Water Treatment: Sedimentation

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Sedimentation/Settling

• Following flocculation, the water then flows

into the settling basins

• Water is nearly quiescent – low flow with

little turbulence little turbulence

• Water resides for at least 3 hours and the

flocs settle out and collect at the bottom.

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Settling in Treatment Train

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Circular Clarifiers

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Type I Settling -- Stokes’ Law

µ

ρρ

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)( −= s

s

gv

where

ν = settling velocityνs = settling velocity

ρs = density of particle (kg/m3)

ρ = density of fluid (kg/m3)

g = gravitational constant (m/s2)

d = particle diameter (m)

µ = dynamic viscosity (Pa�s)5

Overflow rate

cA

Qv =

where

v = overflow v = overflow

rate (m/s)

Q = water flow

(m3/s)

Ac = surface

area (m2)6

Types of Particle Settling

• Type I settling applies to particles that settle with constant velocity -- particles will be removed if v > vs

• If particles flocculate during settling, • If particles flocculate during settling, velocity generally increases – Type II settling

• As particle concentration increases with depth, zone settling occurs

• At bottom of tank compression settling occurs

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SedimentationIdeal sedimentation basins (Type 1) - REGTANGULAR BASINParticle removal is dependent on the

overflow rate, v0

In order for particle to be removed settling

velocity , vs must be sufficient so that it

reaches the bottom during the time the

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water resides in the tank (td).

If Vs = Vo , then

vs > v0 , 100 % particles should be easily removed

vs = v0 , 100 % particles removed

vs < v0 , some fraction of the particles will be removed P =

100 vsv0

Ideal sedimentation basins (Type 2)

UPFLOW CLARIFIER

Settling velocity, vs of the particle to be removed

Velocity of water decreases as the water flows

Sedimentation

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Velocity of water decreases as the water flows

upward (overflow rate = vo, hydraulic surface

loading)

Velocity of the particle remains unchanged

If vs ≥≥≥≥ vo, then 100% of particles remain in tank

If vs < vo, then 0% of particles remain in tank

�Sedimentation basin =

clarifier = settling tank

�Water flows into the

settling basins, where the

flow is almost devoid of

turbulence.

�The water resides here for

time periods ranging from 2

Sedimentation

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time periods ranging from 2

to 8 hours and flocculated

particles settle out as sludge

�The sludge is

mechanically removed

periodically

�Sedimentation basin are

usually rectangular or

circular with either a circular

or upward water flow

pattern.

The key parameters and typical values in the design

of settling tank are:

- surface over flow rate – 20- 35 m3/day/m2

- detention times – 2-8 h

- weir overflow rate – 150 – 300 m3 /day/m

Either rectangular or circular , normally 1 unit

sedimentation system comprises of 2 tanks

Sedimentation

The sedimentation tank be

divided into four zones: inlet,

settling, outlet and sludge

storage

Inlet zone - to evenly distribute

the flow and suspended particles

across the section of the settling

zone*

Sludge storage zone- depends

*

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Sludge storage zone- depends

upon the method of cleaning, the

frequency of cleaning and the

quantity of sludge estimated to be

produced.

Outlet zone- to remove the

settled water from the basin with

out carrying away any of the floc

particlesZones of sedimentation: (a) horizontal flow clarifier , (b)

upflow clarifier

*

*Sedimentation

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*

Example

Calculate the diameter and depth of a circular

clarifier for a design flow of 3800 m3/d and an

overflow rate of 0.00024 m/s and a detention

Sedimentation

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overflow rate of 0.00024 m/s and a detention

time of 3 h. Calculate the weir loading rate by

assuming the total effluent weir is 20 m.

Solution

Calculate the diameter and depth of a circular clarifier for a design flow of 3800 m3/d and an

overflow rate of 0.00024 m/s and a detention time of 3 h. Calculate the weir loading rate by

assuming the total effluent weir is 20 m.

Solution

Volume , V = Q/t

= (3800 m3/d)/ ( 3/24)d

= 475 m3

Q = 3800 m3/d = 0.044 m3/s

Sedimentation

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Surface overflow rate = Q/A

0.00024 m/s = 0.044 m3/s

A m2

Area, A = 183.3 m2

Volume, V = AD

D =V/A

= 475 m3/183.3 m2

= 2.59 m

Diameter = 15.3 m

Weir loading rate = Q/ Lw

= 3800 m3/d

20 m

= 190 m3/day.m ( OK!)

Example

A water treatment plant has a horizontal –flow sedimentation tank with an

overflow rate of 17 m3/d.m2 and wishes to remove particles that have settling

velocities of 0.1 mm/s. What percentage of removal should be expected for each

particle in an ideal sedimentation tank?

Solution

v

Sedimentation

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vs = 0.1 mm/s

v0 = 17 m3/d.m2 = ? mm/s, (do the conversion so, v0= ? mm/s

Note that

vs < v0 , P = 100 vs / vo = 100 ( 0.1)/(0.2) = 50%

Recalculate by considering v0 are 0.2 mm/s and 1 mm/s respectively.

SEDIMENTATIONExample

Determine the surface area of a sedimentation tank . The design

flow is 0.044 m3 /s. Use a design overflow rate of 20 m / day.

Find the depth of the sedimentation for the given overflow rate

and detention time.

Solution:

1) Find the surface area.

First change the flow rate to compatible units.

(0.044 m3/ s)(86,400 s / day) = 3801.6 m3 /day

The surface area is = 3801.6 m3 /day

Tank 1 Tank 2

Flocculation tank

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The surface area is = 3801.6 m3 /day

20 m/ day

=190.m2

2) Find the length to width dimension

Common length-to-width , L: W ratios ( 2:1 < L:W < 5:1 , and

lengths seldom exceed 100 m. A minimum of two tanks is always

provided.

use two tanks, each with a width of 5 m, a total surface area of

190 m2 ,

Length = 190 m2 /(2 tanks)(5 m wide) = 19 m

meet our length-to-width ratio of 3.8 : 1 ( OK!)

Flocculation tank

Se

dim

en

tation

tan

k

Se

dim

en

tation

tan

k

Tank 1

L= 19m,

W = 5 m

Tank 2

L= 19m,

W = 5 m

Q = 3801.6 m3 /day

3) Find the tank depth.

Rule of thumb that the detention time should be 2-8 h.

Assumed the detention time of 120 min

Q = V/t

V = Q t

Sedimentation

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V = Q t

V = (1900.8 m3)(120 min)( d ) = 158 m3

d 1440 min

Depth= 158 m3 =1.684 m,= 1.7 m

95 m2

The final design would then be two tanks, each having the following

dimensions:

5 m wide x 19 m long x 1.7 m deep plus sludge storage depth.

END

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