Turbocharged flow factors

[MOTORMAN377]

Information needed from techies or non-mathamatical challenged motor heads.

Conversion factor for flow rates @ 28" H2O when converting engine to boost.

Example, although any would do :

Small block Chevy...23 degree heads, with 4 valves per cylinder. Intakes:1.65" Exhausts: 1.38" Intake valve area 4.27" Exhaust valve area: 2.99" CFM Intake 320 @ .650" Exhaust 270 @ .650"

These figures are for unmoddified heads, which will be upgraded to 1.72" Intakes and 1.50" Exhausts.

Is there a factor that I can plug into my existing flow rates to show estimated increase in CFM at one pound incremental rates for boost, up to a maximum of say 60" Hg.

Thanks for any help you fellas can supply, and more importantly your patience in researching this matter.


Alan

[deltaf]

Think this might help you:

First work out the pressure ratio.(amount of boost you want to run)

Pressure ratio =14.7+boost (say 5psi) divided by 14.7= 1.34
Approx 34% more air will go into the engine at this 5psi boost level



To calculate the airflow rate of an engine without a turbo-ie;no boost.

Airflow rate = cid x rpm x 0.5 x Ev divided by 1728.

Ev=volumetric efficiency(approx 85% use 0.85 for the formula tho)
1728 converts cubic inches to cubic feet.
0.5 is a constant.

Eg: airflow rate= 302 x 5500 x 0.5 x 0.85 divided by 1728 =408 cfm @5500rpm


Now work out the rate of flow under boost for the same engine.

Airflow rate= pressure ratio x basic engine cfm =
1.34 x 408= 546cfm@5500rpm.

Hope this helps some.





>> Edited by deltaf on Friday 16th January 18:37

[MOTORMAN377]

Yesssssss, deltaf has got hiz sheet to gedder ! Get the refreshments and the munchies...throw on another log...and grab the abacus : > ) watch out world here he comes...see the smoke coming from his ears !

Many thanks, ya made me day and hopefully many others also with your post..........


Alan

[GreenV8S]

Um that doesn't sound right. Basically what you're doing is increasing the air density entering the inlet manifold - this doesn't change the volume flow rate and gas velocities.

[deltaf]

Dont hardly matter...all we want is to get much more air into the cylinder before we set a big fire to it!
However, i am all ears as to what is troubling you mate. Feel free to voice your concerns!

oops, please bear in mind that im less than awake at de momento, as ive had a few drinks and as such im am fuddled beyond mortal comprehenshun...or something like that..



>> Edited by deltaf on Saturday 17th January 22:35

[speedy_thrills]

MOTORMAN377 said:

Conversion factor for flow rates @ 28" H2O when converting engine to boost.

Hi Alan, Here is what I think you need to know.

To calculate the pressure acting on the bottom of a column of water: Pressure = Density of water * Gravitational acceleration * the height of the column of fluid above the said point.

Unfortunately I use SI units so 28 inches = 0.7222m.

So Pressure = 1 (assumed value of fresh water at 4 degrees centigrade) * 9.81 m/sec^2 (gravitational constant of acceleration – only varies a little) * 0.7222

The pressure in your intake manifold is 7.08 Pascals (one Pascal is the force of 1 Newton acting over 1 square meter - Atmospheric pressure is about 101.35 kPa) above or below atmospheric pressure, depending on how you set up your manometer (U – tube as you may call it). Remember that the Manometer only measures pressure absolute to the atmosphere surrounding it.

Edit: And after completely realising I misread the question I do apologise for my irrelevant post.

You just need to factor in some coefficients into the Bernoulli equation and the roughness of your intake manifold. Remember that increasing the flow rate is not always to generate more power with the engine.

Edit for the second time: Although I’m not sure if you are taking into account inefficiencies in the system from what you said above.

>> Edited by speedy_thrills on Thursday 26th May 08:23

>> Edited by speedy_thrills on Thursday 26th May 08:29