Waste water treatment: Settling

Examples are plain settling in water treatment plants and grit settling in wastewater treatment plants.. Type II: Settling of flocculent particles in a dilute suspension: Examples are

Settling of Particles

Settling Types

Type I: Horizontal, discrete, or free settling Examples are plain

settling in water treatment plants and grit settling in wastewater

treatment plants

Type II: Settling of flocculent particles in a dilute suspension:

Examples are settling of flocs after coagulation and flocculation in

water and wastewater treatment plants and grit primary settling in

wastewater treatment plants

Type III: Zone settling or hindered settling Example is the settling of

intermediate concentration of particles where hindernece in the

settling occurs because the particles are close from each other This occurs in the secondary clarifiers at wastewater treatment plants

Type IV: Compression settling where the particles are too close that

settling can occur by compaction This occurs in secondary clarifiers in wastewater treatment plants when the solids at the bottom are

thickened

From MIT OCW

PST AT SST TF

ST

SF

BS

Influent

Effluent AD

BS= Bar screens, GC= Grit chamber, PST= Primary settling tank, AT= Aeration tank, SST= Secondary settling tank, TF= Trickling filter, SS= settling tank, SF= Sand

filter, CT= chlorination tank, AD= Anaerobic digester

CT

GC

Type I Type II Type III & IV

The settling velocity (vs) of discrete particles is given by Stokes Law:

Where, ρs and ρw are the particle and water density, respectively, d is the particle diameter, and µ is water absolute viscosity= 1.01 g/(m.sec)

Once vs is determined, the over-flow rate vo is set to be equal to vs in an ideal reactor However, since reactors deviate from ideality, vo is set at 0.33-0.7 times vs

µ

ρ

−

ρ

=

18

d ) (

g v

2 s

s

Type I: Horizontal, discrete, or free settling

Solution

Will grit particles of 0.2 mm or larger be

collected in a horizontal grit chamber that

is 13.5 m in length? The design flow

width of the chamber is 0.56 m, and the

horizontal velocity is 0.25 m/s?

0.15

13.5m

h

0.6 = h*0.56

h= 1.07 m

s m d

g

) 10 01 1 ( 18

) 10 2

0 )(

1000 2650

( 81

9 18

)

(

3

2 3 2

=

×

×

×

−

=

−

µ ρ ρ

Type II: Settling of flocculent particles in a dilute suspension

The design of these settling tanks usually requires a

batch laboratory test using a column with a depth

similar to that in the treatment plant (> 3m) and a

diameter of 130-205 mm After pouring the waste in

the column, samples are taken at different depths for

different times The samples are tested for

suspended solids and the results are reported as

percentage of the initial suspended solids Lines of

equal percentage of suspended solids are then

drawn as shown in the figure below

>

3m

13-20.5 cm

Sampling ports

The settlement of flocculent particles after coagulation of a surface water with suspended solids concentration of 200 mg/l was determined by a

settling column test The concentration of suspended solids at different

depths and times are listed below Sketch lines representing 50%, 60%, and 70% removals For a depth of 4.0 m, estimate the overall removal for a settling time of 120 min

t (min ( ( Suspended solids concentration (C

1.0

0 1m 2m

3m

4m

Find the percentage removal at each depth and time using :

t (min

1.0

Solution

150 140

120 100

80 60

40 20

3

4

2

1

0

61

31 33 40

48

59 68

52 53 55 65

39 43 49

74

58 62 65 70

53

57

63 67 71 75

61 65

68

Depth, m

Time, min

Insert percent removal on a time-depth chart

150 140

120 100

80 60

40 20

3

4

2

1

0

61

31 33 40

48

59 68

52 53 55 65

39 43 49

74

58 62 65 70

53

57

63 67 71 75

61 65

68

50%

100%

Depth

Time

Draw iso-concentration lines

150 140

120 100

80 60

40 20

3

4

2

1 0

50%

100%

Z 1 =2.7

Z 2 =0.8

Depth

This line corresponds to

61%

Time

% 73

) 4

8 0 )(

70 100

(

) 4

7 2 )(

61 70

( 61

Determine the percent removal (R) at the time needed

Design detention time

By finding Rt at different times, one

can then plot Rt versus t

From this graph one can select the

detention time needed for a certain

percent removal

This value is multiplied by 1.75 for

Desired removal

Needed detention time

For design: Batch settling tests are usually done in a 1-liter graduate

cylinder A settling curve (clarified depth versus time) is generated as shown below

0 0.1 0.2 0.3 0.4

0 10 20 30 40 50 60 70 80

time, min

, m

40cm

cylinder

Type III: Zone settling or hindered settling

Type IV: Compression settling

The area of the final clarifier is then obtained

from the curve as follows:

1 Determine the slope of the hindered settling

for clarification.

2 Determine the area required for

3 Extend the tangents from the hindered

settling region and the compression region

and bisect the angle formed to locate point 1.

4 Draw a tangent to the curve at point 1.

initial height Ho, select a design underflow

7 Determine the area required for thickening:

0 0.1 0.2 0.3 0.4

0 10 20 30 40 50 60 70 80

time, min

, m

0 0.1 0.2 0.3 0.4

0 10 20 30 40 50 60 70 80

time, min

, m 1

tu

Hu

A final clarifier is to be designed for an activated sludge plant treating industrial wastewater having a design flow of 5000 m3/d Batch settling studies have been conducted in the laboratory using acclimated culture

of activated sludge (Co=2,500 mg/L) and a graduated cylinder (0.4m height) The settling curve is as shown below If the design underflow concentration (Cu) is 12,000 mg/L What is the diameter of the settling tank?

Example

0 0.1 0.2 0.3 0.4

0 10 20 30 40 50 60 70 80

time, min

, m

Determine the slope of the hindered settling zone From the graph the

015 0

) 24 60

1 )(

5000

(

m v

Q o

=

×

=

12000

4 0 2500

=

×

=

⇒

=

2 304 4

0

) 35

)(

60 24

1 )(

5000

(

m H

Qt o

Solution

0.1

0.2

0.3

0.4

time, min

, m

1