Paul,
You have often mentioned the gains in horsepower that can be had by
having a big rammed air scoop on top or side of the cowl compared to an
usual inlets next to the spinner.
I am trying to design size such a scoop now to trade airspeed for
horsepower.
As you know in trying to dedicate the exhaust turbine solely for turbo
compound use, I am stuck with a supercharger for boost. The Rotrex is
efficient but to carry three psi of sea level boost at altitudes above
15K it is going to cost me at least 20 to 25 engine hp. My goal is to
reduce this by one half at least.
For all you guys thinking electric supercharger, stop and do the
calculations. I did. Hint.. HUGE amperage, thick heavy cables even at
24 volts and lack of sufficient wing area for PV panels. Did I mention
weight? Ok back to reality. Lets say the scoop is sized at 40 square
inches and allowing for skin drag losses then the supercharger
compressor sees a little more than 5 times the pressure it normally
would without ram effects. Nice. Note at lower altitudes I have enough
boost to not engage the supercharger. At high altitudes if I want the
power, turbo compound and advantages of the jet streams, then the engine
or plane is going to have to work for its lunch.
My assumption is that if I oversize the compressor and make sure it can
handle the volume, then it will work less, and I can run it at its 80%
efficient sweet spot.
For my study I:ve selected model number 30-94. It has a maximum boost
ratio of 2.7 if needed. It is normally used to boost engines wanting
from 250 to 400hp.
The maximum flow is 0.37 kg/sec or 48.84 lbs/min.
So again I am trading airspeed (drag) for work but I suspect the
merits of compressing air and gains in HP. are going to outweigh the
drag of my plane at 140 knots. Especially if the exhaust is designed
well.
Givens:
My chosen airspeed is 140 knots.
The opening to the Rotrex compressor is 76mm or 7 sq. inches.
The P-ported 13B engine is running at 6000 rpms.
At a density altitude of 18K` the STP ambient pressure is 14.95 inches of
Hg or one half that of sea level. In psi terms that is about 7.5psi.
I then need 7.5psi plus the additional 3 psi or 10.5 psi. total of boost
at 18K altitude to equal what the engine would do at sealevel.
The STP temp there is minus 5.14 degrees F. which is good for compressing
air and allows the engine to run cooler. Bad for heat exchanger as the
density is less. Oversize the heat exchangers.
The intake scoop again is 40 square inches ( about 4x10 inches)
Engine air flow needs are estimated as follows:
My rough guess. running E-90 at lean A/F of 14:1 and 0.46 BSFC (probably
more) and 220 hp.
220hp. x 0.46 = 101.2 lbs. of fuel per hour. Divide by 6.4 lbs/gal
(ethanol is heavier) = 15.8 gal./hr. Weight of air is 101.2 lbs x 14
equal to 1416.8 lbs/air per hour and 23.6 lbs per min.
Note* Piston engines produce about 120 hp per 0.1kg/sec of air or 13.2
lbs per min. on gasoline at stoimetric. Double it and you can see my
above calculation are close as gasoline and E-90 have different A/F
ratios.
I need help calculating what pressures and air flows the inlet of the
compressor actually sees. I believe pressure quadruple as speed
squares. I imagine the calculation is done in terms of plane velocity
stated in inches per minutes times the frontal area of the scoop. This
gives cubic inches per minute passing through the opening and we know
how many cubic inches are in a lb of air right? But how many psi of
boost are we picking up at this altitude and speed? One fifth the
volume weight?
I know this is your expertise Paul and I would appreciate any advise on
how to do the proper calculations when you have time. No rush.
Thanks
Doug in Japan.
The intake scoop in theory sees dynamic pressure but that is not what you
will measure in the intake duct as the engine is sucking air through the duct.
What you see at the compressor face will also be a function of how smooth and
how large the scoop area is which you just pointed out.
When the flow into the duct is equal to the forward velocity of the airplane
the drag coef. of a boundary layer bypass type scoop is about .2 based on the frontal
area of the scoop and according to Horner's data. It goes up to .4 if there is little
or no air going in. Surprisingly it also rises ever so slightly when more air is
being pumped in than would be indicated by the forward velocity of the aircraft.
If the scoop is in the slip stream of the prop there will be a mild supercharging
effect. I would add 20% to the air stream velocity at take off and low angle
climb out. 30% for high angle climb out. Much more for static of course. The static
slipstream speed could be as much as 100 MPH. Pito tube dynamic pressure data in that
area would help enormously during development. This type of scoop should help greatly
on getting on the step and out of the water.
BTW dynamic pressure is .0026 times MPH squared. The .0026 is air density
related to MPH.
Here are some pictures of Dave Leonards scoop and others so the new people know
what we are discussing.
--
Paul Lamar ...No rotor no motor.
Paul,
Thanks for your quick reply. The P-38 rad. scoops are sweet. That is
what I modeled my rear heat exchanger inlets on although I stuck the
ears out farther out of the boundary layer. I am amazed at the low drag
co-effecients. 0.2 is quite good. I suspect a rectanglar scoop on top
(like what you see feeding dragsters is way over sized though. Yes I am
counting on putting the prop to work as well.
Just for fun I calculated velocity at inch per mins. 161mph equals
4.293 km/per min.
4293 meters times 39.37 equals 16,9015 inches per min. multipled by the
area of the front face of the scoop and this is the what the compressor
sees. Staying with the aforementioned size for a moment that is
169,015 times 40 or 6,760,616 cubic inches that theortically could flow
into the scoop if the compressor could swallow that much.
One cubic foot of air at STP weighs 0.0807 lbs. Half of that at 18K` or
04035 lbs,
Divide by 1728 (cubic inches in foot) and multiple by total cubic inches
at the front face and I get
157.5 lbs per min. available. In the formula you gave me is the result
at 161 mph is a dynamic pressure (Pd) of 67.39 This seems low as
converted to psi it would be miniscule. Now from here on out I am not
clear what happens. Obviously I don`t need this mass amount of air and
it is just piling up at the inlet and increasing the drag coeff. , but
if there is a 5-1 reduction then the pressure at the compressor inlet
increases perhaps five fold and the flow will reduce correspondingly, I
think maybe:)
Doug in Japan
The air required by a NA 2.6 L rotary at 6000 RPM is 220 cubic feet per minute
or 3.66 cubic feet per second. Lets say you double the HP with a supercharger
so you will need twice the volume of air or 7.32 cubic feet per second.
161 MPH, including the 20% prop slip stream, is 236 feet per second. So the
opening is A X 236 = 7.32... A = 7.32/231 = .032 square foot FA. or 4.56 square inches.
Below that speed you will be sucking air. Above that speed you will start
to compress air. The drag on a .032 square foot scoop at a dynamic pressure
of 161 x 161 X .0026 = 67.4 pounds per square foot so drag is .2 X .032 FA X 67.4 #/sq.ft.
or only .43 pounds of drag. The C sub d here is the .2 number.
HP = .43# X 236 FPS or 101 pound foot per second or only .18 HP.
( One HP is 550 pound foot per second.)
Obviously you want a larger opening. Also obviously a larger opening was chosen
for the P38 and the Me 109.
At 80 MPH you will need twice the opening area. At 40 MPH you will
need four times the area or 18.24 square inches. Or about 4.3 by 4.3 inches.
What a sea plane needs is a variable intake air opening to get maximum supercharge
effect from the prop slip stream at take off without adding a lot of drag at cruise speed.
Paul Lamar ...No rotor no motor.
Paul,
Thanks. You given me some tools where I perhaps come closer to ball park
this.
Several points.
1. I am only looking for a 15% gain in horsepower not 100%. 300 hp with
boosted P-ports is a walk in the park.
2. Air density comes into these equations right? At 18K the density is
one half so that means the volume has to double or the or pressure of
the air has to double to obtain the same mass of oxygen molecules. As I
understand it a large inlet will provide a pressure increase.
3. Next, the chart you posted shows the data at a vol. efficiency of 80%
as it was illustrating a piston engine. When you extrapolated new
curves for a 2600 displacement did you estimate the VE would be over
100%?
4. Finally .Adjustable air inlet. This is great idea. Yep, an adjustable
opening scoop, perhaps with an flapper or butterfly valve installed in
the inlet. Amphibs engines are usually over sized just so they can get
break free of the water suction in a reasonable distance. Like all
airplane it suffers from performance creep as people want more and less
of this or that. But most times we amphib. owners are carrying a weight
penalty to support that extra 50 hp.
An interesting story about the first prototype Seawind which was
running down the water course trying to get airborne without success
with I believe a 200 Hp Lycoming. The pilot gave up and closed the
throttle abruptly and the plane leaped into the air quite high and came
down and continued its journey in submarine mode. The pilot escaped
unhurt and was probably glad his pants got wet if you catch my drift:)
I think the way the stabilizer and prop thrust angle was set-up was
causing a down force on the nose on the take-off attempt but the wing
was already producing lift. This problem was sorted out and now the
Seawind despite its high thrust line behaves like a common front nose
thruster. That is, give it power and the nose comes up.
Off a hard runway 250Hp works fine.
Doug in Japan
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