For the velocity profile for laminar boundary flow $\frac{\upsilon}{U}=sin(\frac{\pi y}{2\delta})$

where u is the velocity at the distance y from the surface of the flat plate and U be the free stream velocity at the boundary layer thickness $\delta$. Obtain an expression for the

boundary layer thickness and the average drag coefficient in terms of the Reynolds Number.

Solution :

The velocity profile is $\dfrac v U=sin\left(\dfrac \pi 2 \dfrac y \delta\right)$

$$ \begin{aligned} \frac{\tau_{0}}{\rho U^{2}} &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \frac{v}{U}\left(1-\frac{v}{U}\right) d y\right]\\ &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \sin \left(\frac{\pi}{2} \frac{y}{\delta}\right)\left[1-\sin \left(\frac{\pi}{2} \frac{y}{\delta}\right)\right] d y\right] \\ &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left[\sin \left(\frac{\pi}{2} \frac{y}{\delta}\right)-\sin ^{2}\left(\frac{\pi}{2} \frac{y}{\delta}\right)\right] d y\right] \\ &=\left[\frac{\partial}{\partial x}\left[\frac{-\cos \frac{\pi y}{2 \delta}}{\frac{\pi}{2 \delta}}\right]-\left[\frac{\frac{\pi y}{2 \delta} \times \frac{1}{2}}{\frac{\pi}{2 \delta}}-\frac{\sin 2\left(\frac{\pi}{2} \frac{y}{\delta}\right)}{4 \times \frac{\pi}{2 \delta}}\right]\right]_{0}^{\delta} \\ &=\frac{\partial}{\partial x}\left[\left(\frac{-\cos \frac{\pi}{2} \frac{\delta}{\delta}}{\frac{\pi}{2 \delta}}+\frac{\cos \frac{\pi}{2} \times \frac{0}{\delta}}{\frac{\pi}{2 \delta}}\right)-\left[\frac{\frac{\pi}{2} \frac{\delta}{\delta} \times \frac{1}{2}}{\frac{\pi}{2 \delta}}-0\right]\right] \\ &=\frac{\partial}{\partial x}\left[\left(0+\frac{1}{\frac{\pi}{2 \delta}}\right)-\frac{\left(\frac{\pi}{4}\right)}{\frac{\pi}{2 \delta}}\right]=\frac{\partial}{\partial x}\left[\frac{2 \delta}{\pi}-\frac{\pi}{4} \times \frac{2 \delta}{\pi}\right]\\ &=\frac{\partial}{\partial x}\left[\frac{2 \delta}{\pi}-\frac{\delta}{2}\right]\\ &=\frac{\partial}{\partial x}\left[\frac{4-\pi}{2 \pi}\right] \delta \\ &=\left(\frac{4-\pi}{2 \pi}\right) \frac{\partial \delta}{\partial x}\\ \therefore \quad \tau_{0} &=\left(\frac{4-\pi}{2 \pi}\right) p U^{2} \frac{\partial \delta}{\partial x} .....(i) \end{aligned} $$

$\tau_{0}$ is also equal $=\mu\left(\frac{d v}{d y}\right)_{a t y=0}$

But

$$ \begin{aligned} v &=U \sin \left(\frac{\pi}{2} \frac{y}{\delta}\right) \\ \left(\frac{d v}{d y}\right) &=U \cos \left(\frac{\pi}{2} \frac{y}{\delta}\right) \times \frac{\pi}{2 \delta} \\ \left(\frac{d v}{d y}\right)_{y=0} &=U \times \frac{\pi}{2 \delta} \cos \left(\frac{\pi}{2} \times \frac{0}{\delta}\right)=\frac{U \pi}{28} \end{aligned} $$

$$\therefore \quad \tau_{0}=\mu\left(\frac{\partial v}{\partial y}\right)_{y=0}=\frac{\mu U \pi}{2 \delta} .....(ii)$$

Equating the two values $\tau_{0}$ given by equations (i) and (ii)

$$\begin{aligned} \left(\frac{4-\pi}{2 \pi}\right) \rho U^{2} \frac{\partial \delta}{\partial x} &=\frac{\mu U \pi}{2 \delta} \Rightarrow \delta \partial \delta=\frac{\mu U \pi}{2} \times \frac{2 \pi}{4-\pi} \times \frac{1}{\rho U^{2}} \partial x \\ \therefore \quad \delta \partial \delta &=\frac{\pi^{2}}{(4-\pi)} \frac{\mu U}{\rho U^{2}}.\partial x \\ &=11.4975 \frac{\mu}{\rho U} \partial x \end{aligned}$$

Integrating, we get $$\quad \frac{\delta^{2}}{2}=11.4975 \frac{\mu}{\rho U} x+C$$

At $x=0,\delta =0$ and hence $C=0$

$$ \begin{aligned} \frac{\delta^{2}}{2} &=11.4975 \frac{\mu}{\rho U} x\\ \delta &=\sqrt{2 \times 11.4975 \frac{\mu}{\rho U} x}=4.795 \sqrt{\frac{\mu}{\rho U}} \times \\ &=4.795 \sqrt{\frac{\mu}{\rho U x}}\\ &=4.795 \sqrt{\frac{\mu}{\rho U x}} \times x \\ &=\frac{4.795 x}{\sqrt{R_{e}}} .....(iii) \end{aligned} $$

(ii) Shear Stress $\left(\tau_{0}\right)$

From equation (ii)

$$ \begin{aligned} \tau_{0} &=\frac{\mu U \pi}{2 \delta}=\frac{\mu U \pi}{\frac{2 \times 4.795 x}{\sqrt{R_{e}}}}=\frac{\mu U \pi \sqrt{R_{e}}}{2 \times 4.795 x} \\ &=\frac{\pi}{2 \times 4.795} \frac{\mu U}{x} \sqrt{R_{e}}\\ &=0.327 \frac{\mu U}{x} \sqrt{R_{e}} \end{aligned} $$

(iii) Drag force $\left({F}_{{b}}\right)$

on one side of the plate is given by

$$ \begin{aligned} F_{D}=\int_{0}^{L} \tau_{0} \times b \times d x &=\int_{0}^{L} 0.327 \frac{\mu U}{x} \sqrt{R_{e}} \times b \times d x \\ &=0.327 \mu U \times b \int_{0}^{L} \frac{1}{x} \sqrt{\frac{\rho U x}{\mu}} d x \\ &=0.327 \mu U \times b \times \sqrt{\frac{\rho U}{\mu}} \int_{0}^{L} x^{-12} d x \\ &=0.327 \mu U \times b \times \sqrt{\frac{\rho U}{\mu}}\left[\frac{x^{1 / 2}}{\frac{1}{2}}\right]_{0}^{L} \\ &=0.327 \times 2 \times \mu U \times b \sqrt{\frac{\rho U}{\mu}} \times \sqrt{L} \\ &=0.655 \times \mu U \times b \times \sqrt{\frac{\rho U L}{\mu}} \end{aligned} $$

(iv) Co-efficient of drag, $C_{D}$

$$ \begin{aligned} C_{D} &=\frac{F_{D}}{\frac{1}{2} \rho A U^{2}}, \text { where } A=b \times L \\ C_{D} &=\frac{0.655 \times \mu U \times b \times \sqrt{\frac{\rho U L}{\mu}}}{\frac{1}{2} \rho U^{2} \times b \times L} \\ &=0.655 \times 2 \times \frac{\mu}{\rho U L} \times \sqrt{\frac{\rho U L}{\mu}}\\ &=1.31×\frac{1}{\sqrt{\rho UL}}=\frac{1.31}{\sqrt{R_e}} \end{aligned} $$