----- Original Message -----
Sent: Sunday, December 21, 2003 11:21
AM
Subject: [FlyRotary] Re: Dumb intake
question
then I'll give it a
shot.
So based on that - it looks like your
length from port to butterfly inlet of TB ranges from 13.1" for 6500 RPM to
11.5 RPM for 7500 RPM. This provides time for the pulse to be generated
by the closing of the port and to arrive after that port opens again after
bouncing back from the throttle body once. So from that Length you would
need to subtract your block to port distance of 2.5" and the length of your
throttle body.
Excellent! This is
just the sort of info I was hoping for, but there's one thing I
still don't understand. When you say the pulse bounces back
from the TB, what is it bouncing off of? I believe the throat
of the TB is straight, with no venturi effect at all, and we'll assume that
the intake runner matches the TB size perfectly. That just leaves
the open butterfly, which has a pretty small
profile. Is it really getting a strong bounce from the open
butterfly? What if you had a slide type opening, like the Ellison TB,
where there is nothing left in the path of the air? Would there just be
no bounce?
Thanks a
bunch,
Rusty (special needs
student)
See, you should have come to class that
day{:>), Rusty
What happens is when a Finite Amplitude
Wave (FAW) hits a change in cross-sectional area (like an opening from your
tube to the atmoshpere or a plenium - or a decrease to a closed tube) energy
is reflected back down the tube. The amount depends on the extent of cross
section change, but opening to the atmosphere reflects most of the energy.
If the cross sectional area it sees is Larger then a FAW wave of the
opposite type is reflected. If the cross sectional area is smaller, a
FAW wave of the same type is reflected.
So in your case, a compression wave is generated
when the intake port closes, the wave travels up the tube until it sees
the atmosphere (at the throat of your throttle body), there a large amount of
the energy (Most of the wave's energy) is reflected back as a expansion wave
(meaning its negative pressure with regard to atmosphere). However, when
it gets to the port the port has closed, so a wave of the same type (Expansion
or negative pressure) is reflected back to the TB throat, again where it sees
a larger cross section. A larger cross section reflects the opposite
type wave, so this time the wave reflected back to the port is a
compression (positive pressure) wave and this when time it arrives,
it finds the intake port now open and helps to shovel more air
into the combustion chamber. Now this is not as desirable as having it
arrive just as the port is closing to overcome reversion - but of some
aid.
Ideally, you would want to take the pulse created
by an opening port (in the rotary because that pulse is stronger) and then
have it bounce back just before that port closes. But, that is a
duration of approx 288 degs which would take at 7500 rpm about .0064 Seconds
which would require a length of approx 46 inches for a two bounce
system. This (I think) is one of the reasons that Mazda interconnected
the two intake ports - it reduces the intake length required because the two
points of interest on the different rotors are approx 93 deg apart rather than
288 deg they are on the same rotor - so the time to travel and therefore
the intake is much shorter.
So you say - lets do a four bounce, well the
problem there is when the pulse comes back to the intake port - it finds it
open (remember in this case, it is from Port opening to port closing so
the port is open each time the pulse arrives - the opposite of the
short runner case). Since the port is opened into the chamber in
this chase, the FAW sees a larger cross sectional area and reverses its type
which sends an expansion wave back to the TB. This reflects back a
compression wave which helps shove a bit more air into the chamber but is
reflected back as a expansion wave. This would continue several times
(2,4) and each time some of the pulse's energy would be lost into the
atmosphere and more loss into the chamber. So by the time the port is in
its closing phase and pushing some of its intake charge back out (Reversion)
the pulse is probably too weak to provide much assistance in overcoming the
reversion.
FWIW
Ed Anderson