[PSIG]

You misunderstand me, and the math. "Calculated" is simple use of energy. You can't get something for nothing, and fuels have specific energy. With the standard 114,000 BTUs in one gallon of unleaded gasoline (petrol), you can calculate how far that will push your vehicle. The BTUs are the total energy content. It has nothing to do with droplets. Pure energy conversion in a perfect environment, that's what you get.

So, that assumes perfect combustion from perfect vapor, etc, in order to extract all of the 114,000 BTUs from that gallon of fuel. If you can do that, then you waste 70 to 80% of that energy in lost heat, due to the inefficiencies of your engine, transmission, gearing, tire drag, aerodynamic drag, and so on. Using optimized values, a gallon of unleaded will produce 44.8 HP per gallon per hour. How fast 44.8 HP will make your car go in one hour will be your miles per gallon.

That's the kind of calculations I'm referring-to. No black magic, no wizardry, no droplets - basic facts using basic energy available, extracted by perfect combustion, calculated against the work that has to be done. What is the absolute maximum you could get if everything was perfect? There is your limit, that you cannot exceed, based on science.

Along the way you have other hurdles, such as combustion limits for lean and rich. This hurdle is a big help in some ways, such as allowing us to have a sparking electric fuel pump in the empty air:vapor space of a fuel tank, and it doesn't blow-up. Facts. If you want an example of how Honda burned mixtures leaner than otherwise burnable, research the CVCC system. BTW - it had tiny little primary head ports, the size of a pencil, fed by a tiny little primary carburetor bore. Sound familiar?

So, in order to avoid contention, arguments, or undesired laughter; I would suggest you move ahead with your project, skip the theory for now, and see how it really works for you. Report the comparative results as you go, relating the theory that applies. Do your thing!

[Casey C]

Thanks, Yeah, people will find out that it's a pretty random number, based on "what generally works", but when you use dry vapor, and no droplets, it loses relevance....it loses a LOT! of relevance.

The more you analyze droplets vs. Vapor, you realize that vapor has never been properly analyzed.
Like spoiled brats we took our engines and liquid fuel, said goodbye to horses and mules, and had no inkling to change it. Went to fancier droplet squirters,
but never investigated the mysterious form of gasoline; called Vapor.
A few stumbled upon it, with outrageous results, yet could never package it for the everyday consumer.

[Casey C]

This is my latest lineup,. Needle will be run off 3 to 1 belt pulleys.

[Casey C]

So, that assumes perfect combustion from perfect vapor, etc, in order to extract all of the 114,000 BTUs from that gallon of fuel.

The question is, what did they burn to get that figure.
I'm sure they did a physical test, to make sure that is a reliable number.
Tell me what you think the chances are that they used pure vapor.
See my point?
Vapor is a whole other animal, that's why I said earlier that reliable data cannot be found in one spot.
Apparently Tom Ogle extracted 5 times the average BTU yield of normal burning mixture.
This shouldn't be too shocking.
No one uses vapor, much less runs tests on its BTU yield.
Like I said from the very beginning, it's a different animal and uncharted, but we do have examples of it's potential, which if one wanted to, could get some revised BTU ratings for gasoline WHEN USED IN ITS VAPOR FORM. (Probably around 600,000 BTU's)
I just seen what was did, and then like you said, went on ahead with effort based on those results.
Thank you.
In regards to laughter, he thought it was funny how I threw AF ratio under the bus. It is kind of funny because it makes no chemical sense when no droplets are present.
I don't need no prodding to go on, thank you; that's what I do.
But I like to have interesting discussions too, in-between, if they are available, so thanks for that too.

[jonbill]

Does petrol vapour have the same chemical reaction with atmospheric air as "droplet" petrol? I think it does.

HC emissions for a healthy car pre catalytic converters were a few hundred ppm. So I think we can agree that effectively *all* the molecules of petrol are burned in the engine with a 'droplet' system. And let's agree that a 'vapour' engine achieves the same.
so both fuel supply types result in all the petrol that goes into the cylinder being burned, right?

If we approximate petrol to be octane, c8h18, then the reaction is:

2 * c8h18 +25 * o2 + 100 * n2 = 16 * co2 + 18 * h2o + 100 n2 + energy

The mass of 2 molecule of octane is (8 * 12) + (18 *1) *2 = 228
The mass of 25 molecules of oxygen is (2 * 16) * 25 = 800
The mass of 100 molecules of nitrogen is (2 * 14) * 100 = 2800

therefore, the complete oxidation of 228 mass units of octane requires 3600 mass units of air.
therefore the air:fuel mass ratio required for complete combustion is 3600/228 = 15.7

As everyone knows, petrol is a mix of different hydrocarbon molecules. Octane (with 8 carbon atoms) is just an approximation. I believe the average number of carbon atoms per molecule in petrol is actually a bit more. If we run the above calculation with an alkane chain with 8.6 carbons, then we get an air:fuel ratio of 14.7 to 1.

So that's where the AFR of 14.7 to 1 comes from. On paper. and I'm not even a chemist.
This applies to vapour and liquid petrol.

Thats the stoichiometric ratio. Of course petrol will happily burn in rather more or rather less air. But varying the air:fuel ratio doesn't vary the amount of energy that is released in the combustion/oxidation of the individual petrol molecules - it just varies the amount of energy released in a cylinder fill (the volume of air being effectively constant and so the amount of fuel must vary).

Oxidation (burning) of one mole (114g) of octane releases 5470 kJ of energy. that's 2hp for 1 hour.

lets say it takes something like 20hp to push my car at a steady 70 mph (I believe it's something like that).
To drive one hour at 70mph then, my car must burn 10 moles of fuel (1.140kg). Irrespective of whether it's a vapour or droplet system. And in that hour I will have driven 70 miles. My car achieves that with 3% TPS, 3500 rpm and an AFR of 14.7:1.

I think you suggested an AFR of 73:1 is doable with a vapour system. I'm pretty sure my engine won't run with that mix. But, let's say I swapped to a vapour engine and run it at 73:1.
At 73:1, that's 1/5th the fuel going in to my engine each cycle(14.7/73 = 0.2), and so only 1/5th the energy being released, and so my engine only produces 4hp for that hour. At the end of my 1 hr journey, I will have used 1/5th the fuel, but I won't have gone 70 miles. Perhaps I will have travelled 20 or 30 miles. so I'll to travel for another several hours with my 4hp to do that 70 mile journey.

Of course, I will have used less fuel to go 70 miles in the vapour engine than I used in my 'droplet' engine to go 70 miles. But this just proves you have to spend more power overcoming wind resistance the faster you go.

[Casey C]

The lean mixture explodes with the same power as the rich mixture. That's what I ve always thought. Perhaps not exactly, but within reason.
The power is determined by how much of the total mixture volume is in the cylinder.
Not by how much fuel per volume of air.
If there is enough fuel in the air to cause an explosion, that's all you need.
If there are no droplets, spontaneous combustion disappears, along with over hot combustion.
That is what is happening in vapor driven cars.
There is virtually zero NOx and 02 as well.
That's just the way it is in actual use.
I'm not a chemist either, but just observe results when it has actually been done.
I already drove my car at 53mph with a single vaporizor channel.
I felt I eventually would need 4 channels for full power.
(Which reminds me, I should probably use smaller channels, unless this single channel is vaporizing the fuel adequately. I need to run a mileage test to find out.)
The carb venturi is about the size of a penny.
I was not happy with the fuel delivery, so I'm setting up a constant variable flow needle valve regulated injection. Thus my use of the Speediuno.
Also advanci.g the timing will be a big help and advantage, as vapor has a slower burn than droplets.
Here's the Venturi I was getting 53 miles per hour.

[jonbill]

No, not at all. The lean mixture burns releasing at most the power available in the fuel. Just like the rich mixture does. If you have 1/5 the fuel, you have 1/5 the power, at most.
But as psig says, do crack on with the practical aspects of making your engine run, then we can come back to why your car won't accelerate.

[Casey C]

Wow, and I thought the virgin birth was mysterious and hard to explain, lol.
BTW the speedy can richen the vapor if needed, but it may not be needed in acceleration because the fuel is so homogenously mixed.
It becomes totally a function of butterfly position.
In a multi port injection car, the ejector fuels only varying segments of air depending on rpm, so they need to squirt more during RPM transition.

[Casey C]

OK let's try this:
Say you're sitting in a room with a lighter.
There's an open 5 gal bucket of gasoline in the corner.
If you flick that lighter, are you more worried about the liquid in that bucket, or the vapor that has filled the room?
Vapor is explosive from 1.4% concentration to 7.2% concentration.
Pieces of the room will fly just as far with 1.4 concentration as they will with a 7.2 concentration.
You could say the same thing with a gasoline mister in the corner of the room.
The vapor in the room is the most powerful explosion because each gasoline molecule is sitting next to an oxygen molecule.
NOT next to 23,000,000,000,000 more gasoline molecules, (the number of molecules in a single droplet. )
See what I'm saying?
Like I said before, vapor is uninvestigated in automotive research but firefighters have some data on gasoline vapor.
Didn't I read this before? Oh yeah, my post on page 1.

[jonbill]

Oh dear.