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Chapter 10

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)

Step-by-Step

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).

Figure 10.14. Example 10.14.7(iii)