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  #31  
Old 08-31-2004, 01:17 AM
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Clutch Cargo Clutch Cargo is offline
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Default Little things count..

....when multiplied by 2000!

I agree with you Mssr MarbleTurtle, it seems like such a small thing, but even 1/2 of a chamberfull 327 cc every 3 revolutions times 2000 = about 23 CFM. That's a pretty hefty pump, better flow than my 5 HP shop compressor. Good thing that isn't the case!
(here is where I am adding my edit) Fortunately MT explained it so even an idiot could understand, and finally, I did.
I have a 13B on my bench and when I rotate it, that last bit of suction on rotor 2 ends about 180 degrees after TDC on rotor 1.
Personally I would like to eliminate the errors on paper first and then go for the anomolies.
For the purpose of mathematical calculations we should have a starting point and an ending point, right?

Since we are talking about the intake volume of air, the place to begin would be the position when rotor one begins to draw air into the engine. The fact that at any time there is air in the part of the chamber in a previous intake to combustion chamber on the other rotor is not a factor in figuring volume of air taken into the engine during a given amount of complete Eshaft rotations. Where it starts and where it finishes taking in the given amount (in the case mentioned, 3 chambers per rotor or 6 1/2 rotations of the eshaft) will give a more correct volumic flow rate number. As mentioned above, 23 CFM is a hefty amount for a " fudge " factor. I for one would not go for an error number that high when figuring boost at numbers less than the "fudge" factor.
So what I learned from MT was that my sample of the rotary "cycle" did not include the "overlap" from rotor B, which is only out of sync with rotor A 60 degrees. I was adding that 1/2 starting charge every 3 cycles and not taking it off at the end.
Since I am not a bank or loan company, I can give credit where credit is due: MT, you did it right!
6000 rotations per minute = 6000 charges@ 1308cc = 277 cfm (or pretty close)

There is that one 1/2 a charge at the end though..............

Some people just strap a turbo on and hope for the best...


Last edited by Clutch Cargo : 09-01-2004 at 09:10 AM. Reason: The stuff I wrote made me look stupid..
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  #32  
Old 08-31-2004, 10:25 AM
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The next fudge factor is engine effeciency. the combustion chanber will not get 100 % filled each time.
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  #33  
Old 08-31-2004, 01:22 PM
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Clutch Cargo Clutch Cargo is offline
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Default there you go again...

Getting technical...

Actually, the chamber DOES get filled each time there is an opening to the outside atmosphere. It is not necessarily filled with a perfect density of air for combustion, but the amount that goes in is more predictable and the "fudge factor" is much smaller. The partial vacuum created by the turning rotor or even a descending piston, is offset by 2116 lbs per square foot (over a ton) of atmoshpheric pressure (at sea level) rushing in to fill the void. Obstruction to the flow may hamper the FLOW, but when the valve closes, the chamber will be full, but not with EXACTLY the same DENSITY of air outside. If there is a supercharger or turbo charger, the density of air will be greater in that finite space, being compressed as it is, and a fuel mixture that has more density will create more power.
But for the purposes of this discussion, the amount of air going into the engine is what is needed for turbo calculations. The density of the air is going to be changing constantly. But once a formula is found and tested, a flow meter (MAF sensor) can be used in real time for fuel / air data.
Getting the initial turbo data concerning the displacement of air in the engine is critical to correct turbo selection.
After that the efficiency of the intake system with regard to air intake velocity, mass, density and flow characteristics are all a matter of post turbo installation tuning.
It starts with the engine displacement. That's why it has to be accurate.

There's an opinion for you.
I could be wrong though.
I was the other day

But I am still alive to enjoy it!
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  #34  
Old 08-31-2004, 05:36 PM
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The combustion chamber won't get filled with 100% fresh air each time.

At idle, exhaust gasses can be more than 70% of the mixture and at high rpm can get down to 10% (90% fresh air) because of the swept volume (the small volume at TDC/BDC) that you can't push push out.

Post script:

Sorry, I'm refering to a rotary with these numbers and swept volume is the total displacement.
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Last edited by mplafleur : 08-31-2004 at 10:09 PM.
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  #35  
Old 08-31-2004, 06:36 PM
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All engines operate at less than 100% intake efficiency. That's just something you work with.

But CC, you are way off on the 6 1/2 rotations! I think you're confusing the half-stoping and half-starting points of the second rotor.

Do this with your bench mounted motor... rotate the eshaft to a position where Rotor A starts to open for an intake charge. Mark this as a starting point. Observe the position of rotor B. It will be somewhere in the middle of accepting/rejecting an intake charge.

Rotate the eshaft 360 degrees... 1 eshaft rotation. Rotor A will now be 1/3 the way from where it started, but since it is a triangle (I know... elipsicle triangle whatever), but it will be at the same point for accepting the new charge on the second lobe of the triangle. Rotor B will be halfway again through the intake/reject process.

To summarize. 1 eshaft rotation, Rotor A accepts 1 full charge, Rotor B gets halfway through accepting 1 charge. (We'll throw out the 1/2 intake at the start on rotor B for now instead of adding 1/2 start and 1/2 end as 1 whole charge... as if this was the start of the engine and no fuel was in rotor B at the halfway point on start.)

Now do 2 eshaft rotations. Rotor A, 2 intake charges. Rotor B, 1 1/2 intake charges. Total charges for 2 eshaft rotations 3 1/2.

3 eshaft rotations. A - 3 charges. B - 2 1/2 charges. Total 5 1/2 charges.

Now the rotor A has come a full 360 degree rotation... the rotors rotate at 1/3 the eshaft rotation. Actually rotor B has come full rotation as well, again in the middle of accepting/rejecting a charge. I think this is where you are seeing the confustion. If you were to stop the engine now you would have 5 1/2 intake charges for 3 eshaft rotations. But the engine doesn't stop here and then start over... it continues the cycle. Keep rotating the eshaft and see what I mean.

4 eshaft rotations. A - 4 charges. B - 3 1/2 charges. Total 7 1/2 charges.

5 eshaft rotations. A - 5 charges. B - 4 1/2 charges. Total 9 1/2 charges.

10 eshaft rotations. A - 10 charges. B -9 1/2 charges. Total 19 1/2 charges.

2000 eshaft rotations. A - 2000 charges. B - 1999 1/2 charges. Total - 3999 1/2 charges.

Do you follow? This is not a fudge factor... this is right. You get 2 intake charges for every eshaft rotation, or 6 intake charges for every rotor rotation... the 2 rotors rotate together... 3 per rotor for every rotor rotation.

If you don't use 2 intake charges per RPM... your calculations are going to be WAAAYYYY off, and this whole exercize will be for naught.
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Last edited by MarbleTurtle : 08-31-2004 at 06:48 PM.
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  #36  
Old 08-31-2004, 08:19 PM
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Quote:
Originally Posted by MarbleTurtle
All engines operate at less than 100% intake efficiency. That's just something you work with.
Ok, that is true for a wide RPM range engine, one that is designed for performance over many RPM ranges, but, it is possible to design an engine to give up to 110% efficiency or more at a designed for RPM. Which, we in the aviation community are using, 95% of my flight time will be at 2400 rpm
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  #37  
Old 08-31-2004, 08:27 PM
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Default Arghhh!

It just dawned on me after all the crumber nunching over the past few days that I remembered... Dust isnt using a rotary... and he even TOLD me as much in a previous post..

Attention to detail..
gotta love it..

Dave
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  #38  
Old 08-31-2004, 08:34 PM
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Sizing a turbo for all engines - see the name of the post, i diverted the atention away from just you silly rotary users or wanna be users to give the whole wide world of aviation a resource, because all the dam info on the net is just nor quite right for aviation, wether it be piton or ratary
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  #39  
Old 08-31-2004, 10:08 PM
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I think by efficiency, we are talking volume efficiency. I don't think we will ever see 100% efficiency because part of the volume will always be remaining charge from the previous cycle.

Obviously with the turbo we are squishing in more air (density).

But wait... what were we trying to find out with this post? Was it how much MAP would blow the Turbo at any given altitude?

Okay... get out the rusty slide rule... 277 cubic feet per minute at 6000 rpm... wait.

Isn't what we are trying to do is not overspeed the Turbo? The exhaust drives the turbine. Without resistance at the impeller, it would continue to spool up to destruction.

The resistance of the impeller is determined by the pressure build up after the impeller and before the intake opening to the cyclind... uh... rotor. Too much resistance, the turbo slows down (unless the exhaust power of the turbine is strong enough to cause cavitation of the impeller... the left side of the turbo map!)

Okay... thinking outloud. Uh... pressure is basically the mass of air in a given volume. Pressure ratio is pressure of the post impeller to pre-impeller.

Hmmm... we expect that since the air is thinner at altitude, the impeller would spin faster to maintain a given pressure post-impeller. But we are not working with set pressures on the turbo map... we are working with ratios.

So if we say our optimum MAP pressure is at 2/1 pressure ratio at sea level... then we go up to some altitude that reduces outside air pressure by... let's just say 25%, (worry about temps later), then the new pressure ratio at altitude to maintain the same MAP pressure (I know, redundant, but we are dealing with 2 maps!), whatever MAP pressure we decided was optimum at sea level, then the pressure ratio goes to 2.667 to one.

2.6667 to 1 at (however many feet gives a 25% drop outside air pressure) altitude generates the same MAP pressure as 2.0 to 1 at sea level. I think 2.667 to 1 is still in the safety zone at 6000 engine RPM for turbo RPM... this will give us X number of cubic feet per minute, which we should try to figure out how much mass of air at Y altitude...

And I don't know if this is right or making any sense so I'm going to sleep before I come up with any more bright ideas that don't look so good in the morning.
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  #40  
Old 08-31-2004, 10:23 PM
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Okay... before I go to sleep, if someone will just give me the pressure at sea level, and at the altitude they want calculated, the turbo map of the turbo they want to use, the engine RPM, CFM at 100% efficiency of the engine at the RPM of choice... um... I think I can figure this out. Maybe...
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  #41  
Old 08-31-2004, 10:46 PM
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Use my example. I'll be first to try it out and give feedback

the pressure at sea level: 30 inches
and at the altitude they want calculated: decreases 2 inches / 1000 ft ???
the turbo map of the turbo they want to use: T04-V2 MAP above
the engine RPM: 4500, 5000, 5500, 6000, 6300
CFM at 100% efficiency of the engine at the RPM of choice... 277@6000

Best advise I've had so far is to keep it at or below 46 MAP at 10,000'

Go MT.
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  #42  
Old 09-01-2004, 12:07 AM
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I modified one of the spreadsheets and this is what I get:

Engine Data
engine rpm 6000 revs/min
displacement 160.0 cu inch
volumetric efficiency 90%
number of turbos 1
compressor efficiency 75%

Ambient Conditions
Altitude at ground 0 feet
local ground baro pressure 29.92 in Hg
ambient temp 90 deg F

Conditions at Plane
Altitude 6000 feet
barometric pressure 14.37 in Hg
Outside air temperature 68.6 deg F

Conditions at Compressor Inlet
Vacuum drawn at inlet 2.0 in Hg
Inlet Pressure 13.71 psia
Inlet density 0.067 lb/ft3

Conditions at Compressor Outlet
outlet pres 12.5 psig
outlet temp 249.7 deg F
P2/P1 1.98
outlet density 0.104 lb/ft3

Conditions at Intercooler Outlet
manifold pressure (desired) 10.0 psig
manifold temp (assume intercooler) 108.6 deg F
manifold density 0.117 lb/ft3
IC pressure drop 2.5 psi

Results, mass and volume flows
compressor air flow 32.6 lb/min, ideal
compressor air flow 29.4 lb/min, actual
compressor air flow 222.0 gm/sec, actual
total engine air flow 222.0 gm/sec, actual

compressor air flow 435.4 ACFM, actual inlet
compressor air flow 283.4 ACFM, actual outlet


I assumed 2.5 psi intercooler and pipping losses. Turbo gives 10 psi boost. Given the barometric pressure and temperature at ground, I calculate temperature and pressure at altitude.
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  #43  
Old 09-01-2004, 09:21 AM
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Clutch Cargo Clutch Cargo is offline
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Default 277 CFM it is...

OK ok you were right MT..
I didn't take the time to get a good sample or segment of the cycle in order to extrapolate the volume calculations to 6000 rpm. I finally arrived at the correct conclusion by ignoring the total per rotation data and took each rotor's volume at one time, then I added the second rotor' data on top of the first ones only, 1/2 of a rev behind. Then the light came on.

At least we have expounded the rotor theory and principle here a little better. Since, so many are planning to use rotory engines, it's all good!

sitting in the corner putting new batteries in my calculator..
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  #44  
Old 09-01-2004, 12:44 PM
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Default Some baseline data.

Quote:
Originally Posted by mplafleur
Given the barometric pressure and temperature at ground, I calculate temperature and pressure at altitude.
There is a table at www.cavalrypilot.com/fm1-301/tab2-1.gif that has standard temp/pressure values calculated much more accurately than the "ballpark" variety.. There is data up to 50,000 ft.

Also, which spreadsheet, and what modifications, so the rest of us can see what you did?

Dave
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  #45  
Old 09-01-2004, 05:15 PM
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OK, i thinks i gots it.

The turbo maps are in corrected LBS of air per minuite, so if you want to compare them for aviation purposes and NOT correct your lbs per minuite to sea level, you must convert the lbs per minuit at sea level to cfm per minuite at sea level and then you have a cfm that will work at any altitude

This may be right, we can check it
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