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 $\theta$

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

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

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

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

$\theta=[\frac{3y^2}{4\rho}-\frac{9y^3}{12\rho ^2}+\frac{3y^5}{20\rho ^4}-\frac{y^4}{8\rho ^3}+\frac{3y^5}{20\rho ^4}-\frac{y^7}{28\rho ^6}]_0^{\rho}$

$\theta=[\frac{3}{4}\rho -\frac{3}{4}\rho +\frac{3}{20}\rho -\frac{1}{8}\rho +\frac{3}{20}\rho -\frac{1}{28}\rho]$

$\theta=\frac{39}{280}\rho$

Now, by Von Karman equation,

$\frac{T_0}{\rho \cdot U^2}=\frac{d\theta}{dx}$................(2)

Putting value of $'\theta'$ in equation (2)

$\frac{T_0}{\rho \cdot U^2}=\frac{d}{dx}(\frac{39}{280}\rho)$

$T_0=\frac{39}{280}\rho \cdot U^2\frac{\partial \rho}{\partial x}$.............(3)

From Newtons law of viscosity

$T_0=u(\frac{\partial u}{\partial y})_{\text{at y=0}}$

From equation (1)

$u=[\frac{3y}{2\rho}-\frac{y^3}{2\rho ^3}]U$

Differentiating above equation w.r.t. y

$\frac{\partial u}{\partial y}=U[\frac{3y}{2\rho}-\frac{3y^2}{3\rho ^3}]$

$(\frac{\partial u}{\partial y})_{\text{at y=0}}=\frac{3U}{2\rho}$...............(5)

Putting above values in equation (4)

$T_0=\frac{3uU}{2\rho}$..............(6)

equating equations (3) & (6)

$\frac{39}{280}\rho \cdot U^2\frac{\partial \rho}{\partial x}=\frac{3uU}{2\rho}$

$\rho \cdot d\rho=\frac{3}{2}u\times \frac{280}{39}\times \frac{dx}{\rho U}$

$\rho \cdot d\rho=\frac{420}{39}\frac{u}{\rho U}dx$

Integrating

$\frac{\rho ^2}{2}=\frac{420}{39}\frac{u}{\rho U}x+C$

at $x=0,\rho=0,C=0$

$\frac{\rho ^2}{2}=\frac{420}{39}\frac{ux}{\rho U}$

$\therefore \rho=\sqrt{\frac{420\times 2}{39}\times \frac{u}{\rho . U}.x}$

$=4.64\sqrt{\frac{u.x}{\rho . U}}$

$=4.64\sqrt{\frac{u.x}{\rho . U}}\frac{x}{x}=4.64\sqrt{\frac{x^2}{\frac{\rho U x}{x}}}$

$\rho=\frac{4.64x}{\sqrt{Rex}}$

(b) $T_0=\frac{3uU}{2\rho}$...............from equation (6)

$T_0=\frac{3uU}{2\times \frac{4.64x}{\sqrt{Rex}}}$

$=0.323\frac{u.U}{x}.\sqrt{Rex}$

local coefficient of drag, $C_D*$

$C_D*=\frac{T_0}{\frac{\rho U^2}{2}}=\frac{0.323\frac{u.U}{x}.\sqrt{Rex}}{\frac{\rho U^2}{2}}$

$=\frac{0.646u.\sqrt{Rex}}{\rho Ux}$

$=\frac{0.646u.\sqrt{Rex}}{\frac{\rho Ux}{u}}$

$C_D*=\frac{0.646}{\sqrt{Rex}}$

(c)$(\frac{\partial u}{\partial y})_{\text{at y=0}}=\frac{3U}{2\rho}$.............positive

Hence flow is attached.