16 replies to this topic

[Hemi_Express2013]

I have this formula (( g/s /1225)x60) x 35.31 and (cid x rpm x ve)/3456 for converting g/s to cfm. Example: 311g/s @ 5500 rpm.

311/1225=.253877551 x 60 = 15.2365306 x 35.31 = 537.9 cfm.

537.9cfm = (345 x 5500) x ve/3456

VE = .97970087(p-ratio?)

If a 5.7 is capable of using 675cfm @5500. Displacement x rpm /2 then 537.9 cfm @ 5500 is 80% of that 675cfm

So if this math is right and given the restrictions. The most this 5.7 could produce is 435 hp. (1.5cfm x hp) = 651 cfm. 80mm TB only flows 641 cfm according to Arringtons test

Just wondering if my direction is right? Any help/ input would be greatly appreciated.

[Hemi_Express2013]

Wow, figured someone would know this stuff and have some input with the calculations. I did a 2nd gear pull on my truck today with my snap-on sulus ultra and was surprised that @ 78.9mph/5034rpm I pulled 563cfm with a VE of 1.12. figured this with 325.608g/s @5034rpm and formula above.

[toofart]

What is it you're asking?  I see a bunch of constants throws up but no source.

[Blown7]

It is pretty simple..... old school...... your in the ball park. Way too many people thinking  that the small (370 Cu In engines)6.1 L need way too much air. The old 350 Cu In engines died on 750 CFM carburetors. My frame of thought is 304 CU In = 500 Holley CFM  AMC stuff

Many people seem to think they need a 90 MM throttle body for a 6.1 L

 

They are way over throttled.

 

Well the Hellcat uses a 92 MM throttle body and thats a 6.2  L supercharged to 2+  atmospheres.(11+ PSI)

[Hemi_Express2013]

What is it you're asking?  I see a bunch of constants throws up but no source.

Just wondering if the formula/math is correct. Converting g/s to cfm, finding VE which would point to the efficency of the system. Just trying to learn the basic math for each parameters calculation in this. Seems to me tuners would easly know this stuff. Sry for not being clearer, but I do have the saved data logs from the snap-on tool. If I can find a way to show them without shop connect program, I'd be glad to post them.
 
 

It is pretty simple..... old school...... your in the ball park. Way too many people thinking  that the small (370 Cu In engines)6.1 L need way too much air. The old 350 Cu In engines died on 750 CFM carburetors. My frame of thought is 304 CU In = 500 Holley CFM  AMC stuff

Many people seem to think they need a 90 MM throttle body for a 6.1 L
 
They are way over throttled.
 
Well the Hellcat uses a 92 MM throttle body and thats a 6.2  L supercharged to 2+  atmospheres.(11+ PSI)

Thanks for the input. I appreciate that.

[toofart]

Ah.  Yep, I agree with B7 and your math.  I did the same calculation a couple years ago (using litres then converting to CFM) and concluded that the stock 80 mm TB is adequate.

[Hemi_Express2013]

Wow, figured someone would know this stuff and have some input with the calculations. I did a 2nd gear pull on my truck today with my snap-on sulus ultra and was surprised that @ 78.9mph/5034rpm I pulled 563cfm with a VE of 1.12. figured this with 325.608g/s @5034rpm and formula above.

So my next question is this given the math from above run,

100% ve airflow(scfm) = displacement x rpm / 3456

Which is saying at 5034 rpm my 5.7 can only pull 502cfm at 100% ve, but I'm pulling 563 cfm. That's 61cfm over 100% VE or 112% efficiency. How is that possible? What am I missing in this equation. I mean I have a ram air system, is it really causing the increased airflow?

[Stricnine]

So my next question is this given the math from above run,

100% ve airflow(scfm) = displacement x rpm / 3456

Which is saying at 5034 rpm my 5.7 can only pull 502cfm at 100% ve, but I'm pulling 563 cfm. That's 61cfm over 100% VE or 112% efficiency. How is that possible? What am I missing in this equation. I mean I have a ram air system, is it really causing the increased airflow?

 

>100% VE can be achieved with intake manifold design, cam timing, exhaust, cylinder head, RPM, etc.

 

One way this can happen: The opening of the intake valve causes a surge of air flow into the cylinder, when the valve closes the momentum of the air causes a rise in pressure at the valve.  That pressure wave travels back up the intake port/runner and when it reaches the "plenum" (open area) at the beginning of the intake runner it tends to cause a reversal of the pressure wave back down the intake runner.  When timed correctly (runner length, valve timing, RPM) the pressure wave arrives during the next period when the intake valve is open and packs a little more air into the cylinder than normal.  This is rather RPM dependent.  This has been called tuned port, and has been around for quite a while.

 

Another is cam timing/phasing (retarding at higher RPM) that takes advantage of the momentum of the air flow relative to the rising piston near the end of the intake valve event.

 

Another thing that can affect the amount of air the computer thinks is going in to the cylinder is intake/exhaust overlap and scavenging. The exhaust pulse is traveling fast, at the end the exhaust port pressure get low enough to start drawing the intake charge out through the exhaust port.  Given the fairly wide LSA on cams used on most builds the overlap is probably not all that large, and minimizes the effect.  Headers will likely be more effective in influencing the overlap effect than the factory iron "log" manifolds.

[Hemi_Express2013]

>100% VE can be achieved with intake manifold design, cam timing, exhaust, cylinder head, RPM, etc.

One way this can happen: The opening of the intake valve causes a surge of air flow into the cylinder, when the valve closes the momentum of the air causes a rise in pressure at the valve. That pressure wave travels back up the intake port/runner and when it reaches the "plenum" (open area) at the beginning of the intake runner it tends to cause a reversal of the pressure wave back down the intake runner. When timed correctly (runner length, valve timing, RPM) the pressure wave arrives during the next period when the intake valve is open and packs a little more air into the cylinder than normal. This is rather RPM dependent. This has been called tuned port, and has been around for quite a while.

Another is cam timing/phasing (retarding at higher RPM) that takes advantage of the momentum of the air flow relative to the rising piston near the end of the intake valve event.

Another thing that can affect the amount of air the computer thinks is going in to the cylinder is intake/exhaust overlap and scavenging. The exhaust pulse is traveling fast, at the end the exhaust port pressure get low enough to start drawing the intake charge out through the exhaust port. Given the fairly wide LSA on cams used on most builds the overlap is probably not all that large, and minimizes the effect. Headers will likely be more effective in influencing the overlap effect than the factory iron "log" manifolds.

I had to read this a couple of times to grasp your information(think headed). Thank you for taking time out to write that, and I appreciate the details of this process you've elaborated on. Good stuff there!

[Blown7]

So my next question is this given the math from above run,

100% ve airflow(scfm) = displacement x rpm / 3456

Which is saying at 5034 rpm my 5.7 can only pull 502cfm at 100% ve, but I'm pulling 563 cfm. That's 61cfm over 100% VE or 112% efficiency. How is that possible? What am I missing in this equation. I mean I have a ram air system, is it really causing the increased airflow?

 

 

Well the formula that I use when I don't use "Rule of Thumb" has more factors like temperature, barometric pressure.( Density)

 

You have to remember all them Gas Laws come into play

 

Alot of things effect the total absolute VE number... so to be 10% off is OK for that formula you use.

[Hemi_Express2013]

Well the formula that I use when I don't use "Rule of Thumb" has more factors like temperature, barometric pressure.( Density)

 

You have to remember all them Gas Laws come into play

 

Alot of things effect the total absolute VE number... so to be 10% off is OK for that formula you use.

Now that is a discussion that has come up a few times in VE calculations. A lot of variables to consider for accuracy. Lot of respect for you guys that have digested this stuff and can put it to work, not something learned over night. I will read as much information as my brain can handle.

[Hemi_Express2013]

So I have been reading as much information on VE as I can find. I have found some very precise mathamatical equations that take into account air density/temps at sea level( I live on a Island) to add to my calculations. I did some changes to my air intake, fabbed my own stuff from 4 intake systems. LMI being the basis of the new air intake. Using my snap-on Ultra, much easier to understand than trying to understand the trinity datalogs,  I did a couple of runs. I had an old max air charge of 328.668 g/s @ 5169rpm. My new maximum is 397.488 g/s @ 4318rpm(roughly 19% increase less 700rpms). AIT was 59*(518.67R)/ that by density of air for a know temp(.0808 lb/ft^3 @ 32f(491.67R) sea level) and outside temps @ 51.8*. MAP @ 1.119 and Baro @ 29.845. SRV Deenergized. So, doing all the calculations at that RPM(4318) and temp, I was pulling 686.5cfm. That's 158% VE. Now I know that's not possible. A stock 6.1 at 6400 rpm @ 100%VE is only pulling 685cfm.

 

So my question is: what is it that I'm not accounting for that's giving me the screwed results. Or is this a point where the duration of my cam is just pulling the intake air straight through, out the exhaust?

[BuckeyeGuy]

Ok,. not a mathmetician here, are you telling me that the 90mm throttle body is too much for my 366 CID NA? 

 

Just asking the question from a math standpoint.  From a dyno standpoint, the answer is a resounding no, it is not too big!.

 

Sorry for the thread jack.   

[toofart]

What is map 1.119 and baro 29.845? I assume 29.845 inHg which is 14.65 psi.

Only thing i can think of is a bad value in your formula, but it's hard to make out your formulas when you pepper them with words

[Cam]

Well let's see, my twin 65's netted me a 30/30 gain almost across the board in relation to a stock 80. Same day, same dyno, but dedicated tuning obviously. Also at the track a solid 2-3 tenths and ~3mph with back to back testing. That's with a excellent set of heads and a wittle itty bitty 214/218 .570 cam.

So math is great and all, but not always money.

[Hemi_Express2013]

What is map 1.119 and baro 29.845? I assume 29.845 inHg which is 14.65 psi.

Only thing i can think of is a bad value in your formula, but it's hard to make out your formulas when you pepper them with words

Sry for that. Yes Map vacuum is 1.119in/Hg and Barometric pressure is 29.845inHg. I only threw those in there for P-ratio, which I thought was an indication of volumetric efficiency? 

 

 

So the equation to reach volumetric efficiency is 3 parts. The first hurdle is to convert a mass flow(in this case g/s) rate into a volumetric flow rate(CFM). Density is inversely proportional to temperature, so in a nut shell, take a known desity of air at a given temp(32*F @ sea level,1 atmosphere) is 0.0808 lb/ft^3 and solve for the ratio between the tempatures. But to do this you cannot use F, it has to be Rankin measurement. To do this you just add 459.67 to the F measurement to get absolute Temp(R*).

 

T1/T2=d2/d1

t1= temperature of air for a known density(32*[email protected]/ft^3

t2=temperature of the intake air measured by the IAT sensor and convert to (*R)

d1=density of air for a known temp(0.0808lb/ft^3 @32*f)

d2=density of the intake air(lb/ft^3)

 

solving for d2 with the given info above =

t1 =  32*F + 459.67(*R) = 491.67®              t1       d2                 491.67R

                                                                      ___ = ___   = d2=  __________  x .0808lb/ft^3  = .07659lb/ft^3

t2 =  59*f + 459.67(*R)=518.67®                  t2       d1                518.67R

 

that is my density of air at temp.  next 3 steps. easier. I'll have to leave you hanging for now sry. something just came up and I gotta go.

[Hemi_Express2013]

Wow, this is much easier on paper. But let me continue. Once air density is figured we can calculate your cars/trucks actual volumetric flow rate with this equation.

AVF= MFk/d2.

AVF= actual volumetric flow rate(ft^3/minute)
MFk= mass flow rate taken from snap on (397.488g/s converted is 52.58lb/min)
d2= we just solved for is .07659lbmin^3

AVF= 686.51259ft^3/min


Now to solve for the theoretical air flow(TAF)

TAF=(ED)(RPM)(VE)/(ES)©
ES= engine stroke 2*1728 in^3/ft^3
C= conversion factor from in^3 to ft^3

345.6495(4318)(1.00)/3456= 431.8618 ft^3

Next equation

VE= (AVF/TAF) * 100%

686.51259/431.8618 * 100%= 158.9658%
VE = 158.9%

So given the math results. Something is happening at this rpm that shouldn't be because no n/a has a ve of 158%. Anyone with a thought/consideration. What am I missing?