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Material/Fluid derivative operator questions

  1. Dec 26, 2012 #1

    The above link shows the material derivative. Which is the derivative that follows a volume of fluid throughout its movement through a fluid. How is this derived from a chain rule? Is the v in that equation the velocity field of the fluid at each point throughout space? Do you act on the position of some arbitrary fluid volume with this operator? Thanks ahead of time.
  2. jcsd
  3. Dec 26, 2012 #2
    Suppose you have a property of interest, say density.

    This property varies in both space and time so f(ρ) = f(x,y,z,t).

    We have to consider what we mean by the partial time derivative [itex]\frac{{\partial f}}{{\partial t}}[/itex]

    This means the time variation of ρ at a fixed point in space (x,y,z t) since we hold x,y,z constant to take the partial derivative with respect to t.

    If we want toconsider 'following' a parcel of fluid then we use the rate of change 'following the fluid' [itex]\frac{{Df}}{{Dt}}[/itex]

    This is derived as follows.

    [tex]\frac{{Df}}{{Dt}} = \frac{{\partial f}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial f}}{{\partial y}}\frac{{dy}}{{dt}} + \frac{{\partial f}}{{\partial z}}\frac{{dz}}{{dt}} + \frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial t}} + u\frac{{\partial f}}{{\partial x}} + v\frac{{\partial f}}{{\partial y}} + w\frac{{\partial f}}{{\partial z}} = \frac{{\partial f}}{{\partial t}} + (u.\nabla )f[/tex]

    Does this help?
  4. Jan 2, 2013 #3
    Thanks for the response. So you are saying that some function of a specific property we are interested in is a function of (x,y,z,t). "f(ρ) = f(x,y,z,t)". I would say that ρ=f(x,y,z,t), but I guess it is more general to state it as you did. Is this correct?
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