air consumption table for 5.9 engine (rpm, P/R, lbs/min)

GOT-Torque

GOT-Torque

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Anyone know how to calculate an air consumption table for a 5.9 (rpm, P/R, lbs/min)?

The table lists the lbs/min of air consumed based upon rpm and pressure ratio (or boost psi). I've seen the tables for many other engines but not this one. Anyone calculated one out or have a copy of one?
 
Diesel Tech

The garrett website has a couple formulas you can use to find your answer.

That is assuming they have calculators in Eastern Nebraska lol
 
Sorry for the delay, my wireless internet doesn't come in to well in my outhouse... One of these days we will get running water...

Thanks for the link, now where did I put that slide ruler...
 
6 cyls
4.02 bore
4.72 stroke
59.9080249152 in2 cyl disp
359.4 in2 engine disp
5.89 liters
3650 rpm
379.625505105 CFM NA
27.9304050302477 lbs/min @ 80F
 
any idea on what to use for volumetric efficiency? IMO, I think this would change based upon rpm, but I don't know much either...

Using a A/F of 16:1 and VE of 70 at 3100 rpms I calculate I need 63.3 lbs/min to make 625 flywheel hp. This would require 52 psi of boost calculated out assuming 180 degree intake temps, 2 psi lost in intercooler, 1 psi lost thru air filter...

My last dyno with a S465 and Mach 4's was 560 wheel hp at 3100 rpms using 50-55 psi boost.

Any input on A/F, VE or intake temps?
 
In my experience, the stock intercooler "charge air cooler" is pretty efficient at cooling intake air temps for a quick blast such as a dyno run. Usually i see 35 - 40* higher IAT than outside incoming air temp. So on a 100* F day, 135-140*F is a realistic number. Obviously the more mass of air that is forced through the stock cooler, the less efficient it will become. Also note that the aluminum cooler is heated by the air and becomes heat soaked and less efficient the longer you try to maintain high temp compressed air flow.
 
As far as Air/Fuel ratio, they make lots power anywhere from 12:1 to 25:1. I've read that 14:1 usually has enough excess fuel to burn 99.99% of available oxygen.

The stoichiometric balance for complete combustion is near 14.63:1.
 
Thanks for the link Brandon, I've got more calculations to add to my spreadsheet now...
 
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