Combustion Analysis: 1.9L Turbo Diesel Engine

Hello everyone!

I've been reading here for a while now, and I finally managed to register.

I also enjoy working with diagrams, and I've come across a question that I can't answer, even after a lot of thought.

Below the diagram, you can see the combustion patterns of a TDI engine (1.9l, PD) for two operating points (red: 1900 rpm, full load; blue: 1000 rpm, full load). You can clearly see the start of the pre-injection and the start of the main injection.

http://www.vflwolfsburg-fanbilder.de/cuwob/Mitglieder/dQB.gif

Now, to my question: Why is the curve so narrow and high at 1000 RPM, but wider at 1900 RPM?

Is it related to the ignition delay? Or to the higher compression, the increased fuel injection, or the increased intake air at 1900 RPM?

Regards,

P.S.: If I've posted in the wrong section, please move it! Thank you!

Simply put: on the x-axis, you'll find the crankshaft position, not seconds or any other unit of time.

Hi,

It is well known that in TDI engines, the injection timing shifts towards later in the combustion cycle. This is because to inject more fuel, the engine can only increase the injection duration, but it cannot start the injection arbitrarily early.

I find it very interesting, but I can't decipher the labeling on the Y-axis.

What does it say?

I interpret the Y-axis as...

Energy rate (J/°K).

m;

Good day!

First of all, thank you for your responses!



@christians

And what does that mean?


@garth.brooks

Does that mean that the start of injection at n = 1900 rpm is 1/min earlier, but because of the longer injection duration, the start of combustion is the same as at n = 1000 rpm?

Why does such a different (softer?) curve result at 1900 rpm?

The y-axis represents the energy rate [J/°KW]!


This engine comes from a dynamometer test bench. How can I relate these test results (full load at 1000 and 1900 RPM) to normal driving conditions in the vehicle?

As Christians already mentioned, the °KW is indicated on the x-axis.
Now, the pressure build-up and thus the energy rate at 1900 min^-1 is "slower" due to the shorter injection time in °KW. However, this becomes consistent again (possibly with a slightly longer injection duration, as more air mass is available at 1900), when you convert the whole thing to a time axis.
Stay cheerful.
OlBe
P.S.: The injection timing is the same for both curves, isn't it?

Hi,

Yes, it looks the same, but should it? The question is whether this refers to the actual 'Grad KW' value, meaning the 0-degree mark, or whether '0-degree' here only represents the start of injection. I would guess the latter. Then it makes sense again.

The curve is lower because, despite the higher pressure, the shorter injection time at 1900 results in less fuel being delivered through the injectors. Consequently, a longer injection time is then required.
As the speed increases, the curve would become flatter and wider.

I also claim that the steepness of the flanks (which corresponds to pressure buildup) of the two curves is approximately the same when plotted on a time axis, and only flattens out when considering the degree-kilowatt relationship. The steepness of the flanks likely represents the closing speed of the valve.

Hello,

'In this diagram, I only see the energy conversion rates related to the crankshaft angle. You can't determine the injection timing from this; you can only see the start of combustion, which is not the same as the start of injection!'

Best, Ben.

Hi,

While I understood the issue with energy, the start of combustion is still related to the start of injection.

But I agree with you, the delay in ignition timing in degrees KW is equivalent to an earlier injection start, which is why fuel is injected earlier.

That would explain the almost identical beginning.