test

When an incompressible fluid flows through a cylindrical tube its velocity relative to the walls changes as a function of the tube radius. In general, this velocity distribution is parabolic: the greatest velocities are achieved at the center of the tube (where R=0) eventually tapering off to 0 at the walls. The parabolic nature of the distribution arises from cylindrical symmetry as well as the fact that the fluid does not move at the walls (the "no-slip" condition).

This gradient in the velocity profile contibutes to the force that a floc experiencing roll-up feels. Flocs actually begin to roll up when the velocity at their edge exposed to the flow exceeds some critical value, which is highly dependent on the floc's diameter, its density, and the capture velocity of the system, among other things.

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Applying a force balance to this:

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$$
\sum

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$$

.

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$$
F_b + F_d - mg\sin (\theta ) = 0
$$

.

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\large
$$

mg\sin \theta + \rho w gV
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\sin \theta + \rho w C_D A
{{V^2 } \over 2}
$$

Solving this equation for the velocity:

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$$
{{V^2 } \over 2}\rho w C_D A

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= \rho p V

g\sin \theta - \rho w gV

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\sin \theta
$$

.

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$$
V^2 = {{2g\sin \theta V_

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} \over {\rho w C_D A

}}(\rho _p - \rho _w )
$$

For the purposes of this approximation, it is assumed the floc's diameter can be modeled as a sphere.

.

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$$
{{V_

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} \over {A_

}} = {{\pi d_p ^3 } \over 6 \over {{

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\over 4}}} =

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d_p
$$

Assuming that the flow through the tube settlers is creeping flow (Re <<1), we can approximate the drag coefficient as 24/Re:

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$$
V = \sqrt {{4 \over {72}}g\sin (\theta )d_p {\mathop

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\nolimits} {{(\rho _p - \rho _w )} \over

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}}
$$

At this velocity a floc that is sitting on the bottom wall of the tube settler should not move. We should call this something like the critical or failure velocity, since velocity values greater than this quantity (at the edge of the floc exposed to the flow) will cause roll-up.

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velocity gradient analysis

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After integration
$$

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{\partial \over {\partial r}}(r{{\partial v_z } \over {\partial r}}) =

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(\partial P} \over {\partial z)r
$$

.
After integration

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$$
r{{\partial v_z } \over {\partial r}} = {1 \over {2\mu }}(\partial P} \over {\partial z)r^2 + c_1
$$

.

This constant of integration should be zero, since we want the velocity of the center of the tube to be finite. Dividing through by r---

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$$
\partial v_z } \over {\partial r = {1 \over {2\mu }}(\partial P} \over {\partial z)r
$$

.

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$$
\partial P} \over {\partial z = {{\Delta P} \over L}
$$

.

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$$
V_

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= R^2 \Delta P} \over {8\mu L = {Q \over {\pi R^2 }}
$$

.

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$$
{{\Delta P} \over L} = {{Q8\mu } \over {\pi R^4 }}
$$

.

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\large
$$
\partial v_z } \over {\partial r = {1 \over {2\mu }}{{Q8\mu } \over {\pi R^4 }}r
$$

.

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\large
$$
\partial v_z } \over {\partial r = {{4Q} \over {\pi R^4 }}r
$$

.

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