Rotary Math
04-19-2007 | 06:07 PM
#53
Registered
Joined: Feb 2004
Posts: 1,802
Likes: 10
From: QLD .au
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)
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.
Last edited by rotarenvy; 04-20-2007 at 04:04 AM.
09-08-2007 | 11:58 AM
#56
Viva Las Vegas
Joined: Mar 2007
Posts: 32
Likes: 0
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!
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!
09-08-2007 | 12:10 PM
#57
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
Crank HP = MFR*60/BSFC*1728 -- something like that, from memory, could be wrong.
MFR = The MASS of air in lb/min
11-25-2007 | 11:17 PM
#58
Registered
Joined: Oct 2007
Posts: 1,574
Likes: 2
From: Boosted...
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......
01-02-2008 | 10:22 AM
#61
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Joined: Aug 2004
Posts: 170
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Ezine site with info on Rotary Volumetric Efficiency explained
05-07-2008 | 03:52 AM
#63
10,000 Revs in the Air!
Joined: Jan 2008
Posts: 233
Likes: 0
From: Tipperary, Ireland
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
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)
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
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.
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.
05-24-2008 | 01:17 AM
#65
AccessPORT Tuner
Joined: Feb 2006
Posts: 27
Likes: 1
From: Athens - Greece
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....
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....
02-23-2012 | 04:36 PM
#70
Voids warranties
Joined: Aug 2011
Posts: 1,632
Likes: 0
From: southern new england
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.
"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.
01-16-2014 | 03:03 AM
#73
Eunos Cosmo T61
Joined: May 2007
Posts: 24
Likes: 0
From: Moscow, Russia
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