Chapter 7, Problem 7.5CP

To determine

For Blasius flat plate relation, devise an iteration method which starts with $f^{11}\left(0\right)=0.2$ ? Compare with table 7.1.

Answer to Problem 7.5CP

Initial value for second derivative is equal to $0.33$ . The values obtained for $\frac{u}{U}$ is almost equal to that of in table 7.1 when initial derivative is equal to $0.33$

Explanation of Solution

Given information:

The Blasius flat plate relation is defined as,

$f^{\text{'}\text{'}\text{'}}+\frac{1}{2}f{f}^{\text{'}\text{'}}=0$

Boundary conditions are,

At $y=0$ ; $f\left(0\right)=f^{\text{'}}\left(0\right)=0$

At $y\to \infty$ ; $f^{\text{'}}\left(\infty \right)\to 1.0$

And,

$\begin{array}{l}{f}^{\text{'}}\left(\eta \right)=\frac{u}{U}\\ \eta =y{\left(\frac{U}{vx}\right)}^{1/2}\end{array}$

The Blasius equation is a third order nonlinear differential equation,

Assume a power series expansion for small values of $\eta$,

$f\left(\eta \right)=A_0+A_1\eta +\frac{A_2}{2!}{\eta }^2+\frac{A_3}{3!}{\eta }^3+\frac{A_4}{4!}{\eta }^4+\mathrm{...}$

Calculation:

According to the explanation given,

$f\left(\eta \right)=A_0+A_1\eta +\frac{A_2}{2!}{\eta }^2+\frac{A_3}{3!}{\eta }^3+\frac{A_4}{4!}{\eta }^4+\mathrm{...}$

Find the first, second and third derivative of above equation,

$f^{\text{'}}\left(\eta \right)=A_1+A_2\eta +\frac{A_3}{2!}{\eta }^2+\frac{A_4}{3!}{\eta }^3+\mathrm{...}$

$f^{\text{'}\text{'}}\left(\eta \right)=A_2+A_3\eta +\frac{A_4}{2!}{\eta }^2+\frac{A_5}{3!}{\eta }^3+\mathrm{...}$

$f^{\text{'}\text{'}\text{'}}\left(\eta \right)=A_3+A_4\eta +\frac{A_5}{2!}{\eta }^2+\frac{A_6}{3!}{\eta }^3+\mathrm{...}$

Apply boundary conditions,

$\begin{array}{l}\eta =0,f^{\text{'}}\left(\eta \right)=0\\ \eta =0,f\left(\eta \right)=0\end{array}$

Therefore,

$\begin{array}{l}{A}_0=0\\ A_1=0\end{array}$

According to $y=0,\frac{{\partial }^{2}u}{\partial y^2}=0$

$\eta =0,f^{\text{'}\text{'}\text{'}}\left(\eta \right)=0$

Therefore we get,

$A_3=0$

Substitute for $f,f^{\text{'}\text{'}},f^{\text{'}\text{'}\text{'}}$ in Blasius equation,

$2\left[A_4\eta +\frac{A_5}{2!}{\eta }^2+\frac{A_6}{3!}{\eta }^3+\mathrm{...}\right]+\left[\frac{A_2}{2!}{\eta }^2+\frac{A_4}{4!}{\eta }^4+\frac{A_5}{5!}{\eta }^5\mathrm{...}\right]\left[A_2+\frac{A_4}{2!}{\eta }^2+\frac{A_5}{3!}{\eta }^3+\mathrm{...}\right]=0$

By equating co-efficient to zero,

$\begin{array}{l}{A}_4=A_6=A_7=0\\ A_5=-\frac{A_2{}^2}{2}\\ A_8=\frac{11}{4}{A}_2{}^3\\ A_{11}=-\frac{375}{8}{A}_2{}^4\end{array}$

For initial condition,

$f^{\text{'}\text{'}}\left(0\right)=0.2$

According to power series expansion,

$f^{\text{'}\text{'}}\left(0\right)=A_2$

According to power series expansion,

$f^{\text{'}}\left(\eta \right)=A_2\eta -\frac{A_2{}^2}{2}\frac{{\eta }^4}{4!}+\frac{11A_2{}^3}{4}\frac{{\eta }^7}{7!}+\mathrm{...}$

Assume, $A_2=0.2$

For $\eta =0.2$

$f^{\text{'}}\left(\eta \right)=$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.0399$

For $\eta =0.4$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.0799$

For $\eta =0.6$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.1198$

Comparing with table 7.1, the obtained values for $f^{\text{'}}\left(\eta \right)=\frac{u}{U}$ is considerably low

Therefore,

Assume, $A_2=0.3$

For $\eta =0.2$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.0599$

For, $\eta =0.4$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.1199$

For, $\eta =0.6$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.1797$

Comparing with table 7.1, the obtained values for $f^{\text{'}}\left(\eta \right)=\frac{u}{U}$ is little low

Therefore,

Assume, $A_2=0.33$

For $\eta =0.2$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.0659$

For, $\eta =0.4$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.1319$

For, $\eta =0.6$

$f^{\text{'}}\left(\eta \right)=\frac{u}{U}=0.1977$

This seems almost equal with the values that in table 7.1

Therefore, initial second derivative is almost equal to $0.33$ and by further iteration; we can find more exact value.

Conclusion:

The initial second derivative is almost equal to $0.33$ and by further iteration.