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

Question

[latex]\frac{D extbf{w}}{Dt}=( extbf{w}. riangledown) extbf{v}+v riangledown^2 extbf{w}[/latex] (10.14.44)

Deduce the following.
(i) If [latex]curl\ extbf{w}= riangledown£[/latex], equation (10.14.44) becomes

[latex]\frac{D extbf{w}}{Dt}=( extbf{w}. riangledown) extbf{v}[/latex] (10.14.45)

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

[latex]\frac{Dw}{Dt}=v riangledown^2w[/latex] (10.14.46)

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

[latex]\frac{\partial w}{\partial t}=v riangledown^2w[/latex] (10.14.47)

Accepted Answer

For the given flow, the Navier-Stokes equation is given by (10.14.6) with [latex] extbf{b}=- riangledown\chi[/latex]. Taking curl on both sides of this equation, we get

[latex]v riangledown^2 extbf{v}-\frac{1}{ ho}+ extbf{b}=\frac{D extbf{v}}{Dt}[/latex] (10.14.6)

[latex]curl\left(\frac{Dv}{Dt} ight)=v riangledown^2 extbf{w}[/latex] (10.14.48)

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

[latex]\frac{D}{Dt}\left(\frac{ extbf{w}}{ ho} ight)=\left(\frac{ extbf{w}}{ ho}. riangledown ight) extbf{v}+\frac{1}{ ho}curl\frac{D extbf{v}}{Dt}[/latex] (8.2.16)
(i) If [latex]curl\ extbf{w}= riangledown£[/latex], then the identity (3.4.27) yields [latex] riangledown^2 extbf{w}= extbf{0}[/latex]. Consequently, equation (10.14.44) reduces to equation (10.14.45).

[latex]curl\ curl\ extbf{u}= riangledown\ div\ extbf{u}- riangledown^2 extbf{v}[/latex] (3.4.27)
(ii) If [latex]v_{3}\equiv 0\ and\ v_{1}\ and\ v_{2}[/latex] are independent of [latex]x_{3}[/latex], we readily find

[latex]w_{1}=w_{2}=0,\ \ w_{3}=v_{2,1}-v_{1,2}[/latex] (10.14.49)

Consequently,

[latex]( extbf{w}. riangledown)= extbf{0}[/latex] (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 [latex]x_{3}[/latex] axis, we have [latex] extbf{w}=v extbf{e}_{0},\ whith\ v=v(R,t)[/latex], where R is the radial distance parallel to the [latex]x_{1}x_{2}\ plane\ and\ extbf{e}_{0}[/latex] is the unit vector in the transverse direction (see Figure 10.14).

From Figure 10.14, we find that

[latex] extbf{e}_{0}=(-sin\ heta) extbf{e}_{1}+(cos\ heta) extbf{e}_{2}=-\frac{x_{2}}{R} extbf{e}_{1}+\frac{x_{1}}{R} extbf{e}_{2}[/latex]

Thus,

[latex] extbf{v}=v extbf{e}_{0}=v[-x_{2} extbf{e}_{1}+x_{1} extbf{e}_{2}][/latex]

so that

[latex]v_{1}=-\frac{v}{R}x_{2},\ \ v_{2}=\frac{v}{R}x_{1},\ \ v_{3}=0[/latex] (10.14.51)

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

[latex]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}[/latex] (10.14.52a)

[latex]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}[/latex] (10.14.52b)

Using these in (10.14.49), we obtain

[latex]w=\frac{v}{R}+\frac{\partial v}{\partial R}[/latex] (10.14.53)

(where [latex]w=w_{3}[/latex]), which yields

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

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

[latex]( extbf{v}. riangledown) extbf{w}=v_{1} extbf{w}_{,1}+v_{2} extbf{w}_{,2}=0[/latex]

Consequently,

[latex]\frac{Dw}{Dt}=\frac{\partial w}{\partial t}[/latex] (10.14.55)

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

Question 10.14.7: For a flow of an incompressible viscous fluid under conserva...

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