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Help A Car Guy Out (Volumetric Efficiency)

  1. Feb 27, 2005 #1
    Any car buffs here?

    Just trying to make sure that I understand (because I am currently confused) about how to convert calculations of air between mass and volume.

    The problem I am trying to solve is calculating the Volumetric Efficiency of an engine at specific points in time. I have a scan tool that will report the following items:

    Flow of Air into the engine in grams/second
    Intake Manifold Absolute Pressure (in Hg or KPA)
    Intake Air Temperature (F or C)
    etc.

    I know the "volume" of my motor, but how do I convert gm/sec into a given volume. Is this just a matter of me understanding the ideal gas law? PV=nRT
    I am just not for sure that I am getting all the measurements converted to the appropriate values.

    For example, assume I have an engine that is 5.7L in size (or ~346 cubic inches). At 100% volumetric efficiency, the motor should consume and expel 5.7 liters of air (although fuel is also pumped into the combustion chamber) every 2 revolutions of the crankshaft.

    Can anyone shed any light on this for me? Theories abound on the car forums, but I am not for sure they really know.

    One popular formula is:

    VE = (3456 x CFM) / (CID x RPM)

    CFM = Cubic Feet per Minute of Air
    CID = Cubic Inch Displacement
    RPM = duh

    Not for sure how this calcuation was developed. The device on most modern cars reports air coming into the motor in terms of g/s. So, if there is an easy conversion of air from g/s to CFM, I can do that (I think).

    If I posted this in the wrong forum, my apologies.
     
    Last edited: Feb 27, 2005
  2. jcsd
  3. Feb 28, 2005 #2

    FredGarvin

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    The mass flow (which you have in g/s) is easily used to calculate the volumetric flow by the relation [tex]\dot{m} = \rho Q[/tex] where [tex]\dot{m}[/tex] is the mass flow rate and [tex]\rho [/tex] is the density of the air entering into the engine. If you keep g/s as your mass flow units, you will have to have density in terms of g/in^3 which will result in your volumetric flow rate being in terms of in^3/sec. Once you know that it's just playing with units after that. It's pretty simple.

    I am still working on the 3456 number in your volumetric efficiency equation. It is undoubtedly a conversion factor, But for what I am still unsure.
     
  4. Feb 28, 2005 #3
    well volumetric efficiency of an engine is the ratio of the actual volume of air at atmospheric pressure and abmiant temp. that gets into the engine on the intake stroke over the total volume at BDC. as the rpm increases it has less time to suck in air so the volumetric efficiency decreases as rpm increases.

    say you have a cylinder that is 40 cubic inches, and on the intake stroke it takes in 35 cubic inches at atmosperic pressure, and at whatever the temperature of the ambiant air outside is. the volumetric efficiency is 35/40 x 100% = 88%.

    ok lets see, in a 4 stroke piston engine, each piston fires every other revolution, so every 2 revolutions, the engine would take in the 5.7L, in your case, at a theoretical 100% volumetric efficiency. so at 2000rpm, at 100% VE, the engine intake flow would be (2000/2) * 5.7 = 5700L per minute = 5700/60 = 95L per second at atmospheric pressure and ambiant temp. you can write this equation to use for any engine speed =>
    Flow = ((RPM/2)*Disp.)/60 = L/s

    now if we can convert this to grams per second, all you would have to do for any engine speed is plug is plug the rpm into the equation, and then read the flow from the intake manifold, divide by the output of the equation, and multiply by 100% to get volumetric efficiency.

    now we can use the ideal gas law PV=nRT. R is just a constant that depends on what units you use. in this case we will use R = 0.08206 and our units will be Liters (L) , atm (atmosphere - 1atm = atmospheric pressure), and Kelvin for temp. we are trying to solve for n which is number of moles.

    n = (PV)/(RT)

    lets say the outside temperature is 20 deg C which is about 65 deg F. this converts to 293K (K = 273 + C --> 273+20). pressure is 1 atm. volume is 95L. if you crunch the numbers you get 3.95 moles.

    ok, but now you need grams, but this post is getting long and i got class, i will come back later and finish it off. ill also summarize it later so you can see all the equations you need right in front of you.

    :bugeye:
     
  5. Feb 28, 2005 #4

    Clausius2

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    Kinda diffucult, isn't it?

    Maybe our friend will understand better the stuff If we show him the equations:

    -Volumetric Efficiency: [tex] \eta_v=\frac{\dot m_a}{\rho_{at} V_c / t_{cycle}}[/tex] where Vc is the volume of the combustion chamber just when the intake valve is completely closed (THIS volume is usually taken as the geometrical maximum volume of the chamber, but it really is not so, because there is a delay crank angle in which the intake valve is opened during the compression stroke). [tex] \rho_{at}=P_{at}/(R_g T_{at})[/tex] is the atmospheric air density, and [tex] t_{cycle} [/tex] is the time to one cycle:

    [tex] t_{cycle}= T/(2n)=2/n[/tex] for a 4 stroke engine, where n is the rpm's.

    So that:

    [tex] \eta_v=\frac{2 \dot m_a}{\rho_{at} V_c n}[/tex]

    Substutute the values into the last formula and you'll get your GLOBAL volumetric efficiency.

    Felix, the volumetric efficiency does not decrease always as the rpm's increase. Depending on how is the charge renovation process designed, and how much the close angles valves are delayed/advanced, the volumetric efficiency usually increases at low rpm's with the rpm's. It is so because the delay crank angle of intake valve closing is optimum to take advance of the inertia of the air which enters into the chamber.

    For high rpms, the volumetric efficiency is inversely proportional to the rpms. It is so because just at the intake there are sonic conditions once some critical rpms are reached. So that, the entering mass flow remains constant above these critical rpms, and ceteris paribus [tex]\eta_v \approx constant/n[/tex].
     
    Last edited: Feb 28, 2005
  6. Feb 28, 2005 #5
    i didnt know any equations for it, i was just trying to derive a basic equation so he could use the g/s reading from his meter to find it. but yea i see what you mean about the VE increasing to a point i just wasnt thinking straight. some engines are designed for a peak torque at a higher rpm range than others (big block v8 vs a rotary engine). the volumetric efficiency increases from idle because as the piston is approching BDC and the intake valve is closing the air is already moving. at slower speeds it is limited to the time the valve is open. when it starts to spin faster, the speed the air is moving helps shove a little more air in than at a slower speed. however, when it gets past a certain point, the time the valve is open is so small that the air velocity isnt enough to help and the VE starts to decrease.....do i have the idea? then with what i said about the different engines, a rotary engine would have its peak VE at a higher rpm than a big block v8?
     
  7. Mar 1, 2005 #6

    Clausius2

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    I think so.

    I'd not dare to afirm that. I don't know much about Wankel engines. But I think the existence of a torque-power peak is not explained only by means of the maximum of the volumetric efficiency. There are several factors like mechanic efficiency, combustion process, combustion efficiency, manifold design....etc.
     
  8. Mar 1, 2005 #7
    Thanks for all the info. A lot to digest so far. I have some more information that I will post to see if you guys can confirm/deny it, but the method is something along the lines of the post by FredGarvin.
     
  9. Mar 1, 2005 #8
    Well, I am not for sure, but I believe it is related to the fact that there are 1728 cubic inches in 1 cubic foot. I guess since the equation is using RPM instead of RPM/2, they double this conversion factor as well.
     
  10. Mar 1, 2005 #9
    I am fairly new to cars, and I always wanted to know a lot more about the way things work and function.

    I would say that the answer to your question is still dependent upon when the valve events actually occur. i.e. when does the intake valve open/close and when does the exhaust valve open/close. Valve events that are good for any motor at 5000 rpm probably won't have the same efficiency at 2000 rpm or 8000 rpm.

    There are so many factors to consider though:
    *Combustion Process (spark, burn rate, combustion chamber design, timing)
    *Cylinder fill charactersitics (cylinder head designers talk about things like "swirl" which is the characteristic of the incoming air to spiral down the cylinder as opposed to just a straight rush downward; otherwise the downward rush of incoming air can eventually adversely impact the piston on its way up)
    *Cam design - particularly the valve events; open/close events, particularly overlap (when both the exhaust and intake valve are open and the exhaust side is helping to "pull" in air)
    *Exhaust flow - back pressure from the exhaust system

    I don't have any empirical evidence (although I am working on it), but my assumption is that the engine will be most "efficient" when producing peak torque.
     
  11. Mar 1, 2005 #10
    BTW, I have some tuning software that allows me to edit the Volumetric Efficiency table within my car's computer, PCM.

    The table itself is supposedly measured in grams of air based on RPM and MAP (Manifold Absolute Pressure).

    My completely stock file looks like:

    [​IMG]

    My version of smoothing this graph looks something like this:

    http://www.ls2gearchatter.com/test/Stock_2001_vette_muti_factor_smoothing.JPG
     
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