Given the velocity distribution in a laminar boundary layer on a flat plate as $\frac{\upsilon}{U}=2(\frac{y}{\delta})-2(\frac{y}{\delta})^3+(\frac{y}{\delta})^4$

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 boundary layer thickness, shear stress, and force on one side of the plate in terms of Reynolds number.

Solution :

The velocity profile, $$\quad \frac{v}{U}=\frac{2 y}{\delta}-\frac{2 y^{3}}{\delta^{3}}+\frac{y^{4}}{\delta^{4}}$$

$$ \frac{\tau_{0}}{\rho U^{2}}=\frac{\partial}{\partial x}\left[\int_{0}^{\delta} \frac{v}{U}\left(1-\frac{v}{U}\right) d y\right] $$

Substituting the given velocity profile in the above equation

$$ \begin{aligned} \frac{\tau_{0}}{\rho U^{2}} & =\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{2 y^{3}}{\delta^{3}}+\frac{y^{4}}{\delta^{4}}\right)\left(1-\left\{\frac{2 y}{\delta}-\frac{2 y^{3}}{\delta^{3}}+\frac{y^{4}}{\delta^{4}}\right\}\right) d y\right] \\ &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{2 y^{3}}{\delta^{3}}+\frac{y^{4}}{8^{4}}\right)\left(1-\frac{2 y}{8}+\frac{2 y^{3}}{\delta^{3}}-\frac{y^{4}}{\delta^{4}}\right) d y\right] \\ &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{4 y^{2}}{\delta^{2}}+\frac{4 y^{4}}{\delta^{4}}-\frac{2 y^{5}}{\delta^{5}}-\frac{2 y^{3}}{\delta^{3}}+\frac{4 y^{4}}{\delta^{4}}-\frac{4 y^{6}}{\delta^{6}}+\frac{2 y^{7}}{\delta^{7}}+\frac{y^{4}}{\delta^{4}}-\frac{2 y^{5}}{\delta^{5}}+\frac{2 y^{7}}{\delta^{7}}-\frac{y^{8}}{\delta^{8}}\right) d y\right] \\ &=\frac{\partial}{\partial x}\left[\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{4 y^{2}}{\delta^{2}}-\frac{2 y^{3}}{\delta^{3}}+\frac{9 y^{4}}{\delta^{4}}-\frac{4 y^{5}}{\delta^{5}}-\frac{4 y^{6}}{\delta^{6}}+\frac{4 y^{7}}{\delta^{7}}-\frac{y^{8}}{\delta^{8}}\right) d y\right] \\ &=\frac{\partial}{\partial x}\left[\frac{2 y^{2}}{2 \delta}-\frac{4 y^{3}}{3 \delta^{2}}-\frac{2 y^{4}}{4 \delta^{3}}+\frac{9 y^{5}}{5 \delta^{4}}-\frac{4 y^{6}}{6 \delta^{5}}-\frac{4 y^{7}}{7 \delta^{6}}+\frac{4 y^{8}}{8 \delta^{7}}-\frac{y^{9}}{9 \delta^{8}}\right]_{0}^{\delta} \\ &=\frac{\partial}{\partial x}\left[\delta-\frac{4}{3} \delta-\frac{1}{2} \delta+\frac{9}{5} \delta-\frac{2}{3} \delta-\frac{4}{7} \delta+\frac{1}{2} \delta-\frac{1}{9} \delta\right] \\ &=\frac{\partial}{\partial x}\left[\frac{315-420+63 \times 9-210-45 \times 4-35}{315}\right] \frac{\partial}{\delta}\\ &=\frac{\partial}{\partial x}\left[\frac{315-420+567-210-180-35}{315}\right] \delta \\ &=\frac{\partial}{\partial x}\left[\frac{882-845}{815}\right] \delta\\ &=\frac{\partial}{\partial x}\left[\frac{37}{315}\right] \delta \\ &=\frac{37}{315} \frac{\partial \delta}{\partial x}\\ \tau_{0} &=\frac{37}{315} \rho U^{2} \frac{\partial \delta}{\partial x} ...(i) \end{aligned} $$

Also shear stress is given by Newton's law of viscosity as

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

where $v=U\left[\frac{2 y}{\delta}-\frac{2 y^{2}}{\delta^{2}}+\frac{y^{4}}{\delta^{4}}\right]$

$ \begin{aligned} \therefore \left(\frac{d v}{d y}\right) &=U\left[\frac{2}{\delta}-\frac{4 y}{\delta^{2}}-\frac{4 y^{3}}{\delta^{4}}\right] \\ \therefore \left(\frac{\partial v}{\partial y}\right)_{y=0} &=U\left[\frac{2}{\delta}-\frac{4}{\delta^{2}}(0)-\frac{4}{\delta^{4}}(0)\right]\\ &=\frac{2 U}{\delta} \end{aligned} $

$$ \tau_{0}=\mu\left(\frac{\partial u}{\partial y}\right)_{y=0}=\mu \times \frac{2 U}{\delta}=\frac{2 U \mu}{\delta} ...(ii) $$

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

$$\begin{aligned} \frac{37}{315} \rho U^{2} \frac{\partial \delta}{\partial x}=\frac{2 U \mu}{\delta} \text { or } \delta \partial \delta=\frac{315}{37} \times \frac{2 U \mu}{\rho U^{2}} \partial x=\frac{630}{37} \frac{\mu}{\rho U} \partial x \end{aligned}$$

On integration, we get

$$\frac{\delta^{2}}{2}=\frac{630}{37} \frac{\mu}{\rho U} x+C$$,

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

$$ \begin{aligned} \frac{\delta^{2}}{2} &=\frac{630}{37} \frac{\mu}{\rho U} x \\ \delta &=\sqrt{\frac{630 \times 2}{37} \frac{\mu}{\rho U}} x=5.84 \sqrt{\frac{\mu x}{\rho U}} \\ &=5.84 \sqrt{\frac{\mu x \times x}{\rho U x}}=5.84 \sqrt{\frac{\mu}{\rho U x}} \times x\\ &=\frac{5.84 x}{\sqrt{R_{e}}} ...(iii) \end{aligned} $$

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

Substituting the value of $\delta$ from (ii) into $(iii)$

$$\begin{aligned} \tau_{0} &=\frac{2 U \mu}{\delta}=\frac{2 U \mu}{\frac{5.84 x}{\sqrt{R_{e}}}}=\frac{2 U \mu}{5.84 x} \sqrt{R_{e}}\\ &=0.34 \frac{U \mu}{x} {\sqrt{R_{e}}} \end{aligned}$$

3) Drag Force $\left(\mathbf{F}_{ \mathbf{D} }\right)$ on one side of the plate :

$$ \begin{aligned} F_{D} &=\int_{0}^{L} \tau_{0} \times b \times d x\\ &=\int_{0}^{L} 0.34 \frac{U \mu}{x} \sqrt{R_{e_{t}}} \times b \times d x\\ &=\int_{0}^{L} 0.34 \frac{U \mu}{\mu} \frac{\sqrt{\rho U \mu}}{\mu} b d x \\ &=0.34 U \mu \sqrt{\frac{\rho U}{\mu}} \times b \int_{0}^{L} x^{-1 / 2} d x\\ &=0.34 U \mu \sqrt{\frac{\rho U}{\mu}} \times b \times\left[\frac{x^{1 / 2}}{1 / 2}\right]_{0}^{L} \\ &=0.34 \times 2 U \mu \sqrt{\frac{\rho U}{\mu}} b \sqrt{L}\\ &=0.68 b \mu U \sqrt{\frac{\rho U L}{\mu}} \end{aligned} $$

4) Drag Co-efficient $\left(\mathrm{C}_{ \mathbf{D} }\right)$

$C_{D}=\frac{F_{D}}{\frac{1}{2} \rho A U^{2}}$,

where $A=b \times L$

$$\begin{aligned} &=\frac{0.68 \times b \times \mu U \times \sqrt{\frac{\rho U L}{\mu}}}{\frac{1}{2} \rho \times b \times L \times U^{2}}\\ &=0.68 \times 2 \frac{\mu}{\rho U L} \times \sqrt{\frac{\rho U L}{\mu}}\\ &=1.36 \times \frac{1}{\sqrt{\frac{\rho U L}{\mu}}} \\ &=1.36 × \frac{1}{\sqrt{R_e}} \end{aligned}$$