Gannet

Obviously, this would be very roughly. But there ought to be a rule of thumb out there.

In my calculations I'm getting something on the order of 1.4 cfm of air to produce 1 rwhp. Does that seem reasonable?

ROCNDAV

iTrader: (100)

According to flow research, the required air flow a muffler needs to attain, in order to keep horsepower loss due to back pressure to a minimum, is 2.2 cfm for every one horsepower. This general rule keeps horsepower losses under one percent. Not sure how much air intake cfm = RWHP.

mrdragster1970

.

I also was taught 2.2 for engine HP. As said, very very rough starting point, and have no idea about rear wheel. We do usually use 75-100 HP loss when making guess'.

.

dmc454

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

Beaflag VonRathburg

iTrader: (176)

I've heard the 2.2 cfm = 1 hp thing also.

I found this earlier when searching for conversion units.

1 HP approx equals 1.45 CFM
1 CFM approx equals 0.0745 lb of air/min
0.108 Lb/min approx equals 1 hp
1 Meter cubed/sec = 35.314 CFS = 2118.867 CFM
1 KG/sec = 132 lbs/min approx equals 1771.812 CFM

TT632

iTrader: (3)

I use 2 cfm/hp and 2.2 for a healthy race engine. You will find 90% of good performance engines will be in this range if you are taking advantage of the cylinder heads capabilities with the right size camshaft, and you have decent compression.

dmc454's "(CFM x .25) x 8= HP potential" which equals 2. Which is at the low end for most high compression race engines and probably about right for a healthy street engine.

chopper

iTrader: (18)

While searching for static compression ratio calculators I found this...
http://falconperformance.sundog.net/compcalculator.asp
is it accurate or even close? I have no idea. You tell Me.

Beaflag VonRathburg

iTrader: (176)

I did some quick messing with it. The cubic inch, compression ratio, and engine sizing parts are correct. The volumetric efficiency I cross referenced agains this VE calculator http://www.ajdesigner.com/phpengine/...efficiency.php and the results also came up accurate.

chopper

iTrader: (18)

Thanks for checking the accuracy beaflag but I thought this might provide the answer for the "how many cfm to make a HP" question. Does it?


Oh, the AJ link seems like a nice one, Thanks!

FASTFATBOY

iTrader: (23)

The 2.0 hp per 1 cfm rule is an estimate, everything has to be "right" for that to happen, VE has to be 100% and better. And that is a flywheel hp rule.

RWHP throws a whole barrel of variables into the pot. Trans, rear end, tires and wheels etc all affect rwhp output.


David

KCS

iTrader: (20)

Superflow says 1.67cfm/1hp

Gannet

Thanks for all the replies.

I get the impression that some folks in this thread are speaking of "cfm" as head flow numbers. That's not what I meant. I meant the actual cfm through the engine.

For example, assume a 427 at 6,000 rpm. The airflow at 100% VE would be 741 cfm. At 90% VE it would be 667 cfm.

The formula for a 4-stroke is: (CI x RPM x .5 x VE)/1728 = CFM

Let's assume 90% VE. If we divide 667 by the 1.4 number I suggested for CFM/RWHP you get 476 rwhp.

Could we expect a well-tuned but still "street" spec engine to make 90% VE? That does not seem unreasonable. Could we expect a well-tuned but street spec 427 to make 476 rwhp? A good one, sure. By "street" tune here I just mean something less than full race, not something with a smooth idle or mild tune.

Part of the point here is that it takes a certain amount of fuel to make a certain amount of power. It consequently takes a certain amount of air to combust that amount of fuel. That being the case, it doesn't matter (within reason) how we get that amount of air. We could do it with a small engine turning many RPM, a large engine turning fewer RPM, or a smaller engine that is supercharged. If all that is true, we ought to be able to work backwards from a given power requirement to a given airflow requirement, and hence to a number of different combinations that will flow the required air, and therefore should make the number. This allows the user to make an apples-to-apples comparison and think about things like cost, reliability, and powerband.

FASTFATBOY

iTrader: (23)

Did you see my post above?

You cannot calculate RWHP like you can FWHP....you just cant. Heres why...

A 500 rwhp engine with:

10 bolt with a 3.42 gear

six speed trans

stock 16 inch wheels and street rubber

WIll make 400 rwhp with:

turbo 400

9 inch rear with 4.11 gear

non locking converter

15 inch wheels with slicks.


There ya go.

No changes to the engine itself. It is still breathing the same air and making the same HP at the flywheel, just not making it to the ground.

David

Gannet

Yeah, I saw your post above. No offense, but I think it's kind of dumb. Everybody knows that certain vehicle factors, which you listed, will cause a typical inertial wheel dyno to "read low". So what? You take that into account, as each of those factors is worth, roughly, a certain percentage of error. Would you have felt better if I had asked about flywheel horsepower, and then applied the typical 15% factor (which would apply for my vehicle)?

Let's try that. I mentioned an original number of, very roughly, 1.4 cfm/rwhp. Multiply that by 1.15 and you get about 1.6 cfm/fwhp. So now, to rephrase the original question:

In my calculations I'm getting something on the order of 1.6 cfm of air to produce 1 fwhp. Does that seem reasonable?

FASTFATBOY

iTrader: (23)

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.

barkingspud

iTrader: (3)

I beg to differ. I believe you CAN in fact calculate FWHP and RWHP. It's just that we aren't privy to the proper mathematics involved to calculate such events. Do you really think that the guys building NASCAR and NHRA cars just "guess" when spending 100k+ on a motor?

FASTFATBOY

iTrader: (23)

Are you KIDDING? They ENGINE dyno and TEST, they dont mathematically GUESS on ANYTHING. Same as a NHRA Pro Stock team. ALot of stuff that was figured "on paper" get thrown in the garbage because they dont make power. Those kinds of operations found that out LONG AGO! They build and test. And if it dont work it gets put in a melt down barrel alot of times.

Furthermore a Nascar team that actually DOES chassis dyno, know the percentage of drivetrain loss because they actually ENGINE dyno the engine and THEN chassis dyno it. They dont do one or the other. Know why? To learn and make the drivetrain have less parasitic losses. And they work pretty much with the same drivetrain all the time with the exception of gear ratios.

Muscle Motorsports in North Carolina has MANY a set of Nascar heads that didnt make power and were sold off.

X.X cfm/HP means nada....again way too many variables.

David

DanO

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

Arfdog

Actually,
BHP = fuel flow / bsfc
= mass air flow / AFR / bsfc

BHP = CFM*multiplier is definitely a valid starting point to try to get HP from a head's CFM rating.



As for Gannet's question,

I get 1.66 CFM = 1 RWHP, using some calculations.

Power is the energy of the fuel per unit time. 1 CF of air, combined with fuel @ rich AFR (13.5) yields 0.006 lb gasoline. That amount of gasoline yields 121 BTU (20,200 BTU/lb gasoline). That is the energy in the combustion chamber. Fuel conversion efficiency in a gasoline engine is 25% and driveline efficiency (not loss) with a manual is 85%. 25% * 85% = 21.2% total fuel energy conversion efficiency to tires.

so 1 CFM @ 13.5:1 AFR = 121 BTU/min * 21% efficiency = 25.6 BTU/min = 0.6 HP

----> 1 CFM = 0.6 RWHP
----> 1.66 CFM = 1 RWHP

And remember the CFM rating is for 1 bank only. So an LS7 (heads flow 379 CFM) flowing 758 CFM total should get 455 rwhp.

Arfdog

I'd say the major factor here is the trans. The auto trans + converter is sucking all that power away. Everything else has similar efficiency, probably within 5% of each other. Gears, rear end, and tires all are for the same vehicle, so have similar construction.