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Aerodynamics HW - fluid particles acceleration based on stream functio

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data
    prove that with flow in a corner, with stream function ψ=Axy, particles are accelerating per [itex]\frac{DV}{Dt}[/itex]=(A2(x2-y2))/r; A=const; r-distance from the center of the corner


    2. Relevant equations

    Vx=U=[itex]\frac{∂ψ}{∂y}[/itex] . . Vy=V=-[itex]\frac{∂ψ}{∂x}[/itex]

    a=[itex]\frac{∂V}{∂t}[/itex]+U[itex]\frac{∂V}{∂x}[/itex]+[itex]\frac{∂V}{∂y}[/itex]

    3. The attempt at a solution

    As per above equations i get velocity components as
    U=Ax and V=-Ay


    then since local acc is 0 acceleration is:

    a=Ax[itex]\frac{A(x-y)}{∂x}[/itex] - Ay[itex]\frac{A(x-y)}{∂y}[/itex]

    finally, as per my calcs, accelerations is:

    a=A2(x+y)

    where did this r come from and also (x2-y2). i was thinking using r2=x2+y2, and using to multiply the whole acceleration expression with r2/(x2+y2), but i am getting nowhere.

    help please
     
  2. jcsd
  3. Mar 4, 2014 #2
    Your equation does not give the acceleration. The acceleration is a vector, and you need to find its components first, before determining its magnitude.

    [tex]a_x=u\frac{∂u}{∂x}+v\frac{∂u}{∂y}[/tex]
    [tex]a_y=u\frac{∂v}{∂x}+v\frac{∂v}{∂y}[/tex]

    Even with this, you still don't match the answer you are trying to prove.

    Chet
     
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