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Posts Tagged ‘coffee’

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)

Steady state solution

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.

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