Water and Wastewater TreatmentSedimentation
SEDIMENTATION (Including worked examples)
Sedimentation concerns the removal by gravitational settling of suspended particles that are denser than water. When chemicals are added to induce or hasten aggregation of the fine and colloidal material the process of coagulation precedes sedimentation. In the case when chemicals are added to throw dissolved impurities out of solution the operation is called chemical precipitation.
The aim of the sedimentation process is to separate solids from water. A number of terms are used including sedimentation and clarification. When the particles have a density comparable with water they can also be separated by flotation. In particular dissolved air flotation is used in the clarification of both domestic and industrial water supplies.
It is useful to classify sedimentation processes in terms of the characteristics and concentration of the solid material. Typically the concentration of solids in water treatment is small so that the suspensions are dilute. The solids concentration is high in wastewater.
When the concentration of particles is not high enough for the settling of individual particles to influence each other the suspension is said to be dilute. This means that the settling velocity of the suspension is the same as that of a single particle. In contrast there is interference of the flow patterns around individual particles and the downwards flux of particles is enough to displace significant amounts of water which moves upwards. The overall result is that there will be a decrease in the settling velocity of the suspension in comparison with a single particle.
Settling of Discrete Particles (Type I Settling)
This is the easiest form of settling to analyse. The forces on a single particle are:
the gravitational force due to its mass,fg = ρpgVp,;
the buoyancy force due to the volume of fluid displaced,fb = -ρwgVp,;
and the drag force due to the hydrodynamic stress, fd = (cDApρwu2)/2.
Here ρp, ρw, g, Vp, u, Apand cDare the density of solid and water, the gravitational acceleration, the particle volume, the settling velocity, the cross-sectional area of the particle and the drag coefficient respectively. A particle at rest is only affected by fgand fb in opposition, as it starts to move the drag force, fd,becomes operational and grows with u2 as the particle accelerates. In a short distance the drag force balances the downward force on the particle which continues to settle at a constant speed, vt. The force balance is
(1)
so that
(2)
for a spherical particle, Vp = (4/3)π(d/2)3, and Ap = π(d/2)2, and Equation (2) is written,
(3
so that on rearranging,
(4)
There are different values of the drag coefficient depending on the flow region,
e.g.(5)
(6)
(7)
where Re = ρwvtd/μ is the particle Reynolds number. When particles are not spherical a modified Reynolds number is sometimes used,
(8)
where φ is the shape factor required to correct for the lack of sphericity. For a perfectly spherical particle φ = 1. Using the drag coefficient for laminar flow Equation (4) is written, (9)
This is called the Stokes Equation. Note it is necessary to solve Equations (4) and (6) simultaneously if the flow is either transitional or turbulent.
Example
Find the settling velocity of a sphere with a diameter of 0.5 mm and a density of 2650 kg m-3 settling in water at 20 0C. (Data ρw ==998.2 kg m-3, μ = 1.002 mPa s).
Firstly estimate
Now check the Reynolds number,
The Reynolds number indicates that the flow is transitional. It is necessary to recalculate the drag coefficient using Equation (6).
Equation (4) can now be used to estimate vt
Using vt = 0.11 m-1 s, recalculate Re, cD and vt. Repeat this procedure until the difference between successive values of vt is sufficiently small.i.e. Re = 35, cD = 1.18, vt = 0.10 m s-1 (c.f. previous value of 0.11 m s-1). It is seldom possible to use the above analysis in the design of settling tanks for both water and wastewater treatment because the size of the particles must be known and the correction factor for sphericity must be determined.
Settling Column Analysis for Type 1 Suspensions
The setting velocity of a dilute suspension can be determined indirectly from the analysis of the mass settlement rate in a settling column. The experiment is simple,Put the suspension in a settling column, ensuring that it is well mixed so that the concentration is uniform throughout.
(ii)Allow the suspension to settle under quiescent conditions and observe the concentration of settling particles at the bottom of the settling column, as a function of time.
Consider what is happening in the column. A particle initially at the surface of the water sinking to the bottom of the column travels a distance Z0 in a time t0 . The average speed of this particle is v0 = Z0/t0. Another particle settling from a height, Zp, reaching the bottom of the column at the same travel time has a settling velocity, vp = Zp/t0. The ratio of the settling velocities vp/v0 is in the same proportion as their respective starting heights,
This is also the fraction of particles of size dp that arrive at the sampling point at the bottom of the column at time t0.
If do is the diameter of the particle that settles through Z0 in time t0, then the settling velocity, vp, of all particles that have a diameter greater than or equal to do will be greater than v0. These particles will arrive at the sample point at the bottom of the column at or before t0. There will be other particles that arrive at the sample point at the bottom of the column at or before t0 that have a diameter less than d0, (and consequently a velocity less than v0), but these particles started settling at some height less than Z0.
If all the particles are randomly distributed throughout the length of the column then the fraction of particles of size dp, with a settling velocity vp, that arrive at the sample point in time t0 will be Zp/Z0 or alternatively vp/v0.
Experimental procedure to determine the removal efficiency of particles that do not flocculate (Type 1 settling)
The procedure used to determine the residence time for from setting column analysis of a Type 1 suspension of non-flocculating particles is now summarised. If C0 is the concentration of the well mixed suspension (note that in practice a settling column is often up to 2 m high) and Ci is the concentration of a sample taken near the bottom of the settling column at time ti, then the mass fraction of particles with settling velocities less than vi = Z0/ti , will be xi = Ci/C0. This is the mass fraction of particles remaining in suspension and 1 xiwill be the mass fraction of settled particles. If this process is repeated several times the mass fraction xi can be plotted as a function of the settling velocity vi. The results of such an experiment are shown in Figure 1.
For a given residence time, t0, the overall percentage removal of particles can be found. In the time, t0, all particles with a settling velocity vtv0 = Z0/t0 will be removed.
From the graph shown in Figure 1 the mass fraction of particles corresponding to a settling speed v0 is x0. This will be the fraction of particles that remains in suspension since particles with speeds lower than v0 cannot settle through the length of the column.Therefore the fraction of particles removed is (1 - x0).
However some particles whose settling velocity is less than v0 will also be removed because they start closer to the bottom of the column and also arrive at the bottom at or before t0.
These remaining particles are removed in the ratio vt/v0 corresponding to the shaded area in Figure 1.
If the relationship between the settling velocity v and the mass fraction x is known then the mass fraction removed by sedimentation is the fraction of fast settling particles plus the fraction of particles that do not need to travel the entire length of the column, that is,
Figure 1Mass fraction of settling particles as a function of their settling velocity.
Figure 2The settling velocity of particles as a function of the mass fraction of unsettled particles.
An example of settling column analysis for a Type 1 suspension.
The following results were obtained from a experiment in a 1.8 m settling column.
Time /(min)06080100130200240420
Concentration / (mg l1)3001891801681561117827
What is the theoretical removal efficiency in a continuous settling basin having a loading rate of 24.7 m day1?
(This is a volumetric flow of 24.7 m3 day-1 per m2 of settling basin area).
Solution
From the results at time t = 0 the initial concentration C0= 300 mg l1.
The mass fraction remaining in suspension at time t is calculated from the ratio of the concentration at time t to C0, i.e.,
The mass fraction remaining in suspension after 60 minutes is 189/300 = 0.63
The mass fraction remaining in suspension after 80 minutes is 180/300 = 0.60
The corresponding setting velocity is calculated from the overall height of the column (1.8 m) divided by the time that the sample was taken, i.e
The settling velocity at 60 minutes is 1.8/60 = 0.030 m min1.
The settling velocity at 80 minutes is 1.8/80 = 0.023 m min1.
The results of the calculations are summarised below:
Time /(min)60 80100130200240420
Mass fraction, xi0.630.600.560.520.370.260.09
Settling velocity vi, /(m min1)0.030.02250.0180.01390.0090.0075 0.00429
The mass fraction, xi, is now plotted as a function of the settling velocity (see Figure (3)).
The value of v0 corresponding to a loading rate of 25 m per day is
The corresponding value of x0 is found either from the graph (see Figure 3) or by using linear interpolation. The residence time is 105 minutes therefore by linear interpolation the concentration at 105 minutes is
mg l1 and the mass fraction x(105) = 166/300 = 0.553
i.e. the mass fraction of solids that remain in suspension is x0 = 0.553. This means that the mass fraction of solids with settling velocities higher than v0 that has settled is (1 - x0) = 0.447.
The next step is to determine to establish the fraction of particles with a settling velocity less than v0 that will have settled in the time t0. The remaining integral has to be evaluated graphically using,
where n is the number of values of xi in the appropriate range, that is all xi below 0.54.
Figure 3Mass fraction as a function of settling velocity
In tabular form the integration proceeds as follows
Time
/ xi / vi / Δxi / Average vi / vi Δxi105 / 0.553 / 0.01741
0.033 / 0.01549 / 0.000511
130 / 0.520 / 0.01385
0.150 / 0.01142 / 0.001713
200 / 0.370 / 0.00900
0.110 / 0.00825 / 0.0009075
240 / 0.260 / 0.00750
0.170 / 0.00589 / 0.001002
420 / 0.090 / 0.00429
0.090 / 0.00214 / 0.0001926
0.000 / 0.00000
To 4 significant figures, the sum of xiv = / 0.004326
The required integral is
Evidently, 25% of the finer solids will be removed as well as the desired fraction.
This means that the mass fraction removed, X, is
(1 – x0) + 0.253 = (1 - 0.553) + 0.253 = 0.447 + 0.253 = 0.70
That is the removal is 70%.
SETTLING COLUMN ANALYSIS FOR FLOCCULATING PARTICLES
(Type II settling)
This analysis is for Type II settling that is a flocculating but not concentrated suspension. Because of the flocculation process, the diameter of particles will be changing throughout the sedimentation process. This means that there is not a unique settling velocity, or even a unique spectrum of settling velocities (as in Type I settlement). Because increase in particle size caused by flocculation varies with time; groups of smaller particles forming larger flocs; it follows that the size and settling velocity will vary with depth. Because of this, in order to obtain realistic information about the settling behaviour of the flocculating suspension, it is necessary to take samples throughout the depth of the settling column at set time intervals. A further complication is arises because the particle’s concentration and settling velocity are also a function of the total depth of settlement (i.e. the maximum depth through which they settle) the settling analysis must be carried out using a column having the same depth as the proposed settling basin. For non-flocculating particles, the overall removal is calculated from the fraction of particles that settle because their settling velocity is equal to or greater than the loading rate plus the fraction of particles with lower settling velocities than the loading rate but will be removed from suspension because they start out closer to the bottom of the tank. The analysis technique for flocculating particles is similar, however in this case the mass fraction of particles remaining has to be calculated as a function of depth and time and presented graphically. If xij is the mass fraction of particles remaining in suspension at depth i and time j, then the percentage of particles removed at that depth and time is
Ways of performing the calculation
(a)By using interpolated values of r
The following calculations are carried out in the table below:
(i)identify the residence time t0 (105 minutes in the example);
(ii)interpolate for the concentration profile at this time;
(iii)convert the concentration profile to r-values (1 c(105)/c0 );
(iv)calculate the values of r;
(v)identify the average depth corresponding to each value of r;
(vi)form the product z r;
(vii)find the sum of all the values of z r;
(viii)divide this by the depth of the column.
Example
The initial concentration of suspended solids is 250 mg l-1. What is the overall removal efficiency for a 3 m deep settling tank with a 1.75 hour residence time?
Data: Concentration of solids observed at various positions in a settling column in mg l-1
Distance belowHeight aboveTime of sampling /(min)
Surface /(m)Bottom /(m)306090120150180
0.52.51338350383023
1.02.018012593655543
1.51.5203150118937058
2.01.02131681351109070
2.50.522018014512310380
3.00.022518815513311395
Solution
Depth / c(90) / c(120) / c(105). / r / r / z(ave) / z r0 / 0 / 0 / 0 / 1.0
0.176 / 0.25 / 0.044
0.5 / 50 / 38 / 44 / 0.824
0.14 / 0.75 / 0.105
1.0 / 93 / 65 / 79 / 0.684
0.106 / 1.25 / 0.1325
1.5 / 118 / 93 / 105.5 / 0.578
0.068 / 1.75 / 0.119
2.0 / 135 / 110 / 122.5 / 0.51
0.046 / 2.25 / 0.1035
2.5 / 145 / 123 / 134 / 0.464
0.040 / 2.75 / 0.11
3.0 / 155 / 133 / 144 / 0.424
Sum = / 0.614
The integral = 0.614/3 = 0.205
Therefore the total removal is 0.424+0.205 = 0.629, i.e. the removal is 63%.
(b)By considering the concentration profile
As there is information about the concentration profile throughout the column the average concentration in suspension at any time is
The mass fraction remaining in suspension must be and the removal = (1 x0(t))
The residence time is 105 minutes construct a table to find the concentration at 105 minutes and do the numerical integration
Depth / 90 min. / 120 min. / 105 min. / C(ave) / dz / C dz0.0 / 0 / 0 / 0
22 / 0.5 / 11
0.5 / 50 / 38 / 44
61.5 / 0.5 / 30.75
1.0 / 93 / 65 / 79
92.25 / 0.5 / 46.125
1.5 / 118 / 93 / 105.5
114 / 0.5 / 57
2.0 / 135 / 110 / 122.5
128.25 / 0.5 / 64.125
2.5 / 145 / 123 / 134
139 / 0.5 / 69.5
3.0 / 155 / 133 / 144
Sum = / 278.5
=278.5/3.0 =92.83
= 92.83/250 = 0.37 so 37% is left in suspension thus 63% is removed
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Mustafa Nasser 2012