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 by $y^2$

$y^2x-x_0y_0y-y_0y^2+y_0^3=0$

$(x-y_0)y^2-x_0\cdot y_0\cdot y+y_0^3=0$

This is required equation of fluid particle in fluid flow.

(ii) Velocity component of Eulerian system:

Let, $u=\frac{dx}{dy}=\text{Velocity component in x-direction}$

$v=\frac{dv}{dy}=\text{Velocity component in y-direction}$

Now, $u=\frac{dx}{dy}$

$=\frac{d}{dt}[x_0e^{-tk}+y_0(1-e^{-2tk})]$

$=-kx_0e^{-tk}+0-(-2k\cdot e^{-2tk}\cdot y_0)$

$=-kx_0e^{-tk}+2ke^{-2tk}y_0$...........(1)

For given equations,

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

$x_0e^{-tk}=x-y_0[1-e^{-2tk}]$

$x_0e^{-tk}=x-\frac{y}{e^{tk}}[1-e^{-2tk}]$

Now putting $y_0$ and $x_0e^{-tk}$ values in equation (1)

$u=-k[x-\frac{y}{e^{tk}}(1-e^{-2tk})]+2k\frac{y}{e^{tk}}\cdot e^{-2tk}$

$=-k[x-\frac{y}{e^{tk}}+\frac{y}{e^{tk}}\cdot e^{-2tk})]+2ky\cdot e^{-3tk}$

$-kx+ky\cdot e^{-tk}+ky\cdot e^{-3tk}$

$u=-kx+ky(e^{-tk}+e^{-3tk})$

$v=\frac{dy}{dt}$

$=\frac{d}{dt}[y_0\cdot e^{tk}]$

$=y_0ke^{tk}$

But $\frac{y}{y_0}=e^{tk}$

$\therefore v=y_0k\frac{y}{y_0}$

$v=ky$