That depends on CAT (charge air temperature). Assuming 14.7:1 stoich for the fuel gives air flow of 0.6125 kg/s for lambda 1.5 and 0.735 for 1.8. This is trapped mass ie the air that stays in the chamber for combustion. If there is any scavenging, that will require additional air flow. Dealing with the trapped mass only for now:Brian Coat wrote:What intake air pressure (abs) and temperature do you estimate Lambda 1.5/1.8 equates to (pre-charge cooler)?gruntguru wrote:Huge guess here. Somewhere between 1.5 and 1.8PlatinumZealot wrote:By the way on the other topic, what lambda ratio do you suppose the Mercedes engines use?
Volume flow = displacement/2 x revolutions per second x VE(trapped) = (1.6 E-3)/2 x 11,000/60 x 1.1 = 0.1613 m3/s.
So for lambda 1.5 (air flow 0.6125 kg/s) the air density in the plenum needs to be 0.6125/0.1613 = 3.8 kg/m3 (DR = 3.8/1.2 = 3.17)
For lambda 1.8 the air density required is 0.735/0.1613 = 4.56 kg/m3. (DR = 4.56/1.2 = 3.8) (DR is Density Ratio - the ratio of charge density to ambient air density)
If the charge air is intercooled back to ambient temperature (not feasible but a useful benchmark) DR = PR so Lambda 1.5 and 1.8 require a PR of 3.17 and 3.8 respectively. Using ideal gas law => P1xV1/T1=P2xV2/T2 so P2/P1=V1/V2 x T2/T1 or
PR = DR x TR (V1/V2 = Density2/Density1 = DR)
So - pick your favourite CAT and plug into the formula.
For example
T1 (ambient) = 293K and T2 (CAT) = 313K and lambda 1.5 gives
PR=DRxTR= 3.17 x 313K/293K = 3.38
A more extreme example.
T1 (ambient) = 293K and T2 (CAT) = 343K and lambda 1.8 gives
PR=DRxTR= 3.8 x 343K/293K = 4.45 (which is approximately 4.45 bar absolute MAP.)