Water-Powered Car: Breaking the Molecular Bond & Burning Hydrogen

[buffordboy23]

buffordboy23, I'm really giving you the benefit of the doubt here. These explanations are getting increasingly implausible. I can understand how earnest people might believe in this. This is the "I want to believe" syndrome. Look at Fleischmann and Pons cold fusion. Do you know how much that website looks like a scam? Flashing gaudy colours, garish typeface, hysterical testimonials. It hurts my eyes. I looks a LOT like a scam. $97 USD for an ebook. Give me a break. Can we drop this?

It is fine to disagree but provide an explanation why. Your current argument is weak; it attacks the website--which I do agree with you about--and me personally, once again, but not the logic of my arguments. What points about my previous arguments are implausible? In hindsight, my initial recommendation of the website probably caused some individuals to doubt the validity of my arguments. The reason I posted the URL to the website in the first place was to provide individuals with a picture of the device, so that they could design one themselves. Notice from my previous posts, that I stated it is not essential to buy the ebook because the device is so simple to build.

 

[Dick]

Because to get the claimed 25-50% fuel saving requires a hugely inefficient electrical generation system. And while I am certainly no expert I would expect that any decent engineer would figure out a way eliminate much beyond essential frictional losses in the drive mechanism. It's not that hard. If nothing else you could power down the field coils.I simply don't find any of the arguments all that compelling. I don't have to disprove them one by one. You've already changed the basis of the argument at least once. And, yes, referring to a scam website for credibility or even a diagram doesn't help. Shall we try the 'injecting hydrogen increases gasoline burning efficiency' one? Sorry, I'm not buying this.

 

[buffordboy23]

Because to get the claimed 25-50% fuel saving requires a hugely inefficient electrical generation system. And while I am certainly no expert I would expect that any decent engineer would figure out a way eliminate much beyond essential frictional losses in the drive mechanism. It's not that hard. If nothing else you could power down the field coils.I simply don't find any of the arguments all that compelling. I don't have to disprove them one by one. You've already changed the basis of the argument at least once. And, yes, referring to a scam website for credibility or even a diagram doesn't help. Shall we try the 'injecting hydrogen increases gasoline burning efficiency' one? Sorry, I'm not buying this.

You make some good points.

The alternator does not have an entirely great efficiency--there are many sources that quote a 50-60% efficiency. Also, not only is hydrogen injected into the fuel system but oxygen as well, since it is one of the products of the electrolysis reactions.

You said that I changed the basis of my argument. I don't recall this but perhaps it is true. When I built and used the device myself, I was skeptical until I saw the results. I must admit that when I made my initial post that I did not entirely understand why such a device worked. It was only when my claim was refuted that I had to research the operation of the systems involved to prove my point. Therefore, with my learning of these subjects, I was probably prone to some minor misunderstandings along the way and conveyed a misunderstanding or two in my statements. Perhaps, I don't have all of the knowledge necessary at the moment to prove my claim. Therefore, the only way of convincing nonbelievers would be to have the support of converts who were also skeptical.

I challenge any users of the forum to build this device and report with their results. It is relatively easy to build and cheap. This is the basic design and the materials that I used, although I believe that there are better materials to use. I used a plastic mason jar as the container--deformation occurred due to heat. For my electrodes, I have used both stainless steel and copper wire that was bent back-and-forth in a shape as depicted in the attachment--the stainless steel has a significantly lower conductivity than copper but the copper electrodes had a build up on them in a few days due to the chemical reactions involved. For the electrolyte, I used baking soda. I ran an output tube from the device into the air filter--these tubes must be sealed well. I connected the device to the car battery via alligator clips. Once attached to the battery, the electrolysis reactions begin to occur--a switch could be used to control the circuit before and after driving. A bracket was used to hold the device during driving. We obtained close to 50 mpg with a 2004 Nissan Sentra driving on country roads at a near constant speed of 50-60 mph.

 

[DaveC426913]

Attaching this electrolysis device requires the regulator to pass more current or pass it more frequently to the field coils, which accesses more of the available mechanical energy from the crankshaft-serpentine belt system RPMs.

If this energy from the alternator/belt mechanism really is wasted as you propose (such that hooking up your device causes no increase in energy requirements) where do you propose the extra energy is going? Heat?

Consider the implication of this: the corollary of your cliam as that hooking up a device (any device, not just yours) that puts a load on the alternator will serve the secondary effect of cooling the engine (since you're now not expelling the extra waste energy-which-is-dumped-as-heat).

You've got to see that this is wrong.

 

[Dick]

I challenge any users of the forum to build this device and report with their results. It is relatively easy to build and cheap. This is the basic design and the materials that I used, although I believe that there are better materials to use. I used a plastic mason jar as the container--deformation occurred due to heat. For my electrodes, I have used both stainless steel and copper wire that was bent back-and-forth in a shape as depicted in the attachment--the stainless steel has a significantly lower conductivity than copper but the copper electrodes had a build up on them in a few days due to the chemical reactions involved. For the electrolyte, I used baking soda. I ran an output tube from the device into the air filter--these tubes must be sealed well. I connected the device to the car battery via alligator clips. Once attached to the battery, the electrolysis reactions begin to occur--a switch could be used to control the circuit before and after driving. A bracket was used to hold the device during driving. We obtained close to 50 mpg with a 2004 Nissan Sentra driving on country roads at a near constant speed of 50-60 mph.

Thanks for sharing the info. I believe that you believe that it works. But I don't think you know how it works. I've been looking for an objective test of such a device (something like Consumer Reports, not someone who is trying to sell me plans to build one), but putting a search term like 'water powered car' into google takes you straight to cootie land. That's not encouraging. I did see a few comments that Mythbusters may have tackled the subject. There must be one somewhere...

 

[buffordboy23]

I found that Mythbusters did do an episode on the topic and the idea failed. Here is a link for a video clip that shows why their test likely failed:



This video claims that they did not use an electrolyte. It also shows the clip of the Mythbuster's electrolysis device producing very little gas and another clip of a homemade device producing a lot of gas. I realize that this does not prove that the electrolysis device actually works but I thought people might find it interesting.

 

[buffordboy23]

Previously, on this forum I have stated that I used the electrolysis device as a gasoline combustion enhancement and reported success in increasing fuel economy. I explained how I thought the device worked, but my ideas were incorrect; particularly, the argument regarding the alternator. Therefore, the refuting arguments from other posters appeared to show that this device could not work and that my results must be due to something else other than the device.

Henceforth, with my basic and limited chemistry background I have been trying to analyze the energy balance among the inputs and outputs, since the totality of the systems and processes involved in such an analysis is more complicated than the arguments of antagonists suggest. Currently, my analysis shows that I am inputting more energy into the system than I am getting out. Although my analysis is deeper and supports the general argument against the viability of the device, I still feel that I am missing key parts in my analysis.

I was curious and decided to search the academic literature to see if their was any publications about this electrolysis device. Surprisingly, I did come across an article that describes the device that I am trying to analyze. Unfortunately, it is not very quantitative, yet my perspective on the issue has become more optimistic. Moreover, the authors cite three sources that support using hydrogen as a combustion enhancement.

For interested readers, the article's citation is listed below.

Z. Dulger, K.R. Ozcelik. Fuel economy improvement by on board electrolytic hydrogen production. International Journal of Hydrogen Energy. 25, (2000), 895-897

Does anyone know the reputation of these authors or how to discern it? Ethics has been an issue in controversial scientific issues.

 

[buffordboy23]

I found another interesting article that should help to clear up the debate over whether or not an on board electrolysis device improves 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.

The graph of their results is posted as an attachment. 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 two 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.


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.

 

[Borek]

I suppose if it works this way it is not because of the energy added in the form of hydrogen, rather the hydrogen presence modifies combustion parameters. This at least doesn't violate nature laws.

Still - car industry have spend about 100 years to optimize the combustion to get as much as possible from the fuel. They may know the trick and rejected it for some reasons.

Attachment pending approval, so can't comment further.

 

[buffordboy23]

I have one more important point about the graph that I forgot. The authors assumed that all of the oxygen and hydrogen produced from the electrolysis device would be present at the time of combustion. Actual systems could come very close to this ideal I believe since the oxygen and hydrogen are produced at different electrodes.

 

[atom888]

I totally agree with Borek about fuel modification. Let's not discuss "free energy" to create unfriendly atmosphere. However, we should not totally reject possibility. I've done some experiment with hydrogen before. It is a strong reaction, complete reaction with no/low heat to the surrounding. It could be that the engine needs oxygen and not hydrogen. Ambient have .. I think 21% oxygen? Electrolysis makes 33% oxygen per break down. If you intake 70%+ gas(from ambient) that not participate in combusion but obsorb heat, then that's a major loss in efficiency. This could be the reason.

 

[Borek]

This oxygen thing can be easily checked. Buffordboy, do you know how much water is electrolized per gallon of gas consumed?

 

[chemisttree]

I have two 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.

A few comments about the attached graph... Each of the dashed/dotted lines represent different equivalence ratios. The graph includes the energy requirement to produce the Brown's gas. Notably, the Prod Energy line is only shown for that portion of the graph that would account for a maximum usage of only about 0.35 mmol/cycle. More than that and the line 'rockets' off the page!

The equivalence ratio is defined as the ratio of the actual fuel/air ratio to the stoichiometric fuel/air ratio. We see in this graph that by varying that ratio between 0.55 (quite lean) and 0.75 (somewhat lean) and then including varying concentrations of Brown's gas, changes in the power output are realized. The magnitude of these changes becomes less and less as we approach a stoichiometric ratio... we only get 3/4ths of the way there in this graph. The 0.75 equivalence ratio line (the uppermost dashed line) shows a modest increase of from 2.5 kW to ~2.7 kW when only 0.2 mmol/cycle of the gas are used. That represents a net increase of only 0.2 kW and to achieve even that level, the electrolysis requires a whopping 1.7 kW of energy which leaves us with a power deficit of roughly 1.5 kW when the unit is operating in that region. I am analyzing this line only because most modern automobile engines operate at or near stoichiometry and this line is the closest to that reality.

If we assume that we have an engine operating at an eqivalence ratio of 0.6 the story is different. This is a quite lean condition but shows the maximum response to injecting Brown's gas in this graph. I estimate that the maximum positive effect of introducing Brown's gas occurs at an injection rate of approximately 0.1 mmol/cycle. This results in an increase of power of ~1.3 KW (1.4 kW at 0.1 concentration of BG - 0.1 kW at 0 concentration of BG). To produce that level of BG, the graph indicates that about 1 kW of energy is required yielding the reported 0.3 kW maximum benefit.

To calculate the Prod Energy line, the author's use an alternator efficiency of 70% (typical values are 55% however, Bosch introduced one that operates at 70%) and an electrolytic efficiency of 30% (small 1kW+ plants can operate at 55%) so the best case scenario could be somewhat more than a 0.3 kW net advantage. Of course, electrolysis plants of 1kW are expensive and anything you could probably come up with in your garage wouldn't be nearly 55% efficient at electrolysis.

Of course none of us drive a car that operates on methane at an equivalence ratio of 0.6! Our cars operate at equivalance ratios of >1 most of the time and we run on gasoline. I suspect that as the molecular weight of the fuel increases and when operated on the other side of stoichiometry that the rather large increases in power (30% in this paper) will vanish into the negative region and it will likely cost money to run the system.

 

[buffordboy23]

It could be that the engine needs oxygen and not hydrogen. Ambient have .. I think 21% oxygen? Electrolysis makes 33% oxygen per break down. If you intake 70%+ gas(from ambient) that not participate in combusion but obsorb heat, then that's a major loss in efficiency. This could be the reason.

Hydrogen is the key input, which is supported by research, and is the only reason that this idea could be plausible. The properties of hydrogen during combustion in the engine causes the flame velocity of the spark ignition to increase dramatically; hydrogen laminar flame velocity is 1.9 m/s, gasoline is 0.37-0.43 m/s according to one of my sources. The result is a more complete burn of the gasoline. Addition of hydrogen also permits the engine to run leaner (with more air), which also increases the thermal efficiency.

Could you please elaborate a little more on the rest of the highlighted portion of your post? I am having trouble following your explanation.

 

[buffordboy23]

This oxygen thing can be easily checked. Buffordboy, do you know how much water is electrolized per gallon of gas consumed?

I can't specify an exact number from the article; they never mentioned it. They did their research using a CFR engine. The kits or e-books sold on the internet supposedly produce 2 grams of hydrogen per hour--I heard this, so I don't know if it is true though.

If it were true, then every second we would be putting about 5 x 10^-4 grams of hydrogen into the engine and also an extra 4 x 10^-3 grams of oxygen; these are really small amounts. If we assume the car usually gets 30 mpg, then the car uses about 1.5 grams of gas per second. If we also assume that the car operates with the stoichiometric ratio (about 15:1), then about 22.5 grams of air is used every second.

It takes about 33 watt-hours to produce 1 gram of diatomic hydrogen, not including inefficiencies. Assuming the inefficiencies with device in the car as discussed above, then we need about 170 watt-hours per gram of hydrogen. This relationship tends to be linear. In truth, it deviates as the temperature of the water goes up, because the entropy increases, reducing the input power we need according to the Gibbs free energy equation.

Suppose we want 1% hydrogen relative to gasoline in terms of mass. Then we need .015 grams/second, or 54 grams per hour, which requires about 9 kilowatt-hours of power.

I believe that once a car is up and running at a steady speed, like 55 mph, we need 20 hp, or about 15 kW, to maintain that speed. I wonder how much extra horsepower could we draw while driving at this speed to produce hydrogen?

Also, is there a limit to the amount of current that a car battery could put out? I would think so, but don't know what it is. This would limit our hydrogen production.

I thought of something else that I haven't been able to answer yet. When we split water, it is approximately at the ambient temperature, but when the products reform back into water during combustion the temperature is very much higher. Does this significantly affect the enthalpy and entropy values of the reaction?

 

[buffordboy23]

To calculate the Prod Energy line, the author's use an alternator efficiency of 70% (typical values are 55% however, Bosch introduced one that operates at 70%) and an electrolytic efficiency of 30% (small 1kW+ plants can operate at 55%) so the best case scenario could be somewhat more than a 0.3 kW net advantage.

Actually, it's the other way around and includes the mechanical efficiency as well. Taken verbatim from page 582, "This production energy was based on a 70% effectiveness of the electrolysis and an overall electric generation and mechanical efficiency of 30%."

 

[buffordboy23]

Of course none of us drive a car that operates on methane at an equivalence ratio of 0.6! Our cars operate at equivalance ratios of >1 most of the time and we run on gasoline. I suspect that as the molecular weight of the fuel increases and when operated on the other side of stoichiometry that the rather large increases in power (30% in this paper) will vanish into the negative region and it will likely cost money to run the system.

This makes sense. In my exploration of this subject, I am learning quite a bit of interesting things about cars.

For example, some car makes and models use a pulsed signal rather than a magnitude-varying current return to the field coil. I think that the frequency of the pulsed signal would be so high that essentially the engine loses power constantly due to electromagnetic drag within the alternator. If the frequency increases more, the drag is essentially the same, but output energy from alternator is increased. This is speculation; I have no mathematics to back it up at this point.

Another example, is that different cars probably operate at different equivalence ratios although the driving conditions are the exact same. More air usually means a better combustion efficiency. I would really like to get some engine data for different car makes for comparison, but haven't came across anything yet. There are a lot of delicate factors involved--perhaps, some vehicles it is possible, others it is not. The large energy needs of the device seem to drown out these small, but possibly important factors. I have no idea. =)

I think some people are getting positive results using the device only because they reprogrammed the ECU to make the car run leaner. With the device attached, they are getting better mileage than before, but they are likely to get better mileage just because of this reprogramming.

If we could electrolyze liquid ammonia, the idea is definitely possible. It takes only 1.55 watt-hours per gram of hydrogen, not including inefficiencies. Good luck with getting this idea to work anytime soon! By the way, ammonia is toxic in large amounts.

 

[buffordboy23]

I came across another interesting article. Here is the citation:

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.

I attached the graph as well.

Basically, a 2% mass fraction of hydrogen relative to gasoline results in a "specific fuel consumption" (sfc) value of about 0.5, where 1.0 represents the sfc at 0% mass fraction.

I posted three citations so far and wanted to comment personally about the authors. For the first citation (Z. Dulger, K.R. Ozcelik), I seriously question the authors' results--article was a technical communication, it lacked quantitative data, and stated that they used a novel bonding material to form the carbon electrode (this novel material was never mentioned). For the second citation (S.O Bade Shrestha, G.A. Karim), I feel that this is a quality article--the authors have many publications on the subject and even stated that the idea of on-board electrolysis does not appear economically advisable. For the third citation, I haven't found anything yet to make me question the authors' claims.

 

[buffordboy23]

I found that Mythbusters did do an episode on the topic and the idea failed.

I can't say the idea works until I have solid scientific evidence that it does. However, I do want to comment on the Mythbusters' episode (see link from original post) that explored the viability of this device, because many people refer to this episode as their source of evidence rather than thermodynamic laws.

With the device attached, the car didn't even start. This doesn't make sense at all unless they were trying to run the car only on the electrolysis products without gasoline. Maybe this was the point of their test but I couldn't tell from the brief video, but if it was, then their test is very different from the claims of this device. You can't run a car only on water.

 

[DaveC426913]

With the device attached, the car didn't even start. This doesn't make sense at all unless they were trying to run the car only on the electrolysis products without gasoline. Maybe this was the point of their test but I couldn't tell from the brief video, but if it was, then their test is very different from the claims of this device.

Did you watch the whole ep? I didn't, but I'll bet money that this was only the first test of many - this one would be the sort of "best case" or "most optimistic" baseline scenario: "what does it do with no gas", then they'll proceed systematically across the scale to "worst case".

 

[buffordboy23]

Did you watch the whole ep?

No, I didn't watch the whole episode as I already mentioned. I only watched the portions of the episode that was presented in this video and others. The main highlights that I saw include testing the device in an isolated context (their production of hydrogen/oxygen as seen from the bubbles is pathetic compared to other units that I have seen), trying to start the vehicle with the device (already discussed), and inputting pure hydrogen into the carburetor (car started, but backfire occurred). Apparently, from posts on other forums, they tried to run a diesel on vegetable oil as well.

Besides, if they did do the appropriate test, I wouldn't be fully convinced because of the dismal production of hydrogen and oxygen as seen in the isolated context and also when attached to the vehicle when trying to start it. The device is not all that it can be. I am not saying that it would work, if it was all that it could be, but that the conclusion would be more convincing.

Perhaps, someone can elaborate more on this episode if they watched it. Were there other tests besides the three that I mentioned?

 

[DaveC426913]

Besides, if they did do the appropriate test, I wouldn't be fully convinced because of the dismal production of hydrogen and oxygen as seen in the isolated context and also when attached to the vehicle when trying to start it.

How can you be sure it could do better? How much gas can be produced given the specs of the rig? Is it not possible that the tests done by the claimants are exaggerated?

(Did you know that on the shopping channel when they show those convcection cookers, they "overclock" them to 220volts so they'll cook faster?)

 

[buffordboy23]

How can you be sure it could do better? How much gas can be produced given the specs of the rig? Is it not possible that the tests done by the claimants are exaggerated?

(Did you know that on the shopping channel when they show those convcection cookers, they "overclock" them to 220volts so they'll cook faster?)

Sure, I agree that there are advertising schemes, but I know this from personal experience. I have built an electrolysis unit myself, hooked it to my car battery, and by comparison I make this judgment.

See my post (#45) for power needs, including the efficiencies, to produce 1 gram of hydrogen--for every gram hydrogen, there are 4 grams oxygen. What size container do you think we would need to store these gases at 25 C and 1 atm? The density of hydrogen is about 0.0899 g/L and oxygen is 1.429 g/L, which equates to about 15 liters.

I am just not seeing it in the video. Besides, I expect that the battery power could operate consistently at about 500 W, which means 3 grams of hydrogen and 12 grams of oxygen, or about 45 liters of combined gas in one hour (or 1 liter every 80 seconds).

 

[buffordboy23]

...to produce 1 gram of hydrogen--for every gram hydrogen, there are 4 grams oxygen.

Actually, it's 8 grams oxygen, so add 3 liter-multiples based on 1 gram of hydrogen to previous figures.

 

[buffordboy23]

Basically, a 2% mass fraction of hydrogen relative to gasoline results in a "specific fuel consumption" (sfc) value of about 0.5, where 1.0 represents the sfc at 0% mass fraction.

This was from post #48. I realized many readers might not understand what sfc is. The discussion about it on wikipedia seems pretty good:

http://en.wikipedia.org/wiki/Specific_fuel_consumption_(shaft_engine)

 

[atom888]

I have to give my curiosity to hydrogen. Hydrogen flame can burn tungsten, which is around 10,000 Kelvin. A normal torch just can't go any higher than some value. What saying is that giving a time interval T, the entropy generate by normal torch is much higher than hydrogen torch. We can surely translate this into efficency without any equation. Another one is that we can extract better work out of higher temperature different. Thought I can't exactly give application to prove this, but we can see it clearly in theory.

Bufferboy, what I was saying about efficiency is that instead of intake ambient air (which contain a lot of nitrogen (78%)) we provide the engine with gases that actually useful (from electrolysis). The exhaust will contain all gasses at some temperature => nitrogen come in at ambient temperature, nitrogen come out with higher temperature taking some of the combustion juice with it.

 

[buffordboy23]

Bufferboy, what I was saying about efficiency is that instead of intake ambient air (which contain a lot of nitrogen (78%)) we provide the engine with gases that actually useful (from electrolysis). The exhaust will contain all gasses at some temperature => nitrogen come in at ambient temperature, nitrogen come out with higher temperature taking some of the combustion juice with it.

I see now.

Unfortunately, we have no choice and must include a large amount of nitrogen because of the amount of air needed to permit the combustion of gasoline. The stoichiometric ratio of air to gasoline is 14.7:1. So assuming a car uses about 1 gram of gasoline per second (which is a decent approximation), it also needs 14.7 grams of air, meaning oxygen comprises only about 3 grams. It is impossible to generate a lot of extra oxygen mass in such a short amount of time to offset the dominant nitrogen mass fraction, but a small yet possibly beneficial amount could be added. I explored this issue before. Based on some calculations and assuming that the car's original equivalence ratio is equal to 1, then the addition of oxygen from electrolysis under ideal conditions would be no better than 0.97 at the vehicle's optimum fuel mileage.

Hydrogen and oxygen will both be useful for the thermal efficiency of combustion. Hydrogen leads to more efficient combustion because of its many inherent characteristics and it also permits lower equivalence ratios (more air) to be used in a vehicle. The added oxygen will further enhance the combustion in conjunction with hydrogen.

Unfortunately, the most recent graph that I posted only looks at hydrogen addition. If it included oxygen, then in general we would see better specific fuel consumption values but I can't specify how much.

I recently did some calculations, which took the associated efficiencies into account, on the number of grams of gasoline needed to produce 1 gram of hydrogen. The result is 8-13 grams depending on different conditions. I realize that some readers may question this low value so I will post some data later.

 

[atom888]

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...

 

[buffordboy23]

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).

 

[atom888]

your calculation is detailed on electrolysis respect to gasoline. I just want to make a quick observation of energy wise in the process.

The final equation I have is an energy balance equation which include .

energy loss due to reduction in nitrogen + hydrogen reaction - enthalphy of water exhaust - energy for electrolysis.

If I assume the energy for electrolysis cancel out hydrogen reaction (100% efficiency), then it just down to energy save by reducing nitrogen and energy loss to increase temperature of water molecules. So far, I do not see a significant shift besides a zero in this equation. If this is the case, then the the process as you descibed, is hydrogen nature of combustion dependant.