Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in boundary layer given by: $\frac{u}{U}=2(\frac{y}{\delta})-(\frac{y}{\delta})^2$

Solution :

Given :

The velocity distribution in boundary layer given by,

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

(i) Displacement thickness $\delta^{*}$

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

Substituting the value of $ \frac{u}{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} $ …