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# Archive for April 25th, 2012

## Compilation of class notes for phy454h1s, continuum mechanics (including exam review and study problems)

Posted by peeterjoot on April 25, 2012

As mentioned in previously I’ve got a compilation of class notes here:

This has been updated now with some problems worked as exam practice and a review section. Included in this document are:

• my lecture notes, often augmented by whatever I felt required to make them understandable to myself
• some solved problems
• some stuff I was playing with on the side
• links to the various mathematica notebooks (some of which have some cool stuff, and can be viewed after download with the free wolfram CDF player).

All of the following, previously posted individually are now also included, and will no longer be maintained independently

Apr 24, 2012 Spin down of coffee in a bottomless cup.

Apr 21, 2012 Continuum mechanics fluids review.

Apr 21, 2012 Continuum mechanics elasticity review.

## Continuum mechanics fluids review.

Posted by peeterjoot on April 25, 2012

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

# Motivation.

Review of key ideas and equations from the fluid dynamics portion of the class.

# Vector displacements.

Those portions of the theory of elasticity that we did cover have the appearance of providing some logical context for the derivation of the Navier-Stokes equation. Our starting point is almost identical, but we now look at displacements that vary with time, forming

\begin{aligned}d\mathbf{x}' = d\mathbf{x} + d\mathbf{u} \delta t.\end{aligned} \hspace{\stretch{1}}(2.1)

We compute a first order Taylor expansion of this differential, defining a symmetric strain and antisymmetric vorticity tensor

\begin{subequations}

\begin{aligned}e_{ij} = \frac{1}{{2}} \left(\frac{\partial {u_i}}{\partial {x_j}} +\frac{\partial {u_j}}{\partial {x_i}} \right).\end{aligned} \hspace{\stretch{1}}(2.2a)

\begin{aligned}\omega_{ij} = \frac{1}{{2}} \left(\frac{\partial {u_i}}{\partial {x_j}}-\frac{\partial {u_j}}{\partial {x_i}} \right)\end{aligned} \hspace{\stretch{1}}(2.2b)

\end{subequations}

Allowing us to write

\begin{aligned}dx_i' = dx_i + e_{ij} dx_j \delta t + \omega_{ij} dx_j \delta t.\end{aligned} \hspace{\stretch{1}}(2.3)

We introduced vector and dual vector forms of the vorticity tensor with

\begin{subequations}

\begin{aligned}\Omega_k = \frac{1}{{2}} \partial_i u_j \epsilon_{i j k}\end{aligned} \hspace{\stretch{1}}(2.4a)

\begin{aligned}\omega_{i j} = -\Omega_k \epsilon_{i j k},\end{aligned} \hspace{\stretch{1}}(2.4b)

\end{subequations}

or

\begin{subequations}

\begin{aligned}\boldsymbol{\omega} = \boldsymbol{\nabla} \times \mathbf{u}\end{aligned} \hspace{\stretch{1}}(2.5a)

\begin{aligned}\boldsymbol{\Omega} = \frac{1}{{2}} (\boldsymbol{\omega})_a \mathbf{e}_a.\end{aligned} \hspace{\stretch{1}}(2.5b)

\end{subequations}

We were then able to put our displacement differential into a partial vector form

\begin{aligned}d\mathbf{x}' = d\mathbf{x} + \left( \mathbf{e}_i (e_{ij} \mathbf{e}_j) \cdot d\mathbf{x} + \boldsymbol{\Omega} \times d\mathbf{x} \right) \delta t.\end{aligned} \hspace{\stretch{1}}(2.6)

# Relative change in volume

We are able to identity the divergence of the displacement as the relative change in volume per unit time in terms of the strain tensor trace (in the basis for which the strain is diagonal at a given point)

\begin{aligned}\frac{dV' - dV}{dV \delta t} = \boldsymbol{\nabla} \cdot \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(3.7)

# Conservation of mass

Utilizing Green’s theorem we argued that

\begin{aligned}\int \left( \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot (\rho \mathbf{u}) \right) dV = 0.\end{aligned} \hspace{\stretch{1}}(4.8)

We were able to relate this to rate of change of density, computing

\begin{aligned}\frac{d\rho}{dt} = \frac{\partial {\rho}}{\partial {t}} + \mathbf{u} \cdot \boldsymbol{\nabla} \rho =- \rho \boldsymbol{\nabla} \cdot \mathbf{u}.\end{aligned} \hspace{\stretch{1}}(4.9)

An important consequence of this is that for incompressible fluids (the only types of fluids considered in this course) the divergence of the displacement $\boldsymbol{\nabla} \cdot \mathbf{u} = 0$.

# Constitutive relation

We consider only Newtonian fluids, for which the stress is linearly related to the strain. We will model fluids as disjoint sets of hydrostatic materials for which the constitutive relation was previously found to be

\begin{aligned}\sigma_{ij} = - p \delta_{ij} + 2 \mu e_{ij}.\end{aligned} \hspace{\stretch{1}}(5.10)

# Conservation of momentum (Navier-Stokes)

As in elasticity, our momentum conservation equation had the form

\begin{aligned}\rho \frac{du_i}{dt} = \frac{\partial {\sigma_{ij}}}{\partial {x_j}} + \rho f_i,\end{aligned} \hspace{\stretch{1}}(6.11)

where $f_i$ are the components of the external (body) forces per unit volume acting on the fluid.

Utilizing the constitutive relation and explicitly evaluating the stress tensor divergence ${\partial {\sigma_{ij}}}/{\partial {x_j}}$ we find

\begin{aligned}\rho \frac{d\mathbf{u}}{dt} =\rho \frac{\partial {\mathbf{u}}}{\partial {t}} + \rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}= -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u}+ \mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u}) + \rho \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(6.12)

Since we treat only incompressible fluids in this course we can decompose this into a pair of equations

\begin{subequations}

\begin{aligned}\rho \frac{\partial {\mathbf{u}}}{\partial {t}} + \rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}= -\boldsymbol{\nabla} p + \mu \boldsymbol{\nabla}^2 \mathbf{u}+ \rho \mathbf{f}.\end{aligned} \hspace{\stretch{1}}(6.13a)

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u} = 0\end{aligned} \hspace{\stretch{1}}(6.13b)

\end{subequations}

# No slip condition

We’ll find in general that we have to solve for our boundary value conditions. One of the important constraints that we have to do so will be a requirement (experimentally motivated) that our velocities match at an interface. This was illustrated with a rocker tank video in class.

This is the no-slip condition, and includes a requirement that the fluid velocity at the boundary of a non-moving surface is zero, and that the fluid velocity on the boundary of a moving surface matches the rate of the surface itself.

For fluids $A$ and $B$ separated at an interface with unit normal $\hat{\mathbf{n}}$ and unit tangent $\hat{\boldsymbol{\tau}}$ we wrote the no-slip condition as

\begin{subequations}

\begin{aligned}\mathbf{u}_A \cdot \hat{\boldsymbol{\tau}} = \mathbf{u}_B \cdot \hat{\boldsymbol{\tau}}\end{aligned} \hspace{\stretch{1}}(7.14a)

\begin{aligned}\mathbf{u}_A \cdot \hat{\mathbf{n}} = \mathbf{u}_B \cdot \hat{\mathbf{n}}.\end{aligned} \hspace{\stretch{1}}(7.14b)

\end{subequations}

For the problems we attempt, it will often be enough to consider only the tangential component of the velocity.

# Traction vector matching at an interface.

As well as matching velocities, we have a force balance requirement at any interface. This will be expressed in terms of the traction vector

\begin{aligned}\mathbf{T} = \mathbf{e}_i \sigma_{ij} n_j = \boldsymbol{\sigma} \cdot \hat{\mathbf{n}}\end{aligned} \hspace{\stretch{1}}(8.15)

where $\hat{\mathbf{n}} = n_j \mathbf{e}_i$ is the normal pointing from the interface into the fluid (so the traction vector represents the force of the interface on the fluid). When that interface is another fluid, we are able to calculate the force of one fluid on the other.

In addition the the constraints provided by the no-slip condition, we’ll often have to constrain our solutions according to the equality of the tangential components of the traction vector

\begin{aligned}{\left.{{\tau_i (\sigma_{ij} n_j)}}\right\vert}_{{A}} ={\left.{{\tau_i (\sigma_{ij} n_j)}}\right\vert}_{{B}},\end{aligned} \hspace{\stretch{1}}(8.16)

We’ll sometimes also have to consider, especially when solving for the pressure, the force balance for the normal component of the traction vector at the interface too

\begin{aligned}{\left.{{n_i (\sigma_{ij} n_j)}}\right\vert}_{{A}} ={\left.{{n_i (\sigma_{ij} n_j)}}\right\vert}_{{B}}.\end{aligned} \hspace{\stretch{1}}(8.17)

As well as having a messy non-linear PDE to start with, our boundary value constraints can be very complicated, making the subject rich and tricky.

## Flux

A number of problems we did asked for the flux rate. A slightly more sensible physical quantity is the mass flux, which adds the density into the mix

\begin{aligned}\int \frac{dm}{dt} = \rho \int \frac{dV}{dt} = \rho \int (\mathbf{u} \cdot \hat{\mathbf{n}}) dA\end{aligned} \hspace{\stretch{1}}(8.18)

# Worked problems from class.

A number of problems were tackled in class

1. channel flow with external pressure gradient
2. shear flow
3. pipe (Poiseuille) flow
4. steady state gravity driven film flow down a slope.

## Channel and shear flow

An example that shows many of the features of the above problems is rectilinear flow problem with a pressure gradient and shearing surface. As a review let’s consider fluid flowing between surfaces at $z = \pm h$, the lower surface moving at velocity $v$ and pressure gradient $dp/dx = -G$ we find that Navier-Stokes for an assumed flow of $\mathbf{u} = u(z) \hat{\mathbf{x}}$ takes the form

\begin{aligned}0 &= \partial_x u + \partial_y (0) + \partial_z (0) \\ u \not{{\partial_x u}} &= - \partial_x p + \mu \partial_{zz} u \\ 0 &= -\partial_y p \\ 0 &= -\partial_z p\end{aligned} \hspace{\stretch{1}}(9.19)

We find that this reduces to

\begin{aligned}\frac{d^2 u}{dz^2} = -\frac{G}{\mu}\end{aligned} \hspace{\stretch{1}}(9.23)

with solution

\begin{aligned}u(z) = \frac{G}{2\mu}(h^2 - z^2) + A (z + h) + B.\end{aligned} \hspace{\stretch{1}}(9.24)

Application of the no-slip velocity matching constraint gives us in short order

\begin{aligned}u(z) = \frac{G}{2\mu}(h^2 - z^2) + v \left( 1 - \frac{1}{{2h}} (z + h) \right).\end{aligned} \hspace{\stretch{1}}(9.25)

With $v = 0$ this is the channel flow solution, and with $G = 0$ this is the shearing flow solution.

Having solved for the velocity at any height, we can also solve for the mass or volume flux through a slice of the channel. For the mass flux $\rho Q$ per unit time (given volume flux $Q$)

\begin{aligned}\int \frac{dm}{dt} =\int \rho \frac{dV}{dt} =\rho (\Delta A) \int \mathbf{u} \cdot \hat{\boldsymbol{\tau}},\end{aligned} \hspace{\stretch{1}}(9.26)

we find

\begin{aligned}\rho Q =\rho (\Delta y) \left( \frac{2 G h^3}{3 \mu} + h v\right).\end{aligned} \hspace{\stretch{1}}(9.27)

We can also calculate the force of the boundaries on the fluid. For example, the force per unit volume of the boundary at $z = \pm h$ on the fluid is found by calculating the tangential component of the traction vector taken with normal $\hat{\mathbf{n}} = \mp \hat{\mathbf{z}}$. That tangent vector is found to be

\begin{aligned}\boldsymbol{\sigma} \cdot (\pm \hat{\mathbf{n}}) = -p \hat{\mathbf{z}} \pm 2 \mu \mathbf{e}_i e_{ij} \delta_{j 3} = - p \hat{\mathbf{z}} \pm \hat{\mathbf{x}} \mu \frac{\partial {u}}{\partial {z}}.\end{aligned} \hspace{\stretch{1}}(9.28)

The tangential component is the $\hat{\mathbf{x}}$ component evaluated at $z = \pm h$, so for the lower and upper interfaces we have

\begin{aligned}{\left.{{(\boldsymbol{\sigma} \cdot \hat{\mathbf{n}}) \cdot \hat{\mathbf{x}}}}\right\vert}_{{z = -h}} &= -G (-h) - \frac{v \mu}{2 h} \\ {\left.{{(\boldsymbol{\sigma} \cdot -\hat{\mathbf{n}}) \cdot \hat{\mathbf{x}}}}\right\vert}_{{z = +h}} &= -G (+h) + \frac{v \mu}{2 h},\end{aligned} \hspace{\stretch{1}}(9.29)

so the force per unit area that the boundary applies to the fluid is

\begin{aligned}\text{force per unit length of lower interface on fluid} &= L \left( G h - \frac{v \mu}{2 h} \right) \\ \text{force per unit length of upper interface on fluid} &= L \left( -G h + \frac{v \mu}{2 h} \right).\end{aligned} \hspace{\stretch{1}}(9.31)

Does the sign of the velocity term make sense? Let’s consider the case where we have a zero pressure gradient and look at the lower interface. This is the force of the interface on the fluid, so the force of the fluid on the interface would have the opposite sign

\begin{aligned}\frac{v \mu}{2 h}.\end{aligned} \hspace{\stretch{1}}(9.33)

This does seem reasonable. Our fluid flowing along with a positive velocity is imparting a force on what it is flowing over in the same direction.

# Hydrostatics.

We covered hydrostatics as a separate topic, where it was argued that the pressure $p$ in a fluid, given atmospheric pressure $p_a$ and height from the surface was

\begin{aligned}p = p_a + \rho g h.\end{aligned} \hspace{\stretch{1}}(10.34)

As noted below in the surface tension problem, this is also a consequence of Navier-Stokes for $\mathbf{u} = 0$ (following from $0 = -\boldsymbol{\nabla} p + \rho \mathbf{g}$).

We noted that replacing the a mass of water with something of equal density would not change the non-dynamics of the situation. We then went on to define Buoyancy force, the difference in weight of the equivalent volume of fluid and the weight of the object.

# Mass conservation through apertures.

It was noted that mass conservation provides a relationship between the flow rates through apertures in a closed pipe, since we must have

\begin{aligned}\rho_1 A_1 v_1 = \rho_2 A_2 v_2,\end{aligned} \hspace{\stretch{1}}(11.35)

and therefore for incompressible fluids

\begin{aligned}A_1 v_1 = A_2 v_2.\end{aligned} \hspace{\stretch{1}}(11.36)

So if $A_1 > A_2$ we must have $v_1 < v_2$.

## Curve for tap discharge.

We can use this to get a rough idea what the curve for water coming out a tap would be. Suppose we measure the volume flux, putting a measuring cup under the tap, and timing how long it takes to fill up. We then measure the radii at different points. This can be done from a photo as in figure (1).

Figure 1: Tap flow measurement.

After making the measurement, we can get an idea of the velocity between two points given a velocity estimate at a point higher in the discharge. For a plain old falling mass, our final velocity at a point measured from where the velocity was originally measured can be found from Newton’s law

\begin{aligned}\Delta v = g t\end{aligned} \hspace{\stretch{1}}(11.37)

\begin{aligned}\Delta z = \frac{1}{{2}} g t^2 + v_0 t\end{aligned} \hspace{\stretch{1}}(11.38)

Solving for $v_f = v_0 + \Delta v$, we find

\begin{aligned}v_f = v_0 \sqrt{ 1 + \frac{2 g \Delta z}{v_0^2} }.\end{aligned} \hspace{\stretch{1}}(11.39)

Mass conservation gives us

\begin{aligned}v_0 \pi R^2 = v_f \pi r^2\end{aligned} \hspace{\stretch{1}}(11.40)

or

\begin{aligned}r(\Delta z) = R \sqrt{ \frac{v_0}{v_f} } = R \left( 1 + \frac{2 g \Delta z}{v_0^2} \right)^{-1/4}.\end{aligned} \hspace{\stretch{1}}(11.41)

For the image above I measured a flow rate of about 250 ml in 10 seconds. With that, plus the measured radii at 0 and $6 \text{cm}$, I calculated that the average fluid velocity was $0.9 \text{m}/\text{s}$, vs a free fall rate increase of $1.3 \text{m}/\text{s}$. Not the best match in the world, but that’s to be expected since the velocity has been considered uniform throughout the stream profile, which would not actually be the case. A proper treatment would also have to treat viscosity and surface tension.

In figure (2) is a plot of the measured radial distance compared to what was computed with 11.41. The blue line is the measured width of the stream as measured, the red is a polynomial curve fitted to the raw data, and the green is the computed curve above.

Figure 2: Comparison of measured stream radii and calculated.

# Bernoulli equation.

With the body force specified in gradient for

\begin{aligned}\mathbf{g} = - \boldsymbol{\nabla} \chi\end{aligned} \hspace{\stretch{1}}(12.42)

and utilizing 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}}(12.43)

we are able to show that the steady state, irrotational, non-viscous Navier-Stokes equation takes the form

\begin{aligned}\boldsymbol{\nabla} \left( \frac{p}{\rho} + \chi + \frac{1}{{2}} \mathbf{u}^2 \right) = 0.\end{aligned} \hspace{\stretch{1}}(12.44)

or

\begin{aligned}\frac{p}{\rho} + \chi + \frac{1}{{2}} \mathbf{u}^2 = \text{constant}\end{aligned} \hspace{\stretch{1}}(12.45)

This is the Bernoulli equation, and the constants introduce the concept of streamline.

FIXME: I think this could probably be used to get a better idea what the tap stream radius is, than the method used above. Consider the streamline along the outermost surface. That way you don’t have to assume that the flow is at the average velocity rate uniformly throughout the stream. Try this later.

# Surface tension.

## Surfaces, normals and tangents.

We reviewed basic surface theory, noting that we can parameterize a surface as in the following example

\begin{aligned}\phi = z - h(x, t) = 0.\end{aligned} \hspace{\stretch{1}}(13.46)

\begin{aligned}\boldsymbol{\nabla} \phi = -\hat{\mathbf{x}} \frac{\partial {h}}{\partial {x}} \hat{\mathbf{z}}.\end{aligned} \hspace{\stretch{1}}(13.47)

Recalling that the gradient is normal to the surface we can compute the unit normal and unit tangent vectors

\begin{subequations}

\begin{aligned}\hat{\mathbf{n}} =\frac{1}{{\sqrt{ 1 + \left( \frac{\partial {h}}{\partial {x}} \right)^2 }}}\left( -\frac{\partial {h}}{\partial {x}}, 1 \right)\end{aligned} \hspace{\stretch{1}}(13.48a)

\begin{aligned}\hat{\boldsymbol{\tau}} =\frac{1}{{\sqrt{ 1 + \left( \frac{\partial {h}}{\partial {x}} \right)^2 }}}\left( 1, \frac{\partial {h}}{\partial {x}} \right)\end{aligned} \hspace{\stretch{1}}(13.48b)

\end{subequations}

## Laplace pressure.

We covered some aspects of this topic in class. [1] covers this topic in typical fairly hard to comprehend) detail, but there’s lots of valuable info there. section 2.4.9-2.4.10 of [2] has small section that’s a bit easier to understand, with less detail. Recommended in that text is the “Surface Tension in Fluid Mechanics” movie which can be found on youtube in three parts \youtubehref{DkEhPltiqmo}, \youtubehref{yiixltf\_HKw}, \youtubehref{5d6efCcwkWs}, which is very interesting and entertaining to watch.

It was argued in class that the traction vector differences at the surfaces between a pair of fluids have the form

\begin{aligned}\mathbf{t}_2 - \mathbf{t}_1 = -\frac{\sigma}{2 R} \hat{\mathbf{n}} - \boldsymbol{\nabla}_I \sigma\end{aligned} \hspace{\stretch{1}}(13.49)

where $\boldsymbol{\nabla}_I = \boldsymbol{\nabla} - \hat{\mathbf{n}} (\hat{\mathbf{n}} \cdot \boldsymbol{\nabla})$ is the tangential (interfacial) gradient, $\sigma$ is the surface tension, a force per unit length value, and $R$ is the radius of curvature.

In static equilibrium where $\mathbf{t} = -p \hat{\mathbf{n}}$ (since $\boldsymbol{\sigma} = 0$ if $\mathbf{u} = 0$), then dotting with $\hat{\mathbf{n}}$ we must then have

\begin{aligned}p_2 - p_1 = \frac{\sigma}{2 R} \end{aligned} \hspace{\stretch{1}}(13.50)

Considering the tangential component of the traction vector difference we find

\begin{aligned}(\mathbf{t}_2 - \mathbf{t}_1) \cdot \hat{\boldsymbol{\tau}} = - \hat{\boldsymbol{\tau}} \cdot \boldsymbol{\nabla}_I \sigma\end{aligned} \hspace{\stretch{1}}(13.51)

If the fluid is static (for example, has none of the creep that we see in the film) then we must have $\boldsymbol{\nabla}_I \sigma = 0$. It’s these gradients that are responsible for capillary flow and other related surface tension driven motion (lots of great examples of that in the film).

## Surface tension for a spherical bubble.

In the film above it is pointed out that the surface tension equation we were shown in class

\begin{aligned}\Delta p = \frac{2 \sigma}{R},\end{aligned} \hspace{\stretch{1}}(13.52)

is only for spherical objects that have a single radius of curvature. This formula can in fact be derived with a simple physical argument, stating that the force generated by the surface tension $\sigma$ along the equator of a bubble (as in figure (3)), in a fluid would be balanced by the difference in pressure times the area of that equatorial cross section. That is

Figure 3: Spherical bubble in liquid.

\begin{aligned}\sigma 2 \pi R = \Delta p \pi R^2\end{aligned} \hspace{\stretch{1}}(13.53)

Observe that we obtain 13.52 after dividing through by the area.

## A sample problem. The meniscus curve.

To get a better feeling for this, let’s look to a worked problem. The most obvious one to try to attempt is the shape of a meniscus of water against a wall. This problem is worked in [1], but it is worth some extra notes. As in the text we’ll work with $z$ axis up, and the fluid up against a wall at $x = 0$ as illustrated in figure (4).

Figure 4: Curvature of fluid against a wall.

The starting point is a variation of what we have in class

\begin{aligned}p_1 - p_2 = \sigma \left( \frac{1}{{R_1}} + \frac{1}{{R_2}} \right),\end{aligned} \hspace{\stretch{1}}(13.54)

where $p_2$ is the atmospheric pressure, $p_1$ is the fluid pressure, and the (signed!) radius of curvatures positive if pointing into medium 1 (the fluid).

For fluid at rest, Navier-Stokes takes the form

\begin{aligned}0 = -\boldsymbol{\nabla} p_1 + \rho \mathbf{g}.\end{aligned} \hspace{\stretch{1}}(13.55)

With $\mathbf{g} = -g \hat{\mathbf{z}}$ we have

\begin{aligned}0 = -\frac{\partial {p_1}}{\partial {z}} - \rho g,\end{aligned} \hspace{\stretch{1}}(13.56)

or

\begin{aligned}p_1 = \text{constant} - \rho g z.\end{aligned} \hspace{\stretch{1}}(13.57)

We have $p_2 = p_a$, the atmospheric pressure, so our pressure difference is

\begin{aligned}p_1 - p_2 = \text{constant} - \rho g z.\end{aligned} \hspace{\stretch{1}}(13.58)

We have then

\begin{aligned}\text{constant} -\frac{\rho g z}{\sigma} = \frac{1}{{R_1}} + \frac{1}{{R_2}}.\end{aligned} \hspace{\stretch{1}}(13.59)

One of our axis of curvature directions is directly along the $y$ axis so that curvature is zero $1/R_1 = 0$. We can fix the constant by noting that at $x = \infty$, $z = 0$, we have no curvature $1/R_2 = 0$. This gives

\begin{aligned}\text{constant} -0 = 0 + 0.\end{aligned} \hspace{\stretch{1}}(13.60)

That leaves just the second curvature to determine. For a curve $z = z(x)$ our absolute curvature, according to [3] is

\begin{aligned}{\left\lvert{\frac{1}{{R_2}}}\right\rvert} = \frac{{\left\lvert{z''}\right\rvert}}{(1 + (z')^2)^{3/2}}.\end{aligned} \hspace{\stretch{1}}(13.61)

Now we have to fix the sign. I didn’t recall any sort of notion of a signed radius of curvature, but there’s a blurb about it on the curvature article above, including a nice illustration of signed radius of curvatures can be found in this wikipedia radius of curvature figure for a Lemniscate. Following that definition for a curve such as $z(x) = (1-x)^2$ we’d have a positive curvature, but the text explicitly points out that the curvatures are will be set positive if pointing into the medium. For us to point the normal into the medium as in the figure, we have to invert the sign, so our equation to solve for $z$ is given by

\begin{aligned}-\frac{\rho g z}{\sigma} = -\frac{z''}{(1 + (z')^2)^{3/2}}.\end{aligned} \hspace{\stretch{1}}(13.62)

The text introduces the capillary constant

\begin{aligned}a = \sqrt{2 \sigma/ g \rho}.\end{aligned} \hspace{\stretch{1}}(13.63)

Using that capillary constant $a$ to tidy up a bit and multiplying by a $z'$ integrating factor we have

\begin{aligned}-\frac{2 z z'}{a^2} = -\frac{z'' z'}{(1 + (z')^2)^{3/2}},\end{aligned} \hspace{\stretch{1}}(13.64)

we can integrate to find

\begin{aligned}A - \frac{z^2}{a^2} = \frac{1}{(1 + (z')^2)^{1/2}}.\end{aligned} \hspace{\stretch{1}}(13.65)

Again for $x = \infty$ we have $z = 0$, $z' = 0$, so $A = 1$. Rearranging we have

\begin{aligned}\int dx = \int dz \left( \frac{1}{{(1 - z^2/a^2)^2}} - 1 \right)^{-1/2}.\end{aligned} \hspace{\stretch{1}}(13.66)

Integrating this with Mathematica I get

\begin{aligned}x - x_0 =\sqrt{2 a^2-z^2} \sgn(a-z)+ \frac{a}{\sqrt{2}} \ln \left(\frac{a \left(2 a-\sqrt{4 a^2-2 z^2} \sgn(a-z)\right)}{z}\right).\end{aligned} \hspace{\stretch{1}}(13.67)

It looks like the constant would have to be fixed numerically. We require at $x = 0$

\begin{aligned}z'(0) = \frac{-\cos\theta}{\sin\theta} = -\cot \theta,\end{aligned} \hspace{\stretch{1}}(13.68)

but we don’t have an explicit function for $z$.

# Non-dimensionality and scaling.

With the variable transformations

\begin{aligned}\mathbf{u} &\rightarrow U \mathbf{u}' \\ p &\rightarrow \rho U^2 p' \\ t &\rightarrow \frac{L}{U} t' \\ \boldsymbol{\nabla} &\rightarrow \frac{1}{{L}} \boldsymbol{\nabla}'\end{aligned} \hspace{\stretch{1}}(14.69)

we can put Navier-Stokes in dimensionless form

\begin{aligned}\frac{\partial {\mathbf{u}'}}{\partial {t'}} + (\mathbf{u}' \cdot \boldsymbol{\nabla}') \mathbf{u}' = \boldsymbol{\nabla}' p' + \frac{1}{{R}} \boldsymbol{\nabla}' \mathbf{u}'.\end{aligned} \hspace{\stretch{1}}(14.73)

Here $R$ is Reynold’s number

\begin{aligned}R = \frac{L U}{\nu}\end{aligned} \hspace{\stretch{1}}(14.74)

A relatively high or low Reynold’s number will effect whether viscous or inertial effects dominate

\begin{aligned}R \sim \frac{{\left\lvert{\text{effect of inertia}}\right\rvert}}{{\left\lvert{\text{effect of viscosity}}\right\rvert}} \sim \frac{ {\left\lvert{\rho ( \mathbf{u} \cdot \boldsymbol{\nabla} ) \mathbf{u} }\right\rvert} }{{\left\lvert{ \mu \boldsymbol{\nabla}^2 \mathbf{u}}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(14.75)

The importance of examining where one of these effects can dominate was clear in the Blassius problem, where doing so allowed for an analytic solution that would not have been possible otherwise.

# Eulerian and Lagrangian.

We defined

1. Lagrangian: the observer is moving with the fluid.
2. Eulerian: the observer is fixed in space, watching the fluid.

# Boundary layers.

## Impulsive flow.

We looked at the time dependent unidirectional flow where

\begin{aligned}\mathbf{u} &= u(y, t) \hat{\mathbf{x}} \\ \frac{\partial {u}}{\partial {t}} &= \nu \frac{\partial^2 {{u}}}{\partial {{y}}^2} \\ u(0, t) &=\left\{\begin{array}{l l}0 & \quad \mbox{forlatex t < 0} \\ U & \quad \mbox{for $t \ge 0$}\end{array}\right.\end{aligned} \hspace{\stretch{1}}(16.76)

and utilized a similarity variable $\eta$ with $u = U f(\eta)$

\begin{aligned}\eta = \frac{y}{2 \sqrt{\nu t}}\end{aligned} \hspace{\stretch{1}}(16.79)

and were able to show that

\begin{aligned}u(y, t) = U_0 \left(1 - \text{erf}\left(\frac{y}{2 \sqrt{\nu t}}\right) \right).\end{aligned} \hspace{\stretch{1}}(16.80)

The aim of this appears to be as an illustration that the boundary layer thickness $\delta$ grows with $\sqrt{\nu t}$.

FIXME: really need to plot 16.80.

## Oscillatory flow.

Another worked problem in the boundary layer topic was the Stokes boundary layer problem with a driving interface of the form

\begin{aligned}U(t) = U_0 e^{i \Omega t}\end{aligned} \hspace{\stretch{1}}(16.81)

with an assumed solution of the form

\begin{aligned}u(y, t) = f(y) e^{i \Omega t},\end{aligned} \hspace{\stretch{1}}(16.82)

we found

\begin{subequations}

\begin{aligned}u(y, t) =U_0 e^{-\lambda y} \cos\left( -i (\lambda y - \Omega t) \right).\end{aligned} \hspace{\stretch{1}}(16.83a)

\begin{aligned}\lambda = \sqrt{\frac{\Omega}{2 \nu}}\end{aligned} \hspace{\stretch{1}}(16.83b)

\end{subequations}

This was a bit more obvious as a boundary layer illustration since we see the exponential drop off with every distance multiple of $\sqrt{\frac{2 \nu}{\Omega}}$.

## Blassius problem (boundary layer thickness in flow over plate).

We examined the scaling off all the terms in the Navier-Stokes equations given a velocity scale $U$, vertical length scale $\delta$ and horizontal length scale $L$. This, and the application of Bernoulli’s theorem allowed us to make construct an approximation for Navier-Stokes in the boundary layer

\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}}(16.84a)

\begin{aligned}\frac{\partial {p}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(16.84b)

\begin{aligned}\frac{\partial {u}}{\partial {x}} + \frac{\partial {v}}{\partial {y}} = 0\end{aligned} \hspace{\stretch{1}}(16.84c)

\end{subequations}

With boundary conditions

\begin{aligned}U(x, 0) &= 0 \\ U(x, \infty) &= U(x) = U_0 \\ V(x, 0) &= 0\end{aligned} \hspace{\stretch{1}}(16.85)

With a similarity variable

\begin{aligned}\eta = \frac{y}{\sqrt{2 \frac{\nu x}{U}}}\end{aligned} \hspace{\stretch{1}}(16.88)

and stream functions

\begin{aligned}u &= \frac{\partial {\psi}}{\partial {y}} \\ v &= -\frac{\partial {\psi}}{\partial {x}}\end{aligned} \hspace{\stretch{1}}(16.89)

and

\begin{aligned}\psi = f(\eta) \sqrt{ 2 \nu x U_0 },\end{aligned} \hspace{\stretch{1}}(16.91)

we were able to show that our velocity dependence was given by the solutions of

\begin{aligned}f''' + f f'' = 0.\end{aligned} \hspace{\stretch{1}}(16.92)

This was done much more clearly in [4] and I worked this problem myself with a hybrid approach (non-dimensionalising as done in class).

FIXME: the end result of this is a plot (a nice one can be found in [5]). That plot ends up being one that’s done in terms of the similarity variable $\eta$. It’s not clear to me how this translates into an actual velocity profile. Should plot these out myself to get a feel for things.

# Singular perturbation theory.

The non-dimensional form of Navier-Stokes had the form

\begin{aligned}(\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u} = -\boldsymbol{\nabla} p + \frac{1}{{\text{Re}}} {\boldsymbol{\nabla}}^2 \mathbf{u}\end{aligned} \hspace{\stretch{1}}(17.93)

where the inverse of Reynold’s number

\begin{aligned}\text{Re} = \frac{U L}{\nu}\end{aligned} \hspace{\stretch{1}}(17.94)

can potentially get very small. That introduces an ill-conditioning into the problems that can make life more interesting.

We looked at a couple of simple LDE systems that had this sort of ill conditioning. One of them was

\begin{aligned}\epsilon \frac{du}{dx} + u = x,\end{aligned} \hspace{\stretch{1}}(17.95)

for which the exact solution was found to be

\begin{aligned}u = (1 + \epsilon) e^{-x/\epsilon} + x - \epsilon\end{aligned} \hspace{\stretch{1}}(17.96)

The rough idea is that we can look in the domain where $x \sim \epsilon$ and far from that. In this example, with $x$ far from the origin we have roughly

\begin{aligned}\epsilon \times 1 + u = x \approx 0 + u\end{aligned} \hspace{\stretch{1}}(17.97)

so we have an asymptotic solution close to $u = x$. Closer to the origin where $x \sim O(\epsilon)$ we can introduce a rescaling $x = \epsilon y$ to find

\begin{aligned}\epsilon \frac{1}{{\epsilon}} \frac{du}{dy} + u = \epsilon y.\end{aligned} \hspace{\stretch{1}}(17.98)

This gives us

\begin{aligned}\frac{du}{dy} + u \approx 0,\end{aligned} \hspace{\stretch{1}}(17.99)

for which we find

\begin{aligned}u \propto e^{-y} = e^{-x/\epsilon}.\end{aligned} \hspace{\stretch{1}}(17.100)

# Stability.

We characterized stability in terms of displacements writing

\begin{aligned}\delta x = e^{(\sigma_R + i \sigma_I) t}\end{aligned} \hspace{\stretch{1}}(18.101)

and defining

• Oscillatory unstability. $\sigma_{\text{R}} = 0, \sigma_{\text{I}} > 0$.
• Marginal unstability. $\sigma_{\text{I}} = 0, \sigma_{\text{R}} > 0$.
• Neutral stability. $\sigma_{\text{I}} = 0, \sigma_{\text{R}} = 0$.

## Thermal stability: Rayleigh-Benard problem.

We considered the Rayleigh-Benard problem, looking at thermal effects in a cavity. Assuming perturbations of the form

\begin{aligned}\mathbf{u} &= \mathbf{u}_\text{base} + \delta \mathbf{u} = 0 + \delta \mathbf{u} \\ p &= p_s + \delta p \\ \rho &= \rho_s + \delta \rho\end{aligned}

and introducing an equation for the base state

\begin{aligned}\boldsymbol{\nabla} p_s = -\rho_s \hat{\mathbf{z}} g\end{aligned} \hspace{\stretch{1}}(18.102)

we found

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \delta \mathbf{u} = -\frac{1}{{\rho_s}} \boldsymbol{\nabla} \delta p - \frac{\delta \rho}{\rho_s} \hat{\mathbf{z}} g\end{aligned} \hspace{\stretch{1}}(18.103)

Operating on this with ${\partial {}}/{\partial {z}} \boldsymbol{\nabla} \cdot ()$ we find

\begin{aligned}\boldsymbol{\nabla}^2 \frac{\partial {\delta p}}{\partial {z}} = -g \frac{\partial^2 {{\delta \rho}}}{\partial {{z}}^2},\end{aligned} \hspace{\stretch{1}}(18.104)

from which we apply back to 18.103 and take just the z component to find

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \delta w = -\frac{1}{{\rho_s}} \frac{\partial {\delta p}}{\partial {z}} - \frac{\delta \rho}{\rho_s} g\end{aligned} \hspace{\stretch{1}}(18.105)

With an assumption that density change and temperature are linearly related

\begin{aligned}\delta \rho = - \rho_s \alpha \delta T,\end{aligned} \hspace{\stretch{1}}(18.106)

and operating with the Laplacian we end up with a relation that follows from the momentum balance equation

\begin{aligned}\left( \frac{\partial {}}{\partial {t}} - \nu \boldsymbol{\nabla}^2 \right) \boldsymbol{\nabla}^2 \delta w=g \alpha \left(\frac{\partial^2 {{}}}{\partial {{x}}^2}+\frac{\partial^2 {{}}}{\partial {{y}}^2}\right)\delta T.\end{aligned} \hspace{\stretch{1}}(18.107)

We also applied our perturbation to the energy balance equation

\begin{aligned}\frac{\partial {T}}{\partial {t}} + (\mathbf{u} \cdot \boldsymbol{\nabla}) T = \kappa \boldsymbol{\nabla}^2 T\end{aligned} \hspace{\stretch{1}}(18.108)

We determined that the base state temperature obeyed

\begin{aligned}\kappa \frac{\partial^2 {{}}}{\partial {{z}}^2} T_s = 0,\end{aligned} \hspace{\stretch{1}}(18.109)

with solution

\begin{aligned}T_s = T_0 - \frac{\Delta T}{d} z.\end{aligned} \hspace{\stretch{1}}(18.110)

This and application of the perturbation gave us

\begin{aligned}\frac{\partial {\delta T}}{\partial {t}} + \delta \mathbf{u} \cdot \boldsymbol{\nabla} T_s = \kappa \boldsymbol{\nabla}^2 \delta T.\end{aligned} \hspace{\stretch{1}}(18.111)

We used this to non-dimensionalize with

\begin{aligned}\begin{array}{l l}x,y,z & \quad \mbox{withlatex d} \\ t & \quad \mbox{with $d^2/\nu$} \\ \delta w & \quad \mbox{with $\kappa/d$} \\ \delta T & \quad \mbox{with $\Delta T$}\end{array}\end{aligned} \hspace{\stretch{1}}(18.112)

And found (primes dropped)

\begin{subequations}

\begin{aligned}\boldsymbol{\nabla}^2 \left( \frac{\partial {}}{\partial {t}} - \boldsymbol{\nabla}^2 \right) \delta w = \mathcal{R} \left( \frac{\partial^2 {{}}}{\partial {{x}}^2} +\frac{\partial^2 {{}}}{\partial {{y}}^2} \right) \delta T\end{aligned} \hspace{\stretch{1}}(18.113a)

\begin{aligned}\left( \text{P}_r \frac{\partial {}}{\partial {t}} - \boldsymbol{\nabla}^2 \right) \delta T = \delta w,\end{aligned} \hspace{\stretch{1}}(18.113b)

\end{subequations}

where we’ve introduced the Rayleigh number and Prandtl number’s
\begin{subequations}

\begin{aligned}\mathcal{R} = \frac{g \alpha \Delta T d^3}{\nu \kappa},\end{aligned} \hspace{\stretch{1}}(18.114a)

\begin{aligned}\text{P}_r = \frac{\nu}{\kappa}\end{aligned} \hspace{\stretch{1}}(18.114b)

\end{subequations}

We were able to construct some approximate solutions for a problem similar to these equations using an assumed solution form

\begin{aligned}\delta w &= w(z) e^{ i ( k_1 x + k_2 y) + \sigma t} \\ \delta T &= \Theta(z) e^{ i ( k_1 x + k_2 y) + \sigma t}\end{aligned} \hspace{\stretch{1}}(18.115)

Using these we are able to show that our PDEs are similar to that of

\begin{aligned}{\left.{{w = D^2 w = D^4 w}}\right\vert}_{{z = 0, 1}} = 0,\end{aligned} \hspace{\stretch{1}}(18.117)

where $D = {\partial {}}/{\partial {z}}$. Using the trig solutions that fall out of this we were able to find the constraint

\begin{aligned}0 =\begin{vmatrix}\left( n^2 \pi^2 + k^2 \right)^2 + \sigma \left( n^2 \pi^2 + k^2\right) & -\mathcal{R} k^2 \\ -1 & n^2 \pi^2 + k^2 + \text{P}_r \sigma\end{vmatrix},\end{aligned} \hspace{\stretch{1}}(18.118)

which for $\sigma = 0$, this gives us the critical value for the Rayleigh number

\begin{aligned}\mathcal{R} = \frac{(k^2 + n^2 \pi^2)^3}{k^2},\end{aligned} \hspace{\stretch{1}}(18.119)

which is the boundary for thermal stability or instability.

The end result was a lot of manipulation for which we didn’t do any sort of applied problems. It looks like a theory that requires a lot of study to do anything useful with, so my expectation is that it won’t be covered in detail on the exam. Having some problems to know why we spent two days on it in class would have been nice.

# References

[1] L.D. Landau and E.M. Lifshitz. A Course in Theoretical Physics-Fluid Mechanics. Pergamon Press Ltd., 1987.

[2] S. Granger. Fluid Mechanics. Dover, New York, 1995.

[3] Wikipedia. Curvature — wikipedia, the free encyclopedia [online]. 2012. [Online; accessed 25-April-2012]. http://en.wikipedia.org/w/index.php?title=Curvature&oldid=488021394.

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

[5] 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.

## Spin down of coffee in a bottomless cup.

Posted by peeterjoot on April 25, 2012

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

# Motivation.

Here’s a variation of a problem outlined in section 2 of [1], which looked at the time evolution of fluid with initial rotational motion, after the (cylindrical) rotation driver stops, later describing this as the spin down of a cup of tea. I’ll work the problem in more detail than in the text, and also make two refinements.

• I drink coffee and not tea.
• I stir my coffee in the interior of the cup and not on the outer edge.

Because of the second point I’ll model my stir stick as a rotating cylinder in the cup and not by somebody spinning the cup itself to stir the tea. This only changes the solution for the steady state part of the problem.

# Guts

We’ll work in cylindrical coordinates following the conventions of figure (1).

Figure 1: Fluid flow in nested cylinders.

We’ll assume a solution that with velocity azimuthal in direction, and both pressure and velocity that are only radially dependent.

\begin{aligned}\mathbf{u} = u(r) \hat{\boldsymbol{\phi}}.\end{aligned} \hspace{\stretch{1}}(2.1)

\begin{aligned}p = p(r)\end{aligned} \hspace{\stretch{1}}(2.2)

Let’s first verify that this meets the non-compressible condition that eliminates the $\mu \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{u})$ term from Navier-Stokes

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{u}&=\left(\hat{\mathbf{r}} \partial_r + \frac{\hat{\boldsymbol{\phi}}}{r} \partial_\phi + \hat{\mathbf{z}} \partial_z\right) \cdot \left(u \hat{\boldsymbol{\phi}}\right) \\ &=\hat{\boldsymbol{\phi}} \cdot\left(\hat{\mathbf{r}} \partial_r u + \frac{\hat{\boldsymbol{\phi}}}{r} \partial_\phi u + \hat{\mathbf{z}} \partial_z u\right) +u\left(\hat{\mathbf{r}} \cdot \partial_r \hat{\boldsymbol{\phi}} + \frac{\hat{\boldsymbol{\phi}}}{r} \cdot \partial_\phi \hat{\boldsymbol{\phi}} + \hat{\mathbf{z}} \cdot \partial_z \hat{\boldsymbol{\phi}}\right) \\ &=\hat{\boldsymbol{\phi}} \cdot \hat{\mathbf{r}} \partial_r u +u\frac{\hat{\boldsymbol{\phi}}}{r} \cdot \left(-\hat{\mathbf{r}}\right) \\ &= 0.\end{aligned}

Good. Now let’s express each of the terms of Navier-Stokes in cylindrical form. Our time dependence is

\begin{aligned}\rho \partial_t u(r, t) \hat{\boldsymbol{\phi}}=\rho \hat{\boldsymbol{\phi}} \partial_t u.\end{aligned} \hspace{\stretch{1}}(2.3)

Our inertial term is

\begin{aligned}\begin{aligned}\rho (\mathbf{u} \cdot \boldsymbol{\nabla}) \mathbf{u}&=\frac{\rho u}{r} \partial_\phi (u \hat{\boldsymbol{\phi}}) \\ &=\frac{\rho u^2}{r} (-\hat{\mathbf{r}}).\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.4)

Our pressure term is

\begin{aligned}-\boldsymbol{\nabla} p=-\hat{\mathbf{r}} \partial_r p,\end{aligned} \hspace{\stretch{1}}(2.5)

and our Laplacian term is

\begin{aligned}\begin{aligned}\mu \boldsymbol{\nabla}^2 \mathbf{u}&=\mu \left( \frac{1}{{r}} \partial_r ( r \partial_r) + \frac{1}{{r^2}} \partial_{\phi\phi} + \partial_{z z}\right)u(r) \hat{\boldsymbol{\phi}} \\ &=\mu \left( \frac{\hat{\boldsymbol{\phi}}}{r} \partial_r ( r \partial_r u) + \frac{-\hat{\mathbf{r}} u}{r^2} \right).\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.6)

Putting things together, we find that Navier-Stokes takes the form

\begin{aligned}\rho \hat{\boldsymbol{\phi}} \partial_t u+\frac{\rho u^2}{r} (-\hat{\mathbf{r}})=-\hat{\mathbf{r}} \partial_r p+\mu \left( \frac{\hat{\boldsymbol{\phi}}}{r} \partial_r ( r \partial_r u) + \frac{-\hat{\boldsymbol{\phi}} u}{r^2} \right),\end{aligned} \hspace{\stretch{1}}(2.7)

which nicely splits into an separate equations for the $\hat{\boldsymbol{\phi}}$ and $\hat{\mathbf{r}}$ directions respectively

\begin{aligned}\frac{1}{{\nu}} \partial_t u=\frac{1}{r} \partial_r ( r \partial_r u) - \frac{u}{r^2}\end{aligned} \hspace{\stretch{1}}(2.8a)

\begin{aligned}\frac{\rho u^2}{r}=\partial_r p.\end{aligned} \hspace{\stretch{1}}(2.8b)

Before $t = 0$ we seek the steady state, the solution of

\begin{aligned}r \partial_r ( r \partial_r u) - u = 0.\end{aligned} \hspace{\stretch{1}}(2.9)

We’ve seen that

\begin{aligned}u(r) = A r + \frac{B}{r}\end{aligned} \hspace{\stretch{1}}(2.10)

is the general solution, and can now fit this to the boundary value constraints. For the interior portion of the cup we have

\begin{aligned}{\left.{{A r + \frac{B}{r}}}\right\vert}_{{r = 0}} = 0\end{aligned} \hspace{\stretch{1}}(2.11)

so $B = 0$ is required. For the interface of the “stir-stick” (moving fast enough that we can consider it having a cylindrical effect) at $r = R_1$ we have

\begin{aligned}A R_1 = \Omega R_1,\end{aligned} \hspace{\stretch{1}}(2.12)

so the interior portion of our steady state coffee velocity is just

\begin{aligned}\mathbf{u} = \Omega r \hat{\boldsymbol{\phi}}.\end{aligned} \hspace{\stretch{1}}(2.13)

Between the cup edge and the stir-stick we have to solve

\begin{aligned}A R_1 + \frac{B}{R_1} &= \Omega R_1 \\ A R_2 + \frac{B}{R_2} &= 0,\end{aligned} \hspace{\stretch{1}}(2.14)

or

\begin{aligned}A R_1^2 + B &= \Omega R_1^2 \\ A R_2^2 + B &= 0.\end{aligned} \hspace{\stretch{1}}(2.16)

Subtracting we find

\begin{aligned}A = -\frac{\Omega R_1^2}{R_2^2 - R_1^2}\end{aligned} \hspace{\stretch{1}}(2.18a)

\begin{aligned}B = \frac{\Omega R_1^2 R_2^2}{R_2^2 - R_1^2},\end{aligned} \hspace{\stretch{1}}(2.18b)

so our steady state coffee flow is

\begin{aligned}\mathbf{u} =\left\{\begin{array}{l l}\Omega r \hat{\boldsymbol{\phi}}& \quad \mbox{latex r \in [0, R_1]} \\ \frac{\Omega R_1^2}{R_2^2 – R_1^2} \left( \frac{R_2^2}{r} -r \right)\hat{\boldsymbol{\phi}}& \quad \mbox{$r \in [R_1, R_2]$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.19)

## Time evolution.

We can use a separation of variables technique with $u(r, t) = R(r) T(t)$ to find

\begin{aligned}\frac{1}{{\nu}} \frac{T'}{T} = \frac{1}{{R}} \left( \frac{1}{r} \partial_r ( r \partial_r R) - \frac{R}{r^2}\right)= -\lambda^2,\end{aligned} \hspace{\stretch{1}}(2.20)

which gives us

\begin{aligned}T \propto e^{-\lambda^2 \nu t},\end{aligned} \hspace{\stretch{1}}(2.21)

and $R$ specified by

\begin{aligned}0 = r^2 \frac{d^2 R}{dr^2} + r \frac{d R}{dr} + R \left( r^2 \lambda^2 - 1 \right).\end{aligned} \hspace{\stretch{1}}(2.22)

Checking [2] (9.1.1) we see that this can be put into the standard form of the Bessel equation if we eliminate the $\lambda$ term. We can do that writing $z = r \lambda$, $\mathcal{R}(z) = R(z/\lambda)$ and noting that $r d/dr = z d/dz$ and $r^2 d^2/dr^2 = z^2 d^2/dz^2$, which gives us

\begin{aligned}0 = z^2 \frac{d^2 \mathcal{R}}{dr^2} + z \frac{d \mathcal{R}}{dr} + \mathcal{R} \left( z^2 - 1 \right).\end{aligned} \hspace{\stretch{1}}(2.23)

The solutions are

\begin{aligned}\mathcal{R}(z) = J_{\pm 1}(z), Y_{\pm 1}(z).\end{aligned} \hspace{\stretch{1}}(2.24)

From (9.1.5) of the handbook we see that the plus and minus variations are linearly dependent since $J_{-1}(z) = -J_1(z)$ and $Y_{-1}(z) = -Y_1(z)$, and from (9.1.8) that $Y_1(z)$ is infinite at the origin, so our general solution has to be of the form

\begin{aligned}\mathbf{u}(r, t) = \hat{\boldsymbol{\phi}} \sum_\lambda c_\lambda e^{-\lambda^2 \nu t} J_{1}(r \lambda).\end{aligned} \hspace{\stretch{1}}(2.25)

In the text, I see that the transformation $\lambda \rightarrow \lambda/a$ (where $a$ was the radius of the cup) is made so that the Bessel function parameter was dimensionless. We can do that too but write

\begin{aligned}\mathbf{u}(r, t) = \hat{\boldsymbol{\phi}} \sum_\lambda c_\lambda e^{-\frac{\lambda^2}{R_2^2} \nu t} J_{1}\left(\lambda \frac{r}{R_2}\right).\end{aligned} \hspace{\stretch{1}}(2.26)

Our boundary value constraint is that we require this to match 2.19 at $t = 0$. Let’s write $R_2 = R$, $R_1 = a R$, $z = r/R$, so that we are working in the unit circle with $z \in [0, 1]$. Our boundary problem can now be expressed as

\begin{aligned}\frac{1}{{\Omega R}} \sum_\lambda c_\lambda J_{1}\left(\lambda z\right)=\left\{\begin{array}{l l}z & \quad \mbox{latex z \in [0, a]} \\ \frac{1}{\frac{R^2}{a^2} – 1} \left( \frac{1}{{z}} – z\right)& \quad \mbox{$z \in [a, 1]$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.27)

Let’s pull the $\Omega R$ factor into $c_\lambda$ and state the problem to be solved as

\begin{aligned}\mathbf{u}(r, t) = \Omega R \hat{\boldsymbol{\phi}} \sum_{i=1}^n c_i e^{-\frac{\lambda_i^2}{R^2} \nu t} J_{1}\left(\lambda_i \frac{r}{R}\right)\end{aligned} \hspace{\stretch{1}}(2.28a)

\begin{aligned}\sum_{i = 1}^n c_i J_{1}\left(\lambda_i z\right) = \phi(z)\end{aligned} \hspace{\stretch{1}}(2.28b)

\begin{aligned}\phi(z) = \left\{\begin{array}{l l}z & \quad \mbox{latex z \in [0, a]} \\ \frac{a^2}{1 – a^2} \left( \frac{1}{{z}} – z\right)& \quad \mbox{$z \in [a, 1]$} \\ \end{array}\right..\end{aligned} \hspace{\stretch{1}}(2.28c)

Looking at section 2.7 of [3] it appears the solutions for $c_i$ can be obtained from

\begin{aligned}c_i = \frac{\int_0^1 r\phi(z) J_1(\lambda_i z) dz}{\int_0^1 r J_1^2(\lambda_i z) dz},\end{aligned} \hspace{\stretch{1}}(2.29)

where $\lambda_i$ are the zeros of $J_1$.

To get a feel for these, a plot of the first few of these fitting functions is shown in figure (2).

Figure 2: First four zero crossing Bessel functions.

Using Mathematica in bottomlessCoffee.cdf, these coefficients were calculated for $a = 0.6$. The $n = 1, 3, 5$ approximations to the fitting function are plotted with a comparison to the steady state velocity profile in figure (3).

Figure 3: Bessel function fitting for the steady state velocity profile for n = 1, 3, 5.

As indicated in the text, the spin down is way too slow to match reality (this can be seen visually in the worksheet by animating it).

# References

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

[2] M. Abramowitz and I.A. Stegun. {\em Handbook of mathematical functions with formulas, graphs, and mathematical tables}, volume 55. Dover publications, 1964.

[3] H. Sagan. Boundary and eigenvalue problems in mathematical physics. Dover Pubns, 1989.