How much boost to run on a 1g

[Chad Giffen]

Here they are:


For the EVO III 16g, has anyone ever had problems on the stock fuel system running one of these turbo's.

Here are some calculations I did...   


Volumetric Flow Rate:

2000cc / 16.39 = 122 cid

VFR = displ * RPM / 3456    =   122 *7000/3456 = 247.1 cfm

*****This is the ideal VFR...  realistically its about 90% of this due to pressure losses going into the engine.

0.90 * VFR
0.90 * 247.10 cfm = 222.4 cfm



Mass Flow Rate

2.703 * Press(amb)  *VFR / (T(amb) +460)  =  2.703 * 14.7 * 222.39 cfm / (78 + 460)  =  16.42 lb/min


Now this mass flow rate is for a naturally aspirated engine....

To aplly to turbo cars..      MFR * Boost Pressure    =      16.42 * (14.7 + 12psi (stock boost)) / 14.7  =  29.83 lb/min


Now for the fuel calculations...

Now if the mass flow rate of air entering the engine must be divided by the number of cylinders....    29.83lb/min  /  4 =  7.46lb/min  per cylinder

convert this from lb/min to lb/hr..   7.46lb/min * 60 min/hr = 447.488 lb/hr


Now...   we know that 1g DSM stock injectos flow 450cc/min which is roughly 41-42 lb/hr of fuel.

Now if we inlclude a duty cycle of no more than 85% for the injectors to be safe (some recommend 80%) we see the new max fuel flow rate is: 

41 lb/hr * .80 = 32.8 lb/hr


Now lets solve for the Air:Fuel Ratio the cylinders will see at 7000rpm....

Air:Fuel Ratio = Air Mass Flow Rate / Inj Flow Rate         so..    447.48lb/hr  /  32.8lb/hr =  13.64 this is actually slightly rich which is good.

If the injectors had a higher duty cycle to get more conservative A/F ratio....


If we assume a 90% duty cycle for the injectors...      447.21 lb/hr  /  (41lb/hr * 0.85)  =  12.83  this is slightly rich...  but much better for cylinder temps.

[daniel Dee]

I didn't notice any power gains past 16 psi with the factory intercooler...

[Graeme H Burvill]

wow, chad, love the math. I like to see tuning math simplified for us non-techs, makes things make sense a lot more. I am wondering what a duty cycle is and why it is mesured in a percentage. I take it this has something to do with the amount of flow utilized in an injector. But I honestly have no idea. Thanks!....maybe i will google this real quick.....

[Graeme H Burvill]

ooh i found this:

An injector in an engine turns on and off very quickly to control the amount of fuel delivered. The amount of time an injector is turned on and delivering fuel is known as the duty cycle. This is measured as a percent, so 50% duty cycle indicates that the injector is held open and held closed for an equal amount of time. When the engine needs more fuel, the time that the injector stays on (its duty cycle) increases so that more fuel can flow into the engine. If an injector stays on all the time, it is said to be static (wide open, or 100% duty cycle). INJECTORS SHOULD NOT GO STATIC IN A RUNNING ENGINE! If an injector is static in a running engine (open 100% of the time), that injector is no longer able to control fuel delivery. It is just “along for the ride”. This could be an indication that the injector is too small for the needs of the engine. Injector duty cycle should usually not exceed 80% in a running engine at any time.

[Graeme H Burvill]

hey chad, wondering if you could explain where all your numbers come from under the volumetric flow rate and mass flow rate? If you have time and feel like writing a brief description next to the numbers that would rock my world. Don't feel obligated though, only if you want to. Cheers man!

[Chad Giffen]

Sure I dont mind.....



Volumetric Flow Rate:      "the physical volume of air your engine can flow (or Air Flow Rate in this case)"

2000cc / 16.39 = 122 cid (cm^3)    
                                                - We know Talons engines are 2.0L (2000 cubic centimeters or cc)
                                                - cid = cubic inches of displacement   (so when a big block guys says
                                                          hes running a chev 350...   the 350 stands for 350 in^3 of volume in all the cylinders)
                                               - Recall 1 cm = 2.54in ....    so (1cm)^3 = (2.54in)^3   ...    1 cm^3 = 16.39in^3
                                               - So we are converting cubic centimeters to cubic inches so we can work with this
                                                 number in upcoming equations.


VFR = displ * RPM / 3456    =   122 *7000/3456 = 247.1 cfm

                                            Volumetric Flow Rate (mesaured in ft^3/min or cubic feet per minute (cfm) =
           
                                            (engine displacement (in^3) * RPM) / 3456

                                            - Now we want the VFR to have units of cfm (cubic feet per minute of air)
                                            - we are inserting the displacement in inches^3 and the RPM...    the 3456 serves as a conversion
                                              factor so we get cfm out of the equation.
                                            - so somehow we must get from...  in^3 * rev/min....     to ft^3/min....  (1ft)^3 = (12in)^3
                                              so...   1 ft^3 = 1756 in^3
                                            - We arent done yet though....    basically what this whole equation is saying is that for 1 rev of the
                                               engine, the volume of all four cylinders in being sucked in. But this is not true....   
                                               air is only sucked in by 2 cylinders for every 1 rev so the amount of air being sucked in by the
                                               engine must be multiplyed by 1/2.
                                            - So to get from in^3 * RPM to ft^3/min going into the engine...   we multiply by 1/1756 to
                                              convert to ft^3 and multiply again by 1/2 to get the actual air going into the engine per rev........
                                             
                                                  1/1756 * 1/2 = 1/3456.....     which is where the conversion factor comes from.

                             
                                  Now we know at any given RPM...   how much air the engine is sucking in per minute.
                                  The reason I input 7000rpm in the equation is obviously cause that is when the fuel system
                                  will be flowing the most fuel and where anyone has to worry about running lean.




Mass Flow Rate

2.703 * Press(amb)  *VFR / (T(amb) +460)  =  2.703 * 14.7 * 247.1 cfm / (78 + 460)  =  18.25 lb/min


This equation is a much more difficult to understand for those who dont know it already.....    first I will start with what everything means...

Mass Flow Rate -->  The mass of air flowing into the engine per minute         ---->  measured in   lb/min
Press(amb)       -->  The ambient (as in right this instant) pressure outside    ---->  known to be 14.7psi at sea level  (atmospheric pressure)
VFR                 -->  Volumetric Flow Rate  (volume of air entering the engine) --->  measured in cfm or ft^3/min
T(amb)            -->  The ambient (as in RIGHT NOW!!!) temperature outside    --->  measured in degrees Farenheit
460                 -->  460 Rankines = 0 deg F...  0 Rankine is when molecules stop moving. A temp in Rankine can be considered the 
                            ABOLUTE TEMP as again at 0 deg Rankine...  everything stops moving. EVERYTHING. After all, temperature is
                            merely a measure of kinetic energy!
2.703              ---> Air density coefficent

Now this equation is not nearly as hard as it looks....    heres why....

Consider 2.703 to be density of air @   1 Degree Rankine @ 1psi...   so it looks something like this....

2.703 * ((Outside Pressure) / (OutsideTemperature))      

now....     the more pressure you have....    the heavier the air is....   the more temp you have...   the more widespread the air molecules which lower the weight of the air. So what is going on here is we are multiplying 2.703 by the ratio of outside pressure over outside temperature to give us the density of air in lb/min.

Now we have the mass of air going in to the engine ...   and to account for how much mass...  we then multiply by the the volumetric flow rate!

so...    (air density) * (volumetric flow rate) = mass flow rate in lb/min!!!


Now this mass flow rate is for a naturally aspirated engine....




Any more questions....    just let me know!

[Graeme H Burvill]

wow, thanks a lot for the response man. I'm gunna read it a few more times and i'll let you know if I have any questions.
Thanks again for sharing man!

[Chad Giffen]

i forgot to take into account Volumetric Efficiency is usually 90%...     the top calcs are now revised...