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Navier Stokes, separation steady/non-steady

  1. May 26, 2007 #1

    I want, for obscur reasons which would lead us too far to explain, to split my flow into two component, one steady and another one non-steady

    [tex]v = v_0 + v'[/tex]

    I'm looking for a simple equation governing the evolution of this non steady components. The complete momentum equation gives

    [tex]\rho\frac{\partial (v_0 + v')}{\partial t} = -\rho\left(\left(v_0 + v'\right)\nabla\right)\left(v_0 + v'\right) - \nabla P + \rho g + \frac{1}{\mu_0}\left(\nabla \times B\right)\times B + \mu\triangle\left(v_0+v'\right) [/tex]

    Assuming that v0 is steady and that both v0 and v' are perturbations of the hydrostatic state, we can get rid of the pressure gradient and the gravity force, leading to the following equation :

    [tex]\rho\frac{\partial v'}{\partial t} = -\rho\left(\left(v_0 + v'\right)\nabla\right)\left(v_0 + v'\right) + \frac{1}{\mu_0}\left(\nabla \times B\right)\times B + \mu\triangle\left(v_0+v'\right) [/tex]

    I would be pleased to get an even easier equation, but I can't find out the hypothesis necessary to get rid of some of the advection terms or the v0 diffusive term.

    Are there some time/space scales which I should compare ?
  2. jcsd
  3. May 26, 2007 #2
    You need to decide if the Vo field is a solution of the Navier-Stokes equation.
    If yes, then the 0-order term disappears completely, of course.
    Then you could develop the equation to first order in v' and drop 2-order terms.
    Usually, the Vo field is a static solution and v' is a time-dependent perturbation.

    Are you studying non-ideal MHD or plasma physics? There are probably many books discussing that, although you will commonly find the effect of resisitivity included instead of the effect of viscosity.
    Last edited: May 26, 2007
  4. May 26, 2007 #3
    Hum I don't think what you say is the solution. You are talking about the perturbation method, where you drop the zero order term and 2nd order terms.

    but here, v' is *not* a perturbation in the way that we don't suppose that v' << v0, it is just a non-steady component. Just saying that the time scale of v0 variations is much much longer than the one for v'... v0 is then considered steady and v' is a deviation from the temporal average.

    I saw something close to what I want to do, in "renolds averaged navier stokes equation", but it seems it is usually used for turbulence modeling.

    In my study, v' doesn't mean turbulence motion at all. I'm looking for some justification of an equation used in solar dynamo modeling, invoquing "malkus-proctor mechanism"

    For solar dynamo modeling, it is usual to solve only the induction equation, with an analytical flow given. The flow consist.. for the sun... of the differential rotation (longitudinal velocity) and a meridional circulation. This is called kinematic dynamo. When one wants to study lorentz force feedback on the flow there is generally two methods, the first consist 'simply' to solve full mhd system by adding the navier stokes equation to your equations... but this is not trivial and needs energy equation etc...

    The second method is what is called "malkus proctor" mechanism, large scale magnetic field feedback on the flow by the following equation :

    [tex]\frac{\partial U'}{\partial t} = \Lambda \left(\nabla \times B\right)\times B + Pm\nabla ^2 U'[/tex] (2)

    with :

    [tex]u=U+U'[/tex] (1)
    [tex]\Lambda[/tex] is called the Elssasser number, Pm is the magnetic Prantl number.

    where U is a steady profile and U' the dime dependant flow field driven by lorentz force.

    This is the equation I want to derive from basic MHD equations for the flow... starting with the basic assumption that you can split the flow field in a steady profile and a time dependant component.

    What I don't really get, is why there is no advection term, or v0 diffusive term.. I could understand the pressure and gravitation miss by assuming perturbations of hydrostatic equilibrium as I said above... but I can't find out a rigourous derivation for the equation.
  5. May 26, 2007 #4
    I realise that maybe prescribing u0 and solving u' might not mean that u0 is steady and u' is not :-/
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