water hydrogen generation fact and fiction

The standard reaction for the electrolysis of water is:
2H2O → 2H2(g) + O2(g)
For this process, 4 moles of electrons take place, therefore the standard free energy change is:
ΔG = -nFε
ΔG = -4*96487*1.229
ΔG = -474330 J

We can expect to extract -474 kJ from the products of the electrolysis and use it to do work.

This is also borne out by looking at the approximate bond energies. 4 H-O bonds must be broken, at ≈460 kJ each, but energy is released when the new bonds form. The O=O bond yields ≈498 kJ and the H-H bonds yield ≈433 kJ each. Our net free energy is:
-4(460000) + 2(433000) + 498000 = -476000 J

So we have to put ≈ 475 kJ into the system to separate the water, and we can get ≈ 475kJ back out in utilizing the free energy.

According to the "Oxyhydrogen" people (more on this later), burning the resulting gases gives off 576 kJ. If the TOTAL energy that can possibly be released to do work is 475 kJ, how are they magically getting 576 kJ? They're generally not.

Now, on to the electrolysis itself. The efficiencies for electrolysis of water are reported as anywhere from 50% to 94%. So, in the best case, we have to put 505 kJ in to be able to get 475 kJ out, and in the worst case we need an input of 950 kJ to be able to get 475 kJ out. Even before factoring in the other losses, we can see from the 50-94% efficiencies, that we will have a net loss of 6-50% of the energy used for electrolysis.

There are thermal losses from heat engines. From Carnot, the MAXIMUM possible efficiency from our engine is 70-75%. The real value is about 25% due to friction and the fact that the combustion is not spontaneously reversible.

So, with the maximum possible theoretical efficiency (which can NOT be obtained in practice) we are now needing anywhere from 675 kJ to 1357 kJ. Even if we allow for the inflated output numbers, and ignore the actual losses we'd encounter, you can NOT produce 576 kJ and use it to keep a process needing 675 kJ to 1357 kJ on-going. And this is before we try to extract any OTHER work from our engine -- we are solely using the output power to try and drive the input power.

In reality, what you end up with is needing to output around 3150 kJ to keep the cycle going, but you are only producing 475 kJ. This is why water-4-gas did not work when Adam and Jamie tried it. The losses completely overwhelm any energies "created". Water is not a fuel -- it is the ashes of hydrogen that has been burned once before.


Now, I know the "newest" schemes do not try to run the car solely off the hydrogen, but call for using it in a "hydro-assist" manner. The claim is that the hydrogen makes the gasoline burn better.

The first thing to note is the losses from above. If we need 3150 kJ to produce two moles of hydrogen gas, and can only get 475 kJ back from the system, then we need to burn an ADDITIONAL 2675 kJ worth of gasoline just to break even!! This isn't looking promising.

But, but, but... the claims are that the hydrogen is a "catalyst" and makes the gasoline burn more efficiently.

So? That claim is just wrong. We know we can't affect the thermal efficiency of the Carnot cycle by very much, so 75% is still going to be "wasted". Even if the hydrogen did act as a "catalyst", there is no more energy to be released -- 99% of the gasoline does undergo combustion. A 300% gain in efficiency would imply that we are now getting 399% of the theoretically-retrievable energy that the gasoline contains (well, actually more than 399% because we also need to cover the losses from the electrolysis). This is just asinine and Carnot says otherwise. Energy can not be created from nothing -- the gasoline can NOT give more than 100% of what it has.

But, but, but... "it's not H2 gas that is produced, it's monatomic hydrogen!"

Ok, let's look at that. The ½H2(g) → H(g) reaction is NOT exothermic until you reach a temperature of about 4000K. At our temperatures you need to INPUT another 800 kJ (4*200 kJ) to get the 2H2(g) into 4H(g). Without even considering the thermal losses from the combustion cycle, we now need an input of about 1600 kJ to our electrolysis system in order to get the claimed 576 kJ of "oxyhydrogen" back out (and that calculation was made in the 19th century and is taken from a 1911 encyclopaedic entry -- wonder why they don't use more modern sources and numbers?). I'm still seeing a major net loss, aren't you?

But, but, but... "the hydrogen DOES improve the efficiency!"

Yes, it is reported that hydrogen can increase the lean limit from 1.7 to 1.85, and it is also reported that hydrogen can reduce the no-load idle consumption of gasoline by up to 50% in small engines.

Let's look at these. First off, these are for volumes of hydrogen that can't possibly be produced by these simple cells. And, the reported lamdas of 1.7 and 1.85 are for natural gas combustion, not gasoline. The lean limit for gasoline is a lamda of about 1.2. Until you get to the lean misfire conditions, leaning your gasoline down WILL reduce consumption. But you also get rough idle and loss of power. Except when idling, you car isn't fully leaned (and even then it's not maxed out) -- in fact, the more demand you're putting on the engine, the richer the computer (or even the carb in non-EFI engines) makes the mixture. By overriding the computer and approaching the lean limit, you can reduce consumption at idle even without hydrogen. But as you place demands on the engine, it can NOT be run this lean. The faster you go, the more engine power must be used to overcome drag, so at highway speeds, even if you are not accelerating, you are not running a lean mixture. The manufacturers do NOT lean the idle down as far as possible, because excessive leaning can burn the spark plugs and pistons and lead to detonation and preignition. Even if it was possible to reduce idle usage of gasoline by up to 50%, the long-term engine damage and the percentage of the time the engine is NOT idling dramatically reduce this savings. Plus, you STILL need extra gasoline to produce the hydrogen in the first place from all the above-mentioned losses.


"Ok, ok... so I can't get such phenomenal gains in mileage from 'hydro-assist.' But what about this PICC stuff?"

Ah, yes... The "pre-ignition catalytic converter". Let's take a look.

The claim is that the PICC will break the isooctane molecule of gasoline down into smaller molecules that will "burn better." Oh, really? When we looked at the energies involved in electrolysis, we noticed that breaking bonds takes energy and forming bonds releases energy. It doesn't matter what happens in-between. If our net result is 2C8H18 + 25O2 → 16CO2 + 18H2O, then we are breaking 37 C-H bonds, 14 C-C bonds and 25 O=O bonds, but re-forming 32 C-O bonds and 36 H-O bonds. Any bonds that form and break in-between are immaterial; we are left with a net change in bonds of about 37800 kJ. But, and this is important, it TAKES energy to make any smaller molecules. We have to put energy in, in order to have the higher bond energies of any smaller molecules. The net change from beginning to end, remains the same: whether you extract 37.8 MJ from the gasoline, or extract 57.8 MJ after putting 20 MJ into the bonds of smaller molecules, is immaterial -- the net change is still 37.8 MJ.

And besides, the PICC sites claim they're making a "plasma". Plasmas do not undergo normal chemical reactions. You won't get combustion in a plasma, nevermind factoring in the large energies necessary to create plasmas in the first place.



Every step of all these schemes consumes far more energy than it could possibly release. There is no magic way to get such phenomenal gains in fuel efficiency out of your existing car.
roofingguy
Senior Member

Registered: 10-28-07
Posts: 4638 Posted 06-25-08 01:58 PM A few points of clarification, since some posters seem to feel the original explanations were intended to be over their heads...

First off, let's take a look at Hess's Law:

"The heat change (ΔH) that accompanies a given chemical reaction is the same whether the reaction occurs in one step or in several steps."

(http://www.science.uwaterloo.ca/~cchieh/cact/c120/hess.html)

What does this mean for w-4-g, oxyhdrogen and PICC? Just what it says. If you start with 2 C8H18 + 25 O2 and end up with 16 CO2 and 18 H2O, there will be the same total change in energy, no matter what path you take to get there. It's like the change in temperatures (or changes in displacements)... if it's 22°C today and was 24°C yesterday, then that is a change of 2 Celsius degrees -- it doesn't matter that it dropped to 16°C overnight... it could have dropped to -40°C overnight and it would *still* be a net change of 2 Celsius degrees from day-to-day. Back to combustion, it doesn't matter what reactions are undergone to get from the isooctane and oxygen to the carbon dioxide and water. You can take many paths to get there (most of which will require the *input* of extra energy, which is subsequently re-released), but the net energy released from that net reaction will always be the same. There is no magic additive to get the gasoline to release more energy than it already does. The standard combustion reaction always releases the same amount of energy. Always.


Yes. That reaction doesn't always take place -- it is the ideal.


But let's look at possible gains using published EPA numbers.

Not all the gasoline is combusted, that's true. But about 99% is, according to EPA estimates... the best gain possible in that regard is about 1%. You can't burn more than the 100% of what's going into the cylinder.

Now, one of the products that the catalytic converter deals with is HxCx unburned hydrocarbons (HC). Ignoring CARB limits and more modern standards, even the 1994 limits on HC emissions were 0.25 g/mile. A quick bit of simple math will tell us that a 25mpg car (2650 grams/gallon for gasoline * 1/25 gallons/mile = 106 grams of gasoline per mile) must only be emitting 0.25g/106g = 0.0024 = 0.24% HC at the maximum. Now, granted most cars will be well under this at the tailpipe, as it is only the upper limit set by Congress, but the catalytic converter is also helping to oxidize the initially higher levels of HC. I would think people would be hard pressed to prove that any modern engine is producing much more than 4% HC, seeing as how engines in the 50's were only producing on the order of 7% HC (2650 g/gallon * 1/15 gallons/mile = 177 g/mile... 13g/mile HC produced / 177 g/mile gasoline used = 7.3%), long before any reductions started being built in to the engine operations and before any exhaust treatments started being employed. So, we're still not finding room for such a major breakthrough in "efficiency". And even then, forming the smaller HxCx hydrocarbons is releasing *some* energy -- just not as much as if complete combustion had taken place for those molecules.

So, the only other place for "better combustion" is for more CO2 to be formed, and less CO. Let's look at that. Again the 1994 standards are 3.4 grams per mile CO, down from 87 grams per mile in the 50's. If the combustion was complete, one gallon of gasoline would produce 8875 grams of CO2. So 8875g CO2/gallon * 1/25 gallon/mile = 355g/mile CO2 produced by the modern 25mpg engine. 8875g/gal * 1/15 gal/mile = 592g/mile CO2 produced by the 1950's engine. The current emission of CO over CO2 is around 1% and the historical value produced was around 20%. Even if the engine still is producing 20% CO and 80% CO2 today (which it isn't), what would be the difference in the energy released? Forming CO2 releases 3.5 times the energy that forming CO does. So, forming 100% CO2 would be releasing 17% more energy (0.2 + 0.8*3.5 = 3, 3(CO/CO2 mix)/3.5(pure CO2) = 1.17).

So the absolute most you could hope for is an overall 20% increase in the combustion efficiency. The actual value you can manage is considerably lower than that, but even assuming the worst of your engine to start with, there just isn't room for more than a 20% gain in the combustion efficiency of gasoline. And oddly enough, the auto makers have tried to release every drop of energy that they can out of the gasoline -- they're not going to leave a 20% gain just sitting there unattended.

Without totally redesigning the car and lowering both the weight and the drag, there is no possible way to do *anything* to the gasoline to double the mileage it produces with an existing car. There is no possible way to even get a 50% gain. 5%, possibly.

The major inefficiency of the internal combustion engine is limited by the theoretical Carnot efficiency. There is very little that can be done to raise the engine's efficiency from 25-30%. Again, this is thermodynamics. So, you can't get the gain by making the gasoline "burn better", and you can't improve on the losses from the engine just by modifying the fuel a little bit. Where do people think these magic gains are coming from?

The only real way you can improve the mileage of an existing vehicle is through more efficient driving.

And again, it takes 105 watts input to the alternator shaft for every 100 watts of electricity it outputs. When you turn on the rear defroster wires or operate power windows, it *does* put more load on the engine. The alternator is never producing surplus electricity. So, your "little bit" of electricity is putting a larger load on the engine. It is. You try to draw more electricity off of it, and it gets harder to turn. This is a regular demonstration at science centers.

So, with the efficiency of the engine and the electrolytic cell, you need the engine to consume about 560 watts (from either gasoline or a gasoline/hydrogen mixture) to produce 100 watts "worth" of hydrogen. See? It takes more energy to make the gas, than what you get back burning it. Always.

So, producing the gas is a net loss of energy, and can't possibly have the effect on combustion that is claimed for it. It doesn't need to be tested. I also don't need to test that I can't fit one gallon of gas into a 1 cup measuring cup. The math says it is not possible. The math says there is no room for it.