Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: buffordboy23 on July 07, 2008, 09:58:05 AM
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You probably already heard about the many claims that you can supplement an automobile engine with a mixture of O2 and H2 to improve fuel economy. Basically, a water electrolysis device is attached to the battery and an output tube from the device permits the gases to flow through the air system into the engine cylinders.
As a result, I am trying to determine the likelihood of this claim quantitatively. The success or failure of this device is ultimately a battle against the First Law of Thermodynamics.
I researched the arguments of many antagonists against the device and found that they are over-simplified. Most arguments state that to split two moles of H20 into H2 and O2 requires an input of 483.6 kJ (the change in enthalpy for the reaction). Moreover, this value in actuality is higher due to the inefficiencies of the electrolysis device and the alternator that supplies the energy for the work. These antagonists say that the most energy we can get back from the combustion of the products is 483.6 kJ, so the device does not work and is a scam.
Here are my thoughts on points that the arguments neglect. First, the energy to split two mole of H20 is not actually 483.6 kJ. Rather it is a lower value since they are neglecting the entropy of the system. Therefore, the gibbs free energy equation, dG = dH - TdS should be the better answer. The value of dG tells us the minimum work that we need to split the same quantity of water. The absorption of heat is also a requirement, but I believe that since the electrolysis device is underneath the hood of the vehicle, it will absorb heat from the engine so that the battery does not have to supply the total heat energy value as well. However, the battery will supply heat to the system due to electromagnetic friction of the conductors so the net result is that the system's temperature will rise over longer time periods, which will tend to reduce the minimum work value necessary to split the water. I assumed that the water in the electrolysis device will be about 50 C, or 323 K after the automobile has been operating for a while. This leads to a value of dG to about 375 kJ/mol. This value in actuality is higher due to the quoted efficiencies of electrolysis (60-80%) and the alternator (50-90%), giving a total efficiency range from 30% to 72%.
Second, it likely that some amount of the products of electrolysis O2 and H2, will reform back into water during their journey to the engine. This value can be obtained from using equations associated with a "plug-flow reactor", which varies depending on the overall volume of the tube from the electrolysis device to the engine cylinders.
Link removed by user. See reply #22 for display of graph.
Third, now only a portion of our original products from the device are available for combustion. I believe that a portion of O2 will react with some of the gasoline molecules that would otherwise be unused during the power stroke combustion. I believe the enthalpy value for iso-octane C8H18 and oxygen is about -1300 kJ (anyone know the entropy value of iso-octane or a good place to find it?) Also, I believe that a substantial amount of the O2 and H2 remaining will react as well; assuming the average engine temperature of 75 C, or 348 K, and taking entropy into account, I find that dG is about -370 kJ/mol (but we don't have our original quantity of products). I have yet to take into consideration the dG for the extra combustion of gasoline, because I don't know how to predict how much of the O2 will react with it. I also wonder that since the reactants are highly compressed under pressure (standard compression ratio 10:1), if this will significantly affect my values as well, specifically the enthalpy of the reactions since this scenario is far from the standard state conditions.
So far, with my rough calculations, I am coming up short with the energy needs for this device to actually work.
My chemistry background is limited, but I would enjoy learning new and applicable concepts to "fully" analyze this problem. Any advice on key ideas that I am neglecting or potential flaws in my thinking or assumptions would be greatly appreciated.
Buffordboy23
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I was curious and decided to search the academic literature to see if their was any publications about this 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
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I have built a system and installed it in one of my vehicles, the increase in efficiency was about 33%(from a maximum of 15mpg to a maximum of 20mpg) i have found that my system can produce about 400ml of the hydrogen oxygen mixture in one minute. although this may be enough to increase the efficiency of the engine, i have an alternative idea of why it increases efficiency.
when the engine and HOD system are both operating correctly(i have had a few problems with the fuse blowing and wires melting, all of which are resolved now) the engine operates at a lower temperature, to my understanding of heat engines, the greater difference between the maximum temperature of the reaction and the lowest temperature of the reaction the more energy is being converted into work. this is why it's efficiency is higher. i believe that the water being produced in the reaction is absorbing more energy than the reaction to form it produces. the only reason i think this is because the engine runs about 10-15C cooler than normal.
not sure if that helps and i don't have the instruments to see if my assumptions are correct.
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to my understanding of heat engines, the greater difference between the maximum temperature of the reaction and the lowest temperature of the reaction the more energy is being converted into work
Can you elaborate? Heat engine efficiency is given by (Th - Tc)/Th - Th stands for temperature of the heat source, Tc for temperature of the cold sink. In general the higher the temperature, the higher the efficiency. What your propose sounds contradictory.
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I have built a system and installed it in one of my vehicles, the increase in efficiency was about 33%(from a maximum of 15mpg to a maximum of 20mpg) i have found that my system can produce about 400ml of the hydrogen oxygen mixture in one minute. although this may be enough to increase the efficiency of the engine, i have an alternative idea of why it increases efficiency.
when the engine and HOD system are both operating correctly(i have had a few problems with the fuse blowing and wires melting, all of which are resolved now) the engine operates at a lower temperature, to my understanding of heat engines, the greater difference between the maximum temperature of the reaction and the lowest temperature of the reaction the more energy is being converted into work. this is why it's efficiency is higher. i believe that the water being produced in the reaction is absorbing more energy than the reaction to form it produces. the only reason i think this is because the engine runs about 10-15C cooler than normal.
not sure if that helps and i don't have the instruments to see if my assumptions are correct.
aveon, where did you find your material/guide to build the device?
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I found my answers and need no further assistance in regards to my post. I researched the academic literature on the subject and found that small amounts of diatomic hydrogen added to gasoline enhances combustion. Essentially, the idea is that hydrogen enhances the flame velocity, which leads to a more complete burn during gasoline combustion. Numerous studies support this result so I don't think this is debatable.
In regards to the success of attaching an electrolytic device and improving fuel economy, the device can work but only within a limited window of engine operating conditions. Overall then, this technology would only benefit those whose engines operate consistently within this range. This statement is debatable and I base it off an article where the authors analyzed the possibility; ultimately, they stated it was not economically advisable.
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.
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I have one more thing to say about the authors' results that I failed to mention previously and is not represented in the graph. The authors assumed a 70% efficiency for the electrolysis device and a 30% efficiency for electric generation and mechanical operation.
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Let's do some sums:
Consider a Faraday Law calculation:
Suppose a car is traveling at 60 mph and using fuel at the rate of 30 mpg. If we assume that to have any influence at all on fuel consumption, it must be producing enough H2 to constitute 1% of the fuel that is being burned, the calculation becomes:
At 60mph, in 1 second it travels 60/3600 miles = 1.66 E-2 miles.
At 30mpg, this is equal to 1.66E-2/30 = 5.6E-4 gal
5.6E-4 gal = 2.1 mL gas
Assuming density of 0.8, 2.1 mL = 1.68 g gas
For 1% = .0168 g H2
Now, what current is required to produce .0168 g H2 in 1 second?
Equation is 2H+ + 2e- = H2
Now, .0168 g H2 = .0084 mol H2 = .0168 mol e-
Charge calc:
.0168 x 96,500 = 1621 coulombs
Now, 1621 C/s = 1621 amps (a ridiculous figure for any automotive system to generate).
If we work on a 1% molar ratio (rather than mass), the number is somewhat smaller (assuming octane as the fuel) at about 14 amps.
Now, in real terms, taking into account the sort of inefficiencies one encounters when trying to make gases electrolytically, I'd imagine that equates to at least 40-50 amps. At 12V, that requires about 600W power, which is a fair drain on an electrical system.
As well as that, before the H2 can increase your fuel efficiency, it must first make up the 600W that is required to make it in the first place. And all this from a mixture in which the H2 molecules number one in every 100.
In terms of the various positive testimonies regarding why this works, I'd say they fall into three categories:
1. Outright lies by people who want you to buy something they are selling.
2. The placebo effect. Those who have put time and money into installing a device like this WANT it to work. Thus, when doing their "test" they inadvertently drive a little lighter on the pedal with their device installed, or use noncomparitive data and so on.
3. People install a device onto an engine that is running too rich. In plumbing the device into their system, they don't seal the joints properly, and air enters the system, thus leaning the mixture and improving the economy.
And you know that something like this is happening when people make these claims for a diesel. At least for petrol, there is a theoretical case for improving efficiency with H2 if you can get enough of it in there. But for a diesel - the autoignition temp for H2 is 571 deg, way above any form of diesel fuel. It thus won't ignite by compression, and will only start to burn after the diesel has ignited - it will therefore actually retard the process.
But finally, there is one overwhelming reason why we know this cannot work - with car companies spending millions - possibly billions - trying to outdo each other for fuel efficiency, no one (that I've heard of) has either installed a device like this or even mentioned it.
It seems to me like the height of naivety to believe that you can just bolt a device onto an engine - any device - and instantly achieve the type of fuel savings that the combined efforts of the world's automotive brains cannot. Even the fabled Toyota Prius, which went to the lengths of using dynamos as brakes in attempt to harness leftover energy, was a failure, with a recent survey by the RACV finding that a small Hyundai Getz was more fuel efficient.
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very in depth and thought out...I think this thread actually sums up the other 3 threads that I had posted questions on.
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just one more thought - couldn't you theoretically attach auxiliary car batteries (store in the trunk perhaps) to obtain the necessary amps/watts to power the system? A bigger alternator -maybe....but possible even?
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just one more thought - couldn't you theoretically attach auxiliary car batteries (store in the trunk perhaps) to obtain the necessary amps/watts to power the system? A bigger alternator -maybe....but possible even?
How would you recharge the batteries? That's the role of the alternator, and ultimately the role of gasoline combustion. A bigger alternator wouldn't be the solution either; argument same as above.
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Now, 1621 C/s = 1621 amps (a ridiculous figure for any automotive system to generate).
If we work on a 1% molar ratio (rather than mass), the number is somewhat smaller (assuming octane as the fuel) at about 14 amps.
I appreciate your analysis. However, I have some concern over the large difference in values among the two methods and the methods assumed.
If you view the graph from the scholarly article that I posted, you will see that the x-axis is labeled "the concentration of (H2 + 1/2 O2) in fuel mixture, mmole/cycle." Therefore, I think one flaw in your analysis is it does not take the produced oxygen from the electrolysis device into consideration--if it did, both of your values would be smaller and the second value would lend greater support to the idea. The second possible flaw is that you are generalizing the results of this study (as I did in a previous post), which focuses on methane combustion, to octane--octane has about 8 times more mass. My knowledge in chemistry is limited but I know that the fuel species are chemically unique. Therefore, I wouldn't be surprised if the concentrations of hydrogen and oxygen relative to octane and the associated engine power output differ from this data. Please don't take offense. All I would like to see is a fair and scientifically-rich argument regarding the viability of this device.
I see two possible options to carry out the analysis. First, we can use your methods using methane gas in our analysis. An important question to consider:
What amount of methane would we need to use to produce the equivalent engine power output of octane?
Second, is it feasible and worthwhile to transform the data into a form based on octane.
At the weblink http://www.process-heating.com/CDA/Articles/Energy_Notes/abdb99d56e268010VgnVCM100000f932a8c0____
it says,
Pure methane has a stoichiometric air-gas ratio of 9.53 to 1 on a volume basis
Unfortunately, I don't have answers to any of these questions, so I'll need assistance to proceed further.
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But finally, there is one overwhelming reason why we know this cannot work - with car companies spending millions - possibly billions - trying to outdo each other for fuel efficiency, no one (that I've heard of) has either installed a device like this or even mentioned it.
This seems like the most reasonable response. There may be an answer, but if it was so simple to improve efficiency, why wouldn't car companies install them? (Please don't tell me something about how they're controlled by oil companies, and it's a conspiracy to make us use more gas).
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Mark: your electrolysis analysis is incomplete. Assuming you need about 2V for electrolysis, you may connect up to 6 cells, thus your current gets multipled by 6 - ie 12V/1A works as 6 times 2V/1A or 2V/6A. Could be it will in pratice not 6*2 but 4*3 or something like that, but if you want to be correct you need to take it into account.
Doesn't mean whole thing becomes feasible after that.
buffordboy23: amount of oxygen is rather neglectable when compared with amount of air entering the engine. You may try to esitimate the ratio.
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buffordboy23: amount of oxygen is rather neglectable when compared with amount of air entering the engine. You may try to esitimate the ratio.
Borek: I agree with your statement about the oxygen being a negligent amount, but this is not how the authors performed their analysis, if you are analyzing my critique of Mark's analysis.
The x-axis shows the concentration of electrolysis products relative to the main fuel methane. The graph says "fuel mixture" which could be ambiguous (does this mixture include air? or do they mean methane with electrolysis products); their reference to figure 10 on page 582 states, "In Fig. 10, the net energy enhancement due to the addition 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."
Using Mark's analysis for gasoline,
1% of 1.68 g gasoline (octane, not methane) = .0168 g total of H2 + 1/2 O2
Obviously, most of this mass total is oxygen, so Mark's mass value for hydrogen is off by about a factor of 10.
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just one more thought - couldn't you theoretically attach auxiliary car batteries (store in the trunk perhaps) to obtain the necessary amps/watts to power the system? A bigger alternator -maybe....but possible even?
How would you recharge the batteries? That's the role of the alternator, and ultimately the role of gasoline combustion. A bigger alternator wouldn't be the solution either; argument same as above.
I disagree. Sometimes you need to spend money to make money. Take a supercharger for example. It's driven directly off of the engine via a belt, yet when its all said and done...it produces MASSIVE amounts of energy for the engine, thus having a positive effect on energy output.
I think the same principle (in theory) could be applied here. the extra drag created by a bigger alternator would have to be compensated by the amount of energy released via the hydrogen combustion....but it's feasible.
I think the main thing is to find that perfect balance - you can get to 100 by many different products... 100x1 50x2 25x4 etc etc etc.....but which yields the greatest area? ...if you catch my drift.. ??? ??? ??? heh
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If we work on a 1% molar ratio (rather than mass), the number is somewhat smaller (assuming octane as the fuel) at about 14 amps.
can you please explain the differences in your figures, as you say molar ratio versus mass? If you calculated it either way, it should compute to the same figures...given that we're looking at 1% of the gasoline taken in
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b34st1y,
In your original post, you said that perhaps auxiliary car batteries or a more powerful alternator would be a solution. These are the ideas that I responded to, not the idea of adding a supercharger because it was never mentioned until now. I have researched the idea of reclaiming some "wasted" energy solely from the alternator, but this scenario is impossible; as engine output increases, the internal rotor of the alternator spins faster within the field coil, and as a result, it must overcome a greater value of electromagnetic "friction."
In regards to adding a turbocharger to the vehicle, I can't offer a decisive refuting argument because I don't know much about its overall efficiencies. I am sure that the fact that it can be powered mechanically makes it an attractive idea for some as a possible solution. However, I wonder if the larger mass of air coupled with the added products of electrolysis will cause the engine to be more susceptible to "knocking."
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Take a supercharger for example. It's driven directly off of the engine via a belt, yet when its all said and done...it produces MASSIVE amounts of energy for the engine, thus having a positive effect on energy output.
It doesn't produce a single joule of energy. It just allows you to prepare more fuel/air mix to be used per engine cycle, so that you can get more energy - but you get more energy using more fuel, not the same amount of fuel.
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its the principle of the matter....trading one form of energy for a more effective form of energy.
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No, it is the same energy, just being created faster.
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ok, so apply that to our situation: we need to create hydrogen (and oxygen) fast enough to compensate enough for at least 1% of the fuel.
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Here is the graph. I have removed the link from my previous post.
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I intend for this to be my last post.
From the graph presented, it is evident that the idea can work for methane but under limited conditions, and ultimately, the results are not desirable overall.
It's difficult to transform this graph into one that focuses on iso-octane, or standard grade gasoline. We would have to assume that since the attached electrolysis device requires so much energy, that this new graph would have to show much larger jumps than shown here for methane with a little concentration of hydrogen and oxygen. I can't speculate if this outcome would be true.
If it were true, I would expect that the vehicle must be run very lean (more air relative to stoichiometric ratio). Depending on the leanness of the fuel/air mixture, power loss may occur. So if this device did work, it could only work for such engine conditions, and benefits would only be seen if the driver consistently drove within such engine conditions.
An alternative way to look at the problem is to question if we can use a substance different than water in the electrolysis device to cut back the energy demand to produce hydrogen on board the vehicle. It appears that we can, but the idea is potentially dangerous and also the design of the electrolysis device has durability issues and decomposition issues with the electrocatalysts. Liquid ammonia consumes 95% less energy than water during electrolysis; ammonia requires 1.55 W-h/g relative to water 33 W-h/g to produce 1 gram of H2. If the hurdles associated with the quick electrocatalyst decomposition can be overcome, it makes good sense that we may see this electrolysis technology in the future.
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has anyone considered the resonance approach to this matter? - Using a defined frequency of current to create a resonance within the water molecule, it's a proven method that uses substantially less energy
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has anyone considered the resonance approach to this matter? - Using a defined frequency of current to create a resonance within the water molecule, it's a proven method that uses substantially less energy
Reference to peer revieved journal please. It'll be against laws of thermodynamics. And these laws are not like speed limit that you either obey or not, they are like a concrete wall - you either stop before hitting, or you break your neck, but you want pass through.
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yes, creating energy would be impossible....but rather try to find a way to transfer energy from a nearby source. Consider the atomic bomb...and the mechanics of the chain reaction.