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Chapter 10

Q. 10.14.2

For steady flow of an incompressible viscous fluid under a conservative body force, prove the following:

\begin{matrix}(i)&\textbf{v}\times \textbf{w}\left(\frac{p}{\rho}+\frac{1}{2}v^2+\chi\right)+v\ curl\ \textbf{w}&(10.14.17)\\ (ii)&(\textbf{v}.\triangledown)\textbf{w}-(\textbf{w}.\triangledown)\textbf{v}=v\triangledown^2\textbf{w}&(10.14.18)\\ (iii)If\\&\textbf{v}=\psi _{,2}e_{1}-\psi_{,1}e_{2}&(10.14.19)\end{matrix}

where ψ = ψ(x_{1},x_{2}), then

\left|\begin{matrix}(\triangledown^2\psi)_{,1}&(\triangledown^2\psi)_{,2}\\ \psi_{,1}&\psi_{,2}\end{matrix}\right|=v\triangledown^4\psi                  (10.14.20)

Step-by-Step

Verified Solution

(i) For an incompressible viscous fluid moving under a conservative force, the Navier-Stokes equation is given by (10.14.9).

v\triangledown^2\textbf{v}-\triangledown\left(\frac{p}{\rho}+\chi+\frac{1}{2}v^2\right)=\frac{\partial \textbf{v}}{\partial t}+\textbf{w}\times \textbf{v}              (10.14.9)
For steady flow \partial \textbf{v}/\partial t=\textbf{0}. Also, for an incompressible fluid,

curl\ \textbf{w}=curl\ curl\ \textbf{v}=grad\ div\ \textbf{v}-\triangledown^2\textbf{v}=-\triangledown^2\textbf{v}                    (10.14.21)

because div v = 0. Using these facts in equation (10.14.9), we obtain (10.14.17).
(ii) From (10.14.21), we get

curl\ curl\ \textbf{w}=-\triangledown^2\textbf{w}                    (10.14.22)

Also, since divv = 0 and div w = 0, we find from identity (3.4.25) that

curl(\textbf{u}\times\textbf{v})=(div\ \textbf{v})\textbf{u}-(div\ \textbf{u})\textbf{v}+(\textbf{v}.\triangledown)\textbf{u}-(\textbf{u}.\triangledown)\textbf{v}               (3.4.25)

curl(\textbf{v}\times \textbf{w})=(\textbf{w}.\triangledown)\textbf{v}-(\textbf{v}.\triangledown)\textbf{w}                      (10.14.23)

Now, taking the curl on both sides of (10.14.17) and using (10.14.22), (10.14.23) and (3.4.16), we obtain (10.14.18).

curl\ \triangledown\phi=\textbf{0}           (3.4.16)
(iii) When v is given by (10.14.19), we get

\textbf{w}=curl\ \textbf{w}=-\triangledown^2\psi\textbf{e}_{3}                  (10.14.24)

so that

(\textbf{v}.\triangledown)\textbf{w}=\left(\psi_{,2}\frac{\partial}{\partial x_{1}}-\psi_{,1}\frac{\partial}{\partial x_{2}}\right)(-\triangledown^2\psi)\textbf{e}_{3}                    (10.14.25)

and

(\textbf{w}.\triangledown)\textbf{v}=0                    (10.14.26)

Using (10.14.24), (10.14.25) and (10.14.26) in (10.14.18), we obtain (10.14.20).