Recent content by kmot

  1. kmot

    I Why is this closed line integral zero?

    Yes. It's a density depending on pressure.
  2. kmot

    I Why is this closed line integral zero?

    g is a generic function of which we know just that it depends only on F(X) grad(F(x)) is gradient of F(x)
  3. kmot

    I Why is this closed line integral zero?

    I've found yet another way to prove this. Convert line integral to surface integral using Stokes theorem: ##\oint{\frac{1}{\rho} \cdot \nabla p \, dl } = \iint_{A}{\nabla \times \left( \frac{1}{\rho} \cdot \nabla p \right) \cdot \vec{n}\, dA } = \iint_{A}{\left(\nabla\frac{1}{\rho}\right) \times...
  4. kmot

    I Why is this closed line integral zero?

    That's wonderful. Thank you. OK, Kelvin theorem doesn't say that per se, but it is implied. It says: For barotropic fluid, i.e. fluid where ##\rho## is a function of only ##p## ##\frac{D\Gamma}{Dt} = \frac{D}{Dt}(\oint \vec{v} \,dl ) = 0## After some manipulations, mainly inserting Newton's...
  5. kmot

    I Why is this closed line integral zero?

    This problem comes from fluid dynamics where Kelvin circulation theorem states, that if density "rho" is a function of only pressure "p", then closed line integral of grad(p) / rho(p) equals zero. It seems so trivial, so that no one ever gives reason for this claim. When trying to solve it...
  6. kmot

    Fluid mechanics - Linearized shallow water equations

    I found the solution. Here are the intermediate steps
  7. kmot

    Fluid mechanics - Linearized shallow water equations

    Hi, In a text describing solution to linearized shallow water equations, I am not able to move forward. It's a 1 dimensional shallow water setup. There is a steady state (velocity) and (height of free surface). On top of this steady state there are u' and h' as disturbances. The goal is to...
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