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Compression, psi and temp.

  1. Mar 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Trying to calculate changes in psi and temperature with a known change of volume.
    One cubic foot of air(14.7 psi) compressed to one tenth its original volume. 10:1 compression ratio.
    Starting temp=70° F or 294° K


    2. Relevant equations
    Don't know the equasion. That is the problem. I am really looking for the correct formula so that I can apply it to my engine building. It helps with boost and compression ratios.


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 28, 2009 #2

    mgb_phys

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    The formula you want is
    PV = nRT
    n is the amountof gas (number of moles) and R is a constant depending on your units.

    But if you don't change the amount of gas you can think of this as
    PV/T = constant
     
  4. Mar 28, 2009 #3
    The gas is a fixed amount. Trapped volume in the engine world.
    Does the constant have a known value. Would you mind walking me through to find the solution.
     
  5. Mar 29, 2009 #4
    Are you looking at an IC engine situation? If so, you want an adiabatic compression description for which
    PV^gamma = const
     
  6. Mar 29, 2009 #5

    mgb_phys

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    If you reduce the volume by 10 then you increase the pressure by 10 - assuming the temperature stays the same,
    In a real situation it's more complicated, the temperature will also go up (like a bike pump getting hot) and then heat will be lost (to the radiator) which will reduce the pressure
     
    Last edited: Mar 30, 2009
  7. Mar 29, 2009 #6
    DrD is right. In PV=nRT there is one too many unknows. You need to specify the conditions of compression: Both constant termperature is common. And the case where no heat is transfered.

    I don't recall the exponent, gamma. However, we do know that air is primarily diatomic molecules, which determines gamma when no heat is transfered to the environment, right?
     
    Last edited: Mar 29, 2009
  8. Mar 30, 2009 #7
    We need to hear from A Atwood as to what his problem really is before we can make any headway on this matter.
     
  9. Mar 31, 2009 #8
    I appreciate the help. What I am calculating, is the psi and temp changes that go on inside of a running engine.

    I found a couple of formulas on your site by doing some searching. I believe they are the correct formuls that I need.

    P2 = P1 (V1/V2) ^y

    for pressure change, and

    T2 = T1 (V1/V2) ^{(y-1)}

    for temp change.

    Y=1.4

    No heat gained or lost from surrounding combustion chamber(adiabatic).

    For a 10:1 engine working at 100% volumetric efficiency, and starting temp of 70° F. I come up with:

    Psi=369
    Temp=896

    Correct??
     
  10. Mar 31, 2009 #9
    The adiabatic compression model works well for the compression stroke, up to the point where combustion begins. At that point, both temperature and pressure rise by some other, undefined rule. Following shortly after TDC, there is an adiabatic expansion phase (the power stroke) until the valves open and the cylinder pressure drops very rapidly at that point.

    The value of gamma depends on what you are compressing. Your 1.4 value is about right for air, but I have seen values as low as 1.35 used. It is hard to predict because what you have is a mixture, not a pure gas (at least that is true in most cases), so the gas properties depend on the mixture and are hard to predict by chemistry.

    The reason for the adiabatic model is that heat transfer is a relatively slow process, and in a single crank cycle, there simply is not time for significant heat transfer to take place.

    PS: your pressure value looks OK, assuming you start with a cylinder pressure at 14.7 psia at the beginning of the stroke.
     
    Last edited: Mar 31, 2009
  11. Apr 1, 2009 #10
    Thanks Dr. D.

    I am only interested in the pressure and temp just prior to ignition @ TDC(top dead center).
    Reason is, knowing these values, helps in determining if the pressure and temperature are signifigant enough to create detonating or pre-ignition. Octane of the fuel requirements as well.

    Is my temp value correct?

    ARN
     
  12. Sep 13, 2009 #11
    disclaimer: i'm an electrical engineer, and am just self-learning the thermo stuff

    i agree on the pressure, though wikipedia has 323 psi (22:1 ratio). i've posted that i disagree and that they don't have any references. i'd love to know if we're right or wp ...

    http://en.wikipedia.org/wiki/Compression_ratio#Compression_ratio_versus_overall_pressure_ratio

    looks like you used 300 C for your calc of temp. 70oF is actually 294 C. but assuming that's what you did, i agree
     
  13. Sep 15, 2009 #12
    For the temp calcs, I had to convert to kelvin, then back to F. All temp calcs must be based off of absolute.

    ARN
     
  14. Sep 15, 2009 #13
    s/C/K/g

    sorry for the type, but the point still stands ... 70oF --> 294 K

    Tc = To * rc^.4
    (300*10^.4 - 273.15) *9/5 + 32 = 896.75 oF
    (294*10^.4 - 273.15) *9/5 + 32 = 869.62 oF

    looks like you used 300K instead of 294K. or did you do something different ?
     
  15. Sep 15, 2009 #14
    Maybe I'm just a little dyslexic, I tried re-calculating and came up with the same result as you.

    70°F with 10.0:1 compression -869°F.

    Not sure what I did wrong the first time.

    Thanks for the correction.

    ARN
     
  16. Jan 21, 2012 #15
    I thought it would be easy Using PV=nRT I get P1V1=nRT1 P2V2=nRT2 (the n and R don't change if you compress or heat it). I rearrange and get V1=nRT1/P1 also T2=P2V2/nR. I also know that V1=(1/10)V2 by combining and canceling I get T2=P2T1/P1. T1=294 P1=14.7. so T2=P2*(294/14.7)

    Long story short I ended up with a relationship between T2 and P2, but there are still too many variables.
     
  17. Jan 21, 2012 #16

    Redbelly98

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    Welcome to PF.
    You have either too many variables, or not enough equations. Try using one of these:
     
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