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Like wtf bro, if you have a piston that's doing 12:1 compression ratio, that's a calculation based on volume. If you compress the air with a compressor, that pressure ratio is factored into the cylinder with the piston at BDC. Where is the calculation for the effective compression ratio where boost is added to the compression ratio at TDC? It doesn't exist, it's a conspiracy man. I mean what is the functional relationship? Is it additive? 14 psi of boost = 2:1 compression ratio, and an engine with a 10:1 geometric compression ratio now has an effective 12:1 geometric compression ratio? That can't be right, because then it would just be easier to use an 12:1 piston and save all the turbo plumbing. Is it multiplicative? 10x2=20? So a 10:1 CR piston with a 2:1 compression ratio from a compressor would effectively have 20:1 compression ratio? Or is it some other non-linear function that can only be gleamed through experimentation?

Like wtf bro, if you have a piston that's doing 12:1 compression ratio, that's a calculation based on volume. If you compress the air with a compressor, that pressure ratio is factored into the cylinder with the piston at BDC. Where is the calculation for the effective compression ratio where boost is added to the compression ratio at TDC? It doesn't exist, it's a conspiracy man. I mean what is the functional relationship? Is it additive? 14 psi of boost = 2:1 compression ratio, and an engine with a 10:1 geometric compression ratio now has an effective 12:1 geometric compression ratio? That can't be right, because then it would just be easier to use an 12:1 piston and save all the turbo plumbing. Is it multiplicative? 10x2=20? So a 10:1 CR piston with a 2:1 compression ratio from a compressor would effectively have 20:1 compression ratio? Or is it some other non-linear function that can only be gleamed through experimentation?

(post is archived)

[–] • 2 pts

You are confused. Compression ratio is the cylinder volume of BDC to TDC. This ratio effects efficiency, not "effectiveness."

Boost is the measure of increase over atmospheric pressure. It doesn't change volumetric compression ratio. As it changes charge density and not physical volume on either end of the stroke/cycle.

[–] • 0 pt

Compression pressure is both the geometric and manifold pressure, this means that both the geometric and atmospheric compression ratios combine to arrive at the cylinder pressure at TDC. Which means there is an effective compression ratio. Compression ratio is compression ratio, it has nothing to do with efficiency and everything to do with the compression of the air.

[–] • 0 pt (edited )

It doesn't change volumetric compression ratio.

I know, I said that, but the density increase increases the effective cylinder pressure, in the same way that the piston itself increases the cylinder pressure as it moves to TDC. So what then becomes the effective compression ratio when boost is factored in? With an axial jet engine the compression ratio is straight forward, the pressure tap reads 147psi at the end of the second spool compressor, that's a 10:1 compression ratio.

With a piston engine, you have volumetric compression of the air in the cylinder and density compression with the turbo. Both are physically compressing the air at a certain ratio, the two events do not exist in isolation, and combine to produce an effective cylinder pressure.

[–] • 1 pt

the pressure tap is psia?

[–] • 0 pt

Ask what you're trying to accomplish. Because you're stating things which are incorrect and asserting I'm the one in error.

[–] • 0 pt

What part is incorrect? What I'm trying to accomplish is stated in the OP.

[–] • 1 pt

Because volumetric efficiency that's why.

[–] • 0 pt

You can assume things and still be in the ballpark, you can assume compression by the piston is adiabatic, when it's very far from it, yet calculation assuming adiabatic conditions gets you in the ballpark. Adding more parameters makes the calculation take longer but you start running into diminishing returns for the parameters you're considering.

[–] • 1 pt

But it's not just compression it's things like cam overlap, manifolds, heads, valves, etc.

This is why the easiest and most accurate way is still a dyno. Like you said you can get close but to perform at the top tier you need real data.

[–] • 0 pt

Volumetric efficiency itself is a measure of how well the geometric compression ratio translates to actual compression of the air. Boost doesn't affect volumetric efficiency, only the density of the fluid. However at certain pressure ratios, the laws of compressibility alter the volumetric efficiency of the engine, because compressibility changes the rules of airflow. When the air approaches sonic velocities(achieved when pressure ratios approach 2:1 between a reservoir and a throat) then a constriction increases pressure, which is the opposite in a subsonic flow.

Think about this when designing intake manifolds, thin long runners are great for low speed operation, because it increases velocity of the air, at higher mass flow, the pressure ratio between the manifold and runner approaches 2:1 and you get sonic choking at the throat. So you design big fat short runners for high mass flow, but then the air doesn't accelerate enough at low speed. So then you try to get the best of both worlds, you use really smooth bell mouth runners to get more flow at low speeds, while still flowing well at higher mass rates.

[–] [deleted] • 1 pt

There is a simple equation for compression ratio of an engine. CR = (Vd + Vc) / Vc

You have all the world's information available to you, why don't you search this out and learn what the fuck you are babbling about.

[–] • 0 pt

I did. Boost pressure is a compression ratio that's added to the geometric compression ratio. There's nothing in the literature regarding the effective compression ratio at top dead center when boost pressure is factored in. Where is boost over atmospheric in your fancy formula.

[–] • 1 pt

Read my reply. This was already explained. You are confused.

[–] • 0 pt

You're confused, or lack knowledge on the subject and cannot answer my question. It's like I'm talking to myself over here.

[–] • 1 pt

If the air going into the cylinder is 2:1 air that's double air bruh.

If the cylinder takes the ten cylinder air and squishes it to one cylinder air, that's ten air bruh.

But if it squishes double air into ten air, that's twenty air BRUH.

[–] • 1 pt

P1V1 = P2V2 P1/T1 = P2/T2 V1/T1 = V2/T2 compression ratio isV2/V1 .. P1' at Standard temp and pressure is 1 atmosphere plus max boost in psi divided by 14.7 figure out the temp rise for ideal .. figure out the after boost temp and pressure and calculate V1' .. this is larger than V1 required to get that unboosted temp and pressure.. then V2/V1'=EFFECTIVE comp ratio

[–] • 0 pt

Thank you sir, I owe you a beer or 12.

[–] • 1 pt

Thanks , I learned something

[–] • 0 pt

You should have never dropped out of school.

[–] • 0 pt

I finished. Train your AI nigger.

[–] • 0 pt

You should check if there is an equation for this. Sometimes you need a computer running a long calculation..

[–] • 0 pt

I have, there isn't. Not in the public domain anyway.