For an incompressible flow \(\rho = {\rm{constant}}\). \(\nabla \cdot V\) is physically the time rate of change of volume of a moving fluid element per unit volume. For an incompressible flow, volume of a fluid element is constant, therefore \[\nabla \cdot V = 0\]This can be also shown from continuity equation. Continuity equation is given as\[\frac{{\partial \rho }}{{\partial t}} + \nabla \cdot \rho V = 0\]Since, for an incompressible flow, \(\rho = {\rm{constant}}\) we have \[\frac{{\partial \rho }}{{\partial t}} = 0\]therefore the continuity equation becomes \[0 + \nabla \cdot \rho V = 0\]\[\nabla \cdot V = \frac{0}{\rho } = 0\]For an irrotational flow, velocity potential \('\phi '\) is defined as \[V = \nabla \phi \]Therefore, a flow which is both incompressible and irrotational the equation can be written as \[\nabla .\left( {\nabla \phi } \right) = 0\]\[{\nabla ^2}\phi = 0\]This equation \({\nabla ^2}\phi = 0\) is called the Laplace's equation.In Cartesian coordinates \[\phi = \phi \left( {x,y,z} \right)\] therefore\[{\nabla ^2}\phi = \frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {z^2}}} = 0\]Also, \[u = \frac{{\partial \psi }}{{\partial y}},v = - \frac{{\partial \psi }}{{\partial x}}\]For an incompressible and irrotational flow \[\frac{{\partial v}}{{\partial x}} - \frac{{\partial u}}{{\partial y}} = 0\]\[\frac{\partial }{{\partial x}}\left( { - \frac{{\partial \psi }}{{\partial x}}} \right) - \frac{\partial }{{\partial y}}\left( {\frac{{\partial \psi }}{{\partial y}}} \right) = 0\]\[ \Rightarrow \frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0\]Therefore, velocity potential as well as stream functions both satisfies Laplace's equation, for an incompressible and irrotational flow.
Example: Prove that velocity potential and stream function for a uniform flow and source flow satisfies Laplace's equation.
Solution:For a uniform flow velocity potential \(\phi = {V_\infty }x\)\[\frac{{\partial \phi }}{{\partial x}} = {V_\infty },\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} = 0\]\[\frac{{\partial \phi }}{{\partial y}} = 0\,;\,\frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0\]Laplace's equation\[\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0 + 0 = 0\]is satisfied. For a uniform flow stream function \(\psi = Vy\,;\)\[\frac{{\partial \psi }}{{\partial x}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} = 0\]\[\frac{{\partial \psi }}{{\partial y}} = V,\frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0\]Therefore, Laplace's equation =\[\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0 + 0 = 0\]is satisfied.
For a source flow , velocity potential \(\phi = \frac{\Lambda }{{2\pi }}\ln r\)\[\frac{{\partial \phi }}{{\partial r}} = \frac{\Lambda }{{2\pi }}\frac{1}{r}\,,\,\frac{{{\partial ^2}\phi }}{{\partial {r^2}}} = - \frac{\Lambda }{{2\pi }}\frac{1}{{{r^2}}}\]\[\frac{{\partial \phi }}{{\partial \theta }} = 0\,,\,\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}} = 0\]Laplace's equation = \[ = \frac{1}{r} \cdot \frac{\partial }{{\partial r}}\left( {r\frac{{\partial \phi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}} + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}}\]\[ = \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {\frac{\Lambda }{{\partial \pi }}} \right] + 0 = 0\]is satisfied.For a source flow stream function \[\psi = \frac{\Lambda }{2} = \theta \,;\,\frac{{\partial \psi }}{{\partial r}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {r^2}}} = 0\]\[\frac{{\partial \psi }}{{\partial \theta }} = \frac{\Lambda }{{2\pi }}\,,\,\frac{{{\partial ^2}\psi }}{{\partial {\phi ^2}}}\, = \,0\]Therefore , Laplace's equation \[\frac{1}{r}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial \psi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\psi }}{{\partial {\theta ^2}}}\]\[ = \frac{1}{r}\frac{\partial }{{\partial r}}\left( 0 \right) + \frac{1}{{{r^2}}}\left( 0 \right) = 0\]is satisfied.
Example: Show that a uniform flow with velocity \({V_\infty }\) is a physically possible incompressible flow and is irrotational.
Solution:For a uniform flow \[{V_\infty } = u = {\rm{constant}}\,\,{\rm{;}}\,\,\mathop V\limits^ \to = {V_\infty }\mathop i\limits^ \to \]Divergence of velocity field\[\nabla \cdot \mathop V\limits^ \to = \frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}}\]\[ = 0 + 0 + 0\]\[ = 0\]Therefore, it shows that it is a physically possible incompressible flow.
Also, curl of velocity field \[\nabla \times \mathop V\limits^ \to = \left| {\begin{array}{*{20}{c}}{\mathop i\limits^ \to }&{\mathop j\limits^ \to }&{\mathop k\limits^ \to }\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\u&v&w\end{array}} \right| = \mathop i\limits^ \to \left( {0 - 0} \right) - \mathop j\limits^ \to \left( {0 - \frac{{\partial u}}{{\partial x}}} \right) + \mathop k\limits^ \to \left( {0 - \frac{{\partial u}}{{\partial y}}} \right)\]\[\nabla \times \mathop V\limits^ \to = 0\]This shows that the flow is irrotational.
Example: For a source flow, show that it is a physically possible incompressible flow except at origin.Also, show that it is irrotational .
Solution: For a source flow, velocity \(\mathop V\limits^ \to = {V_r}\mathop {{e_r}}\limits^ \to = \frac{\Lambda }{{2\pi r}}\mathop {{e_r}}\limits^ \to \)In polar co-ordinates :, the curl of velocity field will be \[\nabla \cdot \mathop V\limits^ \to = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {r{V_r}} \right) + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}\]\[\nabla \cdot \mathop V\limits^ \to = \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {r\frac{\Lambda }{{2\pi r}}} \right] + \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }}\]\[\nabla \cdot \mathop V\limits^ \to = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {\frac{\Lambda }{{2\pi }}} \right) + 0 = 0\]Therefore, the flow is a physically possible incompressible flow, except at origin where r =0.
Also,\[\nabla \times \mathop V\limits^ \to = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to }&{r\mathop {{e_\theta }}\limits^ \to }&{\mathop {{e_z}}\limits^ \to }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{{V_r}}&{r{V_\theta }}&{{V_z}}\end{array}} \right| = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to }&{r\mathop {{e_\theta }}\limits^ \to }&{\mathop {{e_z}}\limits^ \to }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{\frac{\Lambda }{{2\pi r}}}&0&0\end{array}} \right|\]\[\nabla \times \mathop V\limits^ \to = - r\mathop {{e_\theta }}\limits^ \to \left( {\frac{{\partial 0}}{{\partial r}} - \frac{{\partial \Lambda /2\pi r}}{{\partial z}}} \right) + \mathop {{e_z}}\limits^ \to \left( {\frac{{\partial 0}}{{\partial r}} - \frac{{\partial \Lambda /2\pi r}}{{\partial \theta }}} \right) = 0\]Therefore, \(\nabla \times \mathop V\limits^ \to = 0\) (everywhere in the flow).
Example: Prove that velocity potential and stream function for a uniform flow and source flow satisfies Laplace's equation.
Solution:For a uniform flow velocity potential \(\phi = {V_\infty }x\)\[\frac{{\partial \phi }}{{\partial x}} = {V_\infty },\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} = 0\]\[\frac{{\partial \phi }}{{\partial y}} = 0\,;\,\frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0\]Laplace's equation\[\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = 0 + 0 = 0\]is satisfied. For a uniform flow stream function \(\psi = Vy\,;\)\[\frac{{\partial \psi }}{{\partial x}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} = 0\]\[\frac{{\partial \psi }}{{\partial y}} = V,\frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0\]Therefore, Laplace's equation =\[\frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 0 + 0 = 0\]is satisfied.
For a source flow , velocity potential \(\phi = \frac{\Lambda }{{2\pi }}\ln r\)\[\frac{{\partial \phi }}{{\partial r}} = \frac{\Lambda }{{2\pi }}\frac{1}{r}\,,\,\frac{{{\partial ^2}\phi }}{{\partial {r^2}}} = - \frac{\Lambda }{{2\pi }}\frac{1}{{{r^2}}}\]\[\frac{{\partial \phi }}{{\partial \theta }} = 0\,,\,\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}} = 0\]Laplace's equation = \[ = \frac{1}{r} \cdot \frac{\partial }{{\partial r}}\left( {r\frac{{\partial \phi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}} + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\phi }}{{\partial {\theta ^2}}}\]\[ = \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {\frac{\Lambda }{{\partial \pi }}} \right] + 0 = 0\]is satisfied.For a source flow stream function \[\psi = \frac{\Lambda }{2} = \theta \,;\,\frac{{\partial \psi }}{{\partial r}} = 0\,,\,\frac{{{\partial ^2}\psi }}{{\partial {r^2}}} = 0\]\[\frac{{\partial \psi }}{{\partial \theta }} = \frac{\Lambda }{{2\pi }}\,,\,\frac{{{\partial ^2}\psi }}{{\partial {\phi ^2}}}\, = \,0\]Therefore , Laplace's equation \[\frac{1}{r}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial \psi }}{{\partial r}}} \right) + \frac{1}{{{r^2}}}\frac{{{\partial ^2}\psi }}{{\partial {\theta ^2}}}\]\[ = \frac{1}{r}\frac{\partial }{{\partial r}}\left( 0 \right) + \frac{1}{{{r^2}}}\left( 0 \right) = 0\]is satisfied.
Example: Show that a uniform flow with velocity \({V_\infty }\) is a physically possible incompressible flow and is irrotational.
Solution:For a uniform flow \[{V_\infty } = u = {\rm{constant}}\,\,{\rm{;}}\,\,\mathop V\limits^ \to = {V_\infty }\mathop i\limits^ \to \]Divergence of velocity field\[\nabla \cdot \mathop V\limits^ \to = \frac{{\partial u}}{{\partial x}} + \frac{{\partial v}}{{\partial y}} + \frac{{\partial w}}{{\partial z}}\]\[ = 0 + 0 + 0\]\[ = 0\]Therefore, it shows that it is a physically possible incompressible flow.
Also, curl of velocity field \[\nabla \times \mathop V\limits^ \to = \left| {\begin{array}{*{20}{c}}{\mathop i\limits^ \to }&{\mathop j\limits^ \to }&{\mathop k\limits^ \to }\\{\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\u&v&w\end{array}} \right| = \mathop i\limits^ \to \left( {0 - 0} \right) - \mathop j\limits^ \to \left( {0 - \frac{{\partial u}}{{\partial x}}} \right) + \mathop k\limits^ \to \left( {0 - \frac{{\partial u}}{{\partial y}}} \right)\]\[\nabla \times \mathop V\limits^ \to = 0\]This shows that the flow is irrotational.
Example: For a source flow, show that it is a physically possible incompressible flow except at origin.Also, show that it is irrotational .
Solution: For a source flow, velocity \(\mathop V\limits^ \to = {V_r}\mathop {{e_r}}\limits^ \to = \frac{\Lambda }{{2\pi r}}\mathop {{e_r}}\limits^ \to \)In polar co-ordinates :, the curl of velocity field will be \[\nabla \cdot \mathop V\limits^ \to = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {r{V_r}} \right) + \frac{1}{r}\frac{{\partial {V_\theta }}}{{\partial \theta }}\]\[\nabla \cdot \mathop V\limits^ \to = \frac{1}{r}\frac{\partial }{{\partial r}}\left[ {r\frac{\Lambda }{{2\pi r}}} \right] + \frac{1}{r}\frac{{\partial \left( 0 \right)}}{{\partial \theta }}\]\[\nabla \cdot \mathop V\limits^ \to = \frac{1}{r}\frac{\partial }{{\partial r}}\left( {\frac{\Lambda }{{2\pi }}} \right) + 0 = 0\]Therefore, the flow is a physically possible incompressible flow, except at origin where r =0.
Also,\[\nabla \times \mathop V\limits^ \to = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to }&{r\mathop {{e_\theta }}\limits^ \to }&{\mathop {{e_z}}\limits^ \to }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{{V_r}}&{r{V_\theta }}&{{V_z}}\end{array}} \right| = \frac{1}{r}\left| {\begin{array}{*{20}{c}}{\mathop {{e_r}}\limits^ \to }&{r\mathop {{e_\theta }}\limits^ \to }&{\mathop {{e_z}}\limits^ \to }\\{\frac{\partial }{{\partial r}}}&{\frac{\partial }{{\partial \theta }}}&{\frac{\partial }{{\partial z}}}\\{\frac{\Lambda }{{2\pi r}}}&0&0\end{array}} \right|\]\[\nabla \times \mathop V\limits^ \to = - r\mathop {{e_\theta }}\limits^ \to \left( {\frac{{\partial 0}}{{\partial r}} - \frac{{\partial \Lambda /2\pi r}}{{\partial z}}} \right) + \mathop {{e_z}}\limits^ \to \left( {\frac{{\partial 0}}{{\partial r}} - \frac{{\partial \Lambda /2\pi r}}{{\partial \theta }}} \right) = 0\]Therefore, \(\nabla \times \mathop V\limits^ \to = 0\) (everywhere in the flow).
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