## PHY454H1S Continuum Mechanics. Lecture 19: Boundary layers and stream functions. The Blasius problem. Taught by Prof. K. Das.

Posted by peeterjoot on April 4, 2012

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Review. Laminar boundary layer theory.

In the boundary layer we found

- Continuity equation
- momentum equation
- momentum equation

In the inviscid region is a constant in

This will be approximately true in the boundary layer too as illustrated in figure (1).

Starting with

we were able to show that inviscid irrotational incompressible flows are governed by Bernoulli’s equation

In the absence of body forces (or constant potentials), we have

so that our boundary layer equations are

\begin{subequations}

\end{subequations}

With boundary conditions

# Fluid flow over a flat plate (Blasius problem).

Define a similarity variable

Suppose we want

Since we have

or

We can make the transformation

We can introduce stream functions

We can check that this satisfies the continuity equation since we have

Now introduce a similarity variable

Note that we’ve also suddenly assumed that (a constant, which will also kill the term in the N-S equation). This isn’t really justified by anything we’ve done so far, but asking about this in class, it was stated that this is a restriction for this formulation of the Blasius problem.

Also note that this last step requires:

This at least makes sense dimensionally since we have , but where did this definition of come from?

In [] it is mentioned that this is a result of the scaling argument. We did have some scaling arguments that included in the expressions from last lecture, one of which was 2.7

but that doesn’t obviously give us 3.20?

Ah. We argued that

and that the larger of the viscous terms was

If we require that these are the same order of magnitude, as argued in section 8.3 of [2], then we find 3.21.

Regardless, given this change of variables, we can apparently compute

Our boundary conditions are

latex \eta = 0$} \\ f’ = 1 & \quad \mbox{} \\ \end{array}\end{aligned} \hspace{\stretch{1}}(3.25)$

## Deriving the equation of motion.

Attempting to derive 3.24 using the definitions above gets a bit messy. It’s messy enough that I mistakenly thought that we couldn’t possible arrive at a differential equation that has a plain old (non-derivative) in it as in 3.24 above. The algebra involved in taking the derivatives got the better of me. This derivation is treated a different way in [2]. For the purpose of completeness (and because that derivation also leaves out some details), lets do this from start to finish with all the gory details, but following the outline provided in the text.

Instead of pre-determining the form of the similarity variable exactly, we can state it in terms of an unknown and to be determined function of position writing

We still introduce stream our stream functions 3.17, but require that our horizontal velocity component is only a function of our similarity variable

where is to be determined, and is scaled by our characteristic velocity . Observe that, as above, we are assuming that , a constant (which also kills off the term in the Navier-Stokes equation.) Given this form of , we note that

so that

It’s argued in the text that we also want to be a streamline, so that implying that . I don’t honestly follow the rational for that, but it’s certainly convenient to set , so lets do so and see where things go. With

Observe that is necessarily zero with this definition. We can now write

This is like what we had in class, with the exception that instead of a constant relating and we also have a function of . That’s exactly what we need so that we can end up with both and derivatives of in our Navier-Stokes equation.

Now let’s do the mechanical bits, computing all the derivatives. We can compute to start with

We had initially , but , so we’ve now got both and specified in terms of and and their derivatives

We’ve got a bunch of the derivatives that we have to compute

and

and

Our component of Navier-Stokes now takes the form

or (assuming )

Now, it we wish (to make this equation as easy to solve as possible), we can integrate to find the required form of . This gives

It’s argued that we expect

to become singular at , so we should set . This leaves us with

\begin{subequations}

\end{subequations}

and boundary value conditions

\begin{subequations}

\end{subequations}

where 3.38b follows from and 3.37d, and 3.38c follows from the fact that tends to .

## Numeric solution.

We can solve this numerically and find solutions that look like figure (2).

This is the Blasius solution to the problem of fluid flow over a flat plate.

# Singular perturbation theory.

In the boundary layer analysis we’ve assumed that our inertial term and viscous terms were of the same order of magnitude. Lets examine the validity of this assumption

or

or

finally

If we have

and

then

when .

(this is the whole reason that we were able to do the previous analysis).

Our EOM is

with

as

performing a non-dimensionalization we have

or

to force , we can write

so that as we have .

With a very small number modifying the highest degree partial term, we have a class of differential equations that doesn’t end up converging should we attempt a standard perturbation treatment. An example that is analogous is the differential equation

where and . The exact solution of this ill conditioned differential equation is

This is illustrated in figure (3).

Study of this class of problems is called \textit{Singular perturbation theory}.

When we have approximately

but when , we have approximately

# References

[1] Wikipedia. Blasius boundary layer — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 28-March-2012]. http://en.wikipedia.org/w/index.php?title=Blasius_boundary_layer&oldid=480776115.

[2] D.J. Acheson. *Elementary fluid dynamics*. Oxford University Press, USA, 1990.

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