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 …
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