Blasius Equation: Some Explorations Part 3

In the previous post we have solved a modified form of Blasius equation using matlab. But the free pramater that is used for the scaling are not detemined. Here we continue the exploration of solution of the Blasius Equation. The parameter \( \alpha \) does not appear in g formulation, and we can solve for numerically. After solving for g, we choose: 

\begin{equation}\alpha = \sqrt{g’(\infty)} \end{equation}

Then \( f \) will satisfy the proper condition at infinity. With this choice of \(\alpha\), we obtain \( f \) as
\begin{equation} f(\eta) = \frac{1}{\alpha} g\left( \frac{\eta}{\alpha} \right) \end{equation} In the numerical integration to get the solution was carried out to \( \eta \) = 15. As we see from the plot, this is more than we need. The velocity function levels off well before = 10.

We can now determine \(\alpha \):

>> alpha = sqrt(sol.y(2,end))

alpha = 1.444094587085306

Finally we map back to the original function f . Then we check the boundary condition and check for the boundary value of f.

>> f = (deval(sol, ([0 13]./alpha)));

>> f(1,1)/alpha

ans = 0

Fluid Velocity Components

The scaled horizontal velocity component is given by: \[ \frac{u}{U_{\infty}} = f’(\eta) \]

---

>> eta = linspace(0,6, 100);

>> f   = (deval(sol, (eta./alpha)));

>> plot(eta, f(2,:)./alpha^2, 'r','LineWidth', 2.5)

>> xlabel('\eta'); ylabel('u/U_\infty')

>> ylim([0 1.5])

---

The vertical Component is given by: \[ v = \frac{1}{2} \sqrt{\frac{\mu U_{\infty}}{x}} y_F \]
where  \(y_F\) is defined as \[ y_F = \eta f’(\eta) – f(\eta) \]

>> eta = linspace(0,10, 200);

>> f   = (deval(sol, (eta./alpha)));
>> plot(eta, eta.*f(2,:)./alpha^2-f(1,:)./alpha, 'g','LineWidth', 2.5)

>> xlabel('\eta'); ylabel('v_s')

>> ylim([0 2])

We see indeed that there is a constant vertical velocity at the outer limit of the boundary layer. This is caused by the retardation in the boundary layer, which in turn deflects some of the oncoming fluid upward.  At the outer edge of the boundary layer \[ v = \frac{1}{2} \sqrt{\frac{\mu U_{\infty}}{x}}c_{\infty} = \frac{U_{\infty}c_{\infty}}{2 \sqrt{Re_x}} \]

In the next post I will explore the boundary layer thickness.