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Find the displacement thickness, momentum thickness and energy thickness for the velocity distribution in boundary layer given by:

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

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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}

(ii) 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}

(iii) Energy thickness $\delta^{**}$

\begin{aligned} \delta^{* *} &=\int_{0}^{\delta} \frac{u}{U}\left[1-\frac{u^{2}}{U^{2}}\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]^{2}\right) d y \\ &=\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{y^{2}}{\delta^{2}}\right)\left(1-\left[\frac{4 y^{2}}{\delta^{2}}+\frac{y^{4}}{\delta^{4}}-\frac{4 y^{3}}{\delta^{3}}\right]\right) d y \\ &=\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{y^{2}}{\delta^{2}}\right)\left(1-\frac{4 y^{2}}{\delta^{2}}-\frac{y^{4}}{\delta^{4}}+\frac{4 y^{3}}{\delta^{3}}\right) d y \\ &=\int_{0}^{\delta}\left(\frac{2 y}{\delta}-\frac{8 y^{3}}{\delta^{3}}-\frac{2 y^{5}}{\delta^{5}}+\frac{8 y^{4}}{\delta^{4}}-\frac{y^{2}}{\delta^{2}}+\frac{4 y^{4}}{\delta^{4}}+\frac{y^{6}}{\delta^{6}}-\frac{4 y^{5}}{\delta^{5}}\right) d y \\ &=\int_{0}^{0}\left[\frac{2 y}{\delta}-\frac{y^{2}}{\delta^{2}}-\frac{8 y^{3}}{\delta^{3}}+\frac{12 y^{4}}{\delta^{4}}-\frac{6 y^{5}}{\delta^{3}}+\frac{y^{6}}{\delta^{6}}\right] d y \\ &=\left[\frac{2 y^{2}}{2 \delta}-\frac{y^{3}}{3 \delta^{2}}-\frac{8 y^{4}}{4 \delta^{3}}+\frac{12 y^{5}}{5 \delta^{4}}-\frac{6 y^{6}}{6 \delta^{5}}+\frac{y^{7}}{7 \delta^{6}}\right]_{0}^{\delta} \\ &=\frac{\delta^{2}}{\delta}-\frac{\delta^{3}}{3 \delta^{2}}-\frac{2 \delta^{4}}{\delta^{3}}+\frac{12 \delta^{5}}{5 \delta^{4}}-\frac{\delta^{6}}{\delta^{5}}+\frac{\delta^{7}}{7 \delta^{6}}\\ &=\delta-\frac{\delta}{3}-2 \delta+\frac{12}{5} \delta-\delta+\frac{\delta}{7} \\ &=-2 \delta-\frac{\delta}{3}+\frac{12}{5} \delta+\frac{\delta}{7}\\ &=\frac{-210 \delta-35 \delta+252 \delta+15 \delta}{105} \\ &=\frac{-245 \delta+267 \delta}{105}\\ &=\frac{22 \delta}{105} \end{aligned}

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