riff_raff wrote:I haven't done the math, but in general having an exhaust velocity at or above the local airflow velocity would only be beneficial if you were trying to generate thrust from the exhaust flow. One of the main objectives for the exhaust outlet location (besides regulations) is to locate it in an area of relatively low dynamic air pressure to assist exhaust scavenging. I also believe the aero guys like it to be in an area where the varying flow does not adversely affect downforce, but still also helps energize the trailing wake flows.
gambler wrote:I would figure the most true way would be to inject a smoke additive to fuel
and run the engine in the wind tunnel to see where its going.
Ciro Pabon wrote:- "Static" volume of exhaust (yes, I know there is a thread on that specific subject, but I said crude) :
2.4 liters x 18.000 rpm / 4 = 10.800 lt/min
I'll assume an square exhaust with a 10 cm side (Toro Rosso probably has one of these) which gives me an area of 1 square dm.
I'll simplify that even more, assuming the gas is at 1 atm and 20 degrees.
So, it's like 10.000 dm3/1 dm2 = 10.000 dm/min = 1.000 m/min = 60 kph
Kilcoo316 wrote:The volume flow in will be approx dependant on engine displacement and RPM.
0.0024*18000/60 ~ 0.72 m^3 per second
The massflow can then be obtained from this by multiplying by atmospheric air density.
0.72*1.225 ~ 0.882 kg/s
The volume flow out will depend on local density, which is dependant on the air temperature at the exhaust exit point.
(anyone any idea of the temp?)
Vol(out) = 0.882/density
density ~ 101325/(287*temp)
(I'm assuming the exhaust exit is subsonic)
The exhaust exit velocity can then be quickly be calculated from the exhaust exit area.
Vol/Area = Velocity
Kilcoo316 wrote:*Assuming* a subsonic exit - that gives an air density of
P/RT = rho
101325/(287*1273)= 0.277 kg/m^3
So using the massflow rate from earlier, 0.220 kg/s - which gives a volume flow rate of ~ 0.97 m^3/second...
That is ~ 0.48m^3/second per exhaust
Now... how big is an exhaust exit area? Say its 5cm diameter circle, thats then 0.0020m^2
0.48/0.0020 = 244 m/s
Or 880 kph.
What a meal you guys are making of this. This is as simple as things get.
Whatever goes in, has to come out, unless there is somewhere in the cylinder than stores the air....quite unlikely.
Users browsing this forum: 360Spider, Rogerbot [Bot] and 10 guests