# Peeter Joot's Blog.

• ## Archives

 bubba transvere on My letter to the Ontario Energ… Kuba Ober on New faucet installation peeterjoot on Ease of screwing up C string… peeterjoot on Basement electrical now d… peeterjoot on New faucet installation
• ## People not reading this blog: 6,973,738,433 minus:

• 136,375 hits

# Archive for March 26th, 2012

## PHY454H1S Continuum mechanics. Problem Set 2. Two layer flow driven by moving surface. (revisited)

Posted by peeterjoot on March 26, 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.)]

# Problem Q3 (revisited).

I’d produced the following sketches. For a higher viscosity bottom layer $\mu^{(1)} > \mu^{(2)}$, this should look something like figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig4}) whereas for the higher viscosity on the top, these would be roughly flipped as in figure (\ref{fig:continuumProblemSet2:continuumProblemSet2Fig5}).

This superposition can be justified since we have no $(\mathbf{u} \cdot \boldsymbol{\nabla})\mathbf{u}$ term in the Navier-Stokes equations for these systems.

### Exact solutions.

The figures above are kind of rough. It’s not actually hard to solve the system above. After some simplification, I find with Mathematica the following solution

\begin{aligned}u^{(1)}(y) &= -\frac{\mu^{(2)} U -G h^2}{\mu^{(1)}+\mu^{(2)}}-\frac{y \left(G h^2 (\mu^{(2)} -\mu^{(1)}) +2 \mu^{(1)} \mu^{(2)} U \right)}{2 h \mu^{(1)} \left(\mu^{(1)}+\mu^{(2)}\right)}-\frac{G y^2}{2 \mu^{(1)}} \\ u^{(2)}(y) &= -\frac{\mu^{(2)} U -G h^2}{\mu^{(1)}+\mu^{(2)}}-\frac{y \left(G h^2 (\mu^{(2)} -\mu^{(1)}) +2 \mu^{(1)} \mu^{(2)} U \right)}{2 h \mu^{(2)} \left(\mu^{(1)}+\mu^{(2)}\right)}-\frac{G y^2}{2 \mu^{(2)}}.\end{aligned} \hspace{\stretch{1}}(3.49)

Should we wish a more exact plot for any specific values of the viscosities, we could plot exactly with software the vector field described by these velocities.

I suppose it is cheating to use Mathematica and then say that the solution is easy? To make amends for being lazy with my algebra, let’s show that it is easy to do manually too. I’ll do the same problem manually, but generalize it slightly. We can do this easily if we just be a bit marter with our integration constants. Let’s solve the problem for the upper and lower walls moving with velocity $V_2$ and $V_1$ respectively, and let the heights from the interface be $h_2$ and $h_1$ respectively.

We have the same set of differential equations to solve, but now let’s write our solution with the undetermined coefficients expressed as

\begin{aligned}u^{(2)} &= -\frac{G}{2 \mu^{(2)}}\left( h_2^2 - y^2 \right) + \frac{A_2}{\mu^{(2)}} (y - h_2) + B_2 \\ u^{(1)} &= -\frac{G}{2 \mu^{(1)}}\left( h_1^2 - y^2 \right) + \frac{A_1}{\mu^{(1)}} (y + h_1) + B_1.\end{aligned} \hspace{\stretch{1}}(3.55)

Now it’s super easy to match the boundary conditions at $y = -h_1$ and $y = h_2$(the lower and upper walls respectively). Clearly the integration constants $B_1,B_2$ are just the velocities. Matching the tangential component of the traction vectors at $y = 0$ we have

\begin{aligned}A_2 = A_1\end{aligned} \hspace{\stretch{1}}(3.55)

and matching velocities at $y = 0$ gives us

\begin{aligned}-\frac{G}{2 \mu^{(2)}}h_2^2 - \frac{A_2}{\mu^{(2)}} h_2 + V_2 = -\frac{G}{2 \mu^{(1)}}h_1^2 + \frac{A_1}{\mu^{(1)}} h_1 + V_1.\end{aligned} \hspace{\stretch{1}}(3.55)

This gives us

\begin{aligned}u^{(2)} &= \frac{G h_2^2}{2 \mu^{(2)}}\left(\frac{y^2}{h_2^2} - 1 \right) + A \frac{ h_2}{\mu^{(2)}} \left( \frac{y}{h_2} - 1 \right) + V_2 \\ u^{(1)} &= \frac{G h_1^2}{2 \mu^{(1)}}\left(\frac{y^2}{h_1^2} - 1 \right) + A \frac{ h_1}{\mu^{(1)}} \left( \frac{y}{h_1} + 1 \right) + V_1 \\ A &=\frac{V_2 - V_1+ \frac{G h_1^2}{2 \mu^{(1)}}-\frac{G h_2^2}{2 \mu^{(2)}}}{\frac{h_1}{\mu^{(1)}}+\frac{h_2}{\mu^{(2)}}}.\end{aligned} \hspace{\stretch{1}}(3.55)

Plotting this with sliders or animation in Mathematica is a fun way to explore visualizing this. The results vary widely depending on the various parameters. Such a Mathematica notebook can be found in https://raw.github.com/peeterjoot/physicsplay/master/notes/phy454/problemSetIIQ3PlotWithManipulate.cdf. A pair of cached animations with variation of the pressure gradient for $v_1 = 0$, $h_1 = h_2$, showing the superposition of the shear and channel flow solutions can be found here

1. With $\mu^{(1)} > \mu^{(2)}$ fluids with densities of mercury and water in the lower (1) and upper (2) layers respectively.
2. With $\mu^{(2)} > \mu^{(1)}$, this is plotted fluids with densities of water and mercury in the lower (1) and upper (2) layers respectively.