Boost as it relates to relates to port air-flow characteristics
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Boost as it relates to relates to port air-flow characteristics I am inquiring on the differences of velocity, density, and consequently CFM of a boosted engine versus an NA engine.
When the charger is producing boost, is it increasing the density of the air by using the entire intake system (including the cylinder while the intake valve is open) as, in affect, a single container to press against?
If this is true, how is it that a charger does not increase the velocity of a port as stated by numerous intellects?
If this is false, how is it that a charger compresses air? I could imagine the weight of the air itself at the compressor outlet acting like a wall of my "single container", thus offering the charger something to press against, however, simply at high speeds. I liken it to falling into water at a high speed; for mortal purposes, the water might as well be concrete at these speeds.
On an engine with our mild piston speeds, is a charger producing enough compressed air to provide the cylinder at all degrees of crank rotation (or length of stroke) during the intake stroke, a full charge of the compressed air. OR - does the compressed air have time to seek its natural, uncompressed state in the newly immerging volume of the "single container," effectively uncompressing itself and then being compressed again as the charger plays catch-up?
Please shed some light on my dilemma - I cannot sleep and it is driving me crazy.
Mike
PS - I realize all of these things change in accordance with the RPM, so feel free to explain how the system changes with the RPM.
When the charger is producing boost, is it increasing the density of the air by using the entire intake system (including the cylinder while the intake valve is open) as, in affect, a single container to press against?
If this is true, how is it that a charger does not increase the velocity of a port as stated by numerous intellects?
If this is false, how is it that a charger compresses air? I could imagine the weight of the air itself at the compressor outlet acting like a wall of my "single container", thus offering the charger something to press against, however, simply at high speeds. I liken it to falling into water at a high speed; for mortal purposes, the water might as well be concrete at these speeds.
On an engine with our mild piston speeds, is a charger producing enough compressed air to provide the cylinder at all degrees of crank rotation (or length of stroke) during the intake stroke, a full charge of the compressed air. OR - does the compressed air have time to seek its natural, uncompressed state in the newly immerging volume of the "single container," effectively uncompressing itself and then being compressed again as the charger plays catch-up?
Please shed some light on my dilemma - I cannot sleep and it is driving me crazy.
Mike
PS - I realize all of these things change in accordance with the RPM, so feel free to explain how the system changes with the RPM.
are you talking turbocharger or supercharger? a turbo compresses intake air and the speed of the compressor wheel is dependant on the load of the engine not exactly rpm. the turbo is run on exhaust gas so the more load on the motor the higher the exhaust velocity and more air the turbo will compress(boost). a turbocharged motor will move more air then an na engine hence having to add fuel when adding a turbo or upping the boost on a turbo the fuel compensates for the more air the turbo sucks into the motor. not sure exactly what else you want to know. i dont really know much about superchargers, si i cant really help you there.
Last edited by 97blackz28; Feb 4, 2006 at 10:27 PM.
The only way the port velocity would not increase with a higher manifold pressure is if it was already at mach 1 at some point. Velocity of the air mostly depends on the ratio of static pressure (for example at the valve seat)to total pressure (manifold pressure is close enough). With a pressure ratio of 1 there is 0 velocity. With a pressure ratio of about 0.5 the air will reach mach one. At pressure ratios in between, the velocity is between 0 and the speed of sound, and there is a formula to calculate it. But the point is, unless you have mach 1 flow in the port, increased manifold pressure will increase port velocity. Whether most engines see mach 1 in the intake port, i don't know.
If you look at a ve table, you will never see ve decreasing with manifold pressure. It may start to level off, but it won't decrease.
Also, i agree that the increased density of the air has a bigger effect than the increased velocity. I'm just saying velocity does change.
If you look at a ve table, you will never see ve decreasing with manifold pressure. It may start to level off, but it won't decrease.
Also, i agree that the increased density of the air has a bigger effect than the increased velocity. I'm just saying velocity does change.
There is no significant change in velocity or volume in the port, but there is a big change in both at the intake valve. The reason is the air acts like a spring or slinky based on the pressure it sees.
A VE table is just a table and will never change unless you change it. If actual VE changes or not depends on which part of the system you are measuring. The VE of the intake and ports do not change significantly because a fairly stable/constant pressure is maintained keeping roughly the same volume moving through.
As I said though, volume past the intake valve goes way up because of the pressure change. Because of the nature of the boosted system you cannot fill the cylinder enough to equal the boost pressure in the manifold. Even if you started with a naturally aspirated engine capable of 100% VE you couldn't do it.
A VE table is just a table and will never change unless you change it. If actual VE changes or not depends on which part of the system you are measuring. The VE of the intake and ports do not change significantly because a fairly stable/constant pressure is maintained keeping roughly the same volume moving through.
As I said though, volume past the intake valve goes way up because of the pressure change. Because of the nature of the boosted system you cannot fill the cylinder enough to equal the boost pressure in the manifold. Even if you started with a naturally aspirated engine capable of 100% VE you couldn't do it.
Originally Posted by mgray
When the charger is producing boost, is it increasing the density of the air by using the entire intake system (including the cylinder while the intake valve is open) as, in affect, a single container to press against?
If this is true, how is it that a charger does not increase the velocity of a port as stated by numerous intellects?
PS - I realize all of these things change in accordance with the RPM, so feel free to explain how the system changes with the RPM.
If this is true, how is it that a charger does not increase the velocity of a port as stated by numerous intellects?
PS - I realize all of these things change in accordance with the RPM, so feel free to explain how the system changes with the RPM.
Perhaps we could consider the Mass air Flow. If we increase the Mass of the charge then the velocity need not increase with a relative increase in density. There is also the increase in mass from the fueling in-order to maintain the target Air Fuel Ratio. With Higher Densities there would be associated viscosity and momentum influences again resisting increased velocity.
Therefore under increasing pressure we have more mass with no increase or perhaps a decrease in velocity.
With enough pressure the air would condense, and you would not have enough velocity to flow completely past the valve.
but as usual I could be wrong.
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whites10 - are you saying it is totally impossible, or simply with the current charger technology? If I hooked an engine up to a monster tank of compressed air at some ungodly pressure, I cannot see the engine not being able to run. Or, are you saying that even that is impossible? Enlighten me.
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It's possible to equal pressure in the cylinder if the engine is not running or it's running at a very low RPM. When the engine speed goes up there just isn't time to equal the pressure anymore.
The more boost you drive into the intake the bigger the difference in pressure becomes. Boosting isn't about efficiency, it's about acceptable losses for a large total power output.
I wouldn't say impossible. Some configuration like a single cylinder engine being driven by a charger that's actually too small for the engine might do it.
I dismiss it because none of those situations are relavent to an engine running at WOT with an effective booster.
The more boost you drive into the intake the bigger the difference in pressure becomes. Boosting isn't about efficiency, it's about acceptable losses for a large total power output.
I wouldn't say impossible. Some configuration like a single cylinder engine being driven by a charger that's actually too small for the engine might do it.
I dismiss it because none of those situations are relavent to an engine running at WOT with an effective booster.
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Great question. Does anyone out there have flow #'s for heads @ 28"WC vs 14.7 PSI vs 29.5 PSI? That's really what I think you are asking right? A head/port design that flows a given mass NA may not necessarily work well under higher mass flows. 
For those thermodynamic majors, Boyles law states that for air at constant temperature P1 x V1 = P2 x V2, and Charles law that the volume of a gas is directly propostional to its temperature V1/T1 = V2/T2, where temperature = Deg F +460.
If you double the pressure the volume of space the mass of air occupies gets cut in 1/2.
So for a fixed volume (like a cylinder) doubling the pressure (2 x atmospheric) you can get 2 times the amount (mass) of air in it.
This requires that the device pressurizing the cylinder overcome the flow losses (intake and heads) associated with getting twice the air into cylinder and is capable of flowing that mass of air.
This assumes the incoming air temperature is = to the delivered air temperature. Raising the temperature of the intake charge decreases its density and thus mass for a given volume. Thus the use of intercoolers.
Some one out there (probably the turbo manf's) knows these answers.
For those thermodynamic majors, Boyles law states that for air at constant temperature P1 x V1 = P2 x V2, and Charles law that the volume of a gas is directly propostional to its temperature V1/T1 = V2/T2, where temperature = Deg F +460.
If you double the pressure the volume of space the mass of air occupies gets cut in 1/2.
So for a fixed volume (like a cylinder) doubling the pressure (2 x atmospheric) you can get 2 times the amount (mass) of air in it.
This requires that the device pressurizing the cylinder overcome the flow losses (intake and heads) associated with getting twice the air into cylinder and is capable of flowing that mass of air.
This assumes the incoming air temperature is = to the delivered air temperature. Raising the temperature of the intake charge decreases its density and thus mass for a given volume. Thus the use of intercoolers.
Some one out there (probably the turbo manf's) knows these answers.
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We need to remember that velocity is just a vector. In "charging" incoming air you increase the density of a given volume, you can look at it as charging a battery...you are just loading it up with electrons where as here your loading the plenum with air. Just because denstiy increases doesn't mean that velocity increases. When using a "charger" you are increasing the force which acts on the air. As you know when a vaccum is created, air is pushed it because everything flows from higher concentrations to lower concentrations, IT IS NOT SUCKED IN! This pressurization is increasing the density which is increaseing the mass in a constant volume. This increase in mass increases the forces acting on the plenum charge, giving that air a greater push. You're port configuration will dictate the optimum velocity which can dictate CFM (CFM is a bad acronym cause it can be the most mis-leading 3 letters you'll ever come across). In conclusion, in a "charger" you are varying acceleration (don't confuse that with velocity) and mass, which as you know is force.
