Abstract: Water treatment is a process of changing the quality of for potable drinking water & domestic uses. The water treatment plant is designed for the city of Mithi district Tharparkar. The source of water is Naukot canal near Naukot city and the water is supplied to the Mithi city through a big pipeline. The treated water will be supplied to the residents of Mithi city and the treatment plant is designed for 30 years. The treatment units of water treatment plant (WTP) were designed with step-by-step calculations and according to the quality of the water. The future population of the Mithi city from 2025 to 2055 was projected as 449,677.32 and the average discharge for the population of 30 years was estimated as 210,448.8 m3/d. The detailed calculations and drawings were displayed, the results of each unit of WTP were tabulated. It is concluded that it is possible to use this work as a source for other WTP units to build. The removal performance of the WTP units was significantly affected by a variety of factors, such as the age of WTP, repair, economic and political circumstances, technical problems, and water demand.
Keywords: Water treatment plant; design calculation; treatment units; water demand

I.    INTRODUCTION
Water, undubiously is a basic human need. Providing safe and adequate quantities of the same for all rural and urban communities is perhaps one of the most important undertakings, for the public works Dept. Indeed, the well-planned water supply scheme is a prime and vital element of a country's social infrastructures as on this peg hangs the health and wellbeing of its people. Pakistan 2020 population is estimated at 220,892,340 people with growth rate of 2.0% at midyear according to UN data. Pakistan population is equivalent to 2.83% of the total world population and the growth rate is said 2.00% (Pakistan Population-Worldometer). The provisional results of the 2017 Census were presented to the Council of Common Interests on 25 August 2017. According to the results, the total population in Pakistan was 207,906,209, representing a 57% increase in 19 years and population of Mithi city was 219,071 (census 2017). This goes on to say that a very large demand of water supply; for Domestic, Industrial, Firefighting, Public uses, etc.; will have to be in accordance with the rising population. Hence, identification of sources of water supply, there conservation and optimum utilization is of paramount importance.
Tharparkar is one of the most water scarce regions of Pakistan with total population of 1649661 (Cencus-2017). Sometimes, the situation gets worse due to lack of rainfall in the monsoon season. The people of Tharparkar are depending on rainfall because most of the population of Tharparkar is poor they cannot afford the huge cost of water that’s why they drink untreated water. Due protracted low crop production, the food insecurity and malnutrition issue has further worsened as families are forced to sell the goods that they otherwise would have consumed themselves.
Access to clean water has also severely been limited, causing water-borne diseases and compromising the health of the residents. (Pakistan Meteorological Department-PMD). Mithi city is the capital city of Tharparkar, and it does not have any water resources of its own. Water is provided by pipelines from Naukot City, 49 km away. Having its own centralized water treatment plant is the foremost facility that Mithi, Tharparkar need now. Though the RO plant is also there but there are many demerits of RO plants like most of rejected or toxic water discharge on open land near the plant. Many people migrate from other villages towards the Mithi city oftenly.

A.    Objective
The objective of design of water treatment plant is to treat the water and supply it to each and every house, commercial, public places etc. for Mithi. This design of treatment plant is proposed to treat the water up to the desired levels. This water treatment Plant design is proposed for up to 30 years excluding construction and settlement time of the treatment plant.
The plant construction time may be 3 to 4 years, starting from 2021 to 2024 and it will be running from 2025 to 2055and it will provide the treated water to the people of Mithi city. This plant is designed to treat the water and supply it for present and future population as well. Although, Mithi city has the source of water through pipeline from Naukot city and this plant will treat the water. The plant is designed according to the characteristics of water, present and future water demand, present and future population, design period and design flow, and design criteria, design calculation and drawing of each component. This plant is consisting coagulation, flocculation, sedimentation, filtration, disinfection and storage. The water is supplied to the plant through the influent pipe to the primary disinfection, coagulation, flocculation tank, sedimentation tank, filtration system, secondary disinfection, contact tank, clear well and then effluent.

B.    Study Area
The total area of Mithi city is 1,535 km². The source of water for Mithi is ground water and also a pipeline from Naukot to Mithi. The 2020 population of Mithi city is over 250,000, according to National Statistics of GMO.
The design of water treatment plant is proposed at Mithi city, which is 560 meters from RO Plant and 500 meters from Government Polytechnic Institute. The total Perimeter and area of the plant is considered as 0.67 km and 0.03 sq. km. The latitude and longitude of Water Treatment Plant Site are 24°45'30.49"N and 69°47'23.62"E as shown in Figure 1.
Image

Figure 1: Water Treatment Plant Site.

C.    Population Forecasting
There are different methods of population forecasting like arithmetic increase method, geometric increase method, incremental increase method etc. Here, we use the simplified method for population forecasts.
Image
Where
Pn= future population
Po = initial population
R = probable rate of population increases per year
n= number of years considered
Using the formula and by considering population of 2017 census, we will find the population of 2025 and considering 2.10% yearly change (worldometer)
Pn = Po (1 + R) n
P2025 =219,071* (1+2.10%)8
P2025 =258,696.663        (considered as present population)
Similarly using the formula and by considering the above data of 2025, we will find the population of 2055 and also considering 1.86% yearly change (worldometer)
Pn = Po (1 + R) n
P2055 =258,696.663 * (1+1.86%)30
P2055 = 449,677.32         (considered as future population)

D.    Design Period
The plant is considered for 30 years and it will run in 2025 after its completion, because 3 to 4 year will be taking by plant to be constructed, so the construction of the plant will start from 2021 to 2024 that is why I did not consider this construction period with running time of the plant. That is the reason that I considered the population of 2025 as a current population which was projected 258,696.663 and the future population for 30 years was projected 449,677.32.

E.    Types of Demands
There are so many factors involved in demand of water; it is not possible to determine the actual demand. Based on certain empirical formula and thumb rules are employed in determining water demand which may be nearly to the actual demand. Following are the various types of water demand of a city or town.

F.    Domestic Water Demand
The total domestic water consumption may amount to 55 to 60 % of the total water demand. This includes the water required in the houses for cooking, washing, bathing, drinking, gardening and sanitary purpose etc. The domestic demand depends upon the living conditions of consumer such as habits, social status, climate condition etc. As per IS-1172-1993 water requirement for domestic purposes is about 135 liters/day/capita under normal conditions. Table-1 shows the details of water requirement for domestic purpose.
Table-1 Average domestic water consumption
Use of Water    Consumption in
liters/day/per person
Bathing    55
Washing clothes    20
Drinking    5
Cooking    5
Washing of utensils    10
Cleaning of houses    10
Flushing of W.C.    30
Total    135

G.    Industrial and Commercial Water Demand
This consumption includes water used in factories, hotels, offices, hospital etc. The water requirements of industrial needs of a city are generally taken as 50 liter/day/person. This demand depends upon the nature of the city and types of industries. Generally, 20 to 25% of the total water demand may be allowed for industrial water demand. The approximate quantity of water required for industries other than residences as per IS-1172-1993 is given in Table-2.
Table-2 Water supply requirements for building other than residence
Use of Water    Consumption in   liters/day/per person
Industrial    50
Commercial     20
Total    70

H.    Demand for Public Use
Public demand includes the quantity of water required for public utility purpose such as watering of public parks, gardening, sprinkling on roads, use in public fountains etc. As per IS-1172-1993 water requirement for public use is about 20 lpcd.
Table-3 water requirements for Public purpose
Use of Water    Consumption in
liters/day/per person
Public use    20

I.    Compensate Losses Demand
All the water which goes in the distribution pipe does not reach the consumers. Some water is wasted in the pipeline due to leakage, defective pipe joints, faulty valves and fittings. In some cases, quantity of water is lost due to unauthorized and illegal connections. While estimating the total quantity of water some allowances for these losses and wastages should be done. As per IS-1172-1993 water losses/ leakages are about 35 lpcd.
Table- 4 Total averaged water consumption per person per day, month, week, hour and year
Purpose    Consumption in
liters/day/per person
Domestic water demand    135
Industrial    50
Commercial    20
Public use    20
Losses/ leakages    35
Total     260
Consumption in
liters/month/per person    260 * 30 = 7800
Consumption in
liters/week/per person    260 * 7 = 1820
Consumption in
liters/hour/per person    260 / 24 = 10.833
Consumption in
liters/year/per person    260 * 365 = 94900
This is the total water requirement for different purposes, as per IS-1172-1993. This demand is for per person per day, month, week, hour and year. If we want to know the requirement for the whole present and future population then multiply these above values with present and future population.

J.    Present Population Demand
PPD = (Total consumption per person per day) * (Total present population)
         = (260 l/c/d) * (258,696.663capita)
         = 67,261,132.38 l/d    or
         = 67,261.132 m3/d

K.    Future Population Demand
FPD = (Total consumption per person per day) * (Total future population)
        = (260 l/c/d) * (449,677.32 capita)
        = 116916103 l/d         or
        = 116916 m3/d

L.    Maximum Daily Demand
MDD = 1.8 * average daily demand        
MDD = 1.8 * 260 l/c/d                
          = 468 l/c/d or 0.468 m3/c/day
The maximum daily demand factor is 1.8 times the average demand. The maximum daily flow rate is:
1)    The maximum daily flow rate for present community is:
Qm = 1.8 * (67,261.132 m3/d)
= 121,070.0376 m3/d

2)    Similarly, the maximum daily flow rate for future community is:
Qm = 1.8 * (116916 m3/d)
      = 210,448.8 m3/d

M.    Design of Units
The aim of water treatment is to produce and maintain water that is hygienically safe, aesthetically attractive and palatable, in an economical manner. Albeit the treatment of water would achieve the desired quality, the evaluation of its quality should not be confined to the end of the treatment facilities but should be extended to the point of consumer's use. The method of treatment to be employed depends on the characteristics of the raw water and the desired standards of water quality.          

1)    Initial Storage Tank
The initial storage structure is used to store the water for required demand and to avoid the seasonal variation of water supply.

a)    Calculation
The required capacity of the tank is 210,448.8 m3/day.
Let’s find the dimensions for the tank by assuming the height of the tank is 50 meters.
Volume = 210,448.8 m3
V = Area × Height
Area = Volume / Height
Area = 210,448.8 m3/ 50
Area = 4208.976 m2
The tank is rectangular so suppose L= 2B
A = L×B
A = 2B×B
A = 2B2
B2 = A/2
B2 = 4208.976 m2/2
B2 = 2104.488 m2
Taking square root both sides, we get
B = 45.87 m
L = 2B
L = 2×45.87 m
L = 91.75 m
 
Image
Summary
Length     Breath    Height     Area    Volume
91.75 m    45.87 m    50 m    4208.976 m2    210,448.8 m3

⦁    To know the dia, area and velocity for pipe from source to reservoir the length L = 280 meters and assume time T = 3 minutes
Q = V/T
Q = 210,448.8 m3/day
Q = V/T = AL/T
Q = [(Imaged2/4) × L]/T
210,448.8 m3/day × 1day/24 hrs × 1hr/60 min = [(3.14 × d2/4) × 280 m]/3
d = 1.41 m
Area = Imaged2/4 = 3.14 (1.41)2/4
A = 1.57 m2
Q = A × V
210,448.8 m3/day × 1day/24 hrs × 1hr/60 min × 1min/60 sec = 1.71 × V
V = 1.551 m/sec
⦁    To know the dia, area and velocity for pipe from source to reservoir the length L = 280 meters and assume time T = 3 minutes
Q = V/T
Q = 210,448.8 m3/day
Q = V/T = AL/T
Q = [(Imaged2/4) × L]/T
210,448.8 m3/day × 1day/24 hrs × 1hr/60 min = [(3.14 × d2/4) × 50 m]/1.2 min
d = 2.1 m
Area = Imaged2/4 = 3.14 (1.41)2/4
A = 3.46 m2
Q = A × V
210,448.8 m3/day × 1day/24 hrs × 1hr/60 min × 1min/60 sec = 3.46 × V
V = 0.7 m/sec
Table 5. Water characteristics
Parameters    Concentration     WHO standards
pH    7.5    7.5 to 8.5
TDS    900 mg/l    500 to 1000 ppm
Alkanlinity    150 mg/l    20-200 mg/l
Ca    146 mg/l    <60 mg/l
Mg    42 mg/l    < 30 mg/l
Total hardness    104 mg/l    60 -120 mg/l
Na    35 mg/l    sodium should not exceed 30 mg/L
K    9 mg/l    -
F    0.7 mg/l    ± 10% of nominal level (current 0.5 mg/l)
Fe    3.1 mg/l    Not exceeding 0.1 mg/l
TOC    2.0 mg/l    25 ppm
Bromide    not detected    -
Turbidity    20 NTU    Not exceeding 1.5 NTU
Giardia cysts    <1/100 L    4-log (99.99%) reduction or inactivation
Virus    <1/100 L    4-log (99.99%) reduction or inactivation
Cryptosporidium oocysts    1.1–2.0/L    4-log (99.99%) reduction or inactivation

2)    Treatment Units
a)    Primary Disinfection: Considering the characteristics of water given in table 5 and taking help from the example 13-5 from water and wastewater engineering book by Machenzie L. Davis, the primary disinfectant is selected. Ozonator is used as primary disinfectant.
b)    Ozonator Design: Using the example 13-8 from (Water and wastewater engineering, Machenzie book) the ozone disinfection system is designed.
⦁    For drinking water maximum ozone dose concentration is less than 10 mg・min/L
⦁    The pH and temperature are selected for 7.5 and 250C
⦁    Assume t10/t0= 0.65
⦁    Design flow rate = 210,448.8 m3/day
⦁    Assume the time for water to reach the most distant customer at the minimum demand flow rate is 52 hours.
⦁    From bench-scale test data, the second order rate constant was determined to be 3.5 L/mol・s. Assume a transferred dose of 2.0 mg/L.
⦁    Using the EPA’s Ct tables in Appendix D (Water and wastewater engineering, Machenzie book), the Ct to achieve the required log inactivation for each microorganism at a temperature of 250C is
Giardia cysts    0.48 mg.min/L
Viruses    4 log inactivation will occur at the Ct of 0.3 mg・min/L
Cryptosporidium oocysts    4.9 mg・min/L

⦁    The required hydraulic residence time with the bench-scale test dose of 2.0 mg/L transferred dose, the required t 10 is
Ct/C = 4.9 (mg・min/L)/ 2.0 mg・min/L = 2.45 minutes
⦁    Assume the ozone concentration remains constant throughout the contact chamber, the theoretical hydraulic detention time with the assumed t10 / t0 of 0.65 is
t10/t0 = 2.45/t0= 0.65
t0 = 2.45 min /0.65 = 3.769 minutes
⦁    Because ozone generating capacity is expensive and the energy consumption is high, Rakness (2005) recommends an optimized design that takes the decay into account by numerical integration. To perform the numerical integration, the concentration leaving each chamber must be estimated. This requires the decay rate constant in compatible units for Ct calculation. The conversion is
(3.5 L/mol・s) / 48000 mg/mole of ozone = 7.29×10-5 L/mg・s or 4.38×10-3 L/mg・min
⦁    An iterative solution is required. Based on Rakness (2005), assume 10 cells. The spreadsheet solution is shown below. The first trial with 2.0 mg/L achieved the desired Ct of 4.802 mg・min/L. So, no need of any extra cell.
Table 6.
Cell no.    Concentration at cell
influent,
mg/L    HDT,
min    Residual at cell
effluent,
mg/L    t10,
min    Ct,
mg-min/L
1    2.0    0.3769    1.993    0.245    0.488
2    1.993    0.3769    1.986    0.245    0.486
3    1.986    0.3769    1.979    0.245    0.484
4    1.979    0.3769    1.972    0.245    0.483
5    1.972    0.3769    1.965    0.245    0.481
6    1.965    0.3769    1.958    0.245    0.479
7    1.958    0.3769    1.951    0.245    0.478
8    1.951    0.3769    1.944    0.245    0.476
9    1.944    0.3769    1.937    0.245    0.474
10    1.937    0.3769    1.931    0.245    0.473
TOTAL    4.802
1.993 calculated concentration of the influent dose after 0.3769 min using this condorder decay (example 13-3 in water and wastewater engineering by Machenzie book).
 
C = 2.0 mg/L / [1+ (4.38×10-3 L/mg・min × 0.3769) × (2.0 mg/L] = 1.993
3)    Design the Contact Chamber
The volume of the chamber is calculated from the hydraulic detention time and the design flow rate
V =t0×Q= (0.3769) × (210,448.8 m3/d) (1/1440 min/day) = 55.08 m3
Using the Henry and Freeman (1996) optimum ratios, a depth of 6.0 m and an assumed H = 4 L:
L = H/4 = 6/4 = 1.5 meter/cell
V = (H) (Length/cell) (number of cells) (width of cell)
Width of cell = 55.08 m3/ (6 m) (1.5 m/cell) (10 cells)
Width of cell = 0.612 m

II.    COAGULATION
Coagulation describes the effect produced by the addition of a chemical to a colloidal dispersion, resulting in particle destabilization. Operationally, this is achieved by the addition of appropriate chemical and rapid intense mixing for obtaining uniform dispersion of the chemical.

A.    Design Criteria for Alum Dose
Alum required in particular season is given below:
Monsoon = 50 mg/L
Winter = 20 mg/L
Summer = 5 mg/L

B.    Alum Required
The dose of alum can be found from jar test but in our case, we assume 20 mg/L.
Per day alum required = (20 mg/L) × (1g/1000mg) × (1kg/1000g) × 210448.8 m3/d× 1000L/m3
 = 4208.976 kg/d
Consider the amount of alum in a bag is 50kg. So,
Number of bags/day = (4208.976 kg/d)/50kg = 84.18 bags/day

C.    Design of Mechanical Rapid Mix Unit
1)    Flash Mixer: Rapid mixing is and operation by which the coagulant is rapidly and uniformly dispersed throughout the volume of water to create a more or less homogeneous single or multiphase system. This helps in the formation of micro floes and results in proper utilization of chemical coagulant preventing localization of connection and premature formation of hydroxides which lead to less effective utilization of the coagulant. The chemical coagulant is normally introduced at some point of high turbulence in the water. The intensity of mixing is dependent upon the temporal mean velocity gradient ‘G’. This is defined as the rate of change of velocity per unit distance normal to a section. The turbulence and resultant intensity of mixing is based on the rate of power input to the water. Flash mixture is one of the most popular methods in which the chemicals are dispersed. They are mixed by the impeller rotating at high speeds.

D.    Design Criteria for Mechanical Rapid Mix Unit (Radial Impeller)
Detention time = 1 to 10 sec.
Gradient velocity ‘G’ = 600 to 1000 s-1
Ratio of impeller diameter to tank diameter (for radial) = D/T = 0.14-0.5
Ratio of water depth to tank diameter = H/D = 2-4
Efficiency of transfer of motor power to waterpower = 80% or 0.8 for single impeller
Volume of tank = < 8 m3

E.    Assume
Impeller diameter =D = 0.7 meters (depending upon manufacturer)
Power number or Impeller constant (Np) = 5.7 (depending upon manufacturer)
Temperature = 250C
Detention time = 3 seconds
Dynamic viscosity = µ = 0.890 × 10-3 pa•s
G = 800 s-1
H/T = 4

F.    Calculation
Determine the volume of the rapid mix basin
V = Q × t = (2.435 m3/sec) × 3sec = 7.305 m3 < 8 m3   OK
Using the radial impeller guidance, assume H / T =4.0, that is H =4 T. For a round mixing tank
V = [Image (T)2/4] × 4T
T = [(4 × 7.305m3) / (4 ×3.14)]1/3
T = 1.32 meters and H = 4T = 4 × 1.32 = 5.28 meters
The required input waterpower =
P = G2µ Volume=(800 s-1)2 × 0.890 × 10-3 × 7.305m3
P = 569.6 or 570 W
The efficiency of transfer of motor power to waterpower is about 80%, the motor power should be
Motor power = waterpower/0.8 = 570 W/0.8 = 712.5 W
Find geometric ratio = D/T = 0.7/1.32 = 0.53
Although the 0.7 m diameter impeller has a D / T slightly larger than the allowable range, it is satisfactory and, therefore, is selected.
Calculate rotational speed
n = [P/(Np×(D)5 × Image]1/3
n = [712.5 W/ (5.7) (0.7)5(1000kg/m3)]1/3
n = 0.743 rps or 44.62 rpm

Image

G.    Comments
To meet requirements, two rapid mix basins with this design are provided.
Because the average day and minimum flow rates will be less, the detention time at these flows will be longer than 5 s.
To account for variations in water height and wave action, as well as adding a factor of safety in the design volume, the tank is made deeper than the design water depth. This additional depth is called freeboard. The freeboard may vary from 0.45 to 0.60 m.

III.    FLOCCULATION
Flocculation, a gentle mixing stage, increases the particle size from submicroscopic microfloc to visible suspended particles. Microfloc particles collide, causing them to bond to produce larger, visible flocs called pinflocs. Floc size continues to build with additional collisions and interaction with added inorganic polymers (coagulant) or organic polymers. Macroflocs are formed and high molecular weight polymers, called coagulant aids, may be added to help bridge, bind, and strengthen the floc, add weight, and increase settling rate. Once floc has reached it optimum size and strength, water is ready for sedimentation.

A.    Design Criteria for Mechanical Slow Mix unit (Axial Impeller)
Detention time = 20 to 30 minutes
Water depth range = 3 to 5 meters
Maximum tip speed = 2.7 m/s
Ratio of impeller diameter to tank diameter (for radial) = D/T = 0.17-0.4
Ratio of water depth to tank diameter = H/D = 2-4
Velocity of flow = 0.15 to 0.45 m/s
G (high turbidity, solids removal coagulation) = 30 to 80 s-1

B.    Assume
Impeller diameter =D = 3 meters (depending upon manufacturer)
Power number or Impeller constant (Np) = 0.42 (depending upon manufacturer)
Temperature = 250C
Water depth = 4 meters
Detention time = 20minutes
Dynamic viscosity = µ = 0.890 × 10-3 pa•s
Five flocculator compartments with G = 60 s-1, 55 s-1, 50 s-1, 45 s-1, 40 s-1

C.    Calculation
With four flocculation basins, the maximum day flow through each will be
Q = (210448.8 m3/d) × (1/1440 min/d) = 146.145m3/min
Q = (146.145m3/min)/4 = 36.5m3/min
Determine the volume of the flocculation basin
V = Q ×t
V = 36.5m3/min ×20min
V = 730 m3
Each basin is divided into five equal volume compartments or tanks
Vcompartment = 730 m3/5 = 146 m3
With the recommended water depth range of 3 to 5 m, assume water depth of 4.0 m. The surface area is then
Asurface = 146 m3 / 4 = 36 m2
With a rectangular plan, the length equals the width and
L = W = Image36 m2 = 6.0 m
The equivalent diameter exists
T = [(4) × (36 m2)/3.14]1/2
T =7 m
Calculate the required input waterpower for each compartment
P = G2 µ Volume = (60 s-1)2 × (0.890 × 10-3) × 146 m3) =468 W
P = G2 µ Volume = (55 s-1)2 × (0.890 × 10-3) × 146 m3) =393 W
P = G2 µ Volume = (50 s-1)2 × (0.890 × 10-3) × 146 m3) =325 W
P = G2 µ Volume = (45 s-1)2 × (0.890 × 10-3) × 146m3) = 263 W
P = G2 µ Volume = (40 s-1)2 × (0.890 × 10-3) × 146m3) = 207 W
The efficiency of transfer of motor power to waterpower is about 80%, the motor power for each compartment should be
Motor power = waterpower/0.8 = 468 W /0.8 = 585 W
Motor power = waterpower/0.8 = 393W/0.8 = 491 W
Motor power = waterpower/0.8 = 325 W/0.8 = 406 W
Motor power = waterpower/0.8 = 263 W/0.8 = 329 W
Motor power = waterpower/0.8 = 117 W/0.8 = 259 W
Find geometric ratio = D/T = 3/7 = 0.43
Although the 3 m diameter impeller has a D / T slightly larger than the allowable range, is satisfactory and, therefore, is selected.
The tip speed is checked by first computing the rotational speed at G = 60 s-1
n = [P/ (Np×(D)5 × Image]1/3
n = [468 W / (0.42) × (3)5 (1000kg/m3)]1/3
n = 0.0047rps or 0.28 rpm
Tip speed = (rps) × (Image) × (D) = 2.6 m/s
This is less than the design criterion of a maximum tip speed of 2.7 m/s.

D.    Comments
The provision of four flocculation basins meets redundancy requirements.
An additional 0.60 m is added to the water depth as freeboard, so the tank depth is 4.6 m.

IV.    SEDIMENTATION
Sedimentation, also known as settling or clarification is the process of removing solid particles by gravity.

A.    Design Criteria
Sedimentation tank depth = 3 to 5 meters
Reynolds number < 1 [3]
Sludge zone depth = 0.6 to 1 m
Detention time = 2 hrs or 120 min, according to the design criteria described [3]
Diameter of particle = 0.02 mm or 0.00002 m [3]
Dynamic viscosity = µ = 0.890 × 10-3 pa•s
Kinematic viscosity v = 0.893 × 10-6 m2/s (at 250C temperature)
Density of particle Images = 2650 kg/m3
Density of waterImagew = 997.048 kg/m3
Weir loading rate (WLR) = 140 to 320 m3/d•m
f is the Darcy Weisbach factor ranging from 0.02 to 0.03

B.    Assume
Depth of the tank = 5 m
Allowance for sludge depth = 1 m
Freeboard = 0.6 m
Number of tanks =10
Temperature = 250C

C.    Calculation
Q = 2.435 m3/s
Q = (2.435 m3/s)/ (10 tanks) = 0.2435 m3/s/tank or 14.61 m3/min/tank
Detention time t = 2hrs or 120 min
Volume = Q × time
V = (14.61 m3/min) × (120 min) = 1753.2 m3
Depth of the tank = 5 m
Area = (2160 m3)/5 m = 350.64 m2
Diameter of the tank = Image
D = 21.13 m or say 22 m
To check the settling velocity [6]
Vs = [g (Images – Imagew) ds2]/18µ
Vs = [9.81 m/s2 (2650 – 997.048) (0.00002 m)2]/[18 (0.890 × 10-3 pa•s)]
Vs = 4.0 × 10-4 m/s
Reynolds number Re = Vs ds/ v = [(4.0 × 10-4 m/s) (0.00002 m)] / [0.893 × 10-6 m2/s] = 0.008 < 1 (stock’s law is applicable)
Design weir
(90° v-notch) shape weir (50 mm) depth placed (250 mm) center to center, [3]
Weir loading rate (WLR) = Q /Image
WLR = (14.61 m3/min) / (3.14 × 22 m) = 0.21 m3/min•m or 302 m3/day•m (which is within the allowable range) [1,3]
Scoring velocity
Vc = Image
Vc = Image
Vc = 65.7 mm/s or 6.57 cm/s
ImageVs < Vc Ok

Figure 5 Sedimentation tank
Summary
Qdesign    2.435 m3/s
Number of tanks    10
Volume of tank    1753.2 m3
Area of tank    350.64 m2
Depth of tank    5 m
Dia of tank    22 m
Settling velocity    4.0 × 10-4 m/s
Reynolds number    0.008
Weir loading rate    302 m3/day•m
Scouring velocity    6.57 cm/s

V.    FILTRATION
Filtration aims to remove the suspended solids that are not removed in the sedimentation unit or when the removal of these particles takes a long time outside the basin. The rapid sand filter comprises of a bed of sand serving as a single medium granular matrix supported on gravel overlying an under drainage system, the distinctive features of rapid sand filtration as compared to slow sand filtration include careful pre-treatment of raw water to effective flocculate the colloidal particles, use of higher filtration rates and coarser but more uniform filter media to utilize greater depths of filter media to trap influent solids without excessive head loss and back washing of filter bed by reversing the flow direction to clear the entire depth of river. . The removal of particles within a deep granular medium filter such as rapid sand filter occurs primarily within the filter bed and is referred to as depth filtration. Conceptually the removal of particles takes place in two distinct slips as transport and as attachment step. In the first step the impurity particles must be brought from the bulk of the liquid within the pores close to the surface of the medium of the previously deposited solids on the medium. Once the particles come closer to the surface an attachment step is required to retain it on the surface instead of letting it flow down the filter.

A.    Design Criteria
Number of filters = four or more if Q > 8000 m3/d
Filtration rate = 5 to 7.5 m/h for rapid sand filter
Cell width < 6 m
L: W = 2:1 to 4:1
Wash time = more than 10 to 20 minutes
Backwash velocity (minimum) = 37 m/h
Margin of safety = 0.15 to 0.3 m
Backwash conduit velocity = 1.2 to 2.4 m/s
Width of gullet = 0.4 to 2 m
Depth of sand = 0.6 m
Depth of gravel = 0.3 m
Depth of underdrain block = 0.3 m

B.    Assume
Filtration rate = 7 m3/h•m2 or 168 m3/d•m2
Width of a cell = 3 m
Assume a 500 mm diameter of pipe from gullet is to carry the wash water at a velocity of 1.4 m/s
Assume the width of gullet = 0.6 m

C.    Calculation
Number of filters N = 0.0195(Q)0.5 = 0.0195(210448.8 m3/d)0.5 = 8.94 or say 10 filters
Area of filter bed A = Q / Nq = (210448.8 m3/d) / (10) (168 m3/d•m2) = 125.26 m2/filter
As a trial, select the total width of two cells 11.2 m. Each cell has a width of 5.6 m. The length of each cell is then
L = (125.26 m2/filter) / 2 (3 m) = 20.87 or 11.2 m
The L: W = 2:1 which is within the recommended range of 2:1 to 4:1

1)    Backwashing System
The backwash velocity is 864 m/d (example 11-3, Mackenzie book)
(846 m/d) / (24 h/d) = 36 m/h
This is slightly less than recommendation of minimum of 36 m/h
The maximum flow rate of backwash water per trough is at the end trough
(36 m/h) (5.6 m) (1.6 m + 1.33 m) = 591 m3/h
Using the backwash flow rate of 591 m3/h and figure 5, select a trough width dimensions W = 46 cm and Y = 28 cm
Image
Figure.6

From the drawing in the figure, the depth of the trough == 28 cm + (46 cm/2) + 5 cm = 56 cm or 0.56 m
The trough elevation is determined from the backwash expansion calculated (in example 11-3, Mackenzie book)
Height of the weir above undisturbed media = De – D + depth of trough + 0.15 m
Height of the weir above undisturbed media = 0.7 m – 0.5 m + 0.56 m + 0.15 m = 1 m
The volume of backwash water for 15 minutes (0.25 h) of backwash is (36 m/h) (2 cells) (5.6 m) (11.2 m) (0.25 h) = 1128.96 or say 1130 m3
Provide two times of this volume of 2260 m3
Assuming a gravity feed, for the lowest level of water, the backwash tank should be 11 m above the lip of the wash troughs (Kawamura, 2000)
Image

Figure 7 Filtration

Image

Figure 8 Filtration

2)    Gullet Design
Estimate the velocity head in effluent conduit = v2/2g = (1.4 m/s)2 / 2 (9.81 m/s2) = 0.099 m
As shown in the figure 9, estimate the velocity plus entry headloss as
(1.7) (v2/2g) = 1.7 (1.4 m/s)2 / 2 (9.81 m/s2) = 0.168 or 0.17 m
Image

Figure 9
Calculate h
h (m) = depth at distance x in Figure 9
h = diameter of pipe + (velocity head + entrance headloss)
h = 0.5 m + 0.17 m = 0.67 m
Assume 0.6 m as a first trial width b of gullet and estimate depth at upstream end H using Husing equation
H = Image                 Husing equation
Qww = [36 m/h (6 m) (21 m)] / (3600 s/h) = 1.26 m3/s

H = Image= 1.1 m
This is 1.1 m below the bottom of the wash troughs. Using the depth of the trough, the initial estimate of the bottom elevation of the gullet is then
Depth = H + depth of trough
Depth = 1.1 m + 0.56 m = 1.66 m below the lip of the wash trough

VI.    SECONDARY DISINFECTION
Considering the characteristics of water given in table 1 and taking help from the example 13-5 from water and wastewater engineering book by Machenzie L. Davis, the is selected sodium hypochlorite as chlorine secondary disinfectant.

A.    Design Criteria
Residual chlorine = 0.1 to 0.2 mg/L (minimum)
The time required to complete the disinfection performed in a storage tank= 20 to 30 minutes or 0.5 hour.

B.    Calculation
Assume required chlorine = 2.0 mg/L and residual chlorine = 0.2 mg/L
Chlorine demand = required chlorine – residual chlorine
Chlorine demand = 2.0 - 0.2 = 1.8 mg/L
Consumed chlorine =
(2.0 mg/L) × (1000L/m3) ×(210448.8m3/day) ×(1g/1000mg) × (1kg/1000g)
= 420.89 kg/day.

VII.    STORAGE
When the final stages of the treatment process are completed, water can be distributed by high lift pumps to consumers or stored in storage tanks. Thereafter, it can be used as drinking water based on the required household demand.
The time required to complete the disinfection performed in a storage tank is 30 minutes. Storage tank
A.    Design Criteria
D = 4 meter
L = 2 × W
B.    Calculation
Q = V/T
V = Q × T
V = (210448.8 m3/day) × (30 min × (1 hr/60 min) × (1 day/24 hr)
V = 4384.35 m3
A = 4384.35 m3/4 = 1096.08 m2
A = L × W
1096.08 m2 = 2W × W
W = 23.41 m
L = 2 × 23.41 m
L = 46.82 m
Velocity = distance/time
V = (46.82 m/30 min) × (1 min/60 sec)
V = 0.0260 m/s
Qave= 210448.8 m3/day = 8,768.7 m3/hr = 2.435 m3/sec
Using v = 1.5 m/s (Metcalf and eddy 2014)
Q = A × v = (Imaged2/4) × v
A = (2.435 m3/sec)/1.5m/s = 1.623 m2
A = (Imaged2/4)
Imaged2 = Image (1.623 m2 × 4)/3.14
d = 1.438 m
Q = V/T = A × L/T = (Imaged2/4) × L/T
Q/A × L =                     
Q / (Imaged2/4) × L = 1/T                
(2.435 m3/sec) / (Image × 1.4382/4) = L/T
(2.435 m3/sec)/ (3.14 × 1.4382/4) = velocity
v = 1.5 < 2 OK
 
Figure 10 Elevated storage tank

VIII.    CONCLUSION
A typical step-by-step design for WTP units was presented. Procedures and detailed calculations were made. The average discharge of 210,448.8 m3/day and a population of 449,677.32 were used in the design of WTP. The quality and quantity of the surface water source affected the WTP design. Surface water resource such as the water coming from Naukot city to Mithi through a pipeline needs treatment due to the concentration of pollutants. The parameters of each unit and the whole WTP by using the pilot scale should be optimized. Populations should be predicted using various methods to use WTP services without any problems. Based on the obtained calculations and details it is concluded that, the study can be used as a base reference for the future works and to design of any WTP units. Several factors such as age of WTP, maintenance, economic and political situations, technical problems, and water demand had a great impact on the removal efficiency of the WTP units.

REFERENCES
[1]    Davis, M. L. (2010). Water and Wastewater Engineering-Design Principles and Practice. The McGraw Hill Companies.
[2]    Shuokr Qarani Aziz et al., 2019. Step-by-step design and calculations for water treatment plant units. Advances in Environmental Biology 13(8): 1-16. DOI:10.22587/aeb.2019.13.8.1
[3]    WarrenViessman, J.R., & Hammer, M. J. (1985). Water Supply and Pollution Control, 4th edition. Happer &Row, Publishers, NewYork.
[4]    McGhee, T.J. (1991). Water Supply and Sewerage. 6th edition. International Student Edition.
[5]    World Health Organization (1997). Guidelines for Drinking Water Quality. 2nd Edition. Volume 2. Amman, Jordan
[6]    Davis, M. L. (2010). Water and Wastewater Engineering-Design Principles and Practice. The McGraw Hill Companies.
[7]    Punmia, B.C., Jain A. K., & Jain, A. K. (1995). Environmental Engineering-Water Supply Engineering. LAXMI Publication (P) LTD, New Delhi 110002.
[8]    Metcalf & Eddy (2014). Wastewater Engineering: Treatment and Reuse, 5th edition, Inc., Mc Graw-Hill, New York.
[9]    Mukashev, Temirlan. (2015). Water Treatment Facility Design. 10.13140/RG.2.1.1446.3522.
[10]    Design of Water Treatment Plant for Atigre Village Ravindra M. Desai, Dayanand Tavandkar, Arvind Patil, Abhinanadan Kamble, Jayshree Devkhule, Civil Engineering Department, SGI Atigre, Shivaji University.
[11]    Worldometer.
[12]    Metcalf & Eddy (2014). Water Treatment Principles and Design, Third Edition Inc., John Wiley & Sons.
[13]    Kawamura, 2000