Search

Sunday, May 26, 2013

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])


Vel_U

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])

Vel_v


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.

Blasius Equation: Some Explorations Part 2

Solution of Blasius Equation in Matlab

A direct attack on the Blasius equation requires some kind of iteration such as a shooting method, because it is a two-point boundary value problem. Fortunately, there is a reformulation of the problem that avoids an iteration. We begin this reformulation by introducing a new dependent variable :
g(\eta ) = \alpha f(\alpha \eta )
Whereimage is a positive parameter to be determined. As one may show, the equation for g is then the same as that for f:
g(0) = 0,\;g'(0) = 0,\;g''(0) = 1
From the conditions on f , we have g(0) = 0, and

g'(\eta ) = {\alpha ^2}f'(\alpha \eta ) \to {\alpha ^2}

as

\eta  \to \infty  

Instead of imposing the condition on g at infinity , we impose

g''(0) = 1.

Then the problem for g is an initial value problem with initial conditions
g(0) = 0,\;g'(0) = 0,\;g''(0) = 1
In order to carry out this integration numerically using matlab, we need to convert the 3rd order differential equation into a system of three 1st order equations.
{y_1} = g,{y_2} = g',{y_3} = g''
We have the following equivalent system of equations
{y_1}^\prime  = {y_2},\quad {y_2}^\prime  = {y_3},\quad {y_3}^\prime  =  - 0.5{y_1}{y_3}
   1:  >> blas =@(t,f) [f(2); f(3); -0.5*f(1)*f(3)];
   2:  >> options = odeset('RelTol', 1e-10, 'AbsTol', 1e-12);
   3:  >> sol=ode45(blas, [0,15], [0.0 0.0 1.0], options);
   4:  >> plot(sol.x, sol.y(2,:), 'LineWidth', 2.5)
   5:  >> xlabel('\eta'); ylabel('g(\eta)')
   6:  >> plot(sol.x, sol.y(2,:), 'LineWidth', 2.5)
   7:  >> xlabel('\eta'); ylabel('g(\eta)')



blas





Blasius Equation: Some Explorations - Part 1

 

What is Blasius Equation:

Blasuis Equation describes the flow of a fluid over a flat plate.

The x-y coordinate system is chosen so that x is along the plate, and y is perpendicular to the plate. The leading edge of the plate is at x = 0, y = 0. The velocity components - u in the x-direction and v in the y-direction -- are expressed in terms of a stream function (x,y):

image

The fundamental equation which determines y is the x-component of the momentum equation in the boundary layer approximation. For the flat plate there is no pressure gradient, and the boundary layer approximation to the x-momentum equation takes the form

image

Where is the kinematic viscosity. By substituting the first eqaution into the second, we get for the partial differential equation

image

At the wall (y = 0) both velocity components must vanish, and far away from the plate, the horizontal velocity must approach the given free stream velocity image . These conditions translate into the following conditions on image:

image

The absence of a length scale (the plate is semi-infinite in length) suggests a similarity solution, as originally used by Blasius. The solution has the form:

image

The scaleimage is comparable with the boundary layer thickness. This substitution into the equation for  imageleads to the following nonlinear ordinary differential equation forimage :

image

This equation is called the Blasius equation. We will solve it numerically in the next part. Once f is known, the velocity components may be computed as

image