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Tom,
I have no experience with the Ellison, but the
answer may be not in the total air consumed, but in how it is sucked
in. Producing the same HP requires essentially the same air/fuel
regardless - its how it gets there that may make a difference regarding the
Ellison.
The aircraft engine gulps in air in
large chunks. The four large cylinders running at say 2800 rpm and
only two cylinder "suck" each revolution. So there air flow characteristic
is different than a rotary. With the rotary you have six faces (piston
analogs) of less displacement rotating (the rotors not the eccentric shaft) at
approx 2000 rpm (for 6000 rpm eccentric shaft). The rotary sips smaller
chunks of air.
The total amount of air would have to be the same
for both engines (same HP), however, "Average" covers a multitude of difference
in the actual air flow pattern. I see 2 large masses of air in the intake
for the aircraft engine each revolution. The rotary would have 4 smaller
airmass packages ( Yes, the rotary has six faces but only four have come around
in a 720 deg revolution) in the intake. So the interval between the center of
mass for each package is roughly 1/2 that of the Lycoming.
For a specific example let see what numbers may
tell us.
Lets take a Lycoming of 360 CID turning at 2800 rpm
and a rotary of 80 CID with the rotors turning at 2100 rpm (6300 E shaft
). This will have both engines sucking (assuming 100 % Ve for both) approx
291.67 Cubic Feet/Minute. And assuming the same
BSFC they would be producing the same HP.
But, lets see where there are
differences.
1st a 360 CID Lycoming at 2800 rpm has a
period of revolution of 2800/60 = 46.6666 Revs/Sec or a rotation period of
1/rev-sec = 1/ 46.666 = .021428 seconds or 21.428 milliseconds. During
that time its sucking intake air for 2 cylinders in
360 deg of rotation. The rotary however, has its rotors spinning at 2100
rpm (to draw the same amount of air) which gives it a rotation period of 2100/60
= 33.3333 Revs/Sec or a period of 1/35 = 0.02857 seconds or 28.57 ms. The
rotary is also drawing in two chambers of air in 360 deg of
rotation.
Here the rotary e shaft is spinning at 6300 rpm to
give the rotor a rotation rate of 2100 rpm 6300/3 = 2100.
|
Eshaft rpm |
Displacement |
rpm
|
CFM |
|
|
360 |
2800 |
291.67 |
|
|
|
|
|
|
|
80 |
6300 |
291.67 |
|
|
|
|
|
|
6300 |
40 |
2100(rotor) |
291.67 |
So right there we have a difference of approx
25% difference in the rotation time of the pumps pulling in the same average
amount of air. The rotary takes approx 25% more
time than the Lycoming to complete a revolution..
A 360 CID Lycoming (forgetting compression
ratios for this discussion) has 360/4 = 90 cid displacement per cylinder or 180
CID for on rev. The rotary has 40 CID displacement per face and 2
facesx 2 rotors = 4*40 or 160 CID for one rev. So the rotary
has 22% less displacement per revolution and the longer
rotation period.
So if the rotary has less displacement of the
sucking component and must take 25% longer for each revolution. Therefore
the only way it can obtain an equal amount of air is for the intake air to have
a higher velocity than the Lycoming does.
The air velocity of the area in the intake for the
rotary would appear to have to be much higher than the Lycoming. If my
assumptions and calculations are correct that would imply (at least to me) that
to minimize air flow restriction a larger opening would be required on the
rotary compared to the same HP Lycoming. Its not that one is taken in
more air its that the rotary has less time and smaller displacement pump so must
take in the air at a higher velocity.
The fact that the rotary has no valves to
block the flow of air may be one reason that it can over come what would appear
to be less favorable parameters for sucking air. An additional factor that
may play a role is the fact that air mass pulsation in the rotary intake
is less than the Lycoming. This would mean less starting and
stopping of air movement, so the velocity would seem to remain steadier and on
an average higher than for the air pulses for the Lycoming which if you factor
the start/slowing/start of air flow may lower its overall velocity compared to
the rotary.
In summary, while the total air intake in equal for
engines producing equal HP. It is likely that the air flow to the
rotary may be considerably higher in order to ingest the same amount of air
over the same time. This may be why there is a perception that
the Ellison model that may work well for a Lycoming may not work as well
for a rotary.
Well, anyhow, that's my best shot - if its
incorrect perhaps somebody can take it from here, but I think the answer lies in
the different pumping configuration of the two engines.
Best Regards
Ed
I'm under the impression I have an answer.
Isn't there a law of motor performance that says that two motors putting
out the same horsepower are consuming the same amount of air&fuel,
assuming efficiency differences were not significant?
So if you had a 13b and a O-360 putting out the same horsepower for a
single given 1 revolution of the propeller, they should be consuming the same
amount of air and fuel during that 1 propeller revolution. (I THINK chosing 1
propeller rpm is a correct standard)
Bill pointed out that the 13b operates at a higher rpm, and we know that
there's more combustion charges consumed by the 13b to make that 1 prop
rpm.
The difference, the missing piece, each 13b combustion charge consumes a
SMALLER amount of fuel/air than the piston powerplants less frequent
combustion charge. ??? So the 13b burns a smaller
amount more frequently. ???
If this is all true, then the Ellison isn't on the trash heap yet.
Tom
WRJJRS@aol.com wrote:
Group,
I want to remind everyone about how much a priority the large
volume inlets are to us. I believe Ed Anderson was mentioning in one of his
posts how difficult it can be to get a MAP signal in the airbox of one of
our PP engines. This is a perfect indication of why the smaller throttle
bodies used on some of the slow turning engines will kill our HP.
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