A two dimensional flow is described in Langragian system as,

$x=x_0e^{-tk}+y_0(1-e^{-2tk})$ and $y=y_0e^{tk}$

Find:

(i) The equation of fluid particle in flow fluid.

(ii) The velocity component in Eulerian system.

Data:-

$x=x_0e^{-tk}+y_0(1-e^{-2tk})$ and $y=y_0e^{tk}$

Find:

(i) The equation of fluid particle in flow fluid.

(ii) The velocity component in Eulerian system.

Solution:

(i) The equation of fluid particle in flow fluid

We have

$y=y_0e^{tk}$

$\therefore e^{tk}=\frac{y}{y_0}$ and $e^{-tk}=\frac{y_0}{y}$

Also, $x=x_oe^{-tk}+y_0(1-e^{-2tk})$

Now putting $e^{-tk}$ value in above equation

$x=x_0(\frac{y_0}{y})+y_0[1-(\frac{y_0}{y})^2]$

$x=\frac{x_0y_0}{y}+y_0-\frac{y_0^3}{y^2}=0$

Multiplying …