise, the cars won´t move. Again, under my definition.

Of course, in a thermodynamic point of view, under the thermodynamics definition, the Otto engine is maybe 20% efficient and the Diesel engine is only 27% efficient. But this is not the definition I used.

Under the thermodynamics point of view and definition, any system can be, at most, 100% efficient, of course, you are right! But, there are lot of loses, friction, etc.

Sorry if you didn´t read, or didn´t understand the definition I used. Or if I didn´t emphasize my definition enough. Maybe it is my fault.

My intention was just to answer with caution, in simple words (not thermodynamics), to the member of the Forum that asked if someone has tried HHO in a diesel engine.

I did, it works, but ... is it generating more performance? Is it producing more, equal or less energy than that taken from the engine through the alternator-battery? That is the question!

I can say for sure, unless I measure it. Have you measured it? Then give your results.

I hope this time I was clear enough.

May The Force be with all of us

OldBeaver

[chuckm]

We are talking of energy. It doesn´t mater how God put energy inside water, wether by chemistry, thermochemitry, physic-chemestry or whatever means, it is energy.

It most certainly does matter! Thermal energy, as approximated by temperature, is present in any mass with a temp above 0 Kelvin. Whether that energy can be transferred usefully is a different matter. When I say "thermochemical energy", I am referring to the potential energy stored in a material that can be liberated, via a chemical reaction, in the form of heat.

You also wrote that hydrogen combinated with oxygen IS an energy source.
And what is water? From the best of my knoledge, it is a combination of two atoms of hydrogen with one atom of oxygen... or not?

If I phrased that poorly, I apologize. I am simply referring to the fact that hydrogen burns. When hydrogen burns, the chemical reaction (2*H2 + O2 -> 2*H20 + heat energy) produces heat as a by product. If you want to reverse that reaction (dissociate the water into H2 and O2), then you must add energy (2H2O ->electrolysis-> 2*H2 +O2).

You also wrote: "From a physical chemistry standpoint, that means the potential thermochemical energy of water is lower than that of oxygen and hydrogen, at the same pressure and temperature".

I don´t see how it is lower. When I use compression to burn diesel I don´t see why that compression energy will be higher than the energy produced by diesel combusion.

Again, hydrogen, in the presence of oxygen and an ignition source, will produce fire. Fire = energy release. Think of the situation like gravity: I'm standing on a building holding a bowling ball. At the moment I release that bowling ball, it has zero velocity. The kinetic energy of the ball is likewise zero
KE =1/2*m*v^2
However, it has a potential energy equal to the mass times the height times the acceleration due to gravity
PE = m*g*h.
At the moment the ball strikes the ground, its potential energy is zero (because the height equals zero). At the same time, the kinetic energy is at its highest because the velocity is at its highest.
Here's the key to the whole thing: the initial potential energy at the moment I dropped the ball is equal to the kinetic energy of the ball as it hits the pavement.
Now, I want to take the ball back up to the roof. I have to move the ball up there, meaning that I'm putting energy in (Work = Force * distance, where Force = mass * acceleration due to gravity). But due to inefficiency, the work (or change in energy) I expend in bringing the ball back up there is greater than the potential energy the ball acheives at the top of the building (think friction of the elevator, electrical losses in the motor, the energy I expend in walking both my mass and the bowling ball, etc).

NOW, getting back to the HHO system. Hydrogen is the bowling ball at the top of the building. Now exchange the bowling ball's initial potential for the energy stored in the chemical bonds (or if you prefer, the rest energy state of the electrons in a hydrogen molecule). Exchange the bowling ball's final kinetic energy for heat liberated during the chemical reaction. But producing the HHO is more like moving the bowling ball back on top of the building. It takes energy and it is not terribly efficient. Most HHO system run pretty freakin' hot (wasted energy). If you are drawing electrical current from the alternator, then you must also account for the alternator, the belt, and engine inefficiencies. Moving that bowling ball (dissociating water into hydrogen and oxygen) will take more energy than will be released when that same hydrogen is burned in the engine.

Figures are missing. And "movement is proved by moving", as the Greek said.

If you are right, figures will show it. Don't dispair.

If you want figures, I can provide them. If you take 4 grams of hydrogen and combine them with 32 grams of oxygen, forming 36 grams of water, you will release 118 kilocalories of heat. If you could reverse this reaction with 100% efficiency, it would take 118 kcal of energy to do so. But you will not be anywhere near 100% efficient. This is a good reference, though it does have a bunch of biology tacked on the latter part of the article.

[chuckm]

Okay, let me ask this: why don't I hear about people buying a bottle of hydrogen gas and metering it into their intake, regulating the flow according to the amount of diesel (or gas) injected? If you did that, you could throw away the cumbersome HHO generator and the electrical load it puts on your system. Then you could analyze the effects independently of the additional loads on the system. Instead, all I've ever seen are these onboard generators.
I'll even help you design it. I've worked with mass flow controllers before.

[oldbeaver]

Chuckm,

Well, well, good answer. And thank you for your offer on making tests with the car. Maybe I will take your world in the future!

I still have some doubts, as follows. Don´t matter much, anyway.

On the bottle of ready made hydrogen:

Well, they do use it, indeed! See this:

http://www.youtube.com/watch?v=Ykl2PH2B-tM

The car runs on pure hydrogen. The problem is at time they produce hydrogen.

Second, that 128 kcal example. You wrote:

If you take 4 grams of hydrogen and combine them with 32 grams of oxygen, forming 36 grams of water, you will release 118 kilocalories of heat. If you could reverse this reaction with 100% efficiency, it would take 118 kcal of energy to do so.

How do you know that it "costs" the same 128 kcal to reverse the reaction?
It may cost more or less, with some heat coming or leaving, for example.

Think this: when water change its state from liquid to vapor, it takes heat from the environment. What happen in the opposite direction?

What I mean, is that the reaction may not be equal in both senses.

Anyway, my little knowledge on the matter comes to it´s limit about here

And yes, what we are talking about is usable energy.

Even if HHO has a catalist effect, I want to know if that can produce a net improvement or not. In practice.

OldBeaver