Chapter 10
Q. 10.14.4
Q. 10.14.4
Consider a steady motion of an incompressible viscous fluid under a conservative body force. If
H_{0}=\frac{1}{2}v^2+\frac{p}{\rho}+\chi (10.14.33)
prove the following
(i) H_{0} is constant along the field lines of the vector
\textbf{f}=(\textbf{v}\times\textbf{w})\times curl\ \textbf{w} (10.14.34)
(ii) \textbf{v}.\triangledown H_{0}=v(\triangledown^2 H_{0}-\textbf{w}^2) (10.14.35)
Step-by-Step
Verified Solution
For the motion considered, the relation (10.14.17) holds. Using (10.4.33), this relation can be rewritten as
\textbf{v}\times \textbf{w}=\triangledown\left(\frac{p}{\rho}+\frac{1}{2}v^2+\chi\right)+v\ curl\ \textbf{w} (10.14.17)
\triangledown H_{0}=(\textbf{v}\times\textbf{w})-v\ curl\ \textbf{w} (10.14.36)
From (10.14.34) and (10.14.36) we readily see that \textbf{f}.\triangledown H_{0}=0.\ Thus,\ \triangledown H_{0} is orthogonal to f and hence to the field line of f. But \triangledown H_{0} is always orthogonal to the surfaces of constant H_{0}. Hence f must be tangential to a surface of constant H_{0}. That is, H_{0} is constant along the field lines of f.
From (10.14.36), we get
\triangledown^2H_{0}=div(\textbf{v}\times\textbf{w})=\textbf{w}^2-\textbf{v}.\ curl\ \textbf{w} (10.14.37)
and
\textbf{v}.\triangledown H_{0}=-v\textbf{v}.\ curl\ \textbf{w} (10.14.38)
These relations together yield the relation (10.14.35).