Question #149106
If for a two dimensional potential flow, the velocity potential is given by : φ = 4x(3y - 4), determine the velocity at the point (2, 3). Determine also the value of stream function ψ at the point (2, 3).
1
Expert's answer
2020-12-11T09:56:30-0500

In velocity we can write 


u=φx,v=φyu=\dfrac{\partial\varphi}{\partial x}, v=\frac{\partial\varphi}{\partial y}


Given φ=4x(3y4)\varphi=4x(3y-4)

u=φx=12y16,v=φy=12xu=\dfrac{\partial\varphi}{\partial x}=12y-16, v=\dfrac{\partial\varphi}{\partial y}=12x

Calculate the velocity at the point (2, 3)


u=20,v=24u=20, v=24

velocity=u2+v2=(20)2+(24)2=461|velocity|=\sqrt{u^2+v^2 }=\sqrt{(20)^2+(24)^2}=4\sqrt{61}

u=ψy,v=ψxu=\dfrac{\partial\psi}{\partial y}, v=-\dfrac{\partial\psi}{\partial x}

u=ψy=12y16u=\dfrac{\partial\psi}{\partial y}=12y-16

Integrate with respect to yy


ψ=6y216y+g(x)\psi=6y^2-16y+g(x)

ψx=dgdx=12x-\frac{\partial\psi}{\partial x}=-\frac{dg}{d x}=12x

Integrate with respect to xx


g(x)=6x2+Cg(x)=-6x^2+C

Then


ψ(x,y)=6x2+6y216y+C\psi(x,y)=-6x^2+6y^2-16y+C


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