Air / Sand slurry question

I'm not trying to be picky, but why do you mix English and metric measures?

I have worked with fluidized beds, and the best way we have found to determine how much air is necessary to reach the fluidity desired is to experiment, even though there is a scientific method to figure it out.

Air bubble size should be assumed to be somewhere between one particle size and 10 particle sizes. The volume of sand and water is known, and the agitated depth is known to a certain degree, which more or less tells you how much air pressure you need. How much air you need is another question. What's the definition of "suspended?" How did you come up with "diffusers set every 9" apart?"

My suggestion: Get (rent) a big compressor (one capable of delivering more air than you think you'll need) and run some tests. Measure the amount of air it takes to keep the sand "suspended," and then tone down the size of the compressor to that which is needed.

I suggest you use water that is recycled and use a slurry pump that tolerates the grit. This will be a fluidized bed of sand in water. Once it is fluidized an agitator will work to prevent size segregation. The water that goes over the top needs to be collected and pumped into the bottom again. trying to fluidize a mixture of water and sand with air might be a little trucky

Air bubble size should be assumed to be somewhere between one particle size and 10 particle sizes.

Air bubble size is relative to orifice size and psi, small orifice high pressure = high suspension ratio

I agree with this approach, although I would suggest the initial poster get a copy of Octave Levenspeil's text "Fluidization Engineering".

Standard agitators should work fine in this application and will be much more efficient than an air based system in such a shallow tank. What you will need to do is coat the blades in an erosion resistant coating though.

Any agitator manufacturer with any experience in mining should be able to do this with no problem. You will also need to add baffles into the tank, and these should also coated, as should the bottom of the tank under the impeller.

Guys, thanks for your comments thus far...sorry for the mix of Metric and Imperial values, I grew up on both systems and it remains a habit.

I have an agitator in each of the three tanks now. The agitator is a standard design, featuring a 42" impeller with three 18" baffles coming off the walls of the tank. Added to the design I had the shaft extended below the main impeller (which is mounted 2.5' off the tank floor). Off this extension is a mechanically sealed housing which protects a torque converter, wherein the shaft continues through the torque converter and through the bottom of the housing, stopping just a few inches off the tank bottom. On the bottom of this shaft extension is another, relatively flat style paddle.

The reason for the dual paddle system is because the tank must be filled with both the liquid and solid, mix for 40 minutes, then drain out the bottom where it is separated. However, as it's draining, the upper paddle looses contact with the slurry and the solids drop out instantly! The lower paddle was intended to keep the slurry "reasonably fluid" so it could still be pumped out. The design works just fine, except when we simulate a power failure. Under a failure, the solids drop out (as expected), settling over the bottom paddle, but still just under the upper, more aggressive paddle. When power is restored, it was expected that the upper paddle would cause a violent downward mixing, thereby eroding the solids which settled, and finally freeing the lower paddle so it would turn, again, keeping the slurry "fluidized" during pump-out.

Every test we have run, (increasing HP, increasing/decreasing the torque value of the lower agitator, smaller baffles, etc.) we are unable to restore a full slurry where the sand was liberated from the tank bottom.

My next thought to solve this problem is to remove the agitation system and add an air compressor. I think even after a failure, the air would migrate up through the contents and cause the remixing, and I could effectively pump it out.

The calc's I did for air compression is this: 4250lbs of water mixed with 3000lbs of sand (with the sand having a displacement value of 2246lbs against the water). 3000lbs of sand, less the displacement weight equal to water (2246lbs) means I would only need to add another 750lbs of air to balance the sand to zero weight, where it should freely move around. 750lbs of air calculates out to 100 cubic feet per min @ 1PSI.

Am I on crack? I have looked at these numbers to the point where I don't believe myself anymore.