Chapter 10, Problem 113P

The streamwise velocity component of a steady incompressible. laminar. flat plate boundary laver of boundary layer thickness $\delta$is approximated by the sine wave profile of Prob. 10-112. Generate expressions for displacement thickness and momentum thickness as functions of $\delta$, based on this sine wave approximation. Compare the approximate values of $\delta */\delta$and $\theta /\delta$to the values of $\delta */\delta$and $\theta /\delta$ obtained from the Blasius solution.

To determine

The approximate values of $\frac{\theta }{\delta }$ and $\frac{{\delta }^{\ast }}{\delta }$ is with comparison to the values of Blasius result.

Answer to Problem 113P

The approximate values of $\frac{\theta }{\delta }$ and $\frac{{\delta }^{\ast }}{\delta }$ is greater than the values of Blasius result.

Explanation of Solution

Write the expression for the velocity profile of sine wave.

  $u\left(y\right)=U\sin \left(\frac{\pi y}{2\delta }\right)$........ (I)

Write the expression for the displacement thickness.

  ${\delta }^{\ast }={\displaystyle {\int }_0^{\delta }\left(1-\frac{u\left(y\right)}{U}\right)}dy$........ (II)

Here, the boundary layer thickness is $\delta$, the velocity function is $u$ and maximum velocity of the laminar flow over the flat plate is $U$.

Substitute $U\sin \left(\frac{\pi y}{2\delta }\right)$ for $u\left(y\right)$ in Equation (II).

  $\begin{array}{c}{\delta }^{\ast }={\displaystyle {\int }_0^{\delta }\left(1-\frac{U\sin \left(\frac{\pi y}{2\delta }\right)}{U}\right)}dy\\ ={\displaystyle {\int }_0^{\delta }\left(1-\sin \left(\frac{\pi y}{2\delta }\right)\right)}dy\end{array}$........ (III)

Integrate the Equation (III) between the limits $0$ to $\delta$.

  $\begin{array}{c}{\delta }^{\ast }={\left[y-\frac{2\delta \cos \left(\frac{\pi y}{2\delta }\right)}{\pi }\right]}_0^{\delta }\\ {\delta }^{\ast }=\left[\delta -\frac{2\delta \cos \left(\frac{\pi \delta }{2\delta }\right)}{\pi }\right]-\left[0-\frac{2\delta \cos \left(\frac{\pi 0}{2\delta }\right)}{\pi }\right]\end{array}$

  $\begin{array}{c}{\delta }^{\ast }=\left[\delta +\frac{2\delta \cos \left(\frac{\pi }{2}\right)}{\pi }\right]-\left[0+\frac{2\delta \cos \left(0\right)}{\pi }\right]\\ =\left[\delta +0\right]-\left[0+\frac{2\delta }{\pi }\right]\\ =\delta -\frac{2\delta }{\pi }\\ {\delta }^{\ast }=\frac{\delta }{\pi }\left(\pi -2\right)\end{array}$

Thus, the expression for the displacement thickness is $\frac{\delta }{\pi }\left(\pi -2\right)$.

Write the expression for the momentum thickness.

  $\theta ={\displaystyle {\int }_0^{\delta }\frac{u\left(y\right)}{U}\left(1-\frac{u\left(y\right)}{U}\right)}dy$........ (IV)

Substitute $U\sin \left(\frac{\pi y}{2\delta }\right)$ for $u\left(y\right)$ in Equation (III).

  $\begin{array}{c}\theta ={\displaystyle {\int }_0^{\delta }\frac{U\sin \left(\frac{\pi y}{2\delta }\right)}{U}\left(1-\frac{U\sin \left(\frac{\pi y}{2\delta }\right)}{U}\right)}dy\\ ={\displaystyle {\int }_0^{\delta }\sin \left(\frac{\pi y}{2\delta }\right)\left(1-\sin \left(\frac{\pi y}{2\delta }\right)\right)}dy\end{array}$........ (V)

Integrate the Equation (V) between the limit $0$ and $\delta$.

  $\begin{array}{c}\theta ={\displaystyle {\int }_0^{\delta }\left(\sin \left(\frac{\pi y}{2\delta }\right)-{\sin }^2\left(\frac{\pi y}{2\delta }\right)\right)}dy\\ \theta ={\displaystyle {\int }_0^{\delta }\left(\sin \left(\frac{\pi y}{2\delta }\right)-\left(\frac{1-\cos \left(\frac{\pi y}{\delta }\right)}{2}\right)\right)}dy\end{array}$

  $\begin{array}{c}\theta ={\left(\frac{-\cos \left(\frac{\pi y}{2\delta }\right)}{\frac{\pi }{2\delta }}-\frac{y}{2}+\left(\frac{\sin \left(\frac{\pi y}{\delta }\right)}{2}\right)\right)}_0^{\delta }\\ \theta =\left(\begin{array}{l}\frac{-\cos \left(\frac{\pi \delta }{2\delta }\right)}{\frac{\pi }{2\delta }}-\frac{\delta }{2}+\left(\frac{\sin \left(\frac{\pi \delta }{\delta }\right)}{2}\right)+\\ \frac{\cos \left(\frac{\pi 0}{2\delta }\right)}{\frac{\pi }{2\delta }}+\frac{0}{2}-\left(\frac{\sin \left(\frac{\pi 0}{\delta }\right)}{2}\right)\end{array}\right)\\ \theta =\left(\begin{array}{l}\frac{-\cos \left(\frac{\pi }{2}\right)}{\frac{\pi }{2\delta }}-\frac{\delta }{2}+\left(\frac{\sin \left(\pi \right)}{2}\right)+\\ \frac{\cos \left(0\right)}{\frac{\pi }{2\delta }}+\frac{0}{2}-\left(\frac{\sin \left(0\right)}{2}\right)\end{array}\right)\end{array}$

  $\begin{array}{c}\theta =\left(0-\frac{\delta }{2}+0+\frac{2\delta }{\pi }\right)\\ =-\frac{\delta }{2}+\frac{2\delta }{\pi }\\ \theta =\frac{\delta }{\pi }\left(\frac{4-\pi }{2}\right)\end{array}$

Thus, the momentum thickness is $\frac{\delta }{\pi }\left(\frac{4-\pi }{2}\right)$.

Write the expression of displacement thickness ratio to boundary layer thickness.

  $y=\frac{{\delta }^{\ast }}{\delta }$........ (VI)

Substitute $\frac{\delta }{\pi }\left(\pi -2\right)$ for ${\delta }^{\ast }$ in Equation (VI).

  $\begin{array}{c}y=\frac{\frac{\delta }{\pi }\left(\pi -2\right)}{\delta }\\ =\frac{1}{\pi }\left(\pi -2\right)\\ =\left(1-\frac{2}{\pi }\right)\\ =1-0.6366\end{array}$

  $y=0.363$

Write the expression to calculate the ratio of momentum thickness to boundary layer thickness.

  $y^{\prime }=\frac{\theta }{\delta }$........ (VII)

Substitute the $\frac{\delta }{\pi }\left(\frac{4-\pi }{2}\right)$ for $\theta$ in Equation (VII).

  $\begin{array}{c}{y}^{\prime }=\frac{\frac{\delta }{\pi }\left(\frac{4-\pi }{2}\right)}{\delta }\\ =\left(\frac{4-\pi }{2\pi }\right)\\ =\frac{0.86}{6.283}\\ =0.137\end{array}$

Write the expression for boundary layer thickness by blasius solution.

  $\delta =\frac{4.91x}{\sqrt{R_{e_x}}}$........ (VIII)

Here, the location on the flat plate is $x$ and the Reynolds number is $R_{e_x}$.

Write the expression for displacement thickness by blasius solution.

  ${\delta }^{\ast }=\frac{1.72x}{\sqrt{R_{e_x}}}$........ (IX)

Write the expression for the momentum thickness by Blasius solution.

  $\theta =\frac{0.664x}{\sqrt{R_{e_x}}}$........ (X)

Substitute $\frac{1.72x}{\sqrt{R_{e_x}}}$ for ${\delta }^{\ast }$ and $\frac{4.91x}{\sqrt{R_{e_x}}}$ for $\delta$ in Equation (VI).

  $\begin{array}{c}y=\frac{\frac{1.72x}{\sqrt{R_{e_x}}}}{\frac{4.91x}{\sqrt{R_{e_x}}}}\\ =\frac{1.72}{4.91}\\ =0.35\end{array}$

Substitute $\frac{0.664x}{\sqrt{R_{e_x}}}$ for $\theta$ and $\frac{4.91x}{\sqrt{R_{e_x}}}$ for $\delta$ in Equation (VII).

  $\begin{array}{c}{y}^{\prime }=\frac{\frac{0.664x}{\sqrt{R_{e_x}}}}{\frac{4.91x}{\sqrt{R_{e_x}}}}\\ =\frac{0.664}{4.91}\\ =0.135\end{array}$

Conclusion:

The approximate values of $\frac{\theta }{\delta }$ and $\frac{{\delta }^{\ast }}{\delta }$ is greater than the values of Blasius result.