summary ( 500 - 600 words)

· In Water Treatment coagulation is a chemical technique for the destabilization of colloidal particles.

· The size of particles remains between 0.1- 10µm .

· Mostly colloidal particles in natural water are negatively charge and thus remain in suspension in water due to mutual repulsion.

· They have a large ratio of surface area to volume. Now as you increase the surface area of an object it tends to float in water

· Mostly Colloidal particles are negatively charged , this cause particles to remain in suspension

Destabilize the particle by neutralizing the –ve charge

Use some technique to increase particle size so that settlement under gravity can take place.

A chemical which is added to the water during coagulation process in order to destabilized the charge on particle and increase the size of particle to settle under gravity

G is defined as the slope of relative velocity between fluid elements OR rate of change of fluid velocity normal to the direction of velocity

G is a measure of INTENSITY OF MIXING G =(V1-V2)/O

Flocculation results from the velocity differences in the water which causes contact between the moving floc masses. Velocity differences also cause SHEARING STRESSES along planes in the water.

· Measure pH, turbidity and alkalinity of raw water.

· Take 4 or 5 beakers and add 1 litre of sample water to each.

· Add varying doses of coagulants (alum0 in different beakers and rapidly mix at 100-300 rpm for desire time (usually up to 30-60 sec).

· Reduce the speed of machine up to 15-20 rpm and gently mix for 15- 30 min. Stop stirring and settle the sample for at least 30 min.

· Analyze the samples, measure its pH, turbidity and alkalinity.

· If pH is in desired range the beaker corresponding to lowest turbidity give the optimum dose of coagulant (if pH is not in range 4-7 adjust pH and repeat the test).

Rapid DT=30-60 sec

G= 100-300 /sec

Flocculation DT=20-30 min

G=10-100 /sec

It is a process of separation of unstable and destabilized suspended particles from suspension by the force of gravity.

· Settling of coagulated and flocculated water prior to filtration

· Settling of coagulated and flocculated water in a softening plant

· Settling of treated waters in iron and manganese removal plant.

Four basic classifications, depending on the nature

of the solids present in suspension:

· Discrete Settling

· Flocculent Settling

· Zone Settling

· Compression Settling

· DRAG FORCE

· CD is the function of Reynold No., its value decreases as RN increases.

· RN< 1 Laminar flow

· RN > 1000 Turbulent flow

· The inlet or influent zone should provide a smooth transition from the influent flow and should distribute the flow uniformly across the inlet to the tank.

· The normal design includes baffles that gently spread the flow across the total inlet of the tank and prevent short circuiting in the tank.

· The baffle could include a wall across the inlet, perforated with

· holes across the width of the tank.

· Basin inlets should be designed to minimize high flow velocities near the bottom of the tank.

· If high flow velocities are allowed to enter the sludge zone, the sludge could be swept up and out of the tank.

· Sludge is removed for further treatment from the sludge zone

· by scraper or vacuum devices which move along the bottom

The settling zone is the largest portion of the sedimentation basin.

· This zone provides the calm area necessary for the suspended particles to settle.

· The sludge zone, located at the bottom of the tank, provides a storage area for the sludge before it is removed for additional treatment or disposal

· The basin outlet zone should provide a smooth transition from the Sedimentation zone to the outlet from the tank.

· This area of the tank also controls the depth of water in the basin.

· Weirs set at the end of the tank control the overflow rate and prevent the solids from rising to the weirs and leaving the tank before they settle out

· Min. # of tanks Two

· Water depth 3-5m (2m or deep tanks 6.5m also used)

· Hydraulic Detention time selected 8hrs

·

· Overflow rate 20~33m3/m2/day

· weir loading rate ≤250m3/m/day

· Sludge storage 20% extra volume required

Different methods of disinfection are:

Physical (Heat, Sunlight, UV rays)

Chemical (Chemical agents used are chlorine,

iodine, ozone and bromine etc.)

Mechanical (Coagulation + Sedimentation + Filtration)

· 98-99% bacteria are removed by using mechanical methods.

Temperature = 15 oC

Diameter of Tank

Volume = V = t * Q = 30 s x 0.43 m3 / s = 12.9 m3

We design the height to be twice the dia of tank

T = ( 2V / 3.14 )1/3

T = 2 * 12.9 / 3.14 = 2 m

Since, height is twice the diameter of tank

H = 2T = 4 m

Power Input for motor

P = ( 800 )2 * ( 12.9 ) ( 1.139 x 10-3 )

= 9.43 kW

Now for the optimum dia of Impeller , choosing from the table.

Best available Dia = 2.00 m

n = 0.37 rps / 22.2 rpm

So

Best Dia of tank is 3 m due to 0.5 m cover from each side

Discharge = 26.28 m3 / min

Detention Time = 20 min

Values of G as per calculations = G = 40 s-1

Temperature = 15 oC

Now the water will be divided into two basins,

So, Q = 26.28/ 2 = 13.14 m3 / min

Volume = V = t * Q = ( 20 min * 13.14 m3/min ) = 262.8 m3

Since, a Flocculation basin is divided into three compartments , so

V = 262.8 / 3 = 87.6 m3

By Taking Depth of 4m.

Total Surface area of a compartment = A= 87.6/4 = 21.9 m2

As per design criteria, the height above the paddle must be 1m.

4m – 1m = 3m

Minimum Length of Paddle = 3.0m + 1 m = 4m.

Width of compartment = 21.9/4 = 5.475 m

Clearance from sides = 2(0.3)+1 = 1.60 m

So total paddle length = 5.475 – 1.60 = 3.87 m

Each paddle has a width of = Total Width / 2 = 3.87 / 2 = 1.97 m

Now the Dimensions of paddle is 0.15 m x 1.97 m .

For the power:

By calculations, we found Power P = 160 W.

Distance of First paddle board from center = r1 = 0.425 m

Distance of Second paddle board from center = r2 = 0.925 m

Distance of Third paddle board from center = r3 = 1.425 m

And for rpms, n = 0.122 rps / 7 rpm

Motor Power since motor efficiency = 65 %

P = 160 / 0.65 = 246 W

Summary

Water inflow =10000000 MGD = 37854.11 m3 / Day

Overflow rate = 32. 5 m3 / d. m2

Area = As = 37854.11 / 32.5 = 1164 m2

· Select Number of Tanks

Minimum number of tanks shall be 2. For this value of flow, we use 6 tanks.

1164 m2 / 6 = 194 m2 / tank

Trial Width of Tank:

Assuming width is 4 m.

Length = 194 / 4 = 48.5 m

L / W = 48.5 m / 4 m = 12: 1

Acceptable

Selection of Trial Depth:

By calculations it is found to be. 3.6 m (Including cover of 0.6 m + 1 m sludge covering)

L / D = 48.5 / 2 m = 24.25

Accepted

Velocity

Vf = Q / A = 0.43 / 6 x 2 x 4 = 0.0089 m / s

Accepted

Reynolds Number:

Rh = 1

R = (0.0089) ( 1) / 1.140 x 10-6 = 5500

Accepted

For Launders:

At 1/3 of total Length

L = 48.5 / 3 = 16.17 m

Weir Loading rate

WL = 37854.11 / 6 x 3 x 16.17 x 2 = 65 m3 / d . m

Accepted

Summary

QDesign = 37854.11 m3 / day = 0.43 m3 / sec

Number of Tanks = 6

Width of Tanks = 4 m

Length of Each Tank = 48.5 m

L / W = 12 : 1

Depth including Sludge = 3.6 m

L / D = 24. 25

Vf = 0.0089 m /s

Reynolds Number = 5500

Launders = 3 evenly spaced

Length of Launder = 16.17 m

Weir Loading = 65 m3 / d . m

Sludge Collector = Chain and Flight

Specific Gravity of Sand = 2.63

Shape Factor = 0.82

Porosity = 0.45

Temperature = 15 oC

Depth of Sand = 0.73 m

Rate of Loading = 216 m3 / d.m2

For Headloss:

By calculations , we have

Cdf / d = 75025

Reynolds Number

R = Shape factor x dia of particle x rate of loading / viscosity

R = 0.82 x 0.002 x 0.0025 / 1.140 x 10-6

= 3.59

Headloss

HL = (1.067 Va2 D /Ø x g x e4 ) x CDf / dg

= 1.067 x 0.00252 x 0.73 x 75025 / 0.82 x 9.81 x 0.454

= 1.10 m

For Settling Velocity:

Vs = ( 4g ( ps – p )d / 3CDp )1/2

Ps = pw x Sp. G

= 1000 x 2.63 = 2630 kg / m3

Vs = ( 4 x 9.81 x 1630 x 0.002 / 3 x 0.5538 x 1000 ) ½ = 0.277 m /s

For Backwash Velocity

Ee = 0.255

Ee = ( Vb / Vs )0.2247R^0.1

R = Vs x d60 / viscosity = 206535

Vb = 0.046 m /s

Number of Filters

N = 0.0195 ( Q )0.5 = 4

Area of Filter

A = Q / No. of Filters x N x q

Q = Loading rate

N = Number of beds

A = 37854 / 4 x 216 = 87. 6 m2

Dimensions of Cells

Total 2 Cells in one filter

Width of one Cells = 2.6 m

L = 87.6 / 2.6 = 34 m

Detention Time = Volume of Tank / Total Inflow = 217.248 m3 / 0.43 m /s = 505 sec

T = 8.4 min

Summary

Number of Filters 4

Cells in one Unit 2

Length of One Cell 34 m

Width of Cell 2.6 m

Area of one Filter 87. 6 m2

Rate of Filtration 10 m / h

Backwash Velocity 0.046 m /s

Settling Velocity 0.277 m / s

Size of Sand 0.002 mm

Detention Time 8.4 min

Depth of Sand 0.73 m

Depth of water over sand 1.75m

Co efficient of Uniformity 2.8

Headloss 1.10 m

Length of run 20-60 days 12-72 hours Impurities Penetration On Upper Layers

Cost of construction High

Operational cost Medium

Method of cleaning Scrapping upper layer/

Backwashing

Pretreatment Generally Nil

Bacterial removal More effective

Depth of gravel 200-300 mm

The Chlorine demand of a secondary effluent is 8.7mg/L and desired residual is 0.8mg/L. If daily plant flow is 37854 m3/day.

Cl Applied =8.7 mg/L

Cl Residual =0.8 mg/L

ClDEMAND =7.9 mg/L

Daily Inflow = 37854.12 m3 / day

T10 = 100 min

Ct = 200 mg / L

Detention Time = 145 min

Volume of Chamber = 3811 m3

L / W = 40 : 1

Length = 128 m

Width = 3.2 m

Height = 9.6 m

Chemicals = Chlorine along with small

concentration of ozone

Chlorine Demand = 7.9 mg / L