SEDIMENTATION (6)

SEDIMENTATION

Location in the Treatment Plant

• After the source water has been coagulated and flocculated, it is ready for sedimentation.

Objectives of Sedimentation• To separate solids from liquid using the force of gravity. In

sedimentation, only suspended solids (SS) are removed.

• Water Treatment

• Prior to filtration of surface water

• Prior to filtration of coagulated-flocculated water

• After adding lime and soda ash In softening of water

• In iron and manganese removal plants after treating the water

Objectives of Sedimentation

• Wastewater Treatment

• Removal of SS in primary sedimentation basins

• Removal of biological floc in activated sludge processes (final sedimentation basin)

• In tertiary treatment

• Removal of humus after trickling filters (final sedimentation basins)

• Chemical floc removal when coagulation process is used

SEDIMENTATION BASINSShapes−Circular, Rectangular, and square

Circular– 15 to 300 ft (diameter) and 6 to 16 ft (depth) – Typical sizes are 35 to 150 ft (diameter) and 10 to 14 ft (depth)

Square – 35 to 200 ft (width) and 6 to 19 ft (depth)

Rectangular ( depends on sludge removal mechanism) • Freeboard• – 1 to 1.25 ft for circular and square tanks

Types of Settling

• Type I settling (free settling/discrete particle)• Type II settling (settling of flocculated particles)• Type III settling (zone or hindered settling)• Type IV settling (compression settling)

Type I settling (free settling)

Settling of discrete (nonflocculent) particles − Settling of sand particles in grit chamber

In type I settling, a particle will accelerate until the drag force, FD, equals the impelling (due to weight) force, FI; then settling occurs at a constant velocity, Vs.

FI = (ρs - ρ)gV

where;

ρs = particle density

V = particle volume

FD = CDAC ρ (V2s/2)

where;

CD = coefficient of drag

AC = particle cross-sectional area


Assuming spherical particles in a laminar flow regime:

Vs = (ρs − ρ)gd 2

18μwhere:

Vs = settling velocity , m/s

ρs = density of particle, kg/m3

ρ = density of fluid, kg/m3

g = gravitational acceleration, m/s2

d = diameter of particle, mμ = dynamic viscosity, Pa.s

Type I settling (free settling)R = dVs

ѵwhere:

R = Reynold’s numberd = diameter of particle, mVs = settling velocity, m/s

ѵ = kinematic viscosity, m2/s Q Q

What is the settling velocity of a grit particle with a radius of 0.10 mm and a specific gravity of 2.65? The water temperature is 22OC. ρH2O @22OC = 997.774 kg/m3

μH2O@22OC = 0.955 mPa.s.

ѵH2O @22OC = 0.957 μm2/s


Type I settling (free settling)• Design Equations

• The detention time of the particle that enters at point 1 and get removed at point 2 is given by:t = _H_ …. (1)

VO

• The detention time is also equal to the length divided by the horizontal velocity:

t = _L_ …. (2) V • The horizontal velocity is equal flowrate divided by cross-sectional area:

V = _Q_ …. (3) HW

• Combining equations 2 and to eliminate V gives:

t = _LWH_ …. (4) Q

Type I settling (free settling)• Design Equations • Since LWH is the basin volume, Vol, the detention time, t, is equal to the basin volume divided by the flow rate:

t = _Vol_ …. (5) QFrom equation 1 and 4:

LWH = _H_ …. (6) Q VO

Rearranging yields:

VO = Q …. (7)

LW

Or: VO = Q = overflow rate or surface loading (m3/d-m2)

AP

Where Ap is the plan area of the basin

Type I settling (free settling)• Design Equations For an ideal circular settling basin, the horizontal velocity is given by:

V = Q___ 2 ¶rH and

VO = Q = overflow rate or surface loading (m3/d-m2)

AP

The depth an ideal rectangular or circular basins is given by:

H =Vot


Clarifier for Water Treatment

A rectangular sedimentation basin is to be designed for a rapid filtration plant. The flow is 30,300 m3/day, the overflow rate or surface loading is 24.4 m3/d-m2, and the detention time is 6 hours. Two sludge scraper mechanisms for square tanks are to be used in tandem to give a rectangular tank with a length to width ratio of 2:1. Determine the dimensions of the basin.

Solution

The plan area required = (30,000 m3/d) / (24.4 m3/d-m2) = 1242 m2

Since the length (L) is twice the width (W)

Then (2W)(W) = 1242 m2

= 24.9 m and thus L

= 49.8 m

Therefore, the plant dimensions are:

Width = 24.9 m Length = 49.8 m

Since the depth (H) is equal to the settling rate times the detention time

H = (24.4 m3/d-m2)(d/24 )(6 h) = 6.10 m

Depth = 6.10 m


• Percent removal

P = Vs x 100

Vo

Type II settling (settling of flocculated particles)

Particles flocculate during settling; thus they increase in size and settle at a faster velocity.

Examples of Type II settling: −Primary settling of wastewater −Settling of chemically coagulated water and wastewater A batch settling tests are performed to evaluate the

settling characteristics of flocculent suspensions.


Batch settling tests:

− Suspension is poured into column. − Samples are removed from the ports at periodic time intervals. − The samples suspended solids concentrations are determined. − The percent removal is calculated for each sample.

− The percent removal is plotted on a graph as a number vs time and depth.

− Interpolations are made between the plotted points and curves of equal percent removals are drawn as shown in the next figure.



Determining overflow rate and fraction removed:The overflow rates, Vo , are determined for the various settling times (ta, tb, and so on) where the R curves intercept the x-axis. For example, for the curve Rc, the overflow rate is:

VO = H_ x proper conversions

tC

Where:

H = height of column

tc = time defined by intersection of isoconcentration of column


Percent removal versus depth is then plotted. Interpolations are made between these plotted points to construct curves of equal concentration at reasonable percentages, that is, 5 or 10 percent increment. Each intersection point of an isoconcentration line and the bottom of the column defines an overflow rate (VO).


The fractions of solids removed, RT, for the times ta, tb, and so on are then determined. For example, for time tc, the fraction removed would be:

RT = RC + H2 (RD – RC) + H1 ( RE- RD)

H H


A vertical line drawn from ti to intersect all the isoconcentration lines crossing the ti time. The midpoints between isoconcentration lines define heights H1, H2, H3 and so on used to calculate the fraction of solids removed.

The series of overflow rates and removal fractions are used to plot two curves. One of suspended solids removal versus detention time and one of suspended solids removal versus overflow rate. These can be used to size the settling tank.

Eckenfelder (1980) recommends that scale-up factors of 0.65 for overflow rate and 1.75 for detention time be used to design the tank.


Percent Removal

%R = 1 – Ct 100 CO

Where %R = percent removal at one depth and time, %Ct = concentration at time, t, and given depth,mg/L

CO = initial concentration, mg/L

The city of Stillwater is planning to install a new settling tank as an upgrade to their existing water treatment plant. Design a settling tank to remove 65% of the influent suspended solids from their design flow of 0.5 m3/s. A batch-settling test using a 2.0 m column and coagulated water from their existing plant yielded the following data:

Percent removal as a function of time and depthSampling time, min

Depth, m 5 10 20 40 60 90 120

0.5 41 50 60 67 72 73 76

1.0 19 33 45 58 62 70 74

2.0 15 31 38 54 59 63 71


depth, m

Sampling time. min

5 15 25 35 45 55 65 75 85 95 105 115 125

0

0.5

1.0

1.5

2.0

33%

70%

75%

80%

45%

58%

62%


Recalculate the VO per experimental %SS removal:

@ 33% VO = H = 2m 1440 min = 192

m/d tc 15 min d

@ 45% VO = H = 2m 1440 min = 96 m/d

tc 30 min d


% SS removal

overflow rate, m/d

25 50 75 100 125 150 175 200

80

70

60

50

40

30

Vo = 30 m/d


Recalculate the % removal per experimental detention time:@ 15min:RT = 33 + 1.35 45-33 + 0.7 58-45 + 0.55 62-58 + 0.33 70-62

2 2 2 2 + 0.24 75-70 + 0.15 80-75 + 0.04 100-80 2 2 2RT = 49.45 %


@ 30 min:RT = 45 + 0.7 58-45 + 0.55 62-58 + 0.33 70-62

2 2 2 + 0.24 75-70 + 0.15 80-75 + 0.04 100-80 2 2 2 RT = 53.35 %


@ 60 minRT = 61.80%

@ 90 min RT = 64.70%

@120 minRT = 71.38%


% SS removal

detention time, hour

0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.0

80

70

60

50

40

30

T = 1.58 hour


Calculated values based from the graph: tO = ( 1.58 hour)

VO = ( 30 m/d)

Applying the scale-up factors: tO = ( 94.8 min) (1.75) = 165.9 min

VO = ( 30.0 m/d) (0.65) = 19.5 m/d


Comments:1. As implied by the shape of the isoconcentration

lines, and, conceptually, the trajectory of the particles, the settling velocity increases as the particles travel through the tank.

2. The depth of the tank is important because flocculant particles tend to grow in size. Thus, a greater depth facilitates the growth process.

The following test data were gathered to design a settling tank. The initial suspended solids concentration for the test was 20.0 mg/L. Determine the detention time and the overflow rate that will yield 60% removal of suspended solids. The data given are the suspended solids concentration in mg/L.

Sampling time, min

Depth, m 10 20 35 50 70 85

0.5 14.0 10.0 7.0 6.2 5.0 4.0

1.0 15.0 13.0 10.6 8.2 7.0 6.0

1.5 15.4 14.2 12.0 10.0 7.8 7.0

2.0 16.0 14.6 12.6 11.0 9.0 8.0

2.5 17.0 15.0 13.0 11.4 10.0 8.8

% removal @ t = 10 min and H = 0.5 m

% R = 1 – 14 x 100 = 30% 20

Repeat with other concentration valuesDetermine the overflow rate of 60% SS removal form the graph plotted as % SS removal versus overflow rate.Determine the detention time for 60% SS removal from the graph plotted as % SS removal versus detention time

In terms of percent removal

Sampling time, min

Depth, m 10 20 30 35 40 50 60 70 80 85

0.5 30 50 53 65 59 69 70 75 82 80

1.0 25 35 41 47 47 59 59 65 71 70

1.5 23 29 36 40 42 50 53 61 65 65

2.0 20 27 33 37 38 45 49 55 59 60

2.5 15 25 29 35 35 43 45 50 56 56

0

0.5

1.0

1.5

2.0

2.5

10 20 30 40 50 60 70 80 90 100 110

DEPTH

sampling, time, min

30%

46%56%

67%

79%

Recalculate the VO per experimental %SS removal:@ 30%

VO = H = 2.5 m 1440 min = 120 m/d tc 30 min d

@ 46% VO = H = 2.5 m 1440 min = 60 m/d

tc 60 min d@ 56%

VO = H = 2.5 m 1440 min = 40 m/d tc 90 min d

@ 67% VO = H = 2.5 m 1440 min = 30 m/d

tc 120 min d

% SS removal

overflow rate, m/d

20 40 60 80 100 120 140 160

80

70

60

50

40

30

Vo = 36 m/d

@ 30 min:RT = 30 + 2.2 46-30 + 1.13 56-46 + 0.75 67-56

2.5 2.5 2.5 + 0.38 79-67 + 0.07 100-79 2.5 2.5 RT = 54.312 %

@ 60 min:RT = 56.232 %

@ 90 min:RT = 72.712 %

% SS removal

detention time, hour

0.25 0.50 0.75 1.0 1.25 1.50 1.75 2.0

80

70

60

50

40

30

t = 1.15 hour

Calculated values based from the graph: tO = ( 1.15 hour)

VO = ( 36 m/d)

Applying the scale-up factors: tO = ( 69 min) (1.75) = 120.75 min

VO = ( 36.0 m/d) (0.65) = 23.4 m/d

Criteria Used for Horizontal Flow Rectangular Sedimentation Basins

Parameter Range Parameter Range

Inlet Zone Outlet zone

Distance to diffuser wall 2 m Launder length 1/3 – ½ length of basin

Diffuser hole diameter 0.10 – 0.20 m Launder weir loading

140-320 m3/d.m(limit)

Settling Zone

Overflow rate 40-70 m3/d.m2 Width 6 m max per train

Side water depth (SWD) 3-5 m Chain and flight

Length 60 m (max) L:W ≥6:1

Chain and flight L:D 15:1 min

Reynolds number < 20,000 Velocity 0.005-0.018 m/s (mean)

Design the settling tank(s) for the city of Stillwater’s water treatment plant expansion using the design overflow rate found in example #2. The maximum design flow is 0.5 m3/s. assume a water temperature of 10OC.Given:

Vo = 23.4 m/d t = 120.75 minTH2O = 10OC Q = 0.5 m3/s

ѵ = 1.307 x 10 -6 m2/s

Design Parameters• Q design

• Number of tanks• Width of each tank• Length of each tank• L:W• Depth including sludge• L:D• Vf

• Reynold’s number• Launders• Launder’s length• Weir loading• Sludge collector (chain and flight)

• Find the surface area,

AS = 0.5 m3/s (86400 s/d) = 1846.15 m2

23.4 m/d

• Select the number of tanks # tanks = 2 (minimum)

As = 1846 m2 / 4 tanks = 461.5 m2/tank

• Select a trial width for calculationAssume:

W = 6 m (max for chain and flight)

• Check length to width ratioL = 461.5 m2/tank = 76.9 m 6 m/tank

L/W = 76.9 m = 12.82 6 m L/W = 12.8 : 1 (>6:1 acceptable)

• Select a trial depth D = expt’l + allowance + freeboard D = (2.5 m + 1 m + 0.6 m) = 4.1 m

• Check the length to depth ratio: L/D = 76.9 m = 18.76 m 4.1 m L/D = 19:1 (acceptable)• Check the velocity: Vf = Q = 0.5 m3/s

A (4 tanks)(4.1 m depth)(6 m width) Vf = 5.08 x 10 -3 m/s (within the range 0.005 – 0.018 m/s)

• Check Reynold’s Number d = Ax = (4.1 m deep)(6 m wide)

Pw (4.1 m + 6 m + 4.1 m)

d = 1.73 m

R = d Vs = (1.73 m)(5.08 x 10 -3 ) ѵ 1.307 x 10 -6

R = 6733.41 (< 20,000)

• Design the Launders

ȴ Launders = 76.9 m = 25.6 m

3Placed at 1m intervals on center.• Check the weir loading rate Wȴ = 0.5 m3/s (86400 s/d) ( 4 tanks)(3 launders/tank)(25.6 m/launder)(2 sides)

Wȴ = 70.31 m3/d (below 250 m3/d, acceptable)

• Q design = 0.5 m3/s• Number of tanks = 4 tanks• Width of each tank = 6 m• Length of each tank = 76.9 m• L:W = 12.8 : 1• Depth including sludge = 4.7 m• L:D = 19:1• Vf = 5.08 x 10 -3 m/s

• Reynold’s number = 6733.41• Launders = 3 spaced evenly• Launder’s length = 25.6 m• Weir loading = 11 72 m3/d• Sludge collector = chain and flight

Type III settling (zone or hindered settling)

• Is the settling of an intermediate concentration of particles

• The particles are close to each other

• Interparticle forces hinder settling of neighboring particles

• Particles remain in fixed position relative to each other • Mass of particles settle as a zone

Type III settling (zone or hindered settling)

Type IV settling (compression settling)

• Settling of particles that are of high concentration

• Particles touch each other• Settling occurs by compression of the

compacting mass• It occurs in the lower depths of final clarifiers

of activated sludge

Type IV settling (compression settling)

SECONDARY SETTLING

The function of secondary settling tanks that follow trickling filters is to produce a clarified effluent. Secondary settling tanks that follow activated sludge processes also serve the function of thickening to provide a higher solids concentration for either return activated sludge or wasting and subsequent treatment.Things to consider:

design principlesdesign practice

Design principles

• TRICKLING FILTERSTrickling filter solids settling may be classified as Type II flocculant settling. Because the suspended solids loading is low, the overflow rate governs design.

Design principles

• ACTIVATED SLUDGEActivated sludge solids settling classification falls into each of the four types depending on the depth in the clarifier. In the upper water level, discrete floc particles settle(Type I). As the particles sink, they begin to flocculate (Type II). In the lower zones, hindered settling (Type III) and compression settling (Type IV) take place.

Both clarification and thickening are considered in the design of the secondary settling tank for activated sludge systems. Clarification is governed by the settling velocity of the light fluffy particles. Thickening is governed by the mass flux of solids in the zone where hindered settling takes place. It is not possible to make estimates of overflow rates for clarification based on first principles. The irregular nature of activated sludge floc precludes any rational estimate of settling velocity that could be used to select an overflow rate.

The surface area required for thickening may be determined by one of two methods:

solids flux analysis state point analysis

SOLID FLUX ANALYSIS

Zone settling and compression settling occurs in sludge thickeners. As with Type II settling, the methods for analyzing hindered settling require settling tests data. These methods are appropriate for plant expansions or modifications, but have not found use in the design of treatment plants where process sludges from treating the actual water are not available. However, they are useful in explaining the behavior of thickeners and demonstrating design variables.

SOLID FLUX ANALYSIS

Flux is the term used to describe the rate of settling of solids. It is defined as the mass of solids that pass through a horizontal unit area per unit of time (kg/m2.d).

FS = (CU)(ѵ)

= (CS)(zone settling velocity)

Where: FS = solids flux, kg/m2.d

CU = concentration of solids in underflow,

that is, the sludge withdrawal pipe, kg/m3

CS = suspended solids concentration, kg/m3

ѵ = underflow velocity, m/d

SOLID FLUX ANALYSIS

Critical factors in conducting the solids flux analysis:–a cylinder diameter as large as possible, but not less than 20 cm in diameter–An initial height that is, preferably, the same as the thickener, but not less than 1 m–Filling the cylinder from the bottom–Stirring the cylinder very slowly at a speed of 0.5 rpm during the test

SOLID FLUX ANALYSIS

Data from the batch settling curve (Figure 11-4) are used to construct a batch flux curve. Knowing the desired underflow concentration, a line through the desired concentration and tangent to the batch flux curve is constructed. The extension line to the ordinate axis yields the design flux. From the flux and the inflow solids concentration, the surface area may be determined. A safety factor of 0.667 is to be considered.

SOLID FLUX ANALYSIS

A gravity thickener is to be designed to thicken a lime softening sludge from a conventional settling basin. The sludge flow rate is 171.2 m3/d, and the solids mass loading is 4.28 x 10 3 kg/d. the limiting overflow rate to prevent carryover of solids is 0.4 m/h. the thickened sludge should have an underflow solids concentration of 20.0 %. Assume that the sludge yields a batch settling curve as shown in the previous figure. Determine the required surface area and diameter of the thickener for thickening, and verify that the hydraulic loading rate is acceptable.

SOLID FLUX ANALYSISSS, kg/m3 ѵ, m/d FS, kg/m2.d

10 60 600

15 55 825

25 45 1125

30 35 1050

40 14 560

50 7.5 375

60 5.1 306

80 2.6 208

100 1.7 170

150 0.86 129

210 0.43 90.3

SOLID FLUX,

kg/m2.d

1200

1000

800

600

400

200

0 0 50 100 150 200 250

Suspended solids concentration SS, kg/m3

FS design

20% CU

BATCH FLUX CURVE

From Figure 11-420% CU intersecting the Lime graph and projecting it downward to determine subsidence rate, will yield ѵ = 50 m/d

Suspended solids concentration SS, kg/m3

= (0.20 d/m)(20 kg/m3)(50 m/d) = 200 kg/m3

FS DESIGN = (320 kg/m2.d)(0.667) = 213.44 kg/m2.d

AS = 4.28 x 103 kg/d = 20.05 m2

213.44 kg/m2.d

D thickener = 4 (20.05 m2) ½ = 5.05 m Π

Check for the hydraulic loading criterion:

Vo = 171.2 m3/d = 8.54 m/d = 0.36 m/h 20.05 m2

The Vo = 0.36 m/h < 0.4 m/h, is acceptable

Generalization:

• If the hydraulic loading rate is exceeded, then the surface area of the thickener is governed by the hydraulic loading criterion and the surface are and the diameter are recalculated using the overflow rate criterion.

• The thickener design is small and may not be economical because the savings in disposal cost may not be recovered by the end of the design life of the thickener.

STATE POINT ANALYSIS

State point analysis is an extension of solids flux analysis to provide a means to assess different MLSS concentrations and operating conditions relative to the limiting solids concentration.The state point is the intersection of the clarifier overflow solids-flux rate and the underflow solids-flux rate. This analysis accounts for the MLSS concentration, clarifier hydraulic loading, and RAS. It enables an assessment of these operating parameters to determine whether or not the conditions are within the clarifier solids-flux limitations.

The clarifier overflow solids flux isSF = QX

AWhere:

SF = overflow solid flux ratio, kg/m2.dQ = clarifier effluent flow rate, m3/dX = aeration tank MLSS, kg/m3

A = clarifier surface area, m2

State Point AnalysisSOLID FLUX,

kg/m2.

h

10

8

6

4

2

00 2 4 6 8 10 12 14

Solids concentration X, g/L

State pointUnderflow rate operating line

overflow rate operating line

X MLSS

The aeration tank MLSS at any point along the overflow solids-flux line is found by constructing the vertical line to the x-axis. Because the underflow velocity (Ub) can be controlled by controlling the flow rate of return activated sludge, the underflow operating line is used for process control. Ub is defined as

Ub = Qu

AWhere:

Ub = underflow velocity, m/d

Qu = underflow flow rate, m3/d = QR, the

return sludge rate or RAS A = surface area, m2

Other relationship for Ub can be saved to evaluate the clarifier operation:

Ub = SFt – SF

0 - XMLSS

Ub = (Q + QR)(XMLSS /A)

- XMLSS

Where: SFt is the total solids flux.

The underflow operating line is represented as the negative slope of underflow velocity. The underflow solids flux is

SFU = Ub Xi

Where: SFU = solids flux resulting from underflow, kg/m2.d

Xi = solids concentration at a given point, kg/m3

To determine if the operation of the clarifier is within its solid flux limitation, a gravity flux curve is plotted on the state point graph. The gravity solids flux is calculated asSFg = Ci ѵi

Where:SFg = gravity flux, kg/m2.d Ci = initial MLSS concentration, kg/m3

ѵi = initial settling velocity, m/d

Batch settling curve data similar to that in Figure 18-2 are used in the computation.Using Figure 18-3, the plots may be interpreted as follows:•The underflow line at state point A is tangent to the gravity flux curve. This is the limiting solids flux condition.– If the operation is changed to obtain a higher MLSS concentration and the underflow line crosses the lower limb of the gravity flux curve, the limiting solids flux will be exceeded and the clarifier sludge blanket will rise to the effluent weir.

•The underflow line at state point B represents a lower MLSS concentration. This represents an underflow operation relative to the solids loading.

From the state point analysis, it is self evident that the design and behavior of the secondary clarifier is intimately linked to the activated sludge process. The linkage is through control of the MLSS concentration by wasting and the return activated sludge flow rate. Table 18-1 and Figure 18-4 illustrate the use of state point analysis in design and operation of the secondary clarifier.

State Point AnalysisDetermine the limiting solids flux for a 35 m diameter secondary settling tank that is being used to thicken activated sludge. Also, determine whether or not the clarifier is overloaded, critically loaded, or under loaded. The average plant flow rate to the clarifier is 9200 m3/d, the activated sludge MLSS is 3125 mg/L, and the RAS concentration is 10000 mg/L. data from a batch settling analysis is shown below:

Batch Settling Data

MLSS ѵi MLSS ѵi g/m3 m/h g/m3 m/h1000 4.0 5000 0.311500 3.5 6000 0.202000 2.8 7000 0.132500 1.8 8000 0.0943000 1.14 9000 0.074000 0.55

Compute for the Gravity Solids Flux

Ci ѵi SFU Ci ѵi SFU

g/m3 m/h kg/m2.h g/m3 m/h kg/m2.h

1000 4.0 4.00 5000 0.31 1.55

1500 3.5 5.25 6000 0.20 1.20

2000 2.8 5.60 7000 0.13 0.91

2500 1.8 4.50 8000 0.094 0.75

3000 1.14 3.42 9000 0.07 0.63

4000 0.55 2.20

• Plot the gravity solids flux curve (as shown in Figure)

• Add an overflow operating line and MLSS concentration state point at 3125 mg/L

• Determine the clarifier surface area:A = Π(35 m)2 = 962.11 m2

4• Determine the overflow solids flux rate:

SF = (9200 m3/d)(X) = 9.56 m/d X 962.11 m2

• Determine the plotting point for 3125 mg/L MLSS:

SF = (9.56 m/d)(3.125 kg/m3) 1d = 1.24 kg/m3

24 h• Plot the overflow rate operating line through

the points (0,0) and 3125 g/m3, 124 kg/m3)

• Determine the limiting solids flux by drawing a line from 10,000 g/m3 tangent to the gravity flux curve.

• The limiting flux curve is 2.85 kg/m3.h• A plot from 10,000 g/m3 through the state

point indicates a solids flux rate of 1.75 kg/m3.h.

• The underflow line is below the descending line of the gravity flux curve. This implies that the clarifier is underloaded.

SOLID FLUX,

kg/m2.h

6

5

4

3

2

1

00 1000 3000 5000 7000 9000

solids concentration, g/m3

Limiting flux

MLSS

GRAVITY FLUX CURVE

Solids flux rate

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