## PHY454H1S Continuum Mechanics. Lecture 18: Scaling to setup to solve the boundary layer problem. Taught by Prof. K. Das.

Posted by peeterjoot on March 21, 2012

# Disclaimer.

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

# Motivation.

We’ve been talking about impulsively started flow and the Stokes boundary problem.

We’ll now move on to a similar problem, that of fluid flow over a solid body.

# Fluid flow over a solid body.

Consider figure (\ref{fig:continuumL18:continuumL18Fig1}), where we have an illustration of flow over a solid object with a boundary layer of thickness .

We have a couple scales to consider.

- Velocity scale .
- Length scale in the direction .
- Length scale in the direction .

where

As always, we start with the Navier-Stokes equation, restricting ourselves to the steady state case. In coordinates, for incompressible flows, we have our usual momentum, momentum, and continuity equations

\begin{subequations}

\end{subequations}

Let’s look at the scaling of these equations, starting with the continuity equation 3.2c. This is roughly

We require that these have to be of the same order of magnitude.

FIXME: why? This doesn’t make sense to me since in horizontal flow we had , and the two components of the divergence are obviously of different scales.

If these are of the same scale we have

so that

or

Looking at the viscous terms

or

So we can neglect the component of the Laplacian in our momentum equation 3.2a.

How about the inertial terms

Since these are of the same order (in the boundary regions) we cannot neglect either. We also cannot neglect the pressure gradient, since this is what induces the flow.

For the momentum equation we have

We can neglect all the Laplacian terms in the momentum equation.

**Question:** Why compare the magnitude of the viscous terms for the momentum to the magnitude of the same terms in the momentum equation, and not to the LHS of the momentum equation.

**Answer.** That’s a valid point, but our equations are coupled, and contributions from one feed into the other.

We aren’t done yet. For the inertial terms in the momentum equation we have

Note that

We see that both of the momentum inertial terms can be neglected in comparison to the momentum equations.

Putting all of this together, our equations of motion for the boundary flow are now reduced to

\begin{subequations}

\end{subequations}

Let’s try to see what we can do with our pressure term. Let’s start with Navier-Stokes in vector form

Writing the body force as a potential

so that we have

Using the vector identity

we can write

If we consider the non-viscous region of the flow (far from the boundary layer), we can kill the Laplacian term. Again, considering only the steady state, and assuming that we have irrotational flow () in the non-viscous region, we have

or

This is the Bernoulli equation.

Using this we can get an idea of the magnitude of the pressure term. We have

Our equations of motion are finally reduced to

\begin{subequations}

\end{subequations}

## Aside

Observe that if we operate on 3.23 with a divergence operation, then we don’t have to assume non-visous flow but in that case we can only say that (for irrotational flow) we have

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