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  #16  
Old 08-29-2004, 08:08 PM
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Ok, not my spreadsheet, just another item i found bumming around.

I think i have a solution for a set of numbers, it has sorta been verified twice, now if we can work the forumuls and arrive at about what i think i know, whalla
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  #17  
Old 08-29-2004, 08:42 PM
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This is what I think I know

360 cid
25,000 feet
2800 RPM
15.5 psi map
ambient -30
intercooler efficiency 80%
turbo efficiency 75@
engine efficiency 80%
ambient pressure @ 25000 5.7

Results for turbo map lookup

44lbs air min
2.85 pressure ratio

I've sorta verified this twice with professionals, but would like to be able to calculate it and once we can, then we can solve slades problem for flying TOMORROW
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  #18  
Old 08-29-2004, 10:10 PM
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Default crumber nunching..

One more bit of data to throw in the number crunching enigma....

Rotary engines 13B and Renesis deliver 5 1/2 power strokes of 654 cc per 3 revolutions of the eccentric shaft. By the tme you take it to 6000 rpm you get one charge of 1308 cc per eshaft rotation.
A 160 cubic inch reciprocating engine with 4 cylinders will deliver 2 power strokes of about 80 cubic inches (1.31 Litres per revolution).
(information scavenged from metric conversion chart. )
2616 cc is close to 160 cu in for 2 revs of each

The 360 Cu in Lycoming displaces about 180 cu in per revolution (2.94 Litres per rev). The 470 continental (my favorite) displaces about 235 cu in per rev (3.85 Litres per rev) hmmm. Wonder why the HP figures are so high on the rotary?
(hint: The 4 stroke engine needs 2 revolutions to make it's full volume , The Rotary has 2 rotors with a total area of 1308 cc for each power stroke. Since the total displacement is 1308 cc, the area is divided between the 2 rotors and becomes 654cc.
I hope this helps.
Brain wedgie for David (sorry old chap)


I had to edit this a couple of times because my Mazda engineer buddy made me. Accuracy less than 10% or something. Each chamber is 654cc NOT 650cc. One rotor delivers 3 654cc doses per stroke the other just 60 degrees behind in the actual power stroke, misses the complete revolution and gets only 2 whole power doses in during one revolution. The other rotor is just 60 degrees behind and because it overlaps every rotation, you can figure 1308 cc every turn of the crank (unless you turn it just once).
It's all good, it ends up being about 80 cu in per rotation of the crankshaft. That's 100 cu in less than the 360 lyc.

Last edited by Clutch Cargo : 09-01-2004 at 09:39 AM. Reason: for ALL engines...
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  #19  
Old 08-29-2004, 10:58 PM
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Here's part of a post from Ed Anderson on the flyrotary list. Maybe this is a little closer to home:

So a two rotor at 6000 rpm displaces 277 CFM (assuming 100%Ve) - no more, no less regardless of boost. Now the density of that air will change depending on the amount of boost.

Wait, I know someone will say "Ed - at 1.5 pressure ratio (7.5 psi boost at sea level) you actually have approx 1.5*277 = 415 CFM entering the engine". Not true, my friend - you only have 277 CFM entering the engine (at 6000rpm).

However, You DO have 415 CFM entering the inlet to the compessor at 0.076 lbm/Cubic foot. But when it leaves the compressor to feed the engine the flow is reduced to 277 CFM(more or less I am not taking losses into account here) at 6000 rpm, but at a higher air density (approx 1.5* 076lbm/Cubic foot) that is how you get more air into the engine. The engine is a positive displacement pump and will pump essentially the same VOLUME each revolution whether at idle or 6000 rpm. What makes the difference is the density/pressure of the air that it is displacing.

Quote:
slades problem for flying TOMORROW
Maybe not tomorrow, but I might be ready the next day. I'd like to have it flying for sure by next weekend - in case I have to run & hide the plane from the hurricane.
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  #20  
Old 08-29-2004, 11:06 PM
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we should be able to "fake" the program for the program. we know the hp at the rpm 160 at 6000??, we can then put the numbers in at sea level to get the 160 at 0 sea level with no boost and move on from there.

When i put the numbers in for my continental, i get the rated hp +- 2 or 3 percent
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  #21  
Old 08-29-2004, 11:22 PM
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Quote:
we know the hp at the rpm 160 at 6000??,
A little more. I'm told to expect 165HP @ 5200 rpm with no boost. At 6,300 RPM (max prop target), and 8 PSI boost I should have 262 HP.
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  #22  
Old 08-29-2004, 11:39 PM
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Remember that the engine is just an air pump. It will consume air, whose volume is dependant upon RPM, engine displacement and volumetric efficiency. The mass of the air will be dependant on altitude and temperature.

The turbo can compress air and make plenty of it, but the engine will only used a fixed amount.
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  #23  
Old 08-30-2004, 12:18 AM
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Default 2 + 2 = 5

Something doesn't add up here:
277 CFM ???

Say it aint so..
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  #24  
Old 08-30-2004, 02:15 AM
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[quote=Clutch Cargo]One more bit of data to throw in the number crunching enigma....

Rotary engines 13B and Renesis deliver 5 power strokes of 654 cc per revolution..(snip)..so the data for the rotary may be inaccurate when based on a 160 cu in 4 cyl engine.

(snip)

(hint: The 4 stroke engine needs 2 revolutions to make it's full volume , (BIG SNIP)
199.54 per rev 4000 rpm = 798160 cu in per min = 461.9 CFM
I hope this helps.
Brain wedgie for David (sorry old chap)
QUOTE]

Ok.. Thanks everyone for the support.. and Clutch.. I follow you..however, with a "typical" 4 stroke 4 piston engine, you are getting four power stroke for every 2 crankshaft revolutions (instead of 8)... the formulas I have found calculate based on that. Now.. for the Rotary, I follow you on the "5 power strokes" per RPM, but I think we need to clarify something. The Rotors are revolving at one/third the speed of the E-shaft. Remember in Bruce's video, you have to rotate the Eshaft THREE times to get all SIX compression strokes. So.. the 5 power strokes per RPM is ROTOR RPM.. and in actuality is 5 compression strokes per 3 rpm, so a 5/3 ratio..based on the numbers you are tossing at us. Just gives me some more stuff to try and cram into my already full brain.

Dave
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  #25  
Old 08-30-2004, 10:53 AM
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Default Can You Say.. Eureka?

Quote:
Originally Posted by John Slade
So a two rotor at 6000 rpm displaces 277 CFM (assuming 100%Ve) - no more, no less regardless of boost.
Ok... that means that each RPM of the Eshaft moves 80 Cubic Inches. It takes 3 rpms of the eshaft to give all 6 combustion surfaces their share of the load. this gives the rotary 240 CID in those 3 Eshaft rotations to equal what takes a recip 240 CID to do in 2..... incidentally 240 x 2/3=160 so I can see where that value came from, however erroneous it now appears to be.

Plugging in the values for a standard day 15K feet with MAP of 14.9 PSIA I get 44 CFM on the spreadsheet (corrected for PR) that Dust uses.. and 37 CFM on the Javapage I use... fuel flows in the low 30 gpms..and HP of 300 on mine.. 400 on dusts.. things still dont match up.. but I can see where the CFM is coming from now. ALSO.. one thing.. that 2800 rpm dust is talkin about.. HAS to be PROP rpm.. and the engine is running at 2.85 times higher.. at 7980 rpm to generate that kinda flow and power with its displacement. Everything else matches up within 10% of each other on the major numbers

I wouldnt mind bouncing this off a Mazda engineer myself if the opportunity presented itself...

Dave
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  #26  
Old 08-30-2004, 01:18 PM
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Note, my numbers are for a 6 cylinder continental, but the turbos should all work "the same"
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  #27  
Old 08-30-2004, 02:33 PM
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Default ooops

Rotor rpm not "crank" shaft rpm..

Spinning rotors in my head cause data to become faulty.

Made ya think about it though huh?

Trust but verify.

Dust but clarify.

What I meant to say about the 4-banger was that if it is a 160 cubic inch engine (or any 4 stroke standard piston engine, including Continentals) then the advertised cubic inch displacement is the total amount of combustion chamber area taking into consideration all 4 (or 6,8,12, 3,2 or 1) cylinders' area available from the bottom of the compression stroke to the top. It needs 2 full rotations (720 degrees) of the crankshaft to realize the full intake volume advertised by the manufacturer or modifier of the engine. So a typical 160 cubic inch engine will suck in 80 cubic inches per revolution of the crank. At 2800 rpm then, the engine will suck in 224000 cubic inches or 129.6 cubic feet per minute of "air". (my reference to power stroke was actually incorrect as it is the inflow of gas that can be measured more accurately. But I was referring to the individual piston's stroke when not exhausting burned gasses)

Now the Rotary (glasses on) sucks in one full combustion chamber's worth at 654cc (40 cubic inches) during one turn of the "crank" (E) shaft, which is one rotor only and it takes another 1/2 of a turn to get the other rotor's full chamber action involved. At the end of the first full turn of the Eshaft, the second rotor is already almost 1/2 full So it's 1 1/2 turns for 1308 cc or 2616 cc (160 cu in) per 3 crank revolutions. That is 5,232,000 cc per minute at 6000 E shaft RPM or 184.76cubic feet per minute. The rotary needs 6 1/2 turns of the Eccentric shaft to pull in 6 chambers' worth of gas.

That's with my glasses on and empty.

Last edited by Clutch Cargo : 08-30-2004 at 03:19 PM.
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  #28  
Old 08-30-2004, 06:25 PM
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The two rotors work together on every revolution. 3 revolutions of the eshaft create 3 chambers of air for one rotor. 2 rotors working together create 6 chambers of air for 3 eshaft rotations.

Someone could argue that it takes 3 1/2 eshaft rotations to fill six intake charges, but they would be wrong.

It's true that if you start at the exact start of intake for one rotor, the second rotor will need to catch up with a 1/2 eshaft rotation, but this ignores the fact that the 2nd rotor was 1/2 through the first cycle on the start for the 1st rotor.

A test you could work through would be 1 rotor halfway through intake, 1/4 through intake, etc... the closer you get to infinitely small numbers, the closer you get to 3 even rotations. Or you could do the calculus if you are just bored.

For practical purposes, just use 3 eshaft rotations for 6 intake charges.
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  #29  
Old 08-30-2004, 07:05 PM
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is this theoretical cfm for the rotary, then you have to factor in the intake manifold engine efficiency
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  #30  
Old 08-30-2004, 07:20 PM
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Default So.. where does that bring us?

Ok.. so we think we know then that each rev gives us 80 CID, and 3 shaft revs gives us 240 CID.. where a 4 cycle 240 cid would do it in two...At some point I'm gonna go back and re-calc my spreadsheet with the 240 CID numbers and see where they lie...


Part of me wants to say "screw it" and look at someone elses real-time data from a dyno/flowbench (auto if I have to) and work those numbers.. rather than make half assed stabs at developing my own. The only practical/tangible benefit I've derived from this thread and the data in it (so far) is that much of the research was done online from work during slow times.. so I can't say that I've been focusing on theory instead of getting out there and rebuilding the darn engines.

Just got back from walking around an industrial supply store.. seriously have been considering blasting the housings to clean em up.. inside and out.. and I dont feel like paying $250-300 for a portable blasting booth.. so.. might see how good my constructions skills are and see if I can build one for significantly less... I can see now how "getting ready" to do something is sometimes just as or more involved than the actual task....

Got back in the air to maintain currency.. 1.5 hrs in a 172 that has seen many many hours.. trusty.. dependable.. and slow.. I cant wait to do some zooming.. but at the same time I was glad I was in a 172.. cause I was a lil rusty.. and fast moving planes can get out of kilter in a hurry if you arent up to the task.. I need to fly more... especially if I'm gonna be a "test pilot" for this project.

Dave
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