Design of Water Treatment Plant Laboratory Work
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Design of Water Treatment Plant Laboratory Work
DESIGN OF WATER TREATMENT PLANT
NAME OF STUDENT:
ROLL NUMBER:
DEPARTMENT OF CIVIL ENGINEERING
24TH November, 2019
Contents Convectional Surface Water Treatment for Drinking Water 3 Coagulation ( Rapid Mixing) : 4 Flocculation (Slow Mixing) 4 Coagulant: 4 Detention time 5 Speed of Rotation 5 Power Required 5 Velocity Gradient: 5 FLOCCULATION: 5 SEDIMENTATION: 6 Inlet zone 7 SETTLING ZONE 8 SLUDGE ZONE 8 OUTLET ZONE 8 DISINFECTION: 9 CALCULATIONS: 10
Convectional Surface Water Treatment for Drinking Water
- 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.
Speed of Rotation 150 to 1500 rpm
Power Required 2 to 5 kW per m3/min.
Velocity Gradient: 300 – 500 sec-1
FLOCCULATION: It is low mixing process Bring destabilized particles in contact to form flocs. Circular tanks with paddles on vertical shaft. Rectangular tanks with paddles revolving on horizontal shaft. Area of paddle = 10-20% vertical X-sectional area of tank
- 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
Hydraulic Detention time Hours For Small Size Tank 2 hours For Medium Size Tank 5 hours For Large Size Tank 8 hours
- 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.
TOTAL INFLOW = 10000000 GALLONS / DAY = 26.28 m3/min = 0.43 m3 / sec
Temperature = 15 oC
Velocity Gradient: = 800 sec -1
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
Length of Basin 10.24 m Total Basins 2 Total Compartments in a Basin 3 Length of Compartment 4.87 m Width of Compartment 5.47 m Width of a Basin 17 m Power of Motor for one paddle 246 W Length of Paddle 3.87 m Width of Paddle 0.15 m No of Motors required 6 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
According to standard Design criteria of Single Media Rapid Sand Filter, we have the following data:
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
By running through various calculations, we have the following data
T10 = 100 min (By using graph)
T10 / to = 0.7 (from Calculations)
T10 / to = 0.7 = 100 / to = 0.7
V = t0 x Q = 37854.12 m3 / day x 145 min x 1 / 1440 min / day
According to Design Criteria, L > 40W , H = 3W
Using figure for t10 / t0 = 0.7. we find L / W = 40:1
V = L x W x H = 40W x W x 3W = 120 W3
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