ICE Physics question

[DGS49]

Imagine you have a cylindrical vessel with a capacity of one liter. You put a lid on it and seal it. Then you add air, up to a pressure of 12psi. By what factor have you increased the amount of air in the vessel?

So now imagine you have a two-liter internal combustion engine. You turbocharge the engine and the turbocharger provides a maximum of 12psi boost. The engine will now "act like" it has a displacement of greater than two liters...but how much greater? Is it effectively a three liter engine? Four?

That's the question.

 

[JoeMoma]

Ideally, PV=nRT

 

[Mikeoxenormous]

Imagine you have a cylindrical vessel with a capacity of one liter. You put a lid on it and seal it. Then you add air, up to a pressure of 12psi. By what factor have you increased the amount of air in the vessel?

So now imagine you have a two-liter internal combustion engine. You turbocharge the engine and the turbocharger provides a maximum of 12psi boost. The engine will now "act like" it has a displacement of greater than two liters...but how much greater? Is it effectively a three liter engine? Four?

That's the question.

And the answer is Alex...

 

[Likkmee]

Imagine you have a cylindrical vessel with a capacity of one liter. You put a lid on it and seal it. Then you add air, up to a pressure of 12psi. By what factor have you increased the amount of air in the vessel?

So now imagine you have a two-liter internal combustion engine. You turbocharge the engine and the turbocharger provides a maximum of 12psi boost. The engine will now "act like" it has a displacement of greater than two liters...but how much greater? Is it effectively a three liter engine? Four?

That's the question.

Dunno. I do know turbos and supers "effectively increase" the combustion ratio.

 

[evenflow1969]

Imagine you have a cylindrical vessel with a capacity of one liter. You put a lid on it and seal it. Then you add air, up to a pressure of 12psi. By what factor have you increased the amount of air in the vessel?

So now imagine you have a two-liter internal combustion engine. You turbocharge the engine and the turbocharger provides a maximum of 12psi boost. The engine will now "act like" it has a displacement of greater than two liters...but how much greater? Is it effectively a three liter engine? Four?

That's the question.

Lol, that depends on where you are. If you cap a bottle at sea level you would have to pump air out to get to 12 psi. Air pressure at sea lea level is over 14 psi.

 

[percysunshine]

Lol, that depends on where you are. If you cap a bottle at sea level you would have to pump air out to get to 12 psi. Air pressure at sea lea level is over 14 psi.

This is probably an excess pressure experiment. PV=nRT...before and after.

 

[evenflow1969]

This is probably an excess pressure experiment. PV=nRT...before and after.

So ya think he is asking how much more air would be present if you were to add 12 psi to a bottle at sea level making it basically 26 psi?

 

[Crepitus]

Ideally, PV=nRT

"Ideally".

I see what you did there.

 

[Crepitus]

Lol, that depends on where you are. If you cap a bottle at sea level you would have to pump air out to get to 12 psi. Air pressure at sea lea level is over 14 psi.

14.7 to be exact, but that's not what he's asking.

 

[Crepitus]

The easy answer about twice as much. Temperature is gonna make a difference.

Disclaimer:. I'm not an expert on the math, so I could be wrong.

But I don't think I am.

 

[percysunshine]

So ya think he is asking how much more air would be present if you were to add 12 psi to a bottle at sea level making it basically 26 psi?

Well, there is an 'act like' fuzzy question which involves the power delivered by a combustion process at two different pressures and fuel mixtures.

PV=nRT is a static description.

 

[evenflow1969]

The easy answer about twice as much. Temperature is gonna make a difference.

Disclaimer:. I'm not an expert on the math, so I could be wrong.

But I don't think I am.

Nope you are right temperature and pressure would be needed to figure moles and voles I just am not sure what exactly what he asking here.

 

[Crepitus]

Nope you are right temperature and pressure would be needed to figure moles and voles I just am not sure what exactly what he asking here.

I was assuming he wasn't on top of a mountain and room temperature.

 

[evenflow1969]

I was assuming he wasn't on top of a mountain and room temperature.

Well back in class we always assume STP on problems like this. For this forum this person may actually be using this at a current condition other than stp. Who knows what exactly he is trying to accomplish. Would rather not give him answers that would cause damage.

 

[Crepitus]

Well back in class we always assume STP on problems like this. For this forum this person may actually be using this at a current condition other than stp. Who knows what exactly he is trying to accomplish. Would rather not give him answers that would cause damage.

I hope he's not coming to a political forum to figure out his fuel curve or something similar.

 

[percysunshine]

The answer is 42.

 

[evenflow1969]

S

I hope he's not coming to a political forum to figure out his fuel curve or something similar.

Seems to be what he is doing

 

[DGS49]

OK. The difference between a "small" engine and a "big" engine is the displacement of the cylinders - the area of the bore times the stroke, measured in cc's or cubic inches. A four liter engine displaces twice as much volume as a two liter engine. Hence, it will have more power.

Under normal circumstances, the movement of the piston from the top of its throw to the bottom creates a vacuum that draws the air-fuel mixture into the combustion chamber. The pressure in the cylinder at that time is the same as the prevailing atmospheric pressure. The four liter engine will have TWICE AS MUCH air-fuel mixture as the two liter engine. Extraneous factors being equal.

The purpose of turbocharging and supercharging is to charge the cylinder with air-fuel mixture UNDER PRESSURE, so that it is not just the vacuum caused by the moving piston; that force is supplemented, thereby forcing MORE air-fuel mixture into the cylinder - as though it were a bigger engine.

My question is, if the pressure introduced by the turbo is 12psi, how much more air-fuel mixture is introduced into the combustion chamber? My thought is that a virtual volume could be calculated, to wit, the engine gets as much stuff as if it were a three liter engine, or whatever.

That PV=nRT sounds vaguely familiar, although temperature is not a factor in my hypothetical.

"The ideal gas law can also be written and solved in terms of the number of moles of gas: PV = nRT, where n is number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K" Whatever that means.

 

[Crepitus]

S

Seems to be what he is doing

 

[Crepitus]

OK. The difference between a "small" engine and a "big" engine is the displacement of the cylinders - the area of the bore times the stroke, measured in cc's or cubic inches. A four liter engine displaces twice as much volume as a two liter engine. Hence, it will have more power.

Under normal circumstances, the movement of the piston from the top of its throw to the bottom creates a vacuum that draws the air-fuel mixture into the combustion chamber. The pressure in the cylinder at that time is the same as the prevailing atmospheric pressure. The four liter engine will have TWICE AS MUCH air-fuel mixture as the two liter engine. Extraneous factors being equal.

The purpose of turbocharging and supercharging is to charge the cylinder with air-fuel mixture UNDER PRESSURE, so that it is not just the vacuum caused by the moving piston; that force is supplemented, thereby forcing MORE air-fuel mixture into the cylinder - as though it were a bigger engine.

My question is, if the pressure introduced by the turbo is 12psi, how much more air-fuel mixture is introduced into the combustion chamber? My thought is that a virtual volume could be calculated, to wit, the engine gets as much stuff as if it were a three liter engine, or whatever.

That PV=nRT sounds vaguely familiar, although temperature is not a factor in my hypothetical.

"The ideal gas law can also be written and solved in terms of the number of moles of gas: PV = nRT, where n is number of moles and R is the universal gas constant, R = 8.31 J/mol ⋅ K" Whatever that means.

What's the rating on the turbo? What's the displacement of your engine, what's the Boost threshold? Waste gate pressure? Are you intercooling? Methanol or water injection? You aren't putting out enough information.