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Euler equation in Polar coordinates

  1. Sep 26, 2012 #1
    Hello.

    I have 2D Euler equation for fluids. I cant derive it in polar coordinates. I defined functions u(x,y,t) = u'(r, theta, t) and v(x,y,t) = v'(r, theta, t). I started by computing derivatives

    [tex]\frac{\partial u'}{\partial r}=\cos\theta\frac{\partial u}{\partial x}+\sin\theta\frac{\partial u}{\partial y}[/tex]
    and
    [tex]\frac{\partial u'}{\partial \theta}=-r\sin\theta\frac{\partial u}{\partial x}+r\cos\theta\frac{\partial u}{\partial y}[/tex]

    If I express du/dx and du/dy and insert that expressions into Euler Eq. I didnt obtain the right result. my result is

    [tex]\frac{\partial u'}{\partial t}+u'\left(\cos\theta\frac{\partial u'}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u'}{\partial\theta}\right)+v'\left(\sin\theta\frac{\partial u'}{\partial r}+\frac{\cos\theta}{r}\frac{\partial u'}{\partial\theta}\right)=pressure[/tex]
    & similar for v'. Correct result does not contain sin & cos expressions.

    Thanks
     
  2. jcsd
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