Originally posted by turbo buicks
i can see how that makes sense but isnt the gas coming out of the turbine already turbulent. i would think since its only the gas coming out of the turbine housing and not any extra wg gas in the dp then there would be lots of pipe for the little bit of gas to go thru. and since flow is like the pressure difference b/w 2 outlets then if you had tons of psi in the exhaust and none outside the wg the gas would just flow out the fastest cuz its the most psi difference.
The gases coming out of the turbine exducer are actually spinning like a tornado at very high speed[turbines typically spin at anywhere from 60,000-110,000 rpm under boost conditions].
That means that the gases are going like h#ll in a circular direction,but not in a straight or linear direction[the direction you would like it to go].
The hole for the exducer on the turbo is typically from 2.250-2.563 for most modified cars.
The typical D/P diameter is about 3".
So as the gases come out,they expand losing their "cohesion" and their spin to a lesser degree.
That's where the gases become turbulent.
Once the pressure of the gases behind them force them further down the pipe,they tend to align in a unified direction and you get even,straight flow.
I take it by the second part of your comments that you see the exhaust pressure as being lower in the D/P than it is before?
Is that correct?
It should be because energy is being extracted by the turbine wheel[from the exhaust].
The D/P is also larger in diameter than crossover feeding the inlet of the turbo.
Now the pressure and volume drop effected by this can be figured out by running the PV/T formula in SI units[I would figure about 200* F temp drop].
With the drop in pressure and the larger pipe on the outlet side,dumping the W/G output into the pipe shouldn't be a problem unless you've got a *real high* H/P car.
At that I would expect to see a power loss over an outside dump for the W/G because of too much exhaust being handled by too small of a pipe.
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