Skip to main content

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 \(\phi \) .

Comments

Post a Comment

Popular posts from this blog

Importance of Coefficients (Lift and Drag)

Aerodynamic coefficients plays an important role in performance analysis as well as  design of aeroplanes.Coefficient of lift is defined as Lift divided by dynamic pressure, \(L/{q_\infty }S\) where as drag coefficient is defined as Drag divided by dynamic pressure, \(D/{q_\infty }S\) . Coefficient of lift,maximum or \({C_{L,\max }}\) is the determining factor for stalling velocity of aircraft.The higher is the \({C_{L,\max }}\) the lower is the stalling velocity.\[{V_{stall}} = \sqrt {\frac{{2W}}{{{\rho _\infty }S{C_{L,\max }}}}} \] However,\({C_{L,\max }}\) can be increased by the use of mechanical devices like high-lift devices.High lift devices include flaps,slats and slots on the wing. Air plane flying at given altitude with maximum thrust \({T_{\max }}\), the maximum value of \({V_\infty }\) ,corresponds to flight at \({C_{D,\min }}\).\[{V_{\max }} = \sqrt {\frac{{2{T_{\max }}}}{{{\rho _\infty }S{C_{D,\min }}}}} \] The actual value of \({C_L}\) and...
Post a Comment
Read more

Non Lifting flow over a circular cylinder

When there is a superposition of a uniform flow with a doublet (Which is a source-sink pair of flows), a non-lifting flow over a circular cylinder can be analysed. Stream function \('\psi '\) for a uniform flow is \[\psi  = \left( {{V_\infty }r\sin \theta } \right)\left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right)\] velocity field is obtained by differentiating above equation\[{V_r} = \left( {1 - \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\cos \theta \]\[\,{\rm{and}}\,{{\rm{V}}_\theta } =  - \left( {1 + \frac{{{R^2}}}{{{r^2}}}} \right){V_\infty }\sin \theta \] Stagnation points can be obtained by equating \({V_r}\) and \({{\rm{V}}_\theta }\) to zero.Considering an incompressible and inviscid flow pressure coefficient over a circular cylinder is \[{C_p} = 1 - 4{\sin ^2}\theta \]Example: Calculate locations on the surface of a cylinder where surface pressure equals to free-stream pressure considering a non-lifting flow. Solution: Pressure coefficient ...
Post a Comment
Read more

Streamlines

Streamlines: Streamlines represents field lines in a fluid flow. It is a path traced out by massless fluid particles moving with the flow, which is tangential to the instantaneous velocity direction. Different streamlines at same instant in a flow do not intersect or flow across it, because a fluid particle cannot have two different velocities at the same point. For a unsteady flow, streamline pattern is different at different times because the velocity vectors are fluctuating with time both in magnitude and direction. In a fluid flow, a bundle of streamlines is called a streamtube. Walls of an ordinary garden hose form a streamtube for the water flowing through the hose. Differential equation for a streamline in two dimensions is \[vdx - udy = 0\] Question: x and y components of a velocity field are given as \(u = \frac{{4x}}{{\left( {{x^2} + {y^2}} \right)}}\) and \(v = \frac{{4y}}{{\left( {{x^2} + {y^2}} \right)}}\), what is the equation of streamlines.   ...
Post a Comment
Read more