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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 \(\phi \) .

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