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 .
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 .