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Calculations were made using the following assumptions for simplification:
1) When elementary lime particles coagulate, the density of the larger mass stays the same as that of the original particles. This is unlike what happens in flocs, which have a Dfrac of 2.3. (I'm unsure of what you are saying. In the reprecipitation of lime, I'm not sure the fractal dimension stays 3. I would suspect it would behave similiarly to aluminum hydroxide flocs actually. Our assumption is that the solid floc particles are dissolving but not reprecipitating and the original lime is solid with a fractal dimension around 3)
2) Density of lime is 2.211 g/m^3: Particles are uniform.
3) Shape Factor of lime particles = 1: The lime particles are perfectly spherical.
4) Settling velocity = 10 m/day: Given a flow rate of 80 mL/min (as determined by experiment 1). This velocity corresponds to the finer lime particles. (What size does this correspond to?)
CALCULATIONS ANALYSIS
It was assumed that the smallest particle the tube could capture has the same terminal velocity as the capture velocity, and a longer tube can capture smaller sized particles (The relation is shown in figure 1). The tube length at 1.5m has a capture velocity of 10m/day,and the smallest particle it can capture is 0.00135mm. (This is reasonable, can you show how you obtained 1.5 m in the wiki? Can you show alternative lengths and what you would predict? See the plate settler spacing team model and Monroe's CEE 4540 for more information on tube settlers.)
Lime particles will have a larger density than the flocs, which means that their settling velocities will be higher than the assumed 10m/day. Also, it is not neccesary that ALL lime particles settle down - some amount (not determined yet) will have to fall out of the lime feeder to solve the acidity problem. Consequently, the length of the tube needed will be less than 1.5m.
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