Sedimentation

Sedimentation

It is the process of removing solid particles heavier than waterby gravity force.

Sedimentation Sedimentation tanks are either rectangular or circular tanks.

Rectangular Basin

Circular Basin

Sedimentation Sedimentation tanks are either rectangular or circular tanks.

Sedimentation

‐A layer of sludge is formed on the floor of the tank-Typical detention times range from 1 to 10 hours. The longer the detention time, the more expensive the tank is, the better will be the tank's performance.

‐ OFR = 20 – 40 (m3/m2/day)‐ Detention time is 4 hours‐ Tank Depth 3 to 6 meters‐ For rectangular tanks; maximum tank length 40m, Preferable 30m

‐ L:W = 4:1‐ For circular tanks, maximum diameter 40m including the gentle mixing

‐ Average water flow at weir = 450 m3/m/day

Design Criteria of Sedimentation Tank

Sedimentation • Solids collected from the bottom of the tank will be

removed manually by periodically shutting down the tankand washing out the collected sludge, or the tank may becontinuously and mechanically cleaned using a bottomscraper. The effluent from the tank is then filtered.

Circular Sludge Scraper

Weir Overflow• The water velocity within the sedimentation tank is slow in order to 

allow for sedimentation. 

• To remove water from the basin quickly, it is desirable to direct the water into a pipe or small channel for easy transport, which will produce a significantly higher velocity. 

Sedimentation Concept• There are 2 important terms to understand in sedimentation 

zone design:

‐ settling velocity, Vs : The rate at which the particle is settling downward

‐ Velocity at which the tank is designed to operate, called the overflow rate, Vo, the velocity of the liquid rising is the overflow rate.

Overflow rate (Vo) must be less than or equal the settling velocity Vs

Critical settling velocity is the settling velocity of particles which are 100% removed in the basin.

•The percentage of particles removed in  an ideal horizontal sedimentation tank:

P = 100 Vs / Vo 

Since smaller particles have lower settling velocities, if you want to remove smaller particles in the settling basin you have to have a lower overflow rate.

Sedimentation Concept

Sedimentation Concept

• Drag coefficient

• For Laminar Flow

• Overflow rate: Vo

)()/(

2

3

mareasurfacesettlingsmRateFlowRateOverflow

• Settling Velocity: Vs

For flow rate of 120000 m3/d, design the following:‐ Flash mixing tank.‐ Circular sedimentation tank.‐ Rectangular Sedimentation tank & gentle mixing‐ Aluminum sulfate for coagulation for one year if dose = 30ppm and cost of one ton = 250 $‐ Volume of sludge m3/d if turbidity = 60ppm 

SOLUTION:Flash mixing:Hydraulic Retention Time (HRT) = 20-60 seconds, take=40sec.Volume = Q * HRT = 120000*40/(24*60*60) = 55 m3

For circular tank (Assume D=H); then D=4.2m

Example 1

Circular Sedimentation zone: Over Flow Rate (OFR) = 30 m3/m2/dArea surface = 120000 / 30 = 4000 m2

HRT = 3 hoursVolume = 120000 *3/24 = 15000 m3

Depth = volume/area = 15000/4000 = 3.75 m

Gentle mixing zone:HRT = 30 minVolume = 120000*30/24*60 = 2500 m3

Depth = 3.75 – 0.50 = 3.25 mArea = 2500/3.25 = 769 m2

Total area of  sedimentation and gentle mixing  zones  = Area of    sedimentation tank = 4000+769 = 4769 m2

take 6 tanks, the area of tank = 4769/6 = 795m2

Diameter of  sedimentation tank = 32 mArea of one gentle mixing tank = 769/6Diameter of gentle mixing zone = 12.8m

Rectangular Sedimentation zone: Over Flow Rate (OFR) = 30 m3/m2/dArea surface = 120000 / 30 = 4000 m2

Assume L=32 m. L:W=4; then W=8mArea of one tank = 32*8=256 m2

Number of Tanks = 4000/256 = 15.6 Take 16 TanksTherefore Actual Area = 4000/16 = 250 m2 (31.25*8m)HRT = 3 hoursVolume = 120000 *3/24 = 15000 m3

Depth = volume/area = 15000/4000 =  3.75 m

Gentle mixing zone:Number of Tanks = 16HRT = 30 minTotal Volume = 120000*30/24*60 = 2500 m3

Volume of one tank= 2500/16 = 156.25 m2

Width = 8m same as the rectangular tank; Assume 3m depthThen L= 7.75m

Coagulants:Dose = 30 ppm, Q= 120000m3/d Weight of Alum. Sulphate = 30*120000/(1000*1000) = 3.6 ton /d

= 3.6 * 365 = 1314 ton/yearCost of coagulants every year = 1314*250=328500 $

Sludge Volume:

Turbidity of solids = 60ppmEfficiency of sedimentation = 90%Weight of dry solids = 120000*60*0.90/1000*1000 = 6.47 ton/dIf water percentage in the sludge = 98% and dry solids = 2% If specific gravity of sludge = specific gravity of water = 1 ton /m3

Volume of sludge = 6.47 *100/2 = 324 ton/d

In case of Circular Tanks:

Sludge Volume / Tank = 324/6 = 54 m3

Assume Sludge Removal every 4 hrs (6times per day)Sludge to be removed every time = 54/6 = 9m3

In case of RectangularTanks:

Sludge Volume / Tank = 324/16 = 20.25 m3

Assume Sludge Removal every 4 hrs (6times per day)Sludge to be removed every time = 20.25/6 = 3.375m3

Example 2A water treatment plant has a flow rate of 0.6 m3/sec. The settling basin at the plant has an effective settling volume that is 20 m long, 3 m tall and 6 m wide. Will particles that have a settling velocity of 0.004 m/sec be completely removed? If not, what percent of the particles will be removed?

v0 = Q/A = 0.6 m/sec / (20 m x 6 m) = 0.005 m/sec

Since v0 is greater than the settling velocity of the particle of interest, they will not be completely removed.

The percent of particles which will be removed may be found using the following formula:

Percent removed = (vp / v0) 100

= (0.004/0.005) 100 = 80 %

3

Example 3Find the settling velocity for sand particles with a diameter of 0.02 mm; 

What will be the settling velocity for particles with D=0.5mm?