How many CFM of air per RWHP, roughly?
06-08-2010, 08:49 PM
#21
TECH Senior Member
I half agree with a few of you..
You can estimate what your power will be with mathematics.. many programs assist with this but require vast knowledge of fluid dynamics, chemistry, and engines
Companys typically START their engine designs by doing computer simulation then they build the engine test, "calibrate" the computer simulation model, and then make further iterations in the computer simulation to figure out what to try next, etc..
Computer simulation will continue to become more and more important, however, at this point, real world testing is still required for 'very' accurate results.
I do not agree that a volumetric flow rate of air can equal a brake horsepower value.
BHP = Fuel flow x BSFC
BHP = "MASS" air flow x AFR x BSFC
BMEP = IMEP - FMEP
all above are valid equations
BHP = "volume" airflow * multiplier <<--- NOT VALID!!
You'd be better off estimating a BSFC
You can estimate what your power will be with mathematics.. many programs assist with this but require vast knowledge of fluid dynamics, chemistry, and engines
Companys typically START their engine designs by doing computer simulation then they build the engine test, "calibrate" the computer simulation model, and then make further iterations in the computer simulation to figure out what to try next, etc..
Computer simulation will continue to become more and more important, however, at this point, real world testing is still required for 'very' accurate results.
I do not agree that a volumetric flow rate of air can equal a brake horsepower value.
BHP = Fuel flow x BSFC
BHP = "MASS" air flow x AFR x BSFC
BMEP = IMEP - FMEP
all above are valid equations
BHP = "volume" airflow * multiplier <<--- NOT VALID!!
You'd be better off estimating a BSFC
06-09-2010, 08:26 AM
#23
Agreed. Hold a 7.5 locker in one hand and a 9'' locker in the other. Tell me it doesn't take more power to turn that. Same with the tires. Roll a 245/50/16 on a stock snowflake and a 325/50/15 on a prostar with the same initial force and see which one rolls further before stopping.
06-09-2010, 06:09 PM
#24
A simple formula that I was givin was (CFM x .25) x 8= HP potential
Max CFM flow X .25 x 8 ( 330 x .25) X 8 = 660 hp (why not just multiply by 2? its easier than your formula) I understand you are trying to make it work for 8 cyls though.
This is the potential that your heads can make if all other parts are world class parts.
660 x .18= 541 rwhp You are wrong. Did you double check your math? Maybe you meant (660 - (660*.18)) = 541 or in easier form 100% - 18% = 82% then 660*.82 = 541
HP x .18(aprox HP lost from driveline )= rwhp
This was givin to me by 2 Pro engine builders with 45yrs of turning torque wrench. Hope this helps you
Max CFM flow X .25 x 8 ( 330 x .25) X 8 = 660 hp (why not just multiply by 2? its easier than your formula) I understand you are trying to make it work for 8 cyls though.
This is the potential that your heads can make if all other parts are world class parts.
660 x .18= 541 rwhp You are wrong. Did you double check your math? Maybe you meant (660 - (660*.18)) = 541 or in easier form 100% - 18% = 82% then 660*.82 = 541
HP x .18(aprox HP lost from driveline )= rwhp
This was givin to me by 2 Pro engine builders with 45yrs of turning torque wrench. Hope this helps you
06-09-2010, 11:41 PM
#25
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Agreed. Hold a 7.5 locker in one hand and a 9'' locker in the other. Tell me it doesn't take more power to turn that. Same with the tires. Roll a 245/50/16 on a stock snowflake and a 325/50/15 on a prostar with the same initial force and see which one rolls further before stopping.
The power it takes to turn a rear end is something like 10-15 hp at maximum speed. That's a fraction of the power needed to run an auto trans. Same with wheels and tires, driveshafts. Those together amount to another 5-10 hp.
The biggest power draw is the transmission, and a big auto/torque converter combination that can handle big power can draw upwards of 70 hp @ max speed (i've seen the power reading from a drivetrain dyno). That's your main power loss right there.
Last edited by Arfdog; 06-09-2010 at 11:47 PM.
06-10-2010, 09:53 AM
#26
A simple formula that I was givin was (CFM x .25) x 8= HP potential
Max CFM flow X .25 x 8 ( 330 x .25) X 8 = 660 hp
This is the potential that your heads can make if all other parts are world class parts.660 x .18= 541 rwhp
HP x .18(aprox HP lost from driveline )= rwhp
This was givin to me by 2 Pro engine builders with 45yrs of turning torque wrench. Hope this helps you
Max CFM flow X .25 x 8 ( 330 x .25) X 8 = 660 hp
This is the potential that your heads can make if all other parts are world class parts.660 x .18= 541 rwhp
HP x .18(aprox HP lost from driveline )= rwhp
This was givin to me by 2 Pro engine builders with 45yrs of turning torque wrench. Hope this helps you
10-30-2012, 08:06 AM
#27
TECH Enthusiast
Yes, I know it is over two years old - LOL !
******************************************
Peak_HP = Flow_CFM x .257 x Number_of_Cylinders
This is estimated potential Peak HP to expect.
You multiply .87 percent times cam's theoretical max lift , round off to nearest .050" in Flow Test, then see what CFM is at 28 inches
example : .700" Lift cam
.700 Lift times .87 percent = .609" Lift
Flow head at .600" Lift , then take CFM at 28 inches and calculate HP potential with above formula
.257 Factor = for beginning engine builders and engines near 10.0:1 Comp Ratio
.285 Factor = would be for Professional engine builders with wet sump pans, lightweight rotating assemblies, low tension great sealing rings, deep oil pans, etc.
Excellent use of inertia/wave tuning with 9.5 to 11.5:1 Comp Ratios or
11.5 to 13.0:1 CR ranges without fully utilizing inertia/wave tuning effects.
.300 to .310 Factor = Current ProStock Technology with dry sump, unlimited carburetion, Hi Comp Ratio, ultra lightweight rotating assembly, etc, max use of inertia/wave tuning, etc, 14:1 to 17:1 Comp Ratios (usually no better than .3200 efficiency or no worse than .2980 eff %)
******************************************
Peak_HP = Flow_CFM x .257 x Number_of_Cylinders
This is estimated potential Peak HP to expect.
You multiply .87 percent times cam's theoretical max lift , round off to nearest .050" in Flow Test, then see what CFM is at 28 inches
example : .700" Lift cam
.700 Lift times .87 percent = .609" Lift
Flow head at .600" Lift , then take CFM at 28 inches and calculate HP potential with above formula
.257 Factor = for beginning engine builders and engines near 10.0:1 Comp Ratio
.285 Factor = would be for Professional engine builders with wet sump pans, lightweight rotating assemblies, low tension great sealing rings, deep oil pans, etc.
Excellent use of inertia/wave tuning with 9.5 to 11.5:1 Comp Ratios or
11.5 to 13.0:1 CR ranges without fully utilizing inertia/wave tuning effects.
.300 to .310 Factor = Current ProStock Technology with dry sump, unlimited carburetion, Hi Comp Ratio, ultra lightweight rotating assembly, etc, max use of inertia/wave tuning, etc, 14:1 to 17:1 Comp Ratios (usually no better than .3200 efficiency or no worse than .2980 eff %)
10-30-2012, 10:08 AM
#28
Yes, I know it is over two years old - LOL !
******************************************
Peak_HP = Flow_CFM x .257 x Number_of_Cylinders
This is estimated potential Peak HP to expect.
You multiply .87 percent times cam's theoretical max lift , round off to nearest .050" in Flow Test, then see what CFM is at 28 inches
example : .700" Lift cam
.700 Lift times .87 percent = .609" Lift
Flow head at .600" Lift , then take CFM at 28 inches and calculate HP potential with above formula
.257 Factor = for beginning engine builders and engines near 10.0:1 Comp Ratio
.285 Factor = would be for Professional engine builders with wet sump pans, lightweight rotating assemblies, low tension great sealing rings, deep oil pans, etc.
Excellent use of inertia/wave tuning with 9.5 to 11.5:1 Comp Ratios or
11.5 to 13.0:1 CR ranges without fully utilizing inertia/wave tuning effects.
.300 to .310 Factor = Current ProStock Technology with dry sump, unlimited carburetion, Hi Comp Ratio, ultra lightweight rotating assembly, etc, max use of inertia/wave tuning, etc, 14:1 to 17:1 Comp Ratios (usually no better than .3200 efficiency or no worse than .2980 eff %)
******************************************
Peak_HP = Flow_CFM x .257 x Number_of_Cylinders
This is estimated potential Peak HP to expect.
You multiply .87 percent times cam's theoretical max lift , round off to nearest .050" in Flow Test, then see what CFM is at 28 inches
example : .700" Lift cam
.700 Lift times .87 percent = .609" Lift
Flow head at .600" Lift , then take CFM at 28 inches and calculate HP potential with above formula
.257 Factor = for beginning engine builders and engines near 10.0:1 Comp Ratio
.285 Factor = would be for Professional engine builders with wet sump pans, lightweight rotating assemblies, low tension great sealing rings, deep oil pans, etc.
Excellent use of inertia/wave tuning with 9.5 to 11.5:1 Comp Ratios or
11.5 to 13.0:1 CR ranges without fully utilizing inertia/wave tuning effects.
.300 to .310 Factor = Current ProStock Technology with dry sump, unlimited carburetion, Hi Comp Ratio, ultra lightweight rotating assembly, etc, max use of inertia/wave tuning, etc, 14:1 to 17:1 Comp Ratios (usually no better than .3200 efficiency or no worse than .2980 eff %)
This may work at the crank, at the wheel there is no formula to pin down all the variables.
11-23-2012, 07:03 PM
#29
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You cant even calculate FYWHEEL HORSEPOWER like you are trying to do... Too many variables like
VE
BMEP
BSFC
You cant calculate REAR WHEEL HORSEPOWER per CFM with a formula.
If you wanna waste your time trying to do it...knock yourself out. You can build two IDENTICAL combinations and they make different RWHP AND Flywheel HP . WHY? Too many variables.
RWHP Variables are
Flywheel/clutch weight
Driveshaft weight
Rear end rotational weight
Wheel and tire combination
Gear ratio
COnverter effeciency, 2 indentical converters will have different effeciency numbers
Type of trans, turbo 400/350, 4L60E/4L80E all have different effeciency.
FLYWHEEL HORSEPOWER is easier to compute with a formula, but still has many many variables.
If you wanna **** up a rope...be my guest.
VE
BMEP
BSFC
You cant calculate REAR WHEEL HORSEPOWER per CFM with a formula.
If you wanna waste your time trying to do it...knock yourself out. You can build two IDENTICAL combinations and they make different RWHP AND Flywheel HP . WHY? Too many variables.
RWHP Variables are
Flywheel/clutch weight
Driveshaft weight
Rear end rotational weight
Wheel and tire combination
Gear ratio
COnverter effeciency, 2 indentical converters will have different effeciency numbers
Type of trans, turbo 400/350, 4L60E/4L80E all have different effeciency.
FLYWHEEL HORSEPOWER is easier to compute with a formula, but still has many many variables.
If you wanna **** up a rope...be my guest.
11-24-2012, 06:29 PM
#30
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I guarantee Edlebrock, Crane Cams, & every other aftermarket manufacturer out there uses math to design a head, intake, cam, exhaust, etc...
As been stated earlier in this thread, pro engine builders will use math to determine what their goal should be. When their engine dynos a particular number they know right away if there is something wrong & if it's worth tearing it back down to find out what the issue is.
As been stated earlier in this thread, pro engine builders will use math to determine what their goal should be. When their engine dynos a particular number they know right away if there is something wrong & if it's worth tearing it back down to find out what the issue is.
11-30-2012, 05:49 AM
#31
I guarantee Edlebrock, Crane Cams, & every other aftermarket manufacturer out there uses math to design a head, intake, cam, exhaust, etc...
As been stated earlier in this thread, pro engine builders will use math to determine what their goal should be. When their engine dynos a particular number they know right away if there is something wrong & if it's worth tearing it back down to find out what the issue is.
As been stated earlier in this thread, pro engine builders will use math to determine what their goal should be. When their engine dynos a particular number they know right away if there is something wrong & if it's worth tearing it back down to find out what the issue is.
11-18-2013, 09:26 PM
#32
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Apologize in advance for bumping such an old thread, but I came here via google, found some good info and I'd like to ask a follow up question.
I'm trying to reverse engineer an intake limiter, so that different engines which are different sizes get the same amount of air so they'll make approximately the same power.
I see the formulas which calculate what the theoretical max hp are, now what I'm trying to do is determine how to limit the air so the engine is only capable of hitting my set point.
I'm trying to reverse engineer an intake limiter, so that different engines which are different sizes get the same amount of air so they'll make approximately the same power.
I see the formulas which calculate what the theoretical max hp are, now what I'm trying to do is determine how to limit the air so the engine is only capable of hitting my set point.
07-21-2022, 02:41 AM
#34
TECH Apprentice
old but very interessting thread.
why does nobody measure brake specific air consumtion? this would give an indication of an engines efficiency, as opposed to bsfc that is mixture dependent.
why does nobody measure brake specific air consumtion? this would give an indication of an engines efficiency, as opposed to bsfc that is mixture dependent.
07-21-2022, 09:32 AM
#35
TECH Senior Member
Because air is free, and fuel has a cost. Which would you keep track of?
07-21-2022, 09:43 AM
#36
TECH Apprentice
havent we had this discussion somwhere else a few years back?
cost doesnt matter if you are after max power, no? you could compare different engines/set ups as to what they make out of the air mass digested. after all an engine is air limited and not fuel limited.
cost doesnt matter if you are after max power, no? you could compare different engines/set ups as to what they make out of the air mass digested. after all an engine is air limited and not fuel limited.
07-21-2022, 09:58 AM
#37
TECH Senior Member
Fuel parameters, however are far more controllable, due to its liquid state. If you control one of the two, the other's parameters will be close to what is needed.
07-21-2022, 01:11 PM
#38
TECH Apprentice
airmass doesnt depend on density.
here a real world calculation (ls engine, blown, max tq, max hp):
5000 rpm, 535 bhp, 344.6 lbs/h, 11.5 afr → 5.74 lbs/min, 2'605 g/min, 29'959 g/min, 23'045 l/min*, 813 cfm, 1.51 cfm/bhp
6600 rpm, 609 bhp, 442.0 lbs/h, 11.4 afr → 7.37 lbs/min, 3'341 g/min, 38'092 g/min, 29'298 l/min*, 1034 cfm, 1.70 cfm/bhp
* convert at you favourite temp, pressure and humidity. as correction factors used on dyno are unknown, very approximate but comparable back to back. (somebody check calc please.)
im surprised there is such a large variation. what does that tell you?
here a real world calculation (ls engine, blown, max tq, max hp):
5000 rpm, 535 bhp, 344.6 lbs/h, 11.5 afr → 5.74 lbs/min, 2'605 g/min, 29'959 g/min, 23'045 l/min*, 813 cfm, 1.51 cfm/bhp
6600 rpm, 609 bhp, 442.0 lbs/h, 11.4 afr → 7.37 lbs/min, 3'341 g/min, 38'092 g/min, 29'298 l/min*, 1034 cfm, 1.70 cfm/bhp
* convert at you favourite temp, pressure and humidity. as correction factors used on dyno are unknown, very approximate but comparable back to back. (somebody check calc please.)
im surprised there is such a large variation. what does that tell you?
07-21-2022, 03:12 PM
#39
TECH Senior Member
Looks like your example engine is far more efficient at 5000rpm than at 6600rpm.
07-21-2022, 10:52 PM
#40
TECH Apprentice
yes, but not in the conventionl sence of cylinder filling. unfortunately its boosted and corresponding pressure ratios are not given, so we dont know what volumetric efficiency is. however we see that the engine not only fills the cylinder less at hp peak, but also is able to make less out of the air ingested. the difference of 13% certainly reflects friction losses. maybe other factors? do thermodynamics deteriorate with rpm?
edit:
3100 rpm, 228 lbs/h x 0.205 x 11.5/229 bhp = 2.35 cfm/bhp
4000 rpm, 321 x 0.205 x 11.2 = 1.90
4500 rpm, 398 x 0.205 x 11.4 = 2.00
5000 rpm, 365 x 0.205 x 11.5 = 1.60
6800 rpm, 448 X 0.205 x 11.3 = 1.73
apparently this is not so much friction related. as the engine comes on the cam it improves dramatically. (same engine as above set up a bit differently.) looking at the jump 4500→5000 rpm im wondering if the data is correct. ok, from another run i get 1.55 for 4500 rpm.
i have no idea how dependable this data is. would be great if somebody could do this with his own data. published fuel flow is pretty rare, unfortunately.
edit:
3100 rpm, 228 lbs/h x 0.205 x 11.5/229 bhp = 2.35 cfm/bhp
4000 rpm, 321 x 0.205 x 11.2 = 1.90
4500 rpm, 398 x 0.205 x 11.4 = 2.00
5000 rpm, 365 x 0.205 x 11.5 = 1.60
6800 rpm, 448 X 0.205 x 11.3 = 1.73
apparently this is not so much friction related. as the engine comes on the cam it improves dramatically. (same engine as above set up a bit differently.) looking at the jump 4500→5000 rpm im wondering if the data is correct. ok, from another run i get 1.55 for 4500 rpm.
i have no idea how dependable this data is. would be great if somebody could do this with his own data. published fuel flow is pretty rare, unfortunately.
Last edited by Dian; 07-21-2022 at 11:42 PM.