For the Velocity profile for Laminar Boundary Layer : $\frac{u}{U}=\frac{3}{2}(\frac{y}{\delta})-\frac{1}{2}(\frac{y}{\delta})^2$ Determine Boundary layer thickness, Shear stress, Drag force and coefficient of Drag in terms of Reynold’s number.

Solution :

Given :

Velocity distribution,

$$\quad \frac{u}{U}=\frac{3}{2}\left(\frac{y}{\delta}\right)-\frac{1}{2}\left(\frac{y}{\delta}\right)^{3}$$

We have

$$\frac{{t}_{0}}{\rho U^{2}}=\frac{\partial}{\partial x}\left[\int_{0}^{5} \frac{u}{U}\left({I}-\frac{\mu}{U}\right) d y\right]$$

Substituting the value of $\frac{u}{U}=\frac{3}{2}\left(\frac{y}{\mu}\right)-\frac{1}{2}\left(\frac{y}{\mu}\right)^{3}$ in the above equation

$\begin {aligned} \frac{\tau_{0}}{\rho U^{2}} &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left[\frac{3}{2}\left(\frac{y}{\delta}\right)-\frac{1}{2} \left(\frac{y}{\delta}\right)^{3}\right]\left[1-\left\{\frac{3}{2}\left(\frac{y}{\delta}\right)-\frac{1}{2}\left(\frac{y}{\delta} \right)^{3} \right\} \right] d y\right]\\ &=\frac{\partial}{\partial x} \left[\int_{0}^{\delta} \left(\frac{3 y}{2 \delta}-\frac{y^{3}}{2 \delta^{3}} \right)\left(1-\frac{3 y}{2 \delta}+\frac{y^{3}}{2 \delta^{3}}\right) d y\right]\\ &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left(\frac{3 y}{2 \delta}-\frac{9 y^{2}}{4 \delta^{2}}+\frac{3 y^{4}}{4 \delta^{4}}-\frac{y^{3}}{2 \delta^{3}}+\frac{3 y^{4}}{4 \delta^{4}}-\frac{y^{6}}{4 \delta^{6}}\right) d y\right] \\ &=\frac{\partial}{\partial x}\left[\frac{3 y^{2}}{2 \times 2 \delta}-\frac{9 y^{3}}{3 \times 4 \delta^{2}}+\frac{3 y^{5}}{5 \times 4 \delta^{4}}-\frac{y^{4}}{4 \times 2 \delta^{3}}+\frac{3 y^{5}}{5 \times 4 \delta^{4}}-\frac{y^{7}}{7 \times 4 \delta^{6}}\right]_{0}^{\delta}\\ &=\frac{\partial}{\partial x}\left[\frac{3 \delta^{2}}{4 \delta}-\frac{3 \delta^{3}}{4 \delta^{2}}+\frac{3}{20} \frac{\delta^{3}}{\delta^{4}}-\frac{1}{8 \delta^{3}}+\frac{3}{20} \frac{\delta^{3}}{\delta^{4}}-\frac{1}{28} \frac{\delta^{2}}{\delta^{6}}\right]\\ &=\frac{\partial}{\partial x}\left[\frac{3}{4} \delta-\frac{3}{4} \delta+\frac{3}{20} \delta-\frac{1}{8} \delta+\frac{3}{20} \delta-\frac{1}{28} \delta\right]\\ &=\frac{\partial}{\partial x}\left[\frac{6}{20} \delta-\frac{1}{8} \delta-\frac{1}{28} \delta\right]=\frac{\partial \delta}{\partial x}\left[\frac{84-35-10}{280}\right]\\ &=\frac{39}{280} \frac{\partial \delta}{\partial x} \\ \tau_{0} &=\rho U^{2} \times \frac{39}{280} \frac{\partial \delta}{\partial x}\\ &=\frac{39}{280} \rho U^{2} \frac{\partial \delta}{\partial x} .....(i) \end{aligned}$

Also the shear stress is given by,

$\tau_{0}=\mu\left(\frac{d u}{d y}\right)_{z=0}$,

where $u =U\left[\frac{3}{2} \frac{y}{\delta}-\frac{y^{3}}{2 \delta^{3}}\right]$

$ \therefore \quad \frac{d u}{d y}=U\left[\frac{3}{2 \delta}-\frac{3 y^{2}}{20^{3}}\right] $

Hence

$\begin {aligned} \therefore \tau_{0} &=\mu\left(\frac{d u}{d y}\right)_{y=0}\\ &=\mu \frac{3 U}{2 \delta}\\ &=\frac{3}{2} \frac{\mu U}{\delta} .....(ii) \end{aligned}$

Equating the two values of $\tau_{0}$ given by equations $(13.20)$ and $(13.21)$

$ \begin{aligned} \frac{39}{280} \rho U^{2} \frac{\partial \delta}{\partial x} &=\frac{3}{2} \frac{\mu U}{\delta} \\ \delta \partial \delta &=\frac{3}{2} \mu U \times \frac{280}{39} \times \frac{1}{\rho U^{2}} \partial x\\ &=\frac{420}{39} \frac{\mu}{\rho U} \partial x \end{aligned} $

Integrating, we get

$\begin{aligned}\frac{\delta^{2}}{2}=\frac{420}{39} \frac{\mu}{\rho U} x+C \end{aligned}$

where $x=0, \delta=0, \therefore C=0$

$ \begin{aligned} \frac{\delta^{2}}{2} &=\frac{420}{39}. \frac{\mu}{\rho U} x \\ \delta &=\sqrt{\frac{420 \times 2}{39} \frac{\mu}{\rho U}} \\ \delta &=4.64 \sqrt{\frac{\mu x}{\rho U}}\\ &=4.64 \sqrt{\frac{\mu x \times x}{\rho U x}} \\ &=4.64 \sqrt{\frac{\mu}{\rho U x}}\\ &=\frac{4.64 x}{\sqrt{R_{e}}} .....(iii) \end{aligned} $

(i) Shear Stress $(\tau_o)$

Substituting the value of $\delta$ from equation (iii) into equation (ii), we get

$\begin {aligned} \quad \tau_{0} &=\frac{3}{2} \frac{\mu U}{\frac{4.64 x}{\sqrt{R_{e_{x}}}}}\\ &=\frac{3}{9.28} \frac{\mu U \sqrt{R_{e_{x}}}}{x}\\ &=0.323 \frac{\mu U}{x} \sqrt{R_{e_{x}}} \end {aligned}$

(ii) Drag force $(F_D)$

$ \begin{aligned}F_{D} &=\int_{0}^{L} \tau_{0} \times b \times d x \\ &=\int_{0}^{L} 0323 \frac{\mu U}{x} \sqrt{R_{e}} \times b \times d x \\ &=0.323 \int_{0}^{L} \frac{\mu U}{x} \sqrt{\frac{\rho U x}{\mu}} \times b \times d x \\ &=0.323 \mu U \sqrt{\frac{\rho U}{\mu}} \times b \int_{0}^{L} \frac{1}{\sqrt{x}} d x \\ &=0.323 \mu U \sqrt{\frac{\rho U}{\mu}} \times b \int_{0}^{L} x^{-1 / 2} d x \\ &=0.323 \mu U \sqrt{\frac{\rho U}{\mu}} \times b\left[\frac{x^{1 / 2}}{\frac{1}{2}}\right]_{0}^{L} \\&=0.323 \times 2 \mu U \sqrt{\frac{\rho U}{\mu}} \times b[\sqrt{L}] \\ &=0.646 \mu U \sqrt{\frac{\rho U L}{\mu}} \times b \end{aligned} $

(iii) Drag Co-efficient $(C_D)$

$ \begin{aligned} C_{D} &=\frac{F_{D}}{\frac{1}{2} \rho A U^{2}}, \text { where } A=b \times L \\ &=\frac{0.646 \mu U \sqrt{\frac{\rho U L}{\mu}} \times b}{\frac{1}{2} \rho \times b \times L \times U^{2}}\\ &=0.646 \times 2 \times \frac{\mu}{\rho U L} \times \sqrt{\frac{\rho U L}{\mu}}\\ &=\frac{1.292}{\sqrt{\frac{\rho U L}{\mu}}} \\ &=\frac{1.292}{\sqrt{R_{e_{L}}}},\left\{\because \sqrt{\frac{\rho U L}{\mu}}=\sqrt{R_{e_{L}}}\right\} \end{aligned} $