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## Steady state velocity profile of stirred cup of non-bottomless coffee (continued).

Posted by peeterjoot on April 30, 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.)]

# Solving for the flow below the stirring point.

Previously, we found the functional form for an azimuthal flow that has $z$-axis dependence. Attempting apply boundary conditions that including a cylindrical mixing device got us into trouble, since our no-slip condition would simultaneously require zero and non-zero velocity at the point of contact of the mixing interface and the bottom of the vessel (i.e. where the stir stick contacts the bottom of the cup). We can avoid this issue by constraining the mixing to only occur above the bottom of the cup, and then look at the flow that this induces below the mixing point.

Let’s attempt to solve the boundary value problem. We’ll position the mixing cylinder at a height $d$ above the bottom of the cup, and set the radius of that inner cylinder to $s$ as before, with the mixing occurring at an angular velocity of $\Omega$. We now want apply these boundary value constrains to the velocity function we’ve found for the steady state problem. Provided $z < d$, this will have the form

\begin{aligned}u(r, z, 0) = s \Omega \sum C_i J_1(\lambda_i r/R) \sinh(\lambda_i z/R)\end{aligned} \hspace{\stretch{1}}(1.1)

where $\lambda_i$ are the zeros of the order one Bessel function $J_1$. Observe that our boundary value conditions of $u(R, z, 0) = 0$ and $u(r, 0, 0) = 0$ are automatically satisfied. Note that because of the $u(R, z, 0) = 0$ equality here, this form of solution is no good if we are mixing at the edge of the cup, so we require $s \ne R$.

Our remaining boundary value condition $u(s, d, 0) = s \Omega$, means that we have to solve

\begin{aligned}1 = \sum C_i J_1(\lambda_i s/R) \sinh(\lambda_i d/R)\end{aligned} \hspace{\stretch{1}}(1.2)

Having had the same sort of problem in our steady state bottomless coffee problem, we know how to solve this

\begin{aligned}C_i \sinh(\lambda_i d/R) = \frac{\int_0^1 w J_1 (\lambda_i w) dw}{\int_0^1 w J_1^2 (\lambda_i w) dw}.\end{aligned} \hspace{\stretch{1}}(1.3)

So we find that below the point of stirring ($z = d$), our steady state solution for the velocity should be

\begin{aligned}\mathbf{u}(r, z, 0) = s \Omega \hat{\boldsymbol{\phi}} \sum_{i=1}^\infty\frac{\int_0^1 w J_1 (\lambda_i w) dw}{\int_0^1 w J_1^2 (\lambda_i w) dw}J_1(\lambda_i r/R) \frac{\sinh(\lambda_i z/R)}{\sinh(\lambda_i d/R) }\end{aligned} \hspace{\stretch{1}}(1.4)

For a cup size of $R = 5 \text{cm}$, stir radius of $s = 3 \text{cm}$, stir depth $d = 2 \text{cm}$, and angular velocity $\Omega = 2 \pi \text{rad}/\text{s}$, we find

\begin{aligned}\begin{aligned}u(r, z, 0) &=27.1119 J_{1}(7.66341 r) \sinh (7.66341 z)-3.42819 J_{1}(14.0312 r) \sinh (14.0312 z) \\ &+4.97807 J_{1}(20.3469 r) \sinh (20.3469 z)-1.53542 J_{1}(26.6474 r) \sinh (26.6474 z) + \cdots\end{aligned}\end{aligned} \hspace{\stretch{1}}(1.5)

We can plot this velocity field as in figure (1), but it’s hard to see the radial dependence.

Figure 1: Vector field plot of the velocity field below the stir depth.

The radial dependence of the magnitude of the velocity can be seen in figure (2), which plots $u(r, d, 0)$. We see the zero velocity at the edges of the cup as expected, and once we hit the stir height, matching of stir velocity. An animation showing the variation of the radial velocity profile at various depths up to the stir height is available at http://youtu.be/BS8XQdXljSk, and also below.

Figure 2: Radial velocity dependence at the stir height.

Observing this animation we see that the velocity is dominated by the first term in the Bessel series, essentially just scaled by the hyperbolic sine that multiplies it.