Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

 Adam C Scott on avoiding gdb signal noise… Ken on Scotiabank iTrade RESP …… Alan Ball on Oops. Fixing a drill hole in P… Peeter Joot's B… on Stokes theorem in Geometric… Exploring Stokes The… on Stokes theorem in Geometric…

• 290,113

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

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

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 $\delta$.

Flow over a solid object with boundary layer.

We have a couple scales to consider.

• Velocity scale $U$.
• Length scale in the $y$ direction $\delta$.
• Length scale in the $x$ direction $L$.

where

\begin{aligned}L \gg \delta.\end{aligned} \hspace{\stretch{1}}(3.1)

As always, we start with the Navier-Stokes equation, restricting ourselves to the steady state ${\partial {\mathbf{u}}}/{\partial {t}} = 0$ case. In coordinates, for incompressible flows, we have our usual $x$ momentum, $y$ momentum, and continuity equations

\begin{subequations}

\begin{aligned}u \frac{\partial {u}}{\partial {x}} + v \frac{\partial {u}}{\partial {y}} = - \frac{1}{{\rho}} \frac{\partial {p}}{\partial {x}} + \nu \left( \frac{\partial^2 {{u}}}{\partial {{x}}^2}+\frac{\partial^2 {{u}}}{\partial {{y}}^2}\right)\end{aligned} \hspace{\stretch{1}}(3.2a)

\begin{aligned}u \frac{\partial {v}}{\partial {x}} + v \frac{\partial {v}}{\partial {y}} = - \frac{1}{{\rho}} \frac{\partial {p}}{\partial {y}} + \nu \left( \frac{\partial^2 {{v}}}{\partial {{x}}^2}+\frac{\partial^2 {{v}}}{\partial {{y}}^2}\right)\end{aligned} \hspace{\stretch{1}}(3.2b)

\begin{aligned}\frac{\partial {u}}{\partial {x}} +\frac{\partial {v}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(3.2c)

\end{subequations}

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

\begin{aligned}\frac{\partial {u}}{\partial {x}} &\sim \frac{U}{L} \\ \frac{\partial {v}}{\partial {y}} &\sim \frac{v}{\delta}\end{aligned} \hspace{\stretch{1}}(3.3)

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 $v = 0$, and the two components of the divergence are obviously of different scales.

If these are of the same scale we have

\begin{aligned}\frac{U}{L} \sim \frac{v}{\delta}\end{aligned} \hspace{\stretch{1}}(3.5)

so that

\begin{aligned}v \sim \frac{U \delta}{L}\end{aligned} \hspace{\stretch{1}}(3.6)

or

\begin{aligned}v \ll U\end{aligned} \hspace{\stretch{1}}(3.7)

Looking at the viscous terms

\begin{aligned}\nu \frac{\partial^2 {{u}}}{\partial {{x}}^2} &\sim \frac{\nu U}{L^2} \\ \nu \frac{\partial^2 {{u}}}{\partial {{y}}^2} &\sim \frac{\nu U}{\delta^2}\end{aligned} \hspace{\stretch{1}}(3.8)

or

\begin{aligned}\nu \frac{\partial^2 {{u}}}{\partial {{y}}^2} \gg \nu \frac{\partial^2 {{u}}}{\partial {{x}}^2}\end{aligned} \hspace{\stretch{1}}(3.10)

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

How about the inertial terms

\begin{aligned}u \frac{\partial {u}}{\partial {x}} &\sim \frac{U^2}{L} \\ v \frac{\partial {u}}{\partial {y}} &\sim \frac{\delta U}{L} \frac{U}{\delta} \sim \frac{U^2}{L}\end{aligned} \hspace{\stretch{1}}(3.11)

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 $y$ momentum equation we have

\begin{aligned}\nu \frac{\partial^2 {{v}}}{\partial {{x}}^2} &\sim \nu \frac{\delta U}{L} \frac{1}{{L^2}} \sim \nu \frac{\delta U}{L^3} \ll \frac{\nu U}{\delta^2} \\ \nu \frac{\partial^2 {{v}}}{\partial {{y}}^2} &\sim \nu \frac{\delta U}{L} \frac{1}{{\delta^2}} \sim \nu \frac{U}{\delta L} \ll \frac{\nu U}{\delta^2}\end{aligned} \hspace{\stretch{1}}(3.13)

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

Question: Why compare the magnitude of the viscous terms for the $y$ momentum to the magnitude of the same terms in the $x$ momentum equation, and not to the LHS of the $y$ 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 $y$ momentum equation we have

\begin{aligned}u \frac{\partial {v}}{\partial {x}} &\sim \frac{\delta U^2}{L^2} \\ v \frac{\partial {v}}{\partial {y}} &\sim \frac{\delta U}{L} \frac{\delta U}{L} \frac{1}{{\delta}} \sim \frac{\delta U^2}{L^2}\end{aligned} \hspace{\stretch{1}}(3.15)

Note that

\begin{aligned}\frac{\delta U^2}{L^2} = \frac{\delta}{L} \left( \frac{U^2}{L} \right) \ll \frac{U^2}{L}\end{aligned} \hspace{\stretch{1}}(3.17)

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

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

\begin{subequations}

\begin{aligned}u \frac{\partial {u}}{\partial {x}} + v \frac{\partial {u}}{\partial {y}} = - \frac{1}{{\rho}} \frac{\partial {p}}{\partial {x}} + \nu \frac{\partial^2 {{u}}}{\partial {{y}}^2}\end{aligned} \hspace{\stretch{1}}(3.18a)

\begin{aligned}\frac{\partial {p}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(3.18b)

\begin{aligned}\frac{\partial {u}}{\partial {x}} + \frac{\partial {v}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(3.18c)

\end{subequations}

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

\begin{aligned}\frac{\partial {\mathbf{u}}}{\partial {t}} + (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = -\boldsymbol{\nabla} \left( \frac{p}{\rho} \right) + \nu \boldsymbol{\nabla}^2 \mathbf{u} + \mathbf{g}.\end{aligned} \hspace{\stretch{1}}(3.19)

Writing the body force as a potential

\begin{aligned}\mathbf{g} = -\boldsymbol{\nabla} \chi,\end{aligned} \hspace{\stretch{1}}(3.20)

so that we have

\begin{aligned}\frac{\partial {\mathbf{u}}}{\partial {t}} + (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = -\boldsymbol{\nabla} \left( \frac{p}{\rho} + \chi \right) + \nu \boldsymbol{\nabla}^2 \mathbf{u} .\end{aligned} \hspace{\stretch{1}}(3.21)

Using the vector identity

\begin{aligned}(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{u}) + \boldsymbol{\nabla} \left( \frac{1}{{2}} \mathbf{u}^2 \right),\end{aligned} \hspace{\stretch{1}}(3.22)

we can write

\begin{aligned}\frac{\partial {\mathbf{u}}}{\partial {t}} + \boldsymbol{\nabla} \times (\boldsymbol{\nabla} \times \mathbf{u}) + \boldsymbol{\nabla} \left( \frac{1}{{2}} \mathbf{u}^2 \right)= -\boldsymbol{\nabla} \left( \frac{p}{\rho} + \chi \right) + \nu \boldsymbol{\nabla}^2 \mathbf{u} .\end{aligned} \hspace{\stretch{1}}(3.23)

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 ($\boldsymbol{\nabla} \times \mathbf{u} = 0$) in the non-viscous region, we have

\begin{aligned}\boldsymbol{\nabla} \left( \frac{p}{\rho} + \chi + \frac{1}{{2}} \mathbf{u}^2 \right) = 0.\end{aligned} \hspace{\stretch{1}}(3.24)

or

\begin{aligned}\boxed{\frac{p}{\rho} + \chi + \frac{1}{{2}} \mathbf{u}^2 = \text{constant}}\end{aligned} \hspace{\stretch{1}}(3.25)

This is the Bernoulli equation.

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

\begin{aligned}- \frac{\partial {}}{\partial {x}} \left( \frac{p}{\rho} \right) &\sim \frac{\partial {}}{\partial {x}} \left( \frac{1}{{2}} u^2 \right) \\ &\sim \frac{2}{2} u \frac{\partial {u}}{\partial {x}} \\ &\sim U \frac{dU}{dx}\end{aligned}

Our equations of motion are finally reduced to

\begin{subequations}

\begin{aligned}u \frac{\partial {u}}{\partial {x}} + v \frac{\partial {u}}{\partial {y}} = U \frac{dU}{dx} + \nu \frac{\partial^2 {{u}}}{\partial {{y}}^2}\end{aligned} \hspace{\stretch{1}}(3.26a)

\begin{aligned}\frac{\partial {p}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(3.26b)

\begin{aligned}\frac{\partial {u}}{\partial {x}} + \frac{\partial {v}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(3.26c)

\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

\begin{aligned}\boldsymbol{\nabla}^2 \left( \frac{p}{\rho} + \chi + \frac{1}{{2}} \mathbf{u}^2 \right) = 0.\end{aligned} \hspace{\stretch{1}}(3.27)