The chemistry of the reaction, diesel & advance injection

[asdone]

Buond? to all,
My doubt concerns the chemistry of the diesel, once the diesel ? injected how to develop the reaction?
Without a doubt, the peak pressure must be on or after the TDC, which means that c'? a certain delay between the injection and the initiation of the reaction. This delay ? fixed, or the reaction starts immediately and increases of intensity? gradually?

Speaking of the engine, what? the ideal angle for maximum output to where ? appropriate that the reaction reaches its peak pressure? Of course, this dipender? the compression ratio and the displacement, but c'? some online guide or a way to calculate the data values of the advance from the original map?

Please, if you decide to respond to support your intervention with scientific arguments / logic. I'm not interested in things like: "by 2 degrees of advance and you're good to go". I'm sick of try 1000 different configurations at the risk of splitting without getting results.

I want to address the issue in a scientific way!

[sergiot]

as soon as the fuel leaves the injector begins to burn, but the burning lasts for some time even why? the diesel does not come out everything instantly. as you know the injector is open for a certain time, and the last diesel ? still in the injector when the first is down? burning. ? all a progressive phenomenon.

the combustion of the diesel ? very fast, but not instantaneous. as all the phenomena of nature has a progressive development. c'? for? an optimal situation for the development of the energy with the best performance. this optimal time is when the maximum pressure of the cycle is around 10 -20 ? after the upper dead point of the outbreak.
this optimal point for the development of the best pressure peak and little influenced by compression ratio, little displacement, but more? from the bore of the cylinder, meaning that pi? great ? the diameter, the greater the time of expansion of the flame (it should make more? the road to fill the cylinder.)

given this optimal value of the maximum pressure it will go back one time ? the development of the flame and is the point where you have to put the diesel in the cylinder. you go back in time again to take account of the unfolding of the complex operations inside the injector and you will get the position where you have to give the current to the injector. this latest ? the one that we make maps in the maps in advance. the burning time ? then can be translated into degrees engine why? while the combustion develops the engine is running. then you can? translate this time in degrees the engine.
since the combustion time is almost constant but the speed? of the motor variable, equal? of time with the rpm increases, the advance must increase in order to have the optimum point of pressure in the neighborhood of 10-20? the motor shaft.
outside of the optimal point of the pressure the performance ? pi? little, if too much before you can? to prevent the piston rising, if too after it dissipates too much energy in the exhaust.
this last adjustment is also used voluntarily to heat the exhaust to regenerate the dpf.

[asdone]

Good explanation, thanks!

By degrees, the injection timing on the basis of the scheme is achieved that the injection ends always or almost before TDC, which means that the constant that has the most important ? the expansion of the flame (? a constant, right?). Then my goal becomes to quantify how much the flame takes from the centre to reach the walls of the cylinder. It is feasible?

[sergiot]

not ? cos? simple, why? also the flame ? powered by a continuous supply of diesel fuel, which not ? instant.
using the example of the minimum, where the injection time ? minimum of approximately 0,200 msec .
1000 rpm/1' the time to do 1 round engine ? 60000/1000=60 msec , the cio? 60/360 = 0,1666 msec per degree
we know that at 1000 rpm, advance ? about 0? (i.e. at tdc) the electric
to go from the injector to the wall there are about 10?
roughly, we want to 0,1666*10=1,6 msec
then, if the cylinder has a bore of 90 mm and the injector ? at the center, to travel the 45 mm the flame employs 1,6 msec goes to the speed? of 0,045/0,00166 = 271 m/sec = 976 km/h
values with a bit of an approximation, but to give you an idea

[asdone]

Clearly, in this context, the rail pressure influence the time of expansion of the flame? Or the speed? jet ? irrelevant with respect to that of the reaction?

[sergiot]

the pressure of the rail has a double effect:
increases the speed? of the microdroplets out from the injector, but I think that is a little contribution on speed? flame. matters more? the whirling motion of the air in the cylinder
decreases the diameter of the microdroplets, in other words increases the pulverization, and this contributes to the improvement of the quality? of the flame, but not so much on speed? of the flame front.

[ducati83]

sergiot truly without words congratulations...I was able to understand what you have written though things quite complicated...where it would take us a good basis to start....excellent communicator.....
I will follow with close attention to this post...

[asdone]

I tried to make a verification of what was said by sergiot, that still thank you

time propagation = (degrees advance + degrees of the peak of the reaction) / Speed? angular

we have 2 unknowns:

x = time of propagation
y = degree of the peak of the reaction

and some of the data that I obtained from my map times ori (calculated from the first column)

V angle to 4000 rpm = 24000 degrees / second
V angle at 5000 rpm = 30000 degrees / second


We put in the system the following equations:

x=(y+14)/24000
x=(y+20)/30000

the solution:

x=0.001 seconds
y=10 degrees

A little too accurate as the values don't you think? they made coffee table...

[sergiot]

what the mappers, they call in the jargon "advanced injection" in the diesel , or "spark" in the gasoline, for the physical or the engine ? child burning time or time to have the pressure peak. obvious that after the pressure peak in the cylinder pressure ? over, but drops to the time of discharge.
we also take these calculations with the calm resulting from the need? explain in 20 lines that is in the entire volume.

[asdone]

Hehe, but these are the calculations backwards, always work (or almost)

I do not pretend certain to know all the technical background of the calculations and theory that have little to do with the ones I had written, but have served to clarify the ideas!
Thanks sergiot, I have removed many doubts (and I think not only me!) and, above all, you've saved quite a bit of pain to my poor car!