In earlier screencasts introducing us to Blasius solutions, we looked at how to find boundary
layer properties using these solutions. Here, what we're going to do is we're going to find the surface
stress in a boundary layer using the Blasius solutions. Our surface stress, tau sub s,
is the viscosity, mu, dU/dy at y=0. U equals U infinity which is the free stream velocity,
df/d(eta), where f is a function of eta and eta is a similarity variable. Now we need
to find dU/dy. This is the same as dU/d(eta) d(eta)/dy. Eta is defined as y times the square
root of U infinity divided by nu, the kinematic viscosity, times x. This makes d(eta)/dy equal
to this quantity under the square root. So dU/d(eta) is just U inifinity d squared f
/d(eta) squared. So let's put them all together. dU/dy therefore equals U infinity, the square
root of U infinity divided by nu times x, d squared f/ d(eta) squared. In order to solve
for dU/dy, and therefore our shear stress, we're going to have to back to that table
of our boundary layer solutions, and again this was solved using the methods of Blasius.
And if you note, we have a column for d squared f/d(eta) squared, so the question is, which
one of these values do you use? If you recall, we're looking for y = 0, which is the surface
of the plate, and let's go back and see what our definition for eta is. What we realize
is, if y=0, eta=0. Going back to this chart, eta=0, therefore our d squared f/d(eta) squared
equals 0.332. So let's put this all together with our definition for our shear stress.
Again, tau sub s is equal to mu dU/dy, which equals 0.332, that's our d squared f/d(eta)
squared, U infinity times mu, or viscosity, times the square root of U infinity divided
by nu times x. Our kinematic viscosity is the same as the dynamic viscosity, mu, divided
by the density, so we can simplify this even further as 0.332 times U infinity times the
square root of rho U infinity mu divided by x. Let's take a look at a particular situation.
We consider water at 300 Kelvin with a free stream velocity of 2 meters per second, and
what we want to do is find out what the shear stress is at the wall as we go down the plate,
or as a function of x. So when we look these properties up, we find that the density is
997 kilograms per meter cubed, and the viscosity is 855 times 10 to the minus 6 Newton seconds
per meter squared. When we put that into our expression for tau sub s, where again s equals
the surface, this is going to equal 0.87 divided by the square root of x. This is an expression
for us to be able to find the shear stress at the wall, with water flowing at 2 meters
per second at 300 K, anywhere along the length of the flat plate.