Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Air Density, Pressure Differentials, Vacuum ?

  1. Jul 29, 2005 #1
    I am trying to understand some effects of pressure differences and air density particularly as it relates to the good ole american made V-8.

    One of my questions: What is the "speed" of air when a pressure differential is equalized?

    Assume I have a cube that is exactly one cubic foot in size. If the cube is in the room with me, after I close the lid I assume the pressure in the cube is the same as the pressure was in the room (~29.92inHg). If I remove 1 cubic foot of air from the cube (Can I?), what is the pressure in the cube? If I open the lid for 1 millisecond, is the cube instantaneously filled with air? Does it matter how large the opening is that I use to let air into the cube? What if the lid is only opened for 1 nansecond? Is the vacuum completely gone?

    In an engine, there are many restictions to getting air into the cylinder: Air filter, tubing with bends, throttlebody, Intake manifold, cylinder heads, valves, etc. Is the speed of the air related to the pressure differential between the cylinder and the intake manifold? Does piston speed have any impact on the speed of the air?

    Is the cylinder ever not completely full of air? In other words, does the density simply vary based upon the amount of time available to fill the cylinder (RPM), restrictions, density of incoming air, etc?

    I apologize for all the questions and my ignorance. I am just trying to better understand how my engine actually works.

  2. jcsd
  3. Jul 29, 2005 #2


    User Avatar
    Science Advisor
    Gold Member

    Practically, no you can't. But if you could, there would be no air in the cube, and hence no air pressure.

    Very few things happen instantaneously, but yes, this would happen rather quickly.

    The time taken to fill the chamber would indeed depend on the inlet dimensions. Opening the lid for 1 nanosecond might let some air in, but (depending on the opening, and a couple of other things), the vacuum would be decreased. The amount by which the vacuum decreases will depend on how much air got back into the chamber!

    In short, yes. It's essentially the vacuum created by the piston which is sucking the air into the engine (although in truth, it's atmospheric pressure pushing itself into the vacuum).

    To be honest, I'm not sure! It's probably misleading to deal with gases by using statements such as "completely empty" and "completely full". For practical purposes, you can always cram more air into a cylinder and make it "fuller".

    Although leaving the inlet valve open longer, having denser air, or larger ports will allow more air into the cylinder, I doubt that there is enough time in each inlet stroke to allow the cylinder charge and atmosphere to gain a state of equilibrium, setting aside inlet rarefacations due to inlet obstacles, or compressions due to any turbocharging. It won't be far off, however.

    It's late and I've had a few. I might change my mind tomorrow, so read the above with salt! :smile:

    Don't be daft. The ones who don't ask questions are the ignorant ones.
    Last edited: Jul 29, 2005
  4. Jul 29, 2005 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    If the cube had 1 cu ft of air in it, then in theory, you can remove 1 cu ft of air from it. In practice, you can get pretty darn close to that, but not all of it. It just depends on the type of pump and pumping time. But with a good turbo-pump (and a backing pump), you can easily get about 99.999999% of the air out in less than 30 minutes.

    Filled ? Yes. But to what pressure ? Most likely it won't fill to atmospheric pressure in a millisecond.

    It most certainly does. The larger the opening, the lower the impedance (resistance) against airflow.

    By "completely gone" do you mean to ask : "Is the box back at atmospheric pressure" ? It is most certainly not. The RMS speed of an air molecule is about 500 m/s, so if unimpeded, an air molecule will take almost a millisecond to travel across 1 foot. A nanosecond is hardly enough time, even if you take the lid completely off for this time.

    Absolutely ! Generally, you can use the relation [itex]vA = \Delta P / Z [/itex] for any uniform section of the air-flow path (v = air speed, A = cross section area, [itex]\Delta P[/itex] = pressure differential across that section, Z = impedance of that section). The impedance goes typically like [itex]Z = kL/R^n [/itex], where L is the length of the section, R is its radius and n is usually between 3 and 4, but depends on several factors such as whether the flow is laminar or turbulent, ballistic or molecular, etc. The impedance also goes up with every turn you introduce into the air-flow path.

    I'll leave the rest of the engine-related detals to those that actually know something about them. :biggrin:

    Edit : Brewnog got in before I wrote this down...phew !
    Last edited: Jul 29, 2005
  5. Jul 29, 2005 #4
    Ok, so you are both confirming my basic assumptions. I was getting confused regarding descriptions of Volumetric Efficiency. People will discuss a given engine in terms of VE, and use statements like "at point X of Manifold Absolute Pressure and point Y of RPM, the engine is 80% efficient". I don't believe most people understand what they are talking about when making statements like this. To my knowledge, the ideal gas law is used by most computers controlling fuel injected engines. These systems often display the VE of the engine in percentage terms, but this is a representation of Mass Flow compared to a baseline. This baseline is generally:

    100kPa (intake manifold)
    20*C intake air temperature

    and, Flow is determined/calculated by

    cubic ft/min = (Cubic Inches * (RPM / 2) ) / 1728

    In practical terms, I doubt if a small-block chevy mearsuring 350 cubic inches displaces less than 350 cubic inches per engine cycle (2 revolutions). By "practical terms", I mean that the engine has all the standard components and is operated below 8000RPM. I suppose that if one could spin the motor to say 20,000RPM or higher that the time allowed to "fill" the cylinder would overcome the speed at which the air is moving to fill the cylinder.

    It just made me wonder though if the cylinder ever really was "not full" or at what point the engine would stop displacing 350 cubic inches. I realize that there would never really be a "void" within the cylinder (e.g. like there was some sugar cubed size space that was simply void).

    Regarding the four stroke engine, I have been told that if a camshaft were designed to open the intake valve at top dead center (beginning of the intake stroke) and close it at bottom dead center (end of the intake stroke, that the engine could never go above ~2500RPM's. Beynond this RPM, supposedly the time allotted to fill the cylinder would overcome the speed at which the air is moving to fill the cylinder. Any thoughts there?
  6. Jul 29, 2005 #5
    What is RMS?

    Thanks. I deal with some common setups on many engines. I will have to try to figure out how to apply some of that. I have been doing much reading on laminar flow and pressure losses across restrictions particularly related to the intake air tract. However, much of it is still over my head.
  7. Jul 30, 2005 #6


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

  8. Jul 30, 2005 #7


    User Avatar
    Science Advisor
    Gold Member

    It has been discussed here yet. It is true. We have talked yet about angles delays. Sure if you search in PF about valve angle delays you will find the last thread talking about this. An advance in opening and a delay in closing intake valve is needed when operating at medium-high rpms, (or when the Mach number in the intake exceeds 0.6 or so), because it is needed to take into account compressible-inertial behavior of the inlet gas to conserve engine performance. In particular, if such delay-advance is not produced, volumetric efficiency would decay a lot and so engine power.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook