# Aerodynamics HW - fluid particles acceleration based on stream functio

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1. Mar 4, 2014

### leonida

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 $\frac{DV}{Dt}$=(A2(x2-y2))/r; A=const; r-distance from the center of the corner

2. Relevant equations

Vx=U=$\frac{∂ψ}{∂y}$ . . Vy=V=-$\frac{∂ψ}{∂x}$

a=$\frac{∂V}{∂t}$+U$\frac{∂V}{∂x}$+$\frac{∂V}{∂y}$

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$\frac{A(x-y)}{∂x}$ - Ay$\frac{A(x-y)}{∂y}$

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.

2. Mar 4, 2014

### Staff: Mentor

Your equation does not give the acceleration. The acceleration is a vector, and you need to find its components first, before determining its magnitude.

$$a_x=u\frac{∂u}{∂x}+v\frac{∂u}{∂y}$$
$$a_y=u\frac{∂v}{∂x}+v\frac{∂v}{∂y}$$

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

Chet