Interesting exhaust Question.

[Wilsonman02]

I'm sorta in an argument with someone on exhaust flow of turbo charged car. We seem to disagree, anyways.....What is the Average speed of the exhaust gas exiting the turbine of a 500+HP car, i dont need like an average of 525ft per second, more like 500-700, or whatever it might be, i'm just throwing numbers out there.thanks

 

[4geze63]

the turbine wheel spins upto 140 000rpm

 

[gsxtacy]

I am sure there is a way to tell. Its not a fast as a pellet gun though and I have seen some 720 fps pellet guns!!

I would say (guessing) around 550 fps straight out of the turbine wheel exducer.

If its a crush bent/mandrel bent flow argument, might as well give up.

There are restrictions in Crush Bent pipe. Not in mandrel bent.

gsxtacy

 

[aleccolin]

It's completely possible, but you have to fill in some of the blanks.

Pick an rpm. Remember that one revolution of the crankshaft only exhausts half the total displacement of the engine in a 4 stroke (Otto cycle) engine. Multiply that rev number by HALF the displacement of the engine (assuming no wastegate or other leakage) and divide by 60. This will give you a volume flow per second. Pick a turbine housing exit diameter. Halve that number, square it, and multiply by pi. This gives you the cross-sectional area of the turbine housing exit. Now pick a turbine exit length. Multiply this by the cross-sectional area of the exit to give you a volume (simple cylindrical example). So think about it, in order to flow the volume flow per second you first calculated, it will have to flow a certain number of times the volume of the exit per unit time. That, multiplied by the length of the exit, will give you the speed.

Example.

7000 RPM
3.0 Liters
7cm exit diam
3 cm exit length (again, simple cylindrical design here)

A 3/2 liter engine at 7000 rpm must flow 10,500 liters in a minute or 175 liters/sec.

The volume of the exit is (Pi*3.5cm^2)*3cm = 115.45 cc's or 0.11545 liters

The exit must therefore flow it's own volume 1515.8 times in a second for all the exhaust gases the engine expels to exit through it.

Given that, the length of the turbine exit will be traveled the same number of times in a second, or in this case 4547.4 cm/sec or 149.2 ft/sec. May not sound like much, but that equates to 101.7 mph exhaust exit speed.

A 500 hp engine very well may rev higher, or have a greater displacement, either of which would raise the exit velocity, ceteris paribus. Take into account also that larger turbos will have larger exhaust exits and that the exit velocity is inversely proportional to the diameter of the exit. Bigger turbos may have slower exhausts.

This is a very simple example and I probably screwed something up as it's 3 am and I'm still a little drunk, but it should be good enough.

 

[Defiant]

Egad. Interesting question. Of course, it'd change depending on where in the system, but you could start with piston speed. I ain't doin' the math. Keep in mind it'll have a big boost through the exhaust ports, then slow down a bit in the manifold, and on out through the turbine and system.

I've chronographed my RWS 48 at over 1200 fps with lightweight (Prometheus) pellets. They actually break the sound barrier, and sound like a rifle slug going through the air. Bad for the gun, though- not enough load on the mainspring.

 

[Slipstream]

I've chronographed my RWS 48 at over 1200 fps

Doesn't the Model 48 usually do about 1100 fps anyway with standard target pellets?

sorry for the thread hijack.

 

[yoshimitsuspeed]

It's completely possible, but you have to fill in some of the blanks.

Pick an rpm. Remember that one revolution of the crankshaft only exhausts half the total displacement of the engine in a 4 stroke (Otto cycle) engine. Multiply that rev number by HALF the displacement of the engine (assuming no wastegate or other leakage) and divide by 60. This will give you a volume flow per second. Pick a turbine housing exit diameter. Halve that number, square it, and multiply by pi. This gives you the cross-sectional area of the turbine housing exit. Now pick a turbine exit length. Multiply this by the cross-sectional area of the exit to give you a volume (simple cylindrical example). So think about it, in order to flow the volume flow per second you first calculated, it will have to flow a certain number of times the volume of the exit per unit time. That, multiplied by the length of the exit, will give you the speed.

Isn't the exhaust gas still expanding when the exhaust valve opens, and a while afterward? Plus isn't it still under pressure when the exhaust valve opens?
If I am not mistaken (and I could be) those variables would considerably effect the equation.

 

[aleccolin]

That's absolutely true, the ideal gas law would definitely play a role, but in this simplified example I'm just using a "what goes in must come out" sort of example and assuming that the manifold is short enough and the whole system between the head and turbo exit is small enough and hot enough to limit those effects. That's a big assumption, but I would be pretty surprised to see those effects combine to net a several hundred ft/sec change in exit velocity. Not that it's not possible though. In this case, the question is generalized where it really isn't possible to do so, but what else is new

 

[gsxtacy]

550 fps is like Top Fuel dragster flow!!

This is a good question.

Gsxtacy