Chapter 6, Problem 6.1P

Prove that three-dimensional source flow is irrotational.

To determine

To prove:

The velocity having the only component in radial direction & all other components are zero.

Explanation of Solution

Formula used:

The flow can be proved to be irrotational if the cross gradient of velocity is zero.

  $\nabla \times V=0$

Proof:

The flow is said to be irrotational if there will be no net rotation of moving fluid corresponding to chosen one axis at an instant to another.

The rotation of fluid-particle is due to torsion applied by shear force.

For an ideal fluid, there is no shear force due to which it is irrotational.

As the source flow is radially symmetrical flow field and it should be irrotational.

To prove the fluid to be irrotational we take the fluid particle in three directions, there is only one component in the radial direction as per the given problem.

The velocity component in the spherical component is given by:

  $V=V_r{e}_r+V_{\theta }{e}_{\theta }+V_{\varphi }{e}_{\varphi }$

  $e_r$ = unit vector r direction

  $e_{\theta }$ = unit vector in $\theta$ direction

  $e_{\varphi }$ = unit vector in $\varphi$ direction

From equation 2.24 from the textbook we get:

  $\begin{array}{l}\nabla \times V=\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{e}_r& r{e}_{\theta }& (r\sin \varphi )e_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ V_r& r{V}_{\theta }& (r\sin \theta )V_{\varphi }\end{array}\right|\\ \end{array}$

For having only a radial component :

  $V_r=\frac{c}{r^2},V_{\theta }=0\&V_{\varphi }=0$

We get:

  $\begin{array}{l}\nabla \times V=\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{e}_r& r{e}_{\theta }& (r\sin \varphi )e_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ \frac{c}{r^2}& 0& 0\end{array}\right|\\ \end{array}$

As for irrotational flow

Curl of V should be zero:

  $\nabla \times V=0$

Now,

  $\begin{array}{l}\nabla \times V=\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{e}_r& r{e}_{\theta }& (r\sin \varphi )e_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ \frac{c}{r^2}& 0& 0\end{array}\right|=0\\ \end{array}$

Taking Left-hand side:

  $\begin{array}{l}\frac{1}{r^2\sin \theta }\left|\begin{array}{ccc}{e}_r& r{e}_{\theta }& (r\sin \varphi )e_{\varphi }\\ \frac{\partial }{\partial r}& \frac{\partial }{\partial \theta }& \frac{\partial }{\partial \varphi }\\ \frac{c}{r^2}& 0& 0\end{array}\right|\\ =\frac{1}{r^2\sin \theta }\left[e_r(0-0)-r{e}_{\theta }\left\{0-\frac{\partial }{\partial \varphi }\left(\frac{c}{r^2}\right)\right\}+\left(r\sin \theta \right)e_{\varphi }\left\{0-\frac{\partial }{\partial \theta }\right\}\left(\frac{c}{r^2}\right)\right]\\ =\frac{1}{r^2\sin \theta }\left[e_r(0-0)-r{e}_{\theta }\left\{0-0\right\}+\left(r\sin \theta \right)e_{\varphi }\left\{0-0\right\}\right]\\ =\frac{1}{r^2\sin \theta }\left[e_r(0)-r{e}_{\theta }\left\{0\right\}+\left(r\sin \theta \right)e_{\varphi }\left\{0\right\}\right]=0\end{array}$

L.H.S=R.H.S

Hence proved.