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

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

Bernoulli's equation

Euler's equation is given as \[dp =  - \rho vdv\]This equation applies to an incompressible and in-viscid flow where \(\rho  = \,{\rm{constant}}\). In a streamline, in between two points 1 and 2, the above Euler equation can be integrated as \[\int\limits_{{p_1}}^{{p_2}} {dp =  - \rho \int\limits_{{v_1}}^{{v_2}} {vdv} } \]\[{p_2} - {p_1} =  - \rho \left( {\frac{{v_2^2}}{2} - \frac{{v_1^2}}{2}} \right)\]\[ \Rightarrow {p_1} + \frac{1}{2}\rho v_1^2 = {p_2} + \frac{1}{2}\rho v_2^2\]This can be written as\[p + \frac{1}{2}\rho {v^2} = {\rm{constant}}\] for a streamline.For an rotational flow the value of constant is changing from streamline to another.For irrotational flow,the constant is same for all streamlines and \[p + \frac{1}{2}\rho {v^2} = {\rm{constant}}\]throughout the flow. Physical significance of Bernoulli's equation is that when the velocity increases, the pressure decreases  and when the velocity decreases, the pressure increases. ...
Post a Comment
Read more

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