RX8Club.com - Rotary Math

RX8Club.com (https://www.rx8club.com/)

- Series I Tech Garage (https://www.rx8club.com/series-i-tech-garage-22/)

- - Rotary Math (https://www.rx8club.com/series-i-tech-garage-22/rotary-math-106294/)

Bwhahahahahahahahahah

This is some amazing stuff! thanks man :ylsuper:

Quote:

Originally Posted by MazdaManiac

ROTARY MASS FLOW RATE[/b]

MFR = (2.703) x (Pa) x (VFR) ÷ (Ta + 460)
MFR = (2.703) x (14.7) x (369.75) ÷ (85° + 460)
MFR = (14691.683) ÷ (545)
MFR = 26.96 POUNDS per MINUTE (lb/m)

this has been asked before but what does the 2.703 refer to?

MFR = air density lb/cf x CFM

I assume the 2.7 has something to do with density but a clarification on the number since it is some type of conversion factor would be nice.

just looking at the units used, your formula dosn't seem right

MFR lbs/min = (2.703) x (14.7) psi x (369.75) cfm ÷ (85° + 460) deg
lbs/min = psi x cfm / deg

found it. a variation of the ideal gas law adapted for flow.

omg, that's f#^cking hilarious Chicken! :rofl:

MM: truly amazing thread. Way over my head but a great read

Wow, I am amazed! Fan-freakin-tastic! Now I just need to wrap my head around all of this knowledge.. Thanks MM.:banghead:

alright I just have a simple question, I think the answer is "yes"...

In the post where you say: "Working backwards to compute maximum safe boost on the stock fuel system"

It says 5.73psi boost would potentially make 333 or 360 hp.

Does this mean I can simply put a SC or turbo that holds, say, 5psi to redline and safely run stock fuel system making 300+ HP?

I know it also depends on the amount of air being pushed at a certain PSI... I just want to get my brain wrapped around this!

Kind of... 5.73 PSI on an 80 CI motor running at 9000 RPM with perfect Air Temp COULD make that. But as the CFM of air increases so does it's temperature, so density goes down.

Crank HP = MFR*60/BSFC*1728 -- something like that, from memory, could be wrong.
MFR = The MASS of air in lb/min

Quote:

Originally Posted by MazdaManiac (Post 1762434)

I don't know if I'd call that a "myth". I think I'd just call it a completely incoherent thought based on some weird mixture of meth and Jagermeister.

How did you know?
It's Sunday night, I am hitting the Jager.....
No NitroMethane though, waiting for Gainesville (Gatornationals) this spring.
/Shame, It might run in an RX8...... :Drooling_

Forgive me if I missed it but what is the formula for converting CFM to lbs/min?

Look at the quote in post #54.

Ezine site with info on Rotary Volumetric Efficiency explained

Nice write-up.
Fortunately, the legwork was already done for us by our friends down-under.
We have MAF plots that show the actual Ve.

Quote:

Originally Posted by MazdaManiac (Post 1672699)

I wanted to post a thread to which I could reference people for the required math for computing the performance of the Renesis, N/A or FI.
This is just going to be a list of required formulas with the required engine-specific and natural constants already in place.

I will be adding and updating as I compile the info from my video. The constants expressed in the opening are based on published numbers and observed performance data.

For additional reference, here are two spreadsheets:

ROTARY MATH
a practical application of the formulas in this post

INJECTOR MATH
an injector calculation matrix with tables for three staged injectors

These formulas assume the following

Ambient Air Temp (Ta) = 85°F (29.4°C)
Ambient Air Pressure (Pa) = 14.7 PSI (29.98 in/Hg)
Intake Temperature = Ambient
Intercooler Efficiency = 70% ¹
Turbo Compressor Efficiency = 77% ²

{1 Cubic Inch (cid) = 16.39 Cubic Centimeters (cc)}

Renesis Displacement = 1308 cc (80 cid)
Renesis Average Volumetric Efficiency (Ve) = 87%
Renesis Operational Redline = 9200 RPM
Renesis Brake Specific Fuel Consumption (BSFC) = .60 lb/hp/hr (POUNDS per HORSEPOWER per HOUR)
Renesis Total Fuel Injected Volume = 2100 cc (3.33 lb/m)

[[------ NATURALLY ASPIRATED CALCULATIONS ------]]

ROTARY VOLUMETRIC FLOW RATE

VFR = ((DISPLACEMENT) x (RPM) ÷ 1728) x (VOLUMETRIC EFFICIENCY)
VFR = ((80cid) x (9200rpm) ÷ 1728) x (87%)
VFR = (425cfm) x (.87)
VFR = 369.75 cfm

ROTARY MASS FLOW RATE

MFR = (2.703) x (Pa) x (VFR) ÷ (Ta + 460)
MFR = (2.703) x (14.7) x (369.75) ÷ (85° + 460)
MFR = (14691.683) ÷ (545)
MFR = 26.96 POUNDS per MINUTE (lb/m)

POTENTIAL CRANKSHAFT HORSEPOWER ESTIMATE

HPo = (MFR) x (60) ÷ (AFR) x (BSFC)
HPo = (26.96) x (60) ÷ (12) x (.60)
HPo = (1620) ÷ (7.2)
HPo = 225 Hp

ALTERNATE POTENTIAL CRANKSHAFT HORSEPOWER ESTIMATE ³

HPo = (MFR) x (9)
HPo = (27) x (9)
HPo = 243 Hp

REQUIRED FUEL INJECTOR FLOW RATE

IFR = (MFR) ÷ (AFR)
IFR = (27 lb/m) ÷ (12)
IFR = 2.25 lb/m

For cc's, divide by 0.001586:

2.25 lb/m = 1419 cc

[[------ FORCED INDUCTION CALCULATIONS ------]]

Working backwards to compute maximum safe boost on the stock fuel system:

SUPPORTED MASS FLOW RATE

MFRs = (IFR) x (AFR)
MFRs = (3.33) x (12)
MFRs = 39.96 lb/m

SUPPORTED VOLUMETRIC FLOW RATE

VFRs = (85° + 460) x (39.96 lb/m) ÷ (2.703) x (14.7)
VFRs = 548 cfm

DENSITY RATIO

Dr = (VFRs) ÷ (VFR)
Dr = (548) ÷ (369.75)
Dr = 1.48

PRESSURE RATIO

Dr = (Pr) ÷ ((Tin° + 460) ÷ (Tout° + 460))
1.48 = (Pr) ÷ ((85° + 460) ÷ (120° + 460))
1.48 = (Pr) ÷ ((545) ÷ (580))
1.48 = (Pr) ÷ (.94)
Pr = 1.48 x .94
Pr = 1.39

Boost = (Pa x Pr) - Pa
Boost = (14.7 x 1.39) - 14.7
Boost = 5.73 PSI 

POTENTIAL CRANKSHAFT HORSEPOWER ESTIMATE USING SUPPORTED MFR

HPo = (MFR) x (60) ÷ (AFR) x (BSFC)
HPo = (39.96) x (60) ÷ (12) x (.60)
HPo = (2397.6) ÷ (7.2)
HPo = 333 Hp

ALTERNATE POTENTIAL CRANKSHAFT HORSEPOWER ESTIMATE USING SUPPORTED MFR ³

HPo = (MFR) x (9)
HPo = (39.96) x (9)
HPo = 360 Hp

¹ Ieff = (Tout - Tin) ÷ (Tout - Ta)
² Optimal
³ Hotrod geeks use the formula "ten times mass flow equals horsepower" for most piston motors with a fair amount of sucess. Since the rotary is only 90% efficient of the BSFC of a typical piston motor, we use 9 instead of 10 to good effect.
 This assumes (incorrectly) that Ve remains at maximum at redline. This figure represents a best/worst number for boost at redline.

IN SHORT, THE ROTARY IS ONE POWERFUL MOTHER FUCKER!!!!!!!!

Sorry for the somewhat off-topic.
I have the following data on a S/C which I do not understand.
Could someone explain to me how much HP I could make with this??

Theoretical Discharge Rate cc/rev 1460
Maximum RPM (Continuous) ｒｐｍ 10.000
Maximum RPM (Instantaneous) ｒｐｍ 13.000
Maximum Pressure Ratio (Continuous) 1,8
Maximun Pressure Ratio (Instantaneous) 2
Theoretical Discharge Rate at Maximm RPM （Pressure ratio 1.8） m3 /h 675

By converting m3 to cfm, I get 397.29 cfm which is as much as the normal intake flow?
Is it the .8 bar (11.6psi) that makes the difference?
I think I am missing some basic understanding, can someone share some insight?

Thanks in advance....

And part of this nutritious breakfast!

...I am not sure I understand your comments.:confused:

Hmmm...perhaps if I hadn't have slept with my high school math teacher I might have actually learned something and been able to figure some of this out :icon_no2:

All the algebra in the first post kinda makes my head spin. copied this from a recently closed turbo dream thread

"A 2.0L 4-stroke piston motor flows a dramatically different level of air than a 1.3 L rotary."

I pretty much know the ins and outs of how a turbo system works. Unfortunately I don't know much about the rotary engine beyond what I see in the animated videos I see on youtube. Does that statement get explained in the math on the OP?

I certainly dont have any turbo pipe dreams of my own. Im just interested to know what that actually means in layman's terms.

Great XLS!
Could you upgrade it for other EMS with common pulse staging and staging bar by pressure? Like haltech.

Hey funk, the guy that built it has been banned for a while so I don't think it'll get updated.

Quote:

Originally Posted by logalinipoo (Post 4561484)

Hey funk, the guy that built it has been banned for a while so I don't think it'll get updated.

I've read some forum histories and have made my own math last night :)