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Distrubing Results

  1. Jul 5, 2007 #1
    Distrubing Results!!

    I was looking through a proof in my aerodynamics book after I got hung up on something trivial during the derivation. I figured that out, but then I came across something that is very disturbing...

    Look at any proof you want that involves a differential element of size (dx,dy,dz). It is always the case that as you move in the positive (x,y,z) direction, that you have an increase in properties. For example, at one face, you will have a velocity [tex]u[/tex], and at that same face shifted dx to the right, it will have a slightly different velocity [tex]U+\frac{du}{dx} dx [/tex]. Now, this is the standard convention. In fact, this convention is implicit in any formula where you use variational calculus By the chain rule, we have defined: [tex] du= \frac{du}{dx}dx +\frac{du}{dy}dy +\frac{du}{dz}dz [/tex]. So, anywhere you are using the total derivative in a proof, you have stealthily, imposed the condition that flow properties are increasing as you move in the positive (x,y,z) directions, even if you dont draw a differential element in your derivation.

    This is true in the case of the Bernoulli equation, where you take the Navier-Stokes equation and notice that du="du/dx dx...etc etc." and work it out. Nowhere in that derivation did you actually draw a differential free-body diagram and assign a directional increase in properties as you move in the positive (x,y,z) directions, yet you forced yourself into this convention the instant you used the total differential.

    Now lets use the normal convention, making increases in properties as you move in the positive directions and derive Euler's equation.


    Look at the case of an incremental control volume shown above. We clearly see to the left that the flow properties are given by:[tex] \rho ,A,u,p [/tex]

    As per our convention, to the right the flow properties are given by:[tex] \rho + d\rho,A+dA,u+du,p+dp [/tex]

    Now if we use the differential form of the momentum equation we see that:

    [tex]pA + \rho u^2A +pdA = (p+dp)(A+dA) + (\rho+d\rho)(u+du)^2(A+dA)[/tex]

    Expanding this out yields:

    [tex] Adp + Au^2d \rho + \rho u^2dA +2 \rho uAdu=0[/tex]

    Next, take the derivative of the continuity equation:

    [tex] \rho u^2dA+\rho uAdu+Au^2d \rho =0[/tex]

    and subtract it from the expanded equation. You are left with:

    [tex] dp=- \rho udu[/tex] -Q.E.D

    If we take this result and combine it with the continuity equation:

    [tex] \frac{d \rho}{\rho} + \frac{du}{u} + \frac{dA}{A} =0 [/tex]

    and you find, after some manipulation that:

    [tex] \frac{dA}{A}=(M^2-1)\frac{du}{u}[/tex]

    This resulting equation shows us, based on our initial convention, that for subsonic flow M<1, as the area increases +dA, the velocity decreases -du.

    Take this result and itterrate on the first derivation. Now say, to the right the flow properties are given by:[tex] \rho + d\rho,A+dA,u-du,p+dp [/tex]

    Notice I changed it to [tex]u-du[/tex]. Change the u+du terms to u-du terms in all the equations and try and get Eulers equation. You wont, in fact, its impossible. You have changed on term in the same equation, there is no way you could possibly end up at the same result.

    Im trying find an erorr in what Im doing wrong here, but Im not seeing one. Its disturbing becaue I see no reason why I cant assume that the velocity decreases while the area increases, yet get a totally bogus solution.

    Have you ever noticed that every differential element always increases as you move in the direction of dx? Its something I never picked up on until today.
    Last edited: Jul 5, 2007
  2. jcsd
  3. Jul 5, 2007 #2
    Physically, it makes sense to use +(du/dx)dx, because if du/dx is negative, then it will automatically subtract, conversely, if du/dx is positive, it will add.

    On the other hand, -(du/dx)dx will subtract if du/dx is positive, which would be in error with the physical representation of the system since the local gradient is positive.

    I suppose this could be an answer to my question, in a way.....hmmm. It would be a good justification as to why the change in properties is always conventionally taken as positive in the positive (x,y,z) directions.
    Last edited: Jul 5, 2007
  4. Jul 6, 2007 #3


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    Homework Helper

    You answered your own question. Of course, du/dx can have either sign. For example, I could choose to use coordinates that were inverted with respect to yours and then we certainly couldn't both find that u increased as we move "in the positive direction".
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