4 9/25

1. If you put a cup of coffee at the center of a rotating turn table and wait, eventually, the coffee will be executing a “solid body rotation” in which the velocity field is, in cylindrical coordinates:
   
   $$\vec v = {\widehat \imath}_\theta \Omega r$$
   
   
   where $\Omega$ is the angular velocity of the turn table. Draw samples of the streamlines of this flow. Find the vorticity and the strain rate tensor for this flow, using the expressions in appendices B and C. Show that indeed the coffee moves as a solid body; i.e. the fluid particles do not deform, and that for a solid body motion like this, indeed the vorticity is twice the angular velocity.

2. An “ideal vortex flow” is described in cylindrical coordinates by
   
   $$\vec v = \frac{C}{2\pi r} {\widehat \imath}_\theta$$
   
   
   where $C$ is some constant. Draw samples of the streamlines of this flow. In cylindrical coordinates, nabla is given by
   
   $$\nabla = {\widehat \imath}_r \frac{\partial}{\partial r} +... ...ial\theta} + {\widehat \imath}_z \frac{\partial}{\partial z}$$
   
   
   Evaluate
   
   $$\left\vert \begin{array}{ccc} {\widehat \imath}_r & {\wid... ...partial z} \\ v_r & v_\theta & v_z \end{array} \right\vert$$
   
   
   for this flow. Compare the answer with the vorticity $\omega=\nabla\times\vec v$, for which you can find the correct expressions in appendix B. Explain why the determinant does not give the correct result for the vorticity. In particular note that in
   
   $$\nabla\times\vec{v} = \left( {\widehat \imath}_r\frac{\pa... ...hat \imath}_\theta v_\theta + {\widehat \imath}_z v_z \right)$$
   
   
   the unit vectors ${\widehat \imath}_r$ and ${\widehat \imath}_\theta$ are not constants but depend on $\theta$. ( ${\rm d}{\widehat \imath}_r/{\rm d}\theta={\widehat \imath}_\theta$ and ${\rm d}{\widehat \imath}_\theta/{\rm d}\theta=-{\widehat \imath}_r$.)

3. The “circulation“ $\Gamma$ along a closed contour is defined as
   
   $$\Gamma \equiv \oint \vec v \cdot {\rm d}\vec r$$
   
   
   For both the solid body rotation of question 1, and the vortex flow of question 2, find the circulation along the unit circle in the $x,y$-plane. Next, only for the vortex flow, find the circulation along the closed curve consisting of the following segments:
   1. The part of the curve $y=\cosh(x)$, $z=0$ from $x=2$ to $x=-3$;
   2. vertically downward to the curve $y=-\cosh(\cosh(x))$, $z=0$ at $x=-3$;
   3. following the curve $y=-\cosh(\cosh(x))$, $z=0$ from $x=-3$ to the point $x=0$, hence $y=-\cosh(1)$, $z=0$;
   4. in a straight line along the $z$ direction to the point $x=0$, $y=-\cosh(1)$, $z=3.5$;
   5. in a straight line from $x=0$, $y=-\cosh(1)$, $z=3.5$ to the starting point $x=2$, $y=\cosh(2)$, $z=0$.
   Note that in cylindrical coordinates
   
   $${\rm d}\vec r = {\widehat \imath}_r {\rm d}r + {\widehat \i... ... {\widehat \imath}_\theta v_\theta + {\widehat \imath}_z v_z$$
   
   

4. According to the Stokes theorem of Calculus III, you should have
   
   $$\oint \vec v \cdot{\rm d}\vec r = \int\nabla\times\vec v \cdot\vec n{ \rm d}S$$
   
   
   where the second integral is over the inside of the contour. So instead of integrating the circulation $\Gamma$ as you did in question 3, you could have integrated the component of vorticity normal to the circle over the inside of the circle. Show that if you do that integral using the vorticity that you found for solid body rotation in question 1, you do indeed get the same answer as you got in question 3. Fine. But now show that if you do the integral of the vorticity over the inside of the circle for the vortex flow of question 2, you do not get the same answer for the circulation as in question 3. Explain which value is correct. And why the other value is wrong.

5. Write down the worked-out mathematical expressions for the integrals requested in question 5.1. Explain their physical meaning. Don't worry about actually doing the integrations.

   Take the surfaces $S_{I}$, $S_{II}$, $S_{III}$, and $S_{IV}$ to be one unit length in the $z$-direction. (To figure out the correct direction of the normal vector $\vec n$ at a given surface point, note that the control volume in this case is the right half of the region in between two cylinders of radii $r_0$ and $R_0$ and of unit length in the $z$-direction. The vector $\vec n$ is a unit normal vector sticking out of this control volume.)

6. 5.14

7. 5.11

8. 5.12. This question explains why the water stream coming out of a faucet contracts in area shortly below the faucet exit. As always, both mass and momentum conservation are needed.

   The faucet exit velocity is assumed to be of the form of Poisseuille flow:
   

   $$v_z = v_{\rm max} \left(1 - \frac{r^2}{R^2}\right)$$
   
   
   You can assume that the stress tensor at the faucet exit is of the form
   
   $$\bar{\bar\tau} = \left( \begin{array}{ccc} 0 & 0 & \tau_0 r/R 0 & 0 & 0 \tau_0 r/R & 0 & 0 \end{array} \right)$$
   
   
   in other words, much like the strain rate tensor that you derived earlier for Poisseuille flow.

   Take the faucet exit as the entrance of your control volume. Take as exit to your control volume a slighly lower plane at which the radius of the jet has stabilized to $R_2$ and the flow velocity has become uniform (independent of r). For a uniform flow velocity there are no viscous stresses. Gravity and pressure forces can be ignored compared to the high viscous forces in this very viscous fluid.