The control unit EDC16c8 doubts map lambda

[switing63]

Hi all, I have just begun to study the maps contained in the units and they are already? in difficulty, then I would like to ask questions to understand this better, sorry if they are trivial, but keep in mind that I am a beginner who is trying to understand the logic.
At this moment I have before my eyes the map lamdba a 147 140 Hp with ecu EDC16c8 (address 0E9B7A) and the scale of the engine rpm comes up to 3750 rpm/1?
The first question relates to the values we say, out of scale, that is, after this value, for example, 4000 rpm or even in addition to equality? mass d?air the assumed value ? the same that c?? at 3750 rpm?
I mean if all?the intersection between 3750 rpm and 1050 mg I find the value 1,080 seen that the table stops them to 4000 or 4500 worth? always 1,080?
Since? I imagine that this map indicates in a certain sense, the maximum amount? fuel injection for a certain amount? of the air intake to a certain number of laps I would like to ask if the calculation that I am doing ? correct, that is the place to 3750 rpm and 1050 mg 1,080 I find this means that at the most I can inject[1050/(14,5*1,080)]=67,05 mm3 of diesel fuel with a ratio of air-fuel equal to 1050/(67,05*0,84)=18,64:1 or those 67,05 are not mm3 but mg of diesel fuel (which correspond to 67,05/0,84=79,82 mm3)? in this case, the ratio of air to fuel would be equal to eur 15.66 (i.e. 14,5*1,080)? this second solution to me sounds strange to me why? I fear that with a mixture of air and gas/oil cos? close to the stoichiometric ratio, the engine starts to smoke like a steam train.
Thanks for the replies.

[cinqueturbo]

And you'd be a beginner?
Bravo...

[switing63]

Very beginner, sar? a week that I have been mobbed by curiosity? to understand how it works the engine of my car. Thanks for the bravo, I take it as a compliment? I do not know whether to interpret it as the exact answer to the questions I asked myself, what is ? right the first intuition of the out-of-scale, and if in the second case the map as the values of the mm3 and not the mg.

[biela]

Hello

From 3750 the assumed value is the same that there is at 3750 rpm. The ECU gets the last value that can read.
The AFR calculations are in mass or mg.
This map is a Lambda limitation, but the engine is working with higher values. Look at the turbo maps an torque requested map.

[switing63]

Hi
thank you very much for the answer. I continue in English 'cause my english isn't good. You are perfectly right in saying that the engine works with the lambda pi? high, in fact if I'm not wrong to calculate, for example, at 4000 rpm the torque request with the pedal at 100% ? 330 Nm that are for? limited to 311 from the torque limiter, which corresponds to an injection of approximately 60 mm3, which in turn correspond to 50.4 mg of diesel fuel (for diesel, the Italian law provides for a density? between 825 and 845 g / liter, I typically use it in the calculations 0.84) and the turbo pressure ? of 2220 mbar then theoretically we would be already? over 1050mg/cycle (477cc*2,22*1,2928 gr =1369 mg/cycle for? taking into account the intersection of the valves, and then a certain lift of the piston beyond the lower dead point prior to the closure of both, and then some reduction in the volume of the cylinder is available I think that in that situation we would be pi? or less to 1100 mg/cycle, the ratio of the air/fuel ? about 1100/50,4= 21.8 to 1 and the cio? lambda 1,505.
The map I am working during the transient cio? for example, when we say 80 Km/h in the press at the end of the stroke the pedal to accelerate, or even better to speed? pi? low, in such a way to not inject too much fuel in relation to the amount of oxygen available to avoid clouds of black smoke showy.
The one that leaves me very puzzled ? read values of less than 1, type 0,98.
At this moment I don't know what the situation brings the engine to have 660 mg/cycle of air and 1125 RPM, for? if in that situation arrival to inject a maximum of 14,5*0,98 or 1mg of diesel per 14,21 mg (or 46,44 mg/diesel 660 mg) of the air seems to me a situation from the smoke black. If, instead, that 46,44 that results from 660/14,21 were already? mm3, which corresponds to 39 mg then the situation would seem to me the most? acceptable why? by dividing the 660/39 I would have a ratio of air-fuel 16,92 then with a decent excess of air.

[djang01]

12000 rpm there are other maps that do not allow you to smoke why not arriver? i never spread that quantity?, for? the engineers who built the map in question have considered, however, this boundary condition. This happens on almost all engines.