Stream function

Stream function $\left( {{\rm{\psi  = constant}}} \right)$ gives the equation of a streamline and the flow velocity is obtained by differentiating ${\rm{\psi }}$.For a compressible flow $$\rho u = \frac{{\partial \psi }}{{\partial y}}$$ $$\rho v =  - \frac{{\partial \psi }}{{\partial x}}$$ For incompressible flow, $$u = \frac{{\partial \psi }}{{\partial y}}$$ $$v =  - \frac{{\partial \psi }}{{\partial x}}$$ Velocity potential: Velocity potential is defined by $\phi ,\phi  = \phi \left( {x,y,z} \right)$.For an irrotational flow,velocity is given by gradient of $\phi$. $$u = \frac{{\partial \phi }}{{\partial x}},v = \frac{{\partial \phi }}{{\partial y}},w = \frac{{\partial \phi }}{{\partial z}}$$ Stream function is defined for both rotational and irrotational flows,velocity potential is defined for irrotational flows only.However,stream function is defined for two-dimensional flows only,the velocity potential applies to three-dimensional flows also.Since irrotational flow can be described by velocity potential $\phi$,such flow are also called potential flow.

Example: 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 $\psi$. 

Solution:                       $u = 4x = \frac{{\partial \psi }}{{\partial y}}$ 

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

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

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

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

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

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

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

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

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 $\psi$ are perpendicular to lines of constant $\phi$ .