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Compressible Fluid Thermodynamics

  1. Nov 2, 2005 #1
    Hi, Im new to physicsforums, and have no higher education in math or physics; but have much interest.

    I do not think I understand correctly the relationships described under the 'deal gas law' PV=nRT. Specifically, my question is:

    If pressure and temperature are directly proportional when volume and mass are constant, and volume and pressure are inversely proportional when mass is constant, then how exactly does one describe the thermodynamic relationship in say a compression action of ambient air? Such as the compression stroke of an internal combustion engine?

    Ambient air (say 295K) compressed with a 10:1 compression ratio in an engine, should then produce a charge (before ignition) of about 2928*C? This is assuming on heat loss through conduction to the cylinder, but still, how much of this astronomical temp could be lost to the cooling system?

    (273 + 22) * 10 - 22 = 2928*C.

    Am I way off the mark here? This seems like a truly obscene temperature, especially when one considers the autoignition temperature of the air/fuel mixture involved being only around 750*C or so. And the fact that economy engines often use around 10:1 compression ratio with aluminum cylinder liners, heads, and cast aluminum pistons, which should easily melt in such an environment.

    This leads me to believe that my understanding here is fundamentally flawed, and I would appreciate any and all direction regarding this issue.

    Thank you for your time.
     
  2. jcsd
  3. Nov 3, 2005 #2

    Integral

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    I think your conclusion is correct.

    It is not at all clear to me where you got this calculation!
     
  4. Nov 3, 2005 #3
    The equation is derived from:

    273 + 22 is a correction to turn ambient temperature in degrees celsius to kelvins, multiplied by the compression ratio (10), which then has 22 subtracted from it to convert back to degrees celsius.
     
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