Dupa cum spun a 3-a oara teoria ori este gresita ori este aplicata gresit. In cazul de fata teoria este CORECTA DAR APLICATA GRESIT !.
Sa le luam pe rand :
Boost-ul necesar pentru obtinerea 265 cp = 28,6 lb/min
Turatia la care masina dezvolta 265 cp = 3800 rpm ( nu 3600 )
· MAPreq = Manifold Absolute Pressure (psia) required to meet the horsepower target - ce ne intereseaza ( adica pe 16vt )
· Wa = Airflowactual(lb/min) - 28.6 lb/min
· R = Gas Constant = 639.6
· Tm = Intake Manifold Temperature (degrees F) - 18 C = 65 F
· VE = Volumetric Efficiency = 0.92
· N = Engine speed (RPM) = 3800 RPM
· Vd = engine displacement (Cubic Inches, convert from liters to CI by multiplying by 61.02, ex. 2.0 liters * 61.02 = 122 CI)
EXAMPLE:
To continue the example above, let’s consider a 2.0 liter engine with the following description:
· Wa = 28,6 lb/min as previously calculated
· Tm = 65 degrees F
· VE = 92% at peak power
· N = 3800 RPM
· Vd = 2.0 liters * 61.02 = 122 CI
MAP req = ( Wa x R x (460+TM ) / ( Ve x N/2 x Vd )
Deci MAP req = 45.0 psia. ( PRESIUNE ABSOLUTA )
Remember, this is absolute pressure. Subtract atmospheric pressure to get gauge pressure :
45.0 psia – 14.7 psia (at sea level) = 30.3 psig boost
Deci rezultatul este 2.08 BAR.
Alte nelamuriri ai domnule 16vt? Iti recomand sa ramai la vanzarea de casetofoane