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Fluid mechanics: calc the acceleration of a particle at a point.

  1. Nov 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A velocity field is given by [itex]\vec{V}[/itex]= [Ax[itex]^{3}[/itex] + Bxy[itex]^{2}[/itex]][itex]\hat{i}[/itex] + [Ay[itex]^{3}[/itex] + Bx[itex]^{2}[/itex]y][itex]\hat{j}[/itex]; A=0.2 m[itex]^{-2}[/itex]s[itex]^{-1}[/itex], B is a constant, and the
    coordinates are measured in meters. Determine the value and
    units for B if this velocity field is to represent an incompressible
    flow. Calculate the acceleration of a fluid particle at point
    (x, y)=(2, 1). Evaluate the component of particle acceleration
    normal to the velocity vector at this point.

    2. Relevant equations

    u=[itex]\frac{\partial\Psi}{\partial y}[/itex] v=-[itex]\frac{\partial\Psi}{\partial x}[/itex]

    3. The attempt at a solution

    I used the above equation to get value equations u and v, there is something I'm missing, a bit of reasoning that has to be made using the given information to determine a constraint that exists since the fluid is incompressible. This should link the rates of change to each other. But that's where I get stuck, having trouble hanging on to all the concepts..

    u = [2Bxy][itex]\hat{i}[/itex] + [.6y[itex]^{2}[/itex] + Bx[itex]^{2}[/itex]][itex]\hat{j}[/itex]
    v = -[.6x[itex]^{2}[/itex] + By[itex]^{2}[/itex]][itex]\hat{i}[/itex] - [2Bxy][itex]\hat{j}[/itex]

    Assumptions:
    1. Incompressible flow
    2. B is constant
     
    Last edited: Nov 24, 2012
  2. jcsd
  3. Nov 24, 2012 #2

    mfb

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    "Incompressible" links the derivatives of the velocities in both directions.
    This gives you one constraint and allows to calculate B.
     
  4. Nov 24, 2012 #3
    so du = -dv? increases of u lead to decreases of v.
     
  5. Nov 25, 2012 #4

    mfb

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    Not du = -dv, but du/dx = -dv/dy (or the other way round?) should work.
     
  6. Nov 25, 2012 #5
    when I try [itex]\frac{du}{dx}[/itex] = -[itex]\frac{dv}{dy}[/itex] I get

    [2By][itex]\hat{i}[/itex] + [2Bx][itex]\hat{j}[/itex] = [2By][itex]\hat{i}[/itex] + [2Bx][itex]\hat{j}[/itex]

    which doesn't help solve for B, unless I'm messing up somewhere or forgetting something.
    The same happens when I take [itex]\frac{du}{dy}[/itex] = -[itex]\frac{dv}{dy}[/itex]

    [2Bx][itex]\hat{i}[/itex] + [1.2y][itex]\hat{j}[/itex] = [1.2x][itex]\hat{i}[/itex] + [2By][itex]\hat{j}[/itex]
     
  7. Nov 25, 2012 #6

    mfb

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    The second equation can be solved for B.
     
  8. Nov 25, 2012 #7
    I'm getting B = 1.2/2
    does that look right?
     
  9. Nov 26, 2012 #8

    mfb

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    I think so. It solves the equations for i and j at the same time, which looks good.
     
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