$\frac{\upsilon}{U}=2 (\frac{y}{\delta})-2(\frac{y}{\delta})^2$

Where is main stream velocity at boundary layer thickness .Determine the displacement thickness and momentum thickness.

Solution :

Given :

The velocity distribution profile is,

$$\begin{aligned}\frac{v}{U}=2\left(\frac{y}{\delta}\right)-\left(\frac{y}{\delta}\right)^{2}\end{aligned}$$

1) Displacement thickness $\delta^{*}$

$$\begin{aligned}\delta^{*}=\int_{0}^{\delta}\left(1-\frac{v}{U}\right) d y \end{aligned}$$

Substituting the value of $ \frac{v}{U}=2\left(\frac{y}{\delta}\right)-\left(\frac{y}{\delta}\right)^{2},$ we have

$$ \begin{aligned} \delta^{*} &=\int_{0}^{\delta}\left\{1-\left[2\left(\frac{y}{\delta}\right)-\left(\frac{y}{\delta}\right)^{2}\right]\right\} d y \\ &=\int_{0}^{\delta}\left\{1-2\left(\frac{y}{\delta}\right)+\left(\frac{y}{\delta}\right)^{2}\right\} d y\\ &=\left[y-\frac{2 y^{2}}{2 \delta}+\frac{y^{3}}{3 \delta^{2}}\right]_{0}^{\delta} \\ &=\delta-\frac{\delta^{2}}{\delta}+\frac{\delta^{3}}{3 \delta^{2}}\\ &=\delta-\delta+\frac{\delta}{3}\\ &=\frac{\delta}{3} \end{aligned} $$

2) Momentum thickness $\theta$

$$ \begin{aligned} \theta &=\int_{0}^{\delta} \frac{u}{U}\left\{1-\frac{u}{U}\right\} d y \\ &=\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{y^{2}}{\delta^{2} } \right)\left[1-\left(\frac{2 y}{\delta}-\frac{y^{2}}{\delta^{2}} \right) \right] d y \\ &=\int_{0}^{\delta}\left[\frac{2 y}{\delta}-\frac{y^{2}}{\delta^{2}}\right]\left[1-\frac{2 y}{\delta}+\frac{y^{2}}{\delta^{2}}\right] d y \\ &=\int_{0}^{\delta}\left[\frac{2 y}{\delta}-\frac{4 y^{2}}{\delta^{2}}+\frac{2 y^{3}}{\delta^{3}}-\frac{y^{2}}{\delta^{2}}+\frac{2 y^{3}}{\delta^{3}}-\frac{y^{4}}{\delta^{4}}\right] d y \\ &=\int_{0}^{\delta}\left[\frac{2 y}{\delta}-\frac{5 y^{2}}{\delta^{2}}+\frac{4 y^{3}}{\delta^{3}}-\frac{y^{4}}{\delta^{4}}\right] d y\\ &=\left[\frac{2 y^{2}}{2 \delta}-\frac{5 y^{3}}{3 \delta^{2}}+\frac{4 y^{4}}{4 \delta^{3}}-\frac{y^{5}}{5 \delta^{4}}\right]_{0}^{\delta} \\ &=\left[\frac{\delta^{2}}{\delta}-\frac{5 \delta^{3}}{3 \delta^{2}}+\frac{\delta^{4}}{\delta^{3}}-\frac{\delta^{5}}{5 \delta^{4}}\right]\\ &=\delta-\frac{5 \delta}{3}+\delta-\frac{\delta}{5} \\ &=\frac{15 \delta-25 \delta+15 \delta-3 \delta}{15}\\ &=\frac{30 \delta-28 \delta}{15}\\ &=\frac{2 \delta}{15} \end{aligned} $$