For a velocity field of incompressible flow given by u=4x and v=–4y, calculate the stream function and velocity potential.

For a velocity field of incompressible flow given by u = 4x and v =  – 4y, calculate the stream function and velocity potential.Show that lines of constant \phi are perpendicular to lines of constant \psi .

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    u = 4x = \frac{{\partial \psi }}{{\partial y}} 

     

    \Rightarrow \partial \psi  = 4x\partial y

     

    \Rightarrow \int {\partial \psi  = \int {4x\partial y} }

     

    \Rightarrow \psi  = 4xy + f\left( x \right)

     

                                          v =  – 4y = -\frac{{ \partial \psi }}{{\partial x}}

     

    \Rightarrow \frac{{\partial \psi }}{{\partial x}} = 4y

     

    \Rightarrow \partial \psi  = 4y\partial x

     

    \Rightarrow \int {\partial \psi  = \int {4y\partial x} }

     

    \Rightarrow \psi  = 4xy + f\left( y \right)

     

    On comparing these two equations for \psi , stream functions is \psi  = 4xy + {\rm{constant}}.

     

    Also                               u = 4x = \frac{{\partial \phi }}{{\partial x}}

     

    \Rightarrow 4x\partial x = \partial \phi

     

    \Rightarrow \int {\partial \phi  = \int {4x\partial x} }

     

    \Rightarrow \phi  = 4{x^2} + f\left( y \right)

     

                                          v =  – 4y = \frac{{\partial \phi }}{{\partial y}}

     

    \Rightarrow \frac{{\partial \phi }}{{\partial y}} =  – 4y

     

    \Rightarrow \int {\partial \phi  = \int { – 4y\partial y} }

     

    \Rightarrow \phi  =  – 4{y^2} + f\left( x \right)

     

    On comparing the above two equations, f\left( y \right) =  – 4{y^2} and f(x) = 4{x^2}, velocity potential is \phi  = 4\left( {{x^2} – {y^2}} \right).

     

    Now

     

    \psi  = 4xy + {\rm{constant}}, differentiating with respect to x, holding \psi  = {\rm{constant}}

     

    0 = 4x\frac{{dy}}{{dx}} + 4y

     

    \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\psi  = {\rm{constant}}}} =  – \frac{y}{x}

     

    Differentiating \phi  = 4\left( {{x^2} – {y^2}} \right) with respect to x, holding \phi  = {\rm{constant}} 

     

    0 = 2 \times 4x – 2 \times 4 \times y\frac{{dy}}{{dx}}

     

    \Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\phi  = {\rm{constant}}}} = \frac{x}{y}

    On comparing the above, we get {\left( {\frac{{dy}}{{dx}}} \right)_{\psi  = {\rm{constant}}}} =  – {\frac{1}{{\left( {\frac{{dy}}{{dx}}} \right)}}_{\phi  = {\rm{constant}}}}This shows that lines of constant \phi are perpendicular to lines of constant \psi .

    Worldtech Answered on 3rd January 2019.
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