Well,

You can detect the answer from the discussion about what happen in the Primary Sedimentation Tanks

Design factors:

Factor 1: The primary design is the surface loading rate (Q \ A) equal to the velocity (VP = T.H) where T time of H (H) precipitation height. The basin is designed in the case of simple sedimentation in order to take into account the incident of the formation of the blade.

It is difficult to ensure regular distribution of runoff on the entire basin due to insufficient dispersion of energy at the input area and the effect of sludge withdrawal with the outflow at the exit shaft. Therefore, the surface loading rate is calculated with a safety factor of 2.5-1.7.

The second factor, which should be observed in its design, is the time of retention = the actual settling time

It depends on the dimensions of the basin (basin design) and the time of computational deposition.

Scientific references indicate that:

 100% of the suspended materials in the sedimentable wastewater are deposited during the settling time of two hours.

 95% of them deposited during the settling time of one hour and a half.

 90% of which is deposited during the settling time of one hour.

 83% of them were deposited during a settling time of half an hour.

Note that the real precipitation time (actual) is smaller than the arithmetic time and the ratio between the two times relates to the degree of hydraulic effect of the sedimentation basin. The degree of hydraulic effect is related to several factors,

The shape of the entry area, the exit of the water from the basin, the shape of the basin, the specific weight, the temperature of the incoming sewage, the air temperature and the presence of wind at the station site, the water content of the salts and the various degrees of pollution.

The third factor: The design of the primary sedimentation basin is the diameter of the basin, the depth of the water in the basin and the height of the original basin where the depth of water in the basin ranges from 3-5 meters

Factor IV: the slope of the pelvic floor towards the center.

Factor V: The rotation speed of sludge scum ranges from (rpm) to 0.05-0.02

Sincerely

I would say that it is essential to know the density of the particles, their granulometry and the viscocity of the medium in which they are found.

Hello Guilnaz,

the question seems to be not quite clear. It could be understood more in direction of facility processes (e.g. answer Bayan Hussien ) or just dealing with the determination of partcle sizes in a sediment suspension. If so, you possibly could apply the simple Stokes equation. It describes the time of particle sedimentation as a function of particle size (in combination with additional controlling parameters). By the equation you can assess particle diameters. The equation describes the sedimentation in a column of solution of a given height (e.g. water or watery salt solution), and you can rearrange it to find out sedimentation time t (in sec) or R (partcle diameter), and by this you also can do a velocity assessment. Using the equation you will find the time after that you will find only smaller diameters still floating in suspension of a suspension column, i.e. when a diameter greater then X has passed the column and settled at the bottom. (of course, the smaller diameters from lower column heights are also in the bottom sediment). The general equation version is

t = h / (2/9 * ((𝞎s - 𝞎l) / μ) * g * R^2)

with: μ – dynamic viscosity of liquid (kg / (m s)); h – hight of column (in m); R – diameter (m); g – gravity (m / sec^2 (9,81 m/sec^2)); 𝞎l - density of liquid (g / m^3); 𝞎s - density of sample (g / m^3). Checking the parameters you will find many uncertainties. In a first place and striktly speaking the equation is valid for sediment particles with spherical shape only, and for diameters < 20 µm. Concerning μ and 𝞎l: the viscosity and density of a liquid depends on temperature. For some liquids you will find tabulated values in the internet. As an example you will find µ of water at 15' C = 1,138 * 10^-3; at 25' C = 0,89 * 10^-3; so this is quite a difference.Dealing with silicate or carbonate sediments you can assess a rather well defined 𝞎s (about 2.7 – 2.9) of particles; but what about unknown particles like e.g. amorphous hydroxides. This might be in your question a real challenge. Also sedimentologists using the equation have to consider these problems, e.g. in clay mineralogy studies, when particles are not at all spherical, and when the sediment is a mixture containing also higher diameters. Taking in account these constraints and acknowledging that, strictly spoken, the results are operational sedimentologists use the equation quite effectively.

You will find different formulations of the Stokes equation with, however, the same results (hopefully; be aware oft he dimensions). Generally, the particle size determination relies on sedimentation experiments with the studied sediment or suspension. It is done in special cylinders (e.g. Atterberg cylinders; cf. Internet) and carrying out is an additional technique. Finally, you have just to retrieve the supernatatnt suspension and after filtration or centrifugation you can handle a defined grain size. As an example: separation of a grain size < 2 µm takes about 16 to 20 hours in a column of 20 cm height. You can speed up sedimentation by using a centrifuge (increasing g; I would not recommend). Some more summarized instruction on particle separation by sedimentaion you will find e.g. in “D.M. Moore, R.C. Reynolds (1997) X-Ray Diffraction and the Identification and Analysis of Clay Minerals; 2ndedition; 377 pp., ISBN-13 978-0-19-508713-0.”

Good luck

D. Zachmann

I agree with Dr. Vidal,,,for your comment" it is essential to know the density of the particles, their granularity and the viscosity of the medium in which they are found".

Thank you Omar, but I am not Dr. only Magister and you have to say Mg. Rojas.

We use the last name of father and mother, Vidal is my mothers last name.

Interesting....

Just for clarification....

Do you need to calculate it?

Why don't you try to take some experimental data on sedimentation in a graduated cylinder?

Iñigo Legórburu that is also my plan. We have thought about sampling and testing in different time span. But I need to be able to calculate knowing this data as well. right?

Dieter Wolfram Zachmann I know the stockes law, but the problem is that considering stockes law gives the size of particles which managed to sedminet during the specific time. My question is how I should know what is the particle size of the remaining particles which did not manage to sediment yet?

The only data I have is the remaining content of the each component as microgram/liter.

Plus, ofcourse I have the desity of the particles and the density of the liquid they are sedimenting in. But the only real test data I have is the concentration before and after specific time in the sample.

Following answers...

Dear Ms. Mirmoshtaghi,

I would suggest you to look for bibliographic data on:

-Flocculation tests of urban wastewater

-Silty fraction analysis in soils

-Mining flotation technology

All of them use terminal velocity in their design and involve a wide span of particle size and density.

Then, maybe, you can have a first approximation of the terminal velocity of your particles.

Hello Guilnaz,

I am not too sure to understand the problem definitely. However, when you insert a time and the hight in the Stokes equation you will find the particle size still in suspension (size "A"; your advantage: obviously you know density of the particles and the density of the liquid ). If you are interested to find a particle size fraction between "A" and a smaller one ("B") you will just insert the diameter of "B" (and the column height, of course) into the Stokes equation and get the time when only "B" is in suspension. So, when retrieving the sediment you will find a sediment with the size between "A" and "B", and < "B" in suspension.

However, somehow I fear not to have touched the crux of the matter?

Kind egards

D. Zachmann

Flocculation tests of urban wastewater

-Silty fraction analysis in soils

-Mining flotation technology

Iñigo Legórburu is there any specific article you are referring to? In that case could you please send me them?

Thanks

Valuable discussion....

Dear Ms. Mirmoshtaghi,

Sorry for the delay.

The reference for soil particles size analysis by sedimentation is the norm ISO11277. And you can found useful information in the Manuals that US-Department of Agriculture, FAO and UN-Long Range Transboundary Air Pollution have in the Net.

There is, also, a good introduction in Pansu and Gautheyrou's Handbook of Soil Analysis.

Hope this helps,

Flocculation tests of urban wastewater

Let the particle reach its terminal velocity, where the gravitational force works downwards and buoyancy and viscous force work to dim the particles falling. Then from above:

Drag force=6πμUaParticle weight=43πa3gρsParticle buoyance force=43πa3gρw

At terminal velocity there is a balance:

U=43πa3g6πμa (ρs−ρw).

Water with a temperature of 20 °C and a salinity of 34.1 psu has a density ρw = 1024.1 kg m−3. Substituting these values into the expression for U, yields:

U=3×10−4ms−1.

To ensure consistency of the solution with our assumption of slow viscous flow we need to check the value of the Reynolds number:

Re=Uav=0.01.

In such problems, it is always advisable to check the Reynolds number to ensure that this is really small in all the theoretical development mentioned above