Water treatment plant designer document
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Design a water treatment plant.
Link to formula embeded Spreadsheet: Water treatment plant designer
Overview of the spreadsheet:
Link to all autoCAD designs: Water treatment plant design details
Assumptions: Majority of Water to be treated comes from a dam(distance ~10 km)
Since we need to treat river water the basic outline of our filtration system will be like follow:
Current Population: 101005
Design period: 15 years
Design population: Assume geometric population growth with a growth rate of 7.5%
P n = P0(1+r/100)n
P0 = 101005
R = 7.5%
n = 15/10 = 1.5
Pn = 112579
Net water demand (liters/day) = design population x per capita demand
= 112579 x 135 = 15198165 liters/day
Assume 20% water is lost in various treatment processes,
now net water required a day = 1.2 x 15198165 liters/day = 18237798 liters/day
= 18.237798 MLD = 0.211085625 m 3/s
Now summarizing the water treatment steps we need to follow:
Since the influent water is coming from a dam, It will have high organic content. Aeration becomes an important step as bacteria used in water treatment and stabilization are given oxygen by aeration. The bacteria require oxygen for biodegradation to take place. Bacteria in the wastewater use the given oxygen to break down organic stuff that contains carbon to produce carbon dioxide and water. Bacteria cannot biodegrade the incoming organic materials in desired time if there is insufficient oxygen present. Degradation must take place under septic conditions, which are sluggish, unpleasant, and result in incomplete conversions of pollutants, in the absence of dissolved oxygen. Also it will help remove dissolved CO2 and H2S like gasses.
Q = 759.90825 m 3/h
Type of aerator: Diffused aerator
A diffused aerator is highly energy efficient since only air needs to be pumped into the tanks from the diffusers placed at the bottom of tanks. The ability to save energy of 30% to 70% over surface aerators depending on the type of surface aerator used is a big benefit of fine bubble diffusers.
Dimensions of aerator tanks = (l x b x h) = (10 * 8* 5)m3
Plan area of aerator tank: 80 m2
Volume of aerator tank = 400 m3
No of tanks required = 2
No. of diffusers per tank = 80 (1 per 1 m2 of plan area)
Total diffusers required = 2 x 80 = 160
Radius of a diffuser = 0.2 m
Detailed distribution of diffusers in the aeration tank is given in the following diagram:
Detailed Auto-CAD plan of aerator tank showing distribution of diffusers can be found here: ( https://drive.google.com/drive/folders/1u1UNCUCInDnFk1iY33DjOfc5qn0hssFo?usp=sharing)
(All units in ‘cm’)
Diffuse type aerator set-up
Coagulation:
Coagulation is the chemical water treatment process used to remove solids from water, by manipulating electrostatic charges of particles suspended in water. This process introduces small, highly charged molecules into water to destabilize the charges on particles, colloids, or oily materials in suspension. The influent water will have colloidal content as clayey particles. We will use Alum as coagulant and limestone to provide optimum pH for coagulation to occur.
Rapid mixing unit:
Q = 759.90825 m3/h
Assume Mixing time to be 2 min (standard practice)
Volume of mixing basin = (759.90825 x 2)/60 m3 = 25.330275 m3
Generally rapid mixing unit is cubical in shape:
Assume the Rapid mix unit be with side length ‘a’ : a^3 = 25.330275 m3, a = 2.936837 m
Assume a to be nearest integer for ease of construction: a = 3 m
Volume of rapid mix tank = 27 m3
At room temperature T = 20 deg C, Dynamic viscosity will be µ = 1.0016 * 10-3 N.s/m2
Assume G = 800/s
Power required:
P = G2µV = 8002*1.0016 * 10-3*27 = 17.307648 Kwatt
Monthly Alum dosage requirements:
Chemical formula:
Optimal dosage: 20 mg/liter as A study has showed that the removal efficiencies for turbidity and optimum alum dosage were (93% at 20 mg/l, 92% at 20 mg/l, 85% at 30 mg/l, 88% at 30 mg/l and 89.3% at 30 mg/l). And hence we will use 20 mg/l for maximum turbidity removal efficiency.
Q = 759.90825 m3/h = 547133.94 m3/month
Optimal Alum dose = 20 mg/l = 0.02 kg/m3
Alum required = 0.02 x 547133.94 = 10942.6788 kg/month ~ 11,000 kg/month
The pH during coagulation should be maintained in the range of 7-8 for proper flocc formation. This can be achieved by using lime (Ca(OH)2).
The coagulation is generally required only during floods. However to treat water with high efficiency we can perform coagulation year round.
Rapid mix unit design:
Diagram on next page
Sedimentation refers to physical separation of solid concentration from water. The driving force of sedimentation is gravitational force.
Design:
Shape: rectangular sedimentation tank
Q = 759.90825 m3/h
We will divide this flow into 4 tanks. So the each tank will have of 189.98 m3/h
Assumptions:
1) Maximum size of the particle to be removed d = 0.02 mm
2) Expected removal efficiency = 75%
3) Specific gravity G = 2.65 4) Assumed performance of tank = good = n = 1⁄4 5) Dynamic viscosity µ = 1.0016 * 10-3 N.s/m2 (T = 20°C)
6) Kinematic viscosity = 1.00 x 10-6 m2 /s
1) Vertical settling velocity
Settling Velocity Vs = (g x (ρp- ρw) x d2) / (18 x μ)
= (9.81 x 1000 x (2.65 - 1) x (0.02 x 10-3)2) / (18 x 1.0016 x 10-3)
= 3.591253994 x 10 -4 m/s
Vs = 3.59 x 10-4 m/s
Reynolds’s number Re = (ρ x Vs x d) / μ
= (1000 x 3.59 x 10-4 x 0.02 x 10-3) / 1.0016 x10-3
= 7.17 x 10-3 < 1
Hence stoke’s law is applicable and thus computed settling velocity is true.
2)Surface overflow rate
For ideal settling basin and complete removal of particles, we will equate settling velocity to theoretical surface overflow rate.
V0= V S = 3.59 x 10-4x3600x24 = 20.68 m/day
Y / Y0 = 1 − (1 + (n x (V0/(Q/A))))-1/n
As removal efficiency is 75% 🡺 Y / Y0 = 0.75
n = ¼ (for good performance of tank)
🡺 0.75 = 1 − (1 + (0.25 x (V0/(Q/A))))-1/0.75
🡺 V0/(Q/A) = 1.657
🡺 Q/A = V0/ 2.6 = 12.48 m/day
3) Dimensions of tank
Surface area of tank A = Q/(Q/A)= (189.9770625 x 24) / 12.48
A = 365.2298904 m2
Assuming length to width ratio as 3
A = Length x width = 3w x w
🡺 w = √(365.2298904/3) = 11.03373449m
w = 11.03373449 m
🡺 L = 3w = 33.10120347 m
L = 33.10120347 m
Assuming a detention period of 4 hours
🡺 Depth of tank = Q x t / A = 189.9770625 x 4 / (11.03 x 33.10 )
D = 2.08 m
4) Check against resuspension of deposited particles
Velocity that can initiate resuspension of deposited
Particles or scour velocity Vd = √((8 x β / f’) x g x (G -1) x d)
β = 0.04 for uni granular sand f’ = Darcy Weisbach friction factor = 0.03
Vd = √((8 x 0.04 / 0.03) x 9.81 x (2.65 -1) x (0.02 x 10-3)
Vd = 5.8 x 10-2m/s
Horizontal velocity Vh = Q/ (w x D)
= ( 189.9770625 /(60 x 60)) / (11.03373449 x 2.080629954)
Vh = 0.229 x 10-2 m/s < Vd
Hence O.K.
FLOCCULATION
Flocculation is the process of slow mixing in a flocculator so that the particles of the precipitate collide and form a large and dense cluster that will readily settle.
DESIGN
Flocculation Basin Dimensions
Let us assume two trains with three compartments each.
Assumption GT = 4.5 x 104
For first compartment G1 = 50/s
For second compartment G2 = 40/s
For third compartment G3 = 30/s
For fourth compartment G4 = 20/s
For fifth compartment G5 = 10/s
Average compartment G = (50+40+30+20+10)/5 = 30/s
Average Detention Time T = GT/G = (4.5 x 104)/30 = 25 min
T = 25 min
Q = 759.90825 m3/h = 18237.798 m3/day = 12.6651375 m3/min
=> Volume V = 12.6651375 x 25 = 316.6284375 m3 🡺V = 316.63 m3
Let the depth of the tank d = 5 m
Area A = V / d = 316.63 / 5 = 63.3256875 m2 🡺 A = 63.3256875 m2
Total number of compartments = 2 x 5 = 10
Area of one compartment = 63.3256875 / 10 = 6.33 m2
Assume the length of one stage as 3 m
Width of each stage will be
Width = 6.33 /3 = 2.11m
TANK DIMENSIONS
TANK CONFIGURATIONS
Assuming a clear cover of 0.6 m between two wheels of compartment
Power in each Compartment
For first compartment G1 = 50/s
For second compartment G2 = 40/s
For third compartment G3 = 30/s
For fourth compartment G3 = 20/s
For fifth compartment G3 = 10/s
Volume = 316.6284375 m3
Volume in each stage Vs = 316.6284375 / 5 = 63.3256875 m3 = 63.33 m3
At T = 20°C, Dynamic Viscosity μ = 1.0016 x 10-3 N.s/m2
We know, Power P = G2μV
P1 = G12μV = 502x 1.0016 x 10-3 x 63.33= 158.5675215 Watt
P1 = 158.5675215 Watt
P2 = G22μV = 402x 1.0016 x 10-3x 63.33 = 101.4832138 Watt
P2 = 101.4832138 Watt
P3 = G32μV = 302x 1.0016 x 10-3x 63.33 = 57.08430774 Watt
P3 = 57.08430774 Watt
P4 = G2μV = 20 2x 1.0016 x 10-3x 63.33 = 25.37080344 Watt
P4 = 25.37080344 Watt
P5 = G52μV = 102x 1.0016 x 10-3x 63.33 = 6.34270086 Watt
P5 = 6.34270086 Watt
Paddle Design
P = ( Cd x Ap x ρ x Vp3) / 2
For first compartment:
P1 = 158.5675215 Watt
Cd = 1.8 for flat blades
ρ = 1000 kg/m3
Assume actual speed of paddle Vp’ = 0.67 m/s
Vp = Vp’ x 0.75 =0.5 m/s
Assuming width of paddle board as w
Ap = 2 wheels per compartment x 12 boards per wheel x length of paddle board x width of paddle board
= 2 x 12 x L x w = 2 x 12 x 1.30 x w = 31.2w m2
Ap = 31.2 m2
Ap = (P1 x 2) / (Cd x ρ x Vp3)
= (158.5675215 x 2) / (1.8 x 1000 x 0.53) = 1.388556418 m2
🡺 w = 1.39/ 31.2 = 0.066 rev/s w = 4.00 rev/min
For Second, third, fourth, fifth compartment
Calculations had been given in the attached spreadsheet.
CLARIFIER
Clarifier tank is optional as we have already provided a sedimentation tank, but if it is required then sedimentation tank designs can be reused.
Details:
FILTRATION :
Filtration, the process in which solid particles in a liquid or gaseous fluid are removed by the use of a filter medium that permits the fluid to pass through but retains the solid particles.
DESIGN :
Rapid sand Filter:
Q = 18.237798 MLD = 18.24 x 106 liter/day = 759.90825 m3/hr
We will design 4 filter tanks. Same design for all 4 tanks
So, Q for each tank = 189.9770625 m3/hr
Assumptions:
1) Dynamic Viscosity μ = 1.0016 x 10-3 N.s/m2 (T = 20°C)
2) Filter Medium - Uniform Sand of depth 0.67 m
3) Filtering Velocity Va = 5 m/hr
4) Depth of particles dp= 0.6mm
5) Specific Gravity G = 2.65
6) Porosity e = 0.4
7) Shape factor ψ = 0.85
8) backwash water required for backwash = 5 %
9) Time for washing the filter = 1 hr
1) Area of filter:
Design flow rate = (Q + 5% Q) m3/hr = 199.4759156 m3/hr
Design flow rate/day = (Q + 5% Q) x 24
It is assumed that 1 h is lost every day in washing the filter
Q = (Q + 5% Q) x 24/23 m3/hr = 208.1487815 m3/hr
Q = Va x A
=> A = Q / Va = 208.1487815/5 = 41.6297563 m2
Assuming L/B = 1.25 (of a rectangular filter)
Af = L x B = 1.25B2 = 41.6297563 m2
🡺 B = 5.7709449 m (5.8m) 🡺 L = 7.213681126 m (7.25 m)
2) Frictional Head Loss :
Hf = (f’ x L x (1-e) x Va2) / (e3x g x dp) f’ = 1.75 + (150 x (1-e)) / Re Re = (ψ x ρ x Vs x dp) / μ (Vs = Va = 1.4 x 10-3m/s) Re = (0.85 x 1000 x 1.4 x 10-3 x 0.6 x 10-3) / (1.0016 x 10-3) = 0.70 < 1
=> f’ = 1.75 + (150 x (1-0.4)) /0.70 = 129.0121176
=> Hf = (129x0.67x(1-0.4)x(1.4x10 -3)2) /(0.43 x 9.81 x 0.6 x 10-3)
🡺Hf = 265.57 mm
Minimum Thickness for sand bed given by Hudson Formula
l = Va x dp3 x Hf/Bi x 29323 = 5 x2 x 0.63 x 0.265 / 4 x 10-4 x 29323
🡺 l = 0.058m < 0.67m hence ok
3) Estimation of gravel and size gradation
Assume a size gradation of 2 mm at top to 40 mm at the bottom. The depth of gravel layer can be determined by using empirical formula
l = 2.54 k log d
k =constant varies from 10 to 14
Size (mm ) 2 5 10 20 40
Depth (cm) 9.2 21.3 30.5 40 49
Increment (cm) 9.2 12.1 9.2 9.5 9
4) Design of under drainage system:
Plain area of filter = 7.213681126 m x 5.7709449 m
Assumption: Total area of the perforations in all laterals should be 0.5% of the filter area
Total area of perforations =0.5% x 41.6297563 m2 = 2081.487815 cm2
Total cross-sectional area of laterals = 3 x Total area of perforations = 6244.463446 cm2
Area of manifold ( A = πD2/ 4)= 2 x area of laterals
Diameter of central manifold = 126.1327119 cm = 127 cm
Number of laterals ( assume the distance between the laterals = 15 cm) = 7.21 m x 100 (cm/m) / 15 cm= 49 ( for two sides = 98)
So, cross sectional area of one lateral = 6244.463446/ 98 =31.85950738 cm2
Diameter of lateral = ( A =πD2/ 4)= 6.370663993 cm
Length of lateral = 1⁄2 (width of filter – diameter of manifold)
= 1⁄2 (5.7709449 – 1.27) = 2.25 m
Check Length of each lateral / Diameter of lateral < 60
🡺(2.25/0.0623) = 36 < 60
Hence ok
4) Design of back water trough:
Let us assume that the rate of washing of the filter be 60 cm rise/ minute or 0.60 m/minute
The plan area of the filter = 41.6297563 m2
Wash water discharge = 0.60m/min x 60 min/h x 41.6297563
= 1498.671227 m3/hr = 0.416297563 m3/sec
Assume a spacing of 1.6 m for wash water trough which will run parallel
No. of trough = 5.77 / 1.6 = 4 Discharge per unit trough = 0.1040743908 m3/sec
The dimension of a concrete V- bottom trough can be designed by using the empirical formula Q = 1.376 b h3/2
Q = discharge in m3/s (0.1040743908 m3 /sec)
b = width of trough in m
h = water depth in the trough in m
Assume b = 0.4 m; h = 0.33 m
Assume freeboard of 0.1 m, total depth = 0.4 m
Total depth of filter = Depth for under drains + gravel + sand + water depth + free board
= 0.9 + 0.5 + 0.6 + 1.2 + 0.4 = 3.6 m
However instead of going for sand as filter media, we may also opt for anthracite as Sand tends to have a higher uniformity coefficient and therefore, can clog more often when used in a singular media filter configuration. Sand particles tend to be more spherical in shape while anthracite is sharp and angular. Studies have shown that more effective back washes occur with more angular structures.
6) DISINFECTION :
Disinfection describes a process that eliminates many or all pathogenic microorganisms, except bacterial spores, on inanimate objects (Tables 1 and 2). In health-care settings, objects usually are disinfected by liquid chemicals or wet pasteurization.
Chlorination, ozone, ultraviolet light, and chloramines are primary methods for disinfection.
We will go with chlorine disinfection.
DESIGN :
Assuming Chlorine Dosage = 0.3 mg/L = 0.3 x 10-3 kg/m3
=> C x Q = 0.3 X 10-3 x 18237.798 = 5.4713394 kg/day
Consumed chlorine = 5.5 kg/day
The end
