Static pressure in a multi phase suspension

I have a headscratcher: Suppose you have a suspension, say sand dispersed in water. Looked at as a bulk, over the whole volume, the density would be higher than water. Do we expect a higher pressure near the botom of the vessel because of this?

Two possible scenarios com to mind: Sand (or another solid) is dispersed and sinking to the ground, sand is dispersed and a mixer addes energy and keeps the sand dispersed. When I try to solve this headscratcher, My thinking goes like this:

• With sinking particles, there's an upward force acting on the particles (friction according to Hookes/Kaskas equation)
• action, reaction, etc. - there's a force acting on the water
• hence there's a higher pressure at the bottom of a suspension, however ...
• in a non equlibrium case, the pressure does not correspond to the bulk density of water + solid, but is somewhat smaller
• In an equilibrium case (mixer), the pressure should be equivalent to pressure according to the bulk density

In most real life cases, a solid will either one or all of ...

• settle quickly
• be not much more dense than the fluid
• be of low percentage of the total volume

...so the effect will be really small

This is my line of thinking so far. If wrong, I'd like to be shown where. If correct, I want confirmation. Either way, I expect answers to be backed up by some authoritative source.

p.s. If my thinking is correct, you could make a nice experiment out of my bullet points 4 & 5.

A settling particle causes pressure disturbance due to its motion, and this is certainly felt everywhere in the fluid, and in particular at the walls of the container (including its bottom), which can be measured with sufficiently sensitive pressure sensors. For a single small sphere of radius $a$ settling at constant velocity $\mathbf{U}$, the disturbance in pressure (i.e. deviation from hydrostatic pressure of the undisturbed fluid) at location $\mathbf{x}$ from its center, is :$$p=\frac{3\mu}{2a} \frac{\mathbf{U}\cdot\mathbf{x}}{r^3},\quad r\equiv |\mathbf{x}|$$ in which $\mu$ is fluid viscosity. Although this result is for an unbounded fluid and for a single particle, you may apply it to the case of a dilute suspension in a finite container as an approximation; pressure disturbance at the wall is predominantly due to particles close to the wall (pressure disturbance decays rapidly as $\sim1/r^2$). Above result is for low Reynolds number flow which is good enough for particles in a suspension. The settling speed is given by $$\mathbf{U}=\frac{2\Delta \rho g a^2}{9\mu}\hat{\mathbf{U}}$$ in which $g$ is gravitational acceleration, $\Delta \rho$ is the density difference between particle of the dispersed phase and water, and $\hat{\mathbf{U}}$ is the unit vector along the direction of settling. Substituting above expression into the first one for $p$ we get: $$p=\frac{\Delta\rho g a}{3r^3}\hat{\mathbf{U}}\cdot\mathbf{x}$$ Above formula shows that, smaller the particle smaller is the pressure disturbance induced by its motion, which for practical purposes may be negligible.
• @mart Yes. $\mathbf{x}$ is the position vector of the point w.r.t. sphere centre where pressure is to be known. $r$ is just the radial distance to that point. – Deep May 25 '18 at 15:48