Marc Wiese wrote:
Ah so right Tom!
2 rotor = 80 cu. 3 rotor= 120 cu in (as spec by Mazda...). Mass
flow
>from calcs 20B=41.3 lbs min; mass flow from 13 b =27.3 lbs min. So
actually 41.3/27.3=1.5128205
You add more of course than just a rotor, or displacement!
Marc Wiese
Is this 80 cu per rotor, or per rotor face.?
Royce
Now you are getting into a real can of worms :-)
Considered as a two cycle it is 1.3 liter but it is not a two
cycle.
It is equivalent, as far as the volume of air it pumps, to a 2.6
liter
4 cycle 4 cylinder engine. However the volumetric efficiency is at
least 20% higher than any piston engine. No poppet valves to get
in
the
way of the intake ports. Therefore in terms of turbo charger
selection I would use about a 3 liter equivalency.
Paul Lamar
Paul,
Can I clarify the airflow calculation that I uploaded last week in
that
spreadsheet? That tried to work out the airflow more analytically.
If you read Corky Bell or Hugh MacInnes, the standard formula for a
four
stroke piston engine's airflow is airflow rate = ((cubic inch
displacement)
X (rpm) X (volumetric efficiency) X (0.5))/1728. The 1728 divisor
converts
cubic inches to cubic feet.
What's the 0.5 in the equation for? Bell writes, "The .5 is due to
the
fact
that a four-stroke-cycle engine fill sits cylinders only on one-half
the
revolutions." (p.27)
That's not true as far as rotary engine airflow is concerned because
rotary
rpms have mixture flowing every revolution. So rotary racers who
post to
the
net leave out the 0.5 when they calculate airflow; they double the
airflow,
as you did in suggesting a 2.6 liter piston equivalence.
Then the question of volumetric efficiency is another issue. Bell
gives
a
small-block Ford as an example in one of his problems and uses a
volumetric
efficiency coefficient of 0.85. In discussions on this list some
time
ago,
when Paul Yaw was building up his super-turbo-something for someone,
he
used
1.07. I'm told that rotary tuner HKS recommends using 1.1 as the
volumetric
efficiency coefficient. Some racers on listserve message I've read
suggest
1.0. Beats me if there is an exact right number but the bottom line
is
that
the volumetric efficiency also boosts the airflow. Your estimate of
a
20%
better volumetric efficiency is right in there with those other
guys.
So, to the best of my ability to tell, knowledgeable people sizing
turbos
would use around 278 cubic feet per minute of airflow for naturally
aspirated two rotor 13B engines running at 6000 rpm. I can't imagine
doing
anything besides adding 50% when you go to a three-rotor 20B because
that's
what you've done to the displacement.
If your comment, Paul, that you'd use a 3 liter piston equivalency
was
intended to be your recommendation for a 20B engine, it seems to me
that
would be too conservative, though it probably is about right for a
13B.
Of course, if one had a different prop reduction ratio and wanted to
optimize at a different rpm, that would affect your airflow
calculation
and
turbo sizing.
Barry Gardner
Wheaton, IL
I was referring to a 13B two rotor.
This chart is part of a Mazda tech paper that can be downloaded from:
http://www.thecarricos.com/ACRE/
Paul Lamar
It looks like at STP, air =.08071 lb/cf. Wonder why Jeff used a .9
volumetric efficency? Anyone?
Marc Wiese
I have not seen the spread sheet. Ar you sure it is for a rotary?
Jeff built some UAV sequential turbo charge Rotax 914's that made it to 57,000 feet.
Perhaps it was for a rotary and he was being conservative.
Paul Lamar
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