Chapter 10, Problem 97P

The stramwise velocity component of steady, incompressible, laminar, flat plate boundary layer thickness $\delta$ is approximated by the simple linear expression, $u=Uy/\delta$ for $y<\delta$ , and u=U for $y>\delta$ (Fig. 10-97). Generate expressions for displacement thickness momentum thickness as functions of $\delta$ , based on this linear approximation, Compare the approximate values of $\delta */\delta$ and $\theta /\delta$ to the $\delta */\delta$ and $\theta /\delta$ obtained from the Blasius solution.

Answer: 0.500, 0.167

FIGURE P10-97

To determine

The comparison between different ratio when approximation method and blasius methods are used simultaneously.

Answer to Problem 97P

The approximation value for $\frac{{\delta }^{*}}{\delta }$ is $0.5$.

The approximation value for $\frac{\theta }{\delta }$ is $0.167$.

The blasius solution for $\frac{{\delta }^{*}}{\delta }$ is $0.35$.

The blasius solution for $\frac{\theta }{\delta }$ is $0.135$.

Explanation of Solution

Given information:

Write the expression for velocity profile if $y<\delta$.

  $u=\frac{Uy}{\delta }$  ...... (I)

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

Write the expression for velocity profile if $y>\delta$.

  $u=U$  ...... (II)

Here, maximum velocity of the laminar flow over the flat plate is $U$.

Write the expression for displacement thickness.

  ${\delta }^{*}={\displaystyle \underset{0}{\overset{\delta }{\int }}\left(1-\frac{u_{y<\delta }}{U}\right)}dy+{\displaystyle \underset{\delta }{\overset{\infty }{\int }}\left(1-\frac{u_{y>\delta }}{U}\right)}dy$  ...... (III)

Write the expression for momentum thickness.

  $\theta ={\displaystyle \underset{0}{\overset{\delta }{\int }}\frac{u_{y<\delta }}{U}\left(1-\frac{u_{y<\delta }}{U}\right)dy}+{\displaystyle \underset{\delta }{\overset{\infty }{\int }}\frac{u_{y>\delta }}{U}\left(1-\frac{u_{y>\delta }}{U}\right)}dy$  ...... (IV)

Write the expression for Blasius solution for boundary layer thickness.

  $\delta =\frac{4.91x}{\sqrt{{\mathrm{Re}}_x}}$  ...... (V)

Here, any location on flat plate is $x$, Reynolds number is ${\mathrm{Re}}_x$.

Write the expression for Blasius solution for displacement thickness.

  ${\delta }^{*}=\frac{1.72x}{\sqrt{{\mathrm{Re}}_x}}$  ...... (VI)

Write the expression for Blasius solution for momentum thickness.

  $\theta =\frac{0.664x}{\sqrt{{\mathrm{Re}}_x}}$  ...... (VII)

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

  $R=\frac{{\delta }^{*}}{\delta }$  ...... (VIII)

Write the expression for momentum thickness to boundary layer thickness.

  $S=\frac{\theta }{\delta }$  ...... (IX)

Calculation:

Substitute $\frac{Uy}{\delta }$ for $u_{y<\delta }$, and $U$ for $u_{y>\delta }$ in Equation (III).

  $\begin{array}{c}{\delta }^{*}={\displaystyle \underset{0}{\overset{\delta }{\int }}\left(1-\frac{\left(\frac{Uy}{\delta }\right)}{U}\right)}dy+{\displaystyle \underset{\delta }{\overset{\infty }{\int }}\left(1-\frac{U}{U}\right)}dy\\ ={\displaystyle \underset{0}{\overset{\delta }{\int }}\left(1-\frac{y}{\delta }\right)}dy+{\displaystyle \underset{\delta }{\overset{\infty }{\int }}\left(0\right)}dy\\ ={\left[y-\frac{y^2}{2\delta }\right]}_0^{\delta }\\ =\delta -\frac{{\delta }^2}{2\delta }\end{array}$

  ${\delta }^{*}=\frac{\delta }{2}$

The displacement thickness ${\delta }^{*}$ is $\frac{\delta }{2}$.

Substitute $\frac{Uy}{\delta }$ for $u_{y<\delta }$, and $U$ for $u_{y>\delta }$ in Equation (IV).

  $\begin{array}{c}\theta ={\displaystyle \underset{0}{\overset{\delta }{\int }}\frac{\frac{Uy}{\delta }}{U}\left(1-\frac{\frac{Uy}{\delta }}{U}\right)dy}+{\displaystyle \underset{\delta }{\overset{\infty }{\int }}\frac{U}{U}\left(1-\frac{U}{U}\right)}dy\\ ={\displaystyle \underset{0}{\overset{\delta }{\int }}\frac{y}{\delta }\left(1-\frac{y}{\delta }\right)dy}+{\displaystyle \underset{\delta }{\overset{\infty }{\int }}\frac{U}{U}}\left(0\right)dy\\ ={\displaystyle \underset{0}{\overset{\delta }{\int }}\frac{y}{\delta }\left(1-\frac{y}{\delta }\right)dy}+0\end{array}$

  $\begin{array}{c}\theta ={\left[\frac{y^2}{2\delta }-\frac{y^3}{3{\delta }^2}\right]}_0^{\delta }\\ =\left(\frac{{\delta }^2}{2\delta }-\frac{{\delta }^3}{3{\delta }^2}\right)-0\\ =\frac{\delta }{2}-\frac{\delta }{3}\\ =\frac{\delta }{6}\end{array}$

The momentum thickness $\theta$ is $\frac{\delta }{6}$.

Substitute $\frac{\delta }{2}$ for ${\delta }^{*}$ in Equation (VIII).

  $\begin{array}{c}R=\frac{\left(\frac{\delta }{2}\right)}{\delta }\\ =\frac{\delta }{2\delta }\\ =\frac{1}{2}\\ 0.5\end{array}$

Ratio of displacement thickness to boundary layer thickness is $0.5$.

Substitute $\frac{\delta }{6}$ for $\theta$ in Equation (IX).

  $\begin{array}{c}S=\frac{\left(\frac{\delta }{6}\right)}{\delta }\\ =\frac{\delta }{6\delta }\\ =\frac{1}{6}\\ =0.167\end{array}$

Ratio of displacement thickness to boundary layer thickness is $0.167$.

Divide Equation (VI) with Equation (V) when blasius solution is considered.

  $\begin{array}{c}\frac{{\delta }^{*}}{\delta }=\frac{\left(\frac{1.72x}{\sqrt{{\mathrm{Re}}_x}}\right)}{\left(\frac{4.91x}{\sqrt{{\mathrm{Re}}_x}}\right)}\\ =\frac{1.72}{4.91}\\ =0.35\end{array}$

Divide Equation (VII) with Equation (V) when blasius solution is considered.

  $\begin{array}{c}\frac{\theta }{\delta }=\frac{\left(\frac{0.664x}{\sqrt{{\mathrm{Re}}_x}}\right)}{\left(\frac{4.91x}{\sqrt{{\mathrm{Re}}_x}}\right)}\\ =\frac{0.664}{4.91}\\ =0.135\end{array}$

Conclusion:

The approximation value for $\frac{{\delta }^{*}}{\delta }$ is $0.5$.

The approximation value for $\frac{\theta }{\delta }$ is $0.167$.

The blasius solution for $\frac{{\delta }^{*}}{\delta }$ is $0.35$.

The blasius solution for $\frac{\theta }{\delta }$ is $0.135$.