BUT, my reasoning is that if SVO works in the big engines, why not in our little guys? After all, the big ones use kero and also the SVOs.
Well as I pointed out before there is one major difference. The in cylinder temperatures of full size diesels is WELL above what is needed to vaporize and ignite SVO.
Somehow, the big atomization device handles the thicker oil/fuel.
The fuel is forced under pressure, minumum 3000PSI, up to 30,000PSI through an atomizer(injector). That's why they get away with it.
And about power/RPM. If the volatiles are only 22%, and only those are burning, surely we would now only be see something like 22% of the RPM over a stock fuel? Or at least far less than 90% RPM.
That's not logical. The engine at full throttle breathes the same amount of air whether is is being driven by fuel or another source of power like if you were motoring it with an electric motor for example, regardless of the amount of fuel or type of fuel. So what the difference? Take for example water injection in engines. It doesn't reduce the amount of power the engine makes, or can make, even though it is inert. The fuel has to burn to make power. To burn it needs air. We have airflow through the engine. and that is basically fixed for any carb, muffler, port timing, and RPM. We know that fuel burns within a certain air to fuel ratio range. If we increase the inert portion of the fuel(oil in this case) we need to open the needle to get an acceptable air to fuel ratio for the engine to run. The combustion process happens in a similar manner, but the incompressible liquid oil doesn't burn and just takes up space and absorbs heat in the combustion chamber, and is ejected with the exhaust. That heat the oil absorbs ends up as lost power output, I think.
I don't doubt that some of the oil is burning. There will be some oil droplets in the fuel air mixture small enough to vaporize during combustion and burn. This is why in post 353 fuel consumption will be the only way I will believe such a claim.
So power is basically determined by airflow. Why? the amount of available oxygen determines how much fuel you can burn.
Fuel A: (Methanol)
Mass Air/Fuel ratio = 6.4:1
Heat of combustion of fuel = 22.7 MJ/Kg
Fuel B: (Kerosene)
Mass Air/Fuel ratio = 14.4:1
Heat of combustion of fuel = 42.8 MJ/Kg
Say we have an engine that pumps 1 Kg of air per hour.
This means that on fuel A 1/6.4 of a Kg of fuel would be burned for 3.55MJ of energy input At 12% efficiency this is .425MJ per hour or .158HP per hour output.
On fuel B we can only burn 1/14.4Kg of fuel for 2.97MJ of energy input at 15% efficiency this is .445MJ per hour or .165HP per hour output.
These figures are rough and our engines always run far richer, but that just means that more fuel is used, but there is no more fuel burned than the oxygen allows. If a mixture of a fuel contains an inert component that just means that more mixture needs to be pumped into the engine per hour, but the air/fuel ratio of the burning components must be the same.
To clarify the inert portion of the fuel mix. If the above engine was fed a mix of 50% fuel B and 50% oil then it would consume twice as much mix, but would make roughly the same amount of power. It's the same as if you run the engine rich. You pump a lot more fuel through the engine and get no more power. The unburnt fuel is inert in this case, and is just dumped out the exhaust.
Andy, if the engine gets better than ~3.7 minutes per ounce of this mix, some of the corn oil is burning.