Basic criteria for map

[Vega25]

Buonsera all guys and happy Easter.
I would like to begin this discussion by asking to analyze the basic concepts of any map to be processed.
I have documented a lot here and there, but there are always a lot of doubt about some of the concepts that I consider the "key".
I don't want to open this discussion to know specifically what to touch on a map, but in order to understand if the reasoning is pu? go well or not.
From what I've read, all the maps are linked to each other. Unthinkable only change the map to leave the other. For example..If you increase the flow rate of the diesel, I can not increase the turbo pressure...
By making these arguments, I came to the conclusion that the basis of everything, I could put The AFR as a starting point.
This is because? I would like to impostarmi a AFR optimal to the combustion, and therefore, all of the changes that will go? to make the flow rate fuel and air, will be made on the basis of the AFR impostatomi initially.
I ask this because? following other discussions, I noticed that when you start to make changes considering only the last AFR and using it as a factor to correct for the changes already? made. But wouldn't it be more? makes sense to do as I have written before?
You will find that as reasoning can be deemed to have been sensible, or am I mistaken?

Thanks to anyone who wants to tell me a basis on which to reason. :D

[mate89]

I'm sorry if I break but can you explain to me well cos'? AFR?

[djang01]

If you increase the flow rate of diesel fuel, you need more? air, and how much air? You build on the ratio of air and fuel and then once you know how much you need do a conversion in boost pressure

[switing63]

AFR ? a stands for AIR FUEL RATIO, or ratio of air to fuel. For an Otto motor, that runs on gasoline this relationship ? fixed ? equal to 14.7/1, meaning that for every gram of gasoline need, 14.7 pounds of air to burn completely, in engines with Diesel cycle, the ratio to the theoretical air fuel ? similar (14,5/1) to that of the engines eight? precisely ? theoretical why? really? the engine runs extremely lean and gi? reaching values close to 18/1 begins to smoke a lot. For example, in the 1.9 jtd 16v 150 Hp (just why? I'm studying a bit of that) looking at the maps you see that are injected into the 63,75 mm3 of diesel (which corresponds to approximately 63,75*0.84=53,55 mg) to the regime of maximum power where the amount? air should be about 1100 mg per cycle and then with an AFR equal to 1100/53,55=20,5/1 very far from the stoichiometric ratio to the theoretical
Hello Vega25, I am a newbie and I'm trying to learn myself how to read and interpret the values of the maps in the ecu, then take it all with the benefit of inventory to what I say, for? according to me what you say, or rely solely on the maps lambda to do your mapping does not ? feasible for two reasons, the first that if I understand well the operation of those maps ? during transients that control and limit the diesel during a phase of acceleration and not for example to speed? constant and in the second place, why? as I said before, diesel engines do not have a air/fuel ratio fixed as the gasoline, and then ? impossible to say what ? the optimal value. If there is an optimal value, we pretend 18/1 much to shoot a number, at this point, it was sufficient to read each cycle the value of the MAF sensor (Mass Air Flow) divide it by that famous 18 of the example, and I knew already? how much fuel to inject without asking other problems.
Of course, as I said this ? my opinion, for? saw that I still know little to see if some user pi? expert there can? teach some thing.

[Vega25]

[switing63]

Always premising that I know little, if I have understood the logic of operation of a ecu EDC16 all part from the map pedal or driver's wish, for example if I want to travel at 100 Km/h constant in the gear you want? the gas pedal of say 25% (I throw them a percentage on the case), this map on one axis there are the engine rpm and on the other a percentage and then crossing 2000 rpm and 25% pressure on the pedal avr? as a result, a value of the cup in Nm request that we pretend is 170 Nm, at this point you looked at a map "Torque limiter", which says that at 2000 rpm the max torque request should not exceed, say, 350 Nm, as 170 ? lower the request is left as is, and passing for a map conversion torque/diesel where in the a-axis there are the engine rpm and the other of the pair crossing the given 2000 rpm 170 Nm I provide? the quantity? to inject let's say for example 31 mm3, this sar? the quantity? that is injected to speed? constant, in this case, I believe that the maps lamdba not be verified because? I am asking considerably less power than I could get at 2000 rpm with the pedal pressed at 100%, then very little fuel in relation to the available air. Conversely, if 100 Km/h I want to go to 130 Km/h it happens that from the 25% you want? the pedal at the bottom of the stroke (not always but ? only an example) here redoing the turn of the first ask? for example, through the crossroads pedal/percentage 390 Nm why? cos? says the map the pedal, but the torque limiter abbasser? the request to 350 Nm that is always in the in the map conversion will correspond to 74 mm3 of fuel. It is obvious that if in the instant x I was injecting 31mm3, and the next revolution of the crankshaft will have to? inject 74mm3 cio? pi? double wants us to be a check that there is enough air to burn, then in these circumstances will have? check through the map lamdba not to exceed a certain amount of diesel to avoid smoking like a chimney. Then the 74 mm3 them, I will have the injection x+1 but gradually according to the increase of the amount? of air gets to the cylinders.

[Vega25]

Your reasoning does not seem to me wrong, on the contrary....For? at this point, I ask you: How to intervene maps lambda?

[switing63]

It is always a hypothesis of my own, then wait for trigger input from someone who knows. I believe that the map lambda, to intervene more? or less so?. Always referring to the example, at time x the request ? 31 mm3 of diesel to the next round let's call it x+1 of the motor shaft and the request goes to 74 mm3, at this point, the ecu looks at the data that comes from the MAF sensor and see that the engine breathes in that moment, suppose 700mg of the air and from the map smoke has a value of 1.08 at the intersection between 2000 rpm and 700 mg/cycle, and then makes a calculation type 700/(14,5*1,08)=44,69 that I think they are already mm3 and not mg of diesel? on this I have big doubts, that inietter? let's say the 44,69 mm3, then this will result in? an increase of the combustion gases that impact on the turbine will spin more? quickly and this pomper? pi? the air in the cylinder, and so on to the next round x+2 the sensora maf reading? 800mg/cycle, and then inietter? even more? diesel finch? arriver? the target value of 74mm3. I simplified it a lot, but I believe that pi? whether or not the concept is this.

[mate89]

Good explanation, thanks!

[Vega25]

For which not ? it's wrong to think that if the basis of your calculations put a relationship stoichiometric AFR, map, lambda avr? just to be correct no?

[switing63]

According to me the lambda corrects for a lot, the problem ? to understand how, I at the time that changes in the dark more? deep.
A Diesel cycle engine ? unthinkable that it can be run at a stoichiometric ratio why? fumerebbe like a train to the coal, must always work in excess of air to the minimum air/fuel ratio pu? to get even 100 to 1, then the ultra skinny.
As I have already? expressed in other post that I opened to ask for enlightenment on the interpretation of the maps the lambda, I do not think that the calculation is based on a stoichiometric ratio, let's call it “classic” of a gasoline engine cio? 14.7 to 1 why? c’? something that I did not go back, that is, in some points of the map lambda (I I always refer to the EDC16c8) read values of 0,97 cio? even lower than 1, then in a rich fuel mixture that if gi? a Diesel goes bad in 1 let alone to 0.97, so I hypothesized that there may be at the base is the classic 14,7 but, for example, the value obtained is gi? calculated in mm3 and not in mg.
I try to explain better with an example, if 1250 rpm and with a deal? of air of 660 mg I 0,99 value (14,7*0.99=14,55), I think that the deal? maximum injectable is not equal to 660/(14,7*0.99)=45,35 mg of 54 mm3 of diesel oil, i.e. an AFR of 14,55 but for example is already? 45,35 calculated in mm3 then the weight would be only 38,1 mg, and then the AFR would be equal to 660/38,1=17,33 gi? pi? reasonable for a Diesel.
Or are they completely out of the way, and the “magic number” to be applied to the formula may well be another (for example, 18 (many consider it to be the limit beyond which they will not go to avoid smoking), and then always referring to the example of the first if he was 18 I would have 660/(18*0,99)=37,04 mg that are 44mm3.
Of course, mine are all assumptions we would like someone who has done some tests with diagnostic tools to be able to see in the circumstance the quantity? actually injected.

[SandroMarciano]

Hi, the afr set to edc16c8 refers to the stoichiometric typical of a diesel engine, which is 14.5/1 (1000 points on the map lambda).
The AFR minimum for a car powered diesel pu? also get up to 12/1, a value that guarantees the maximum performance. That said, the logic of the edc16c8 "allows" the afr from switing mentioned only for a short period of time (immediately after the request of the pi? power on the part of the driver), when immediately after the turbo sees? to go in pressure and the operation will return? of new excess air.
I hope I was clear :)