The velocity profile within a laminar boundary layer over a fiat plate is given by

$\frac{u}{U}=sin (\frac{\pi y}{2\rho})$

where, 'U' is the mean stream velocity and $'\rho'$ is the boundary layer thickness. Determine

(1) Displaced thickness

(2) Momentum thickness

To find:-

$\rho *=?$ $\theta=?$

Solution:-

(1) Displacement thickness

$\rho *=\int _0^\rho (1-\frac{u}{U})dy$

$=\int _0^\rho (1-sin (\frac{\pi y}{2\rho}))dy$

$=[y-[\frac{(\frac{\pi y}{2\rho})}{\frac{\pi}{2\rho}}]]_0^\rho$

$=[y+\frac{2\rho}{\pi}cos(\frac{\pi y}{2\rho})]_0^\rho$

$=[(\rho + \frac{2\rho}{\pi}cos(\frac{\pi y}{2\rho}))-(0+\frac{2\rho}{\pi}cos(0))]$

$=[\rho+0-0-\frac{2\rho}{\pi}]=\rho-\frac{2\rho}{\pi}$

$\therefore \rho *=(\frac{\pi -2}{\pi})\rho$

(ii) Momentum thickness

$=\theta=\int _0^\rho \frac{u}{U}[1-\frac{u}{U}]dy$

$=\int _0^\rho sin(\frac{\pi y}{2\rho})[1-sin(\frac{\pi y}{2\rho})]dy$

$=\int _0^\rho [sin(\frac{\pi y}{2\rho})-sin ^2(\frac{\pi y}{2\rho})]dy$

$=\int _0^\rho [sin(\frac{\pi y}{2\rho})-[\frac{1-cos(\frac{\pi y}{2\rho})}{2}]]dy$

$=\int _0^\rho [sin(\frac{\pi y}{2\rho})-\frac{1}{2}+\frac{cos(\frac{\pi y}{2\rho})}{2}]dy$

$=[\frac{-cos(\frac{\pi y}{2\rho})}{\frac{\pi}{2\rho}}-\frac{y}{2}+\frac{sin(\frac{\pi y}{\rho})}{\frac{\pi}{\rho}}]_0^\rho$

$=[\frac{-cos(\frac{\pi y}{2\rho})}{\frac{\pi}{2\rho}}-\frac{\rho}{2}+\frac{sin(\frac{\pi y}{\rho})}{\frac{\pi}{\rho}}]-[\frac{-cos(0)}{\frac{\pi}{2\rho}}-0+\frac{sin(0)}{\frac{\pi}{\rho}}]$

$=[(0-\frac{\rho}{2}+0)-(\frac{2\rho}{\pi}-0+0)]$

$=-\frac{\rho}{2}+\frac{2\rho}{\pi}=\frac{2\rho}{\pi}-\frac{\rho}{2}$

$\theta=(\frac{4-\pi}{2\pi})\rho$