ObjectiveObjective

In this lesson we will answer the In this lesson we will answer the following questions:following questions:

How does sedimentation fit into the How does sedimentation fit into the water treatment process? water treatment process?

5.6 SEDIMENTATION5.6 SEDIMENTATION

To remove the suspended To remove the suspended material from water by the action material from water by the action of gravityof gravity

Sedimentation tanks can be Sedimentation tanks can be rectangular, square or circular in rectangular, square or circular in shape. The most common types shape. The most common types are rectangular, and circular with are rectangular, and circular with centre feed.centre feed.

SEDIMENTATION TANK

Sedimentation TankSedimentation Tank

Types of particleTypes of particle

a. Discrete / individual particlea. Discrete / individual particle

-size, velocity are constant during the -size, velocity are constant during the settlingsettling

b. Flocculate particleb. Flocculate particle

- size, velocity are changing during the - size, velocity are changing during the settlingsettling

- the particles flocculate and grow - the particles flocculate and grow bigger in size bigger in size

Types of sedimentationTypes of sedimentation

Type 1 sedimentationType 1 sedimentation

- particles concentration is - particles concentration is very very lowlow

- settle as - settle as individualindividual particles particles

- example : sand and grit material - example : sand and grit material removal in wastewater treatment removal in wastewater treatment processprocess

Type 2 sedimentationType 2 sedimentation

- particles concentration is - particles concentration is lowlow

- particles - particles flocculateflocculate during during settlingsettling

- example : particles removal in - example : particles removal in sedimentation tanksedimentation tank

Type 3 sedimentation / zone Type 3 sedimentation / zone sedimentationsedimentation

- particles concentration is - particles concentration is highhigh

- particles tend to settle as a - particles tend to settle as a massmass and form a layer called “blanket”and form a layer called “blanket”

- distinct clear zone and sludge - distinct clear zone and sludge zone are presentzone are present

- example : secondary - example : secondary sedimentation in wastewater sedimentation in wastewater treatment planttreatment plant

Sedimentation ConceptsSedimentation Concepts

ss - settling velocity - settling velocityoo - over flow rate - over flow rate

oo = = = =

Where Where Q Q = flow rate= flow rateAs As = surface area= surface areaHH = depth of water= depth of watertt = detention time= detention time

As

Q

t

H

If If ss > > oo , particles will completely , particles will completely settlesettle

If If ss < < oo , particles , particles do not settledo not settle unless unless the particles are at h level when the particles are at h level when entering the sedimentation tank, whereentering the sedimentation tank, where

h = h = ss t t

To get the effective of sedimentation tank, To get the effective of sedimentation tank,

oo <<< <<< ss. This can be achieved by . This can be achieved by increasing increasing

the area of the tank (the area of the tank (oo = Q/As = Q/As))

Type 2 Sedimentation AnalysisType 2 Sedimentation Analysis

To determine the criteria of the To determine the criteria of the particles and the effectiveness of particles and the effectiveness of the tankthe tank

Using settling columnUsing settling column diameter of the column is not diameter of the column is not

important but the depth of the important but the depth of the water is same as the actual water water is same as the actual water depthdepth

occur in steady stateoccur in steady state

MethodMethod record the initial concentration of the record the initial concentration of the

suspended solids, Cosuspended solids, Co withdrawn the sample at every sample withdrawn the sample at every sample

ports at selected time intervals, Ctports at selected time intervals, Ct calculate the percent of removal for calculate the percent of removal for

every sample point and sampling timeevery sample point and sampling time

plot the %R at depth versus time graphplot the %R at depth versus time graph

100% xC

CCR

o

to

From the analysis, can determineFrom the analysis, can determine

the effectiveness of the tank at the effectiveness of the tank at selected timeselected time

oo = = HH tt

As = As = QQ oo

Relationship between analysis Relationship between analysis (column) and actual condition (column) and actual condition (tank)(tank)

- Scale-up factors of - Scale-up factors of 0.65 for 0.65 for overflow rateoverflow rate and and 1.75 for 1.75 for detention timedetention time to be used to to be used to design the tank.design the tank.

Example 1

Given: Given: Co = 400 mg/L, Overflow rate = 2.7 mCo = 400 mg/L, Overflow rate = 2.7 m33/m/m22.hour.hourFind the effectiveness of settling column.Find the effectiveness of settling column.

Time (t) 5 10 20 40 60 90 120

Depth, m Percent Removal (%R)

0.6 41 50 60 67 72 73 76

1.2 19 33 45 58 62 70 74

1.8 15 31 38 54 59 63 71

SolutionSolution1. 1. oo = = HH t t ii

2. 2. Plot the graph (%R at depth (H) Plot the graph (%R at depth (H) vs Time (t))vs Time (t))

.min401

.min60

./7.2

8.123

hr

xhrmm

mHt

o

3. The effectiveness of settling column3. The effectiveness of settling column

22

)(

2

)( 433322211 RR

H

HRR

H

HRR

H

HRT

%6.662

)5460(

8.1

79.0

2

)6070(

8.1

55.0

2

)70100(

8.1

46.0

Example 2Example 2

  

Determine the surface area of Determine the surface area of settling tank for 0.5 msettling tank for 0.5 m33/s and find /s and find the depth of the clarifier for the the depth of the clarifier for the overflow rate of 32.5 m/d and overflow rate of 32.5 m/d and detention time of 95 min. A detention time of 95 min. A minimum of two tanks is provided, minimum of two tanks is provided, each with a width of 12 m.each with a width of 12 m.

Solution :Solution :

Example 3Example 3

A rectangular sedimentation A rectangular sedimentation tank is to be designed for a flow tank is to be designed for a flow of 20 million litre per day using a of 20 million litre per day using a 2:1 length-width ratio and 2:1 length-width ratio and overflow rate of 24 moverflow rate of 24 m33/m/m22.day. .day. the tank is to be 2 m deep. the tank is to be 2 m deep. Determine the dimensions for Determine the dimensions for the tank and the detention time.the tank and the detention time.

SolutionSolution

DimensionsDimensionsQ = AVQ = AV = 833.33 = 833.33

mm22

L:W = 2:1L:W = 2:1

A = L x W = 2W x W = 2WA = L x W = 2W x W = 2W22 = 833.33 = 833.33

W = 20.41mW = 20.41m

L = 2W = 2(20.41) = 40.82 mL = 2W = 2(20.41) = 40.82 m

daymm

daym

V

QA

./24

/2000023

3

Detention timeDetention time

t = = t = =

= 120 min = 0.0833 day= 120 min = 0.0833 day

oV

Hdaymm

m

./24

223

An ideal rectangular sedimentation tank An ideal rectangular sedimentation tank illustration the settling of discrete particlesillustration the settling of discrete particles

Long rectangular settling basinLong rectangular settling basin