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The table below lists the average bond energies that you would need to determine reaction enthalpies

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

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

Solutions

Expert Solution

Solution :

The velocity distribution in boundary layer given by

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

1.  Displacement thickness ( \( \delta^* \))

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

Substituting \( \frac{u}{U}=2\left(\frac{y}{\delta}\right)-\left(\frac{y}{\delta}\right)^{2} \) in above equation, we get 

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

2.  Momentum thickness \( (\theta) \)

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

3.  Energy thickness \( \theta^{**} \)

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


\( \)we calculated Displacement thickness, momentum thickness and energy thickness. 

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