What would it take to make 1.44 lbs-ft per cid ?

273 said
Were measuring cfm but ok mass flow rate, M = r x v x a right ?

So and NA engines Air density varies but say it's static for a moment at atmosphere psi 14.7 & we add a turbo and double it to 29.4 psi overall so the density is twice that of an NA engine so per cubic foot it's got about twice the oxygen and fuel molecules why makes about double the power.

So density is r in the formula so in the turbo r would be twice the amount than in an NA engine in this example so doubling the density would double the mass flow rate with Velocity and Area still being the same turbo vs boosted, NA 100r x 1v x 1a = 100m, Boosted 200r x 1v x 1a = 200m.

Is this wrong ? Seems right.

I totally agree, the velocity in the engine probably way higher and same with overall air flow, but we see some correlations between the bench and what works in the engine.

Trust me I understand the concept. You squeeze more air in a given volume, now move it along at a fixed velocity. You're flowing more air but the velocity is constant.

But pressure and velocity arent constant--at all-- so the formulas break down. I'm trying to explain that it's not as simple as plugging it into a formula. Those formulas also assume a laminar flow which is absolutely not the case.

I also think it's helpful to remember...

1. Power production is a result of burn rate of air/fuel (mass flow)

2. Pressure is a result of flow restriction

Anyway, I still understand the need to have a certain amount of flow velocity. I can see how it's helpful for increasing volumetric efficiency.