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Turbocharger: transient behaviour

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Xplode

Automotive
Jan 10, 2008
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Dear,

I have a question regarding the transient behaviour of a turbocharger. Are the following reasonings correct?

1) Assuming the engine is working at constant torque and you want to increase the engine speed.

I thought, first of all, you increase the air-fuel-ratio. This causes an increase in the exhaust temperature and engine power, accelerating the load. To keep the torque constant, the boost pressure will decrease. As there is more exhaust power present at the entrance of the turbine, the turbocharger will accelerate, increasing the volumetric flow rate and boost pressure. Eventually the air-fuel-ratio has to be decreased slightly to limit the increase of the exhaust temperature.

2) Assuming the engine is working at constant power, but the load increases and the rpm decreases.

As the rpm decreases, you move towards slightly higher boost pressure and lower volumetric flow rate in the compressormap (assuming constant compressor speed). To keep the power steady, the air-fuel-ratio should be increased.
Then, what happens to the turbocharger?
I thought: OR, as a cause of the lower flow rate, there is less exhaust power present to the turbine and it decelerates.
OR, as the air-fuel-ratio is increased, the exhaust temperature increases and more exhaust power is present to the turbine.

Thank you for your time and help!

Yours sincerely,
Xplode
 
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For situation 1:
Increase fueling, which decreases the air-fuel-ratio initially. Turbo begins to spin up. As you achieve the desired engine speed, you tweak fueling back. Since your condition is "constant torque" your final fueling (per engine stroke, not per unit time) will be about the same as your beginning fueling, with a higher engine speed and probably higher turbocharger speed. Realize that with "constant torque" if you wanted a final higher engine speed you need to either decrease the load or grab a lower gear, or you can't maintain the new engine speed with the same torque.

For situation 2: Not sure how you want your engine to respond here - please clarify if neither of the following describes your situation. Initially, the engine speed and turbocharger will both begin to slow.
possibility 1: If the engine is maintaining a torque setting, fueling will stay the same but the engine and turbo will be running slower and exhaust temperature will rise.
possibility 2: If the engine is maintaining a speed setting, fueling will increase until the engine speeds back up (or the max torque is reached). The steady state for the constant engine speed will be a higher fueling level (per stroke and time in this case) and probably higher turbocharger speed with the same initial engine speed (unless torque limited). Exhaust temperature will be higher.
 
These questions sound VERY academic. Do you have a real world work related problem or is this a student assignment or thesis or the like?

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Are we talking gasoline or diesel here.

If we're talking gasoline, you simply open the throttle more, which puts more mass flow through the engine, which makes more exhaust, which spins the turbo more.

If we're talking diesel, yes it is going richer initially, but so what? Diesels don't much care about their air/fuel ratio as long as it's within boundaries imposed by emissions.
 
Thanks for the help!

I don't have any experience with turbocharging, but we're planning a research where they might appear as a solution.
Trying to get it all clear, I asked myself some questions and got a clarifying answer now. Thanks

Regards
 
There are many papers on this. Try Googling something like:

"transient behaviour" turbocharger diesel

It is a fairly complicated issue, specially when things like VGT and EGR controllers are included.

Do a search on "transient overboost" too. It's what makes the Ford diesels feel quick.

- Steve
 
The turbocharger has absolutely no idea of what sort of an engine is connected to it, or crankshaft rpm, or engine load.

A turbo only sees mass flow, pressure ratio, and turbine inlet temperature. The turbine responds accordingly.
 
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