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## Q. 10.14.7

For a flow of an incompressible viscous fluid under conservative body force, show that the vorticity equation is given by

$\frac{D\textbf{w}}{Dt}=(\textbf{w}.\triangledown)\textbf{v}+v\triangledown^2\textbf{w}$                  (10.14.44)

Deduce the following.
(i) If $curl\ \textbf{w}=\triangledown£$, equation (10.14.44) becomes

$\frac{D\textbf{w}}{Dt}=(\textbf{w}.\triangledown)\textbf{v}$                      (10.14.45)

(ii) If the motion is two-dimensional (where $v_{3}\equiv 0\ and\ v_{1}\ and\ v_{2}$ are independent of $x_{3}$), equation (10.14.44) reduces to

$\frac{Dw}{Dt}=v\triangledown^2w$                  (10.14.46)

where $w\equiv w_{3}$.
(iii) If the motion is two-dimensional and in circles with centers on $x_{3}$ axis, equation (10.14.46) reduces to

$\frac{\partial w}{\partial t}=v\triangledown^2w$                    (10.14.47)

## Verified Solution

For the given flow, the Navier-Stokes equation is given by (10.14.6) with $\textbf{b}=-\triangledown\chi$. Taking curl on both sides of this equation, we get

$v\triangledown^2\textbf{v}-\frac{1}{\rho}+\textbf{b}=\frac{D\textbf{v}}{Dt}$              (10.14.6)

$curl\left(\frac{Dv}{Dt}\right)=v\triangledown^2\textbf{w}$                (10.14.48)

Using this expression in Beltrami’s vorticity equation (8.2.16), we obtain the equation (10.14.44).

$\frac{D}{Dt}\left(\frac{\textbf{w}}{\rho}\right)=\left(\frac{\textbf{w}}{\rho}.\triangledown\right)\textbf{v}+\frac{1}{\rho}curl\frac{D\textbf{v}}{Dt}$              (8.2.16)
(i) If $curl\ \textbf{w}=\triangledown£$, then the identity (3.4.27) yields $\triangledown^2\textbf{w}=\textbf{0}$. Consequently, equation (10.14.44) reduces to equation (10.14.45).

$curl\ curl\ \textbf{u}=\triangledown\ div\ \textbf{u}-\triangledown^2\textbf{v}$          (3.4.27)
(ii) If $v_{3}\equiv 0\ and\ v_{1}\ and\ v_{2}$ are independent of $x_{3}$, we readily find

$w_{1}=w_{2}=0,\ \ w_{3}=v_{2,1}-v_{1,2}$                (10.14.49)

Consequently,

$(\textbf{w}.\triangledown)=\textbf{0}$                      (10.14.50)

and equation (10.14.44) reduces to equation (10.14.46).
(iii) If the motion is two-dimensional and in circles with centers on the $x_{3}$ axis, we have $\textbf{w}=v\textbf{e}_{0},\ whith\ v=v(R,t)$, where R is the radial distance parallel to the $x_{1}x_{2}\ plane\ and\ \textbf{e}_{0}$ is the unit vector in the transverse direction (see Figure 10.14).

From Figure 10.14, we find that

$\textbf{e}_{0}=(-sin\ \theta)\textbf{e}_{1}+(cos\ \theta)\textbf{e}_{2}=-\frac{x_{2}}{R}\textbf{e}_{1}+\frac{x_{1}}{R}\textbf{e}_{2}$

Thus,

$\textbf{v}=v\textbf{e}_{0}=v[-x_{2}\textbf{e}_{1}+x_{1}\textbf{e}_{2}]$

so that

$v_{1}=-\frac{v}{R}x_{2},\ \ v_{2}=\frac{v}{R}x_{1},\ \ v_{3}=0$                    (10.14.51)

These give, on noting that $v=v(R,t)\ and\ R_{,i}=x_{i}/R,i=1,2,$

$v_{1,2}=-\frac{v}{R}-\frac{1}{R}\frac{\partial v}{\partial R}R_{,2}x_{2}+\frac{v}{R^2}R_{,2}x_{2}=-\frac{v}{R}-\frac{1}{R^2}\frac{\partial v}{\partial R}x^2_{2}+\frac{v}{R^3}x^2_{2}$                      (10.14.52a)

$v_{2,1}=\frac{v}{R}+\frac{1}{R}\frac{\partial v}{\partial R}R_{,1}x_{1}-\frac{v}{R^2}R_{,1}x_{1}=\frac{v}{R}+\frac{1}{R^2}\frac{\partial v}{\partial R}x^2_{1}-\frac{v}{R^3}x^2_{1}$                    (10.14.52b)

Using these in (10.14.49), we obtain

$w=\frac{v}{R}+\frac{\partial v}{\partial R}$                    (10.14.53)

(where $w=w_{3}$), which yields

$w_{,i}=\left[\frac{\partial^2v}{\partial R^2}-\frac{1}{R}\frac{\partial v}{\partial R}-\frac{v}{R^2}\right]\frac{x_{i}}{R},\ \ i=1,2$                      (10.14.54)

From (10.14.51), (10.14.53) and (10.14.54) we find that

$(\textbf{v}.\triangledown)\textbf{w}=v_{1}\textbf{w}_{,1}+v_{2}\textbf{w}_{,2}=0$

Consequently,

$\frac{Dw}{Dt}=\frac{\partial w}{\partial t}$                  (10.14.55)

Substituting this in (10.14.46), we obtain (10.14.47).