Determine,

(a) Boundary layer thickness

(b) Local coefficient of drug

(c) Check whether the flow is attached or not.

Data:-

$f(\eta)=\frac{3n}{2}-\frac{n^3}{2}$; $\eta =\frac{y}{\rho}$ & $f(\eta)=\frac{u}{U}$

To find:-

(a) $\rho =?$

(b) $C_D*=?$

(c) To check whether the flow is attached or not=?

Solution:-

(a) From given data,

$\frac{u}{U}=(\frac{3}{2}\frac{y}{\rho})-\frac{1}{2}(\frac{y}{\rho})^3$

$\frac{u}{U}=\frac{3y}{2\rho}-\frac{y^3}{2\rho ^3}$..................(1)

Momentum Thickness $\theta$

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

$=\int _0^{\rho}\frac{3y}{2\rho}-\frac{y^3}{2\rho ^3}[1-(\frac{3y}{2\rho}-\frac{y^3}{2\rho ^3})]dy$

$=\int _0^{\rho}\frac{3y}{2\rho}-\frac{y^3}{2\rho ^3}[1-\frac{3y}{2\rho}+\frac{y^3}{2\rho ^3}]dy$

$=\int _0^{\rho}[\frac{3y}{2\rho}-\frac{9y^2}{4\rho ^2}+\frac{3y^4}{4\rho ^4}-\frac{1}{2}\frac{y^3}{\rho ^3}+\frac{3y^4}{4\rho ^4}-\frac{1}{4}\frac{y^6}{\rho …