MrsDrPoe: The Blasius Boundary Layer Solution

It's Thesis Tuesday on the blog, and today we'll be looking at the Blasius Boundary Layer solution.

Often when discussing boundary layer flow over a flat plate, the Blasius boundary layer solution is examined.  This solution is based on the fact that, even though the boundary layer profile changes along the plate, its general shape remains the same.  To capture this relationship, Blasius defined a similarity variable, eta = y*sqrt(Uinfinity/(2*nu*x)).  We can define a stream function for the flow based on the volumetric flow rate between streamlines: psi = sqrt(2*nu*Uinfinity*x)*f(eta).  Using our stream function/velocity relations and the product rule, we can see that:

u = d/dy(psi) = Uinfinity*f'(eta)

v = -d/dx(psi) = sqrt((nu*Uinfinity)/(4*x))*(eta*f'(eta) - f(eta)).

Since we now have our velocity components in terms of x and y, we can plug these expressions into the boundary layer momentum equation to obtain (with MUCH manipulation): 

f''' + f*f'' = 0.

The boundary conditions for this differential equation are f = f' = 0 at eta = 0 and f' -> 1 as eta -> infinity.  Check out the figure here.

From the computed solution, it can be seen that u = 0.99*Uinfinity when eta = 5.0.  This gives us: delta = 5*sqrt(nu*x/Uinfinity) = 5*x/sqrt(Rex), where Rex is the local Reynolds number at any x-location along the plate (Rex = Uinfinity*x/nu).  Furthermore, since we now know the velocity profile in the boundary layer, we can calculate the shear stress at any x-location along the plate by tauw = 0.332*Uinfinity^(3/2)*sqrt(rho*mu/x).  We can also determine the friction coefficient at a particular x-location Cf = 0.664/sqrt(Rex).

Next week we'll look at an example problem involving these equations.  Happy studying!