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Math Help - dP/dz = -kg

  1. #1
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    Question dP/dz = -kg

    Hi,

    this is physics, but it's the math that is a problem to me. Problem is added in pdf-file. Can anyone quide me from EQ1 to EQ2 and further to EQ3?

    If I use a computer program it gives me arctan...

    Thanks!

    - Trond
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  2. #2
    Super Member Aryth's Avatar
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    To get from equation 1 to equation 2, you must differentiate the mean pressure function with respect to z:

    \frac{\partial}{\partial{z}} p(z) = \frac{0\left(1 + \left(\frac{z}{h}\right)^2\right) - p_0\left(0 + \left(\frac{2zh^2 - 0z^2}{h^4}\right)\right)}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}

    \frac{\partial}{\partial{z}} p(z) = \frac{-p_0\frac{2z}{h^2}}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}

    \frac{\partial}{\partial{z}} p(z) = -2\frac{p_0z}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2h^2}

    That's to EQ2.

    All they do for EQ3 is just substitute p_0 for P, and since I'm not too familiar with this topic, I'd just go with it.
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  3. #3
    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Aryth View Post
    To get from equation 1 to equation 2, you must differentiate the mean pressure function with respect to z:

    \frac{\partial}{\partial{z}} p(z) = \frac{0\left(1 + \left(\frac{z}{h}\right)^2\right) - p_0\left(0 + \left(\frac{2zh^2 - 0z^2}{h^4}\right)\right)}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}

    \frac{\partial}{\partial{z}} p(z) = \frac{-p_0\frac{2z}{h^2}}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2}

    \frac{\partial}{\partial{z}} p(z) = -2\frac{p_0z}{\left(1 + \left(\frac{z}{h}\right)^2\right)^2h^2}

    That's to EQ2.

    All they do for EQ3 is just substitute p_0 for P, and since I'm not too familiar with this topic, I'd just go with it.
    I also find strange that g appears in EQ 1 and not in EQ 2. But it wouldn't be constant anyway... I don't really know what happened to g in the derivation.
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  4. #4
    Super Member Aryth's Avatar
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    I believe you are right. I think they are 2 separate equations derived independently and not one from the other.
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  5. #5
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    EQ 2 and 3 are only the left side of EQ 1. Further I set the solution of the diff like right side of EQ 1. So g will be back in EQ 4.

    Thanks, this helped me further on. (By the way, P is pressure. And the next steps will end up with a function of temperature with respect to z (after substitude with a gas law). z is height from sea level).
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