Water Supply

Bangalore Institute of Technology

Water supply and Sanitation project

WATER SUPPLY AND SANITATION PROJECT

CONTENTS

1

1. Introduction

2

2. Objectives of planned water supply

4

3. Factors to be considered for water supply

4

4. Water quality standard for drinking water

6

5. Design period

7

6. Units of water treatment plant

8

7. Units of sewage treatment plant

9

8. Design of water treatment plant

11

9. Flow diagram of water treatment plant

26

10.Flow 10.Flow diagram of water supply scheme

28

11.Design 11.Design of sewage treatment units

29

12.Flow 12.Flow diagram of sewage treatment plant

34

13.Bibliography

35

Extensive Survey Project

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1. INTRODUCTION PLACE: - MELKOTE

In preparing schemes of water supply the duties of engineer are to collect all the information that may prove useful for preparing plan and all the estimates of the project.

Before actual construction of water supply work is taken up in it is necessary to prepare plan and the full schemes and also design the various component of the project. The proper planning will ensure an a n economical and efficient functional scheme, which will serve the various objectives in view efficient and with minimum m inimum expenditure and recurring operational troubles

A WATER SUPPLY SCHEMES ESSENTIAL CONSISITS OF:

1) 2) 3) 4)

Works for collection of water. Works for conveyance of water. Works for treatment and purification of water and, Works for distribution of water to the consumers.

In planning good water supply schemes a compromise between efficiency and economic is generally decided and the schemes possessing at least defects and professional advice and guidance is very essential for the s chemes.

In the case of rural water supply schemes, Since the commutations are generally small and poor and hence the various element of the scheme have to be necessarily simple and comparatively cheaper requiring minimum skilled supervision and maintenance .In the case, the sources are generally ground water leading to special problems like fluorides, Iron, Sailing etc., One should take of all these the problems while designing.

NECESSITY OF PLANNED WATER SUPPLY

Water is a chemical compound and may occur in a liquid form or in a solid form or in a gaseous form .All these three form of water are extremely useful to man, providing the luxuries and comforts, in addition to fulfilling his basic necessity of life .No life can

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exist without water since water is essential for life as air. It has been estimated that two thirds of human body is constituted c onstituted of water.

From the public health point of view, it is necessary that the water required for their need must be invariably free from all type of impurities whether suspended or dissolved in water and no risk should occur to the health of the public as a result of any water contamination.

A public water supply or water works system has to be both from the point of view of providing an adequate and reliable supply of water catering to all the public need ensuring that the supply so made are not only portable but also als o fully protected against every infection which might otherwise pollute water and cause outbreak of disease.

The planned water supply scheme should not only help in supplying wholesome water to the people for drinking cooking, bathing etc.., so as to keep disease away and there by promoting better health but it should a lso help in supplying water for fountains, garden etc.., thus helping in maintaining better sanitation and beautification of surrounding thereby reducing environmental pollution .The schemes s hould therefore help in promoting wealth and welfare of the entire humanity as a whole.

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2. OBJECTIVES OF PLANNED WATER SUPPLY

1) 2) 3) 4)

To supply safe and wholesome water to consumers: To supply water in adequate quantity, and To make water easily available to the consumer so as to encourage Personnel and base hold clean lines.

3. FACTOR TO BE CONSIDERED FOR WATER SUPPLY PROJECT The various factors should be taken into consideration before designing any

water supply

project is:

1. 2. 3. 4. 5. 6.

Population forecast. Per capita supply. Selection of source. Quality requirement. Distribution system. Design period

I.POPULATION FORECAST The recent population of a region is obtained by recent census figure and perspective population as a result of growth, development, educational and industrial expansion etc.., are estimated by followed methods. i). ARITHMETICAL INCREASE METHOD: This method is based upon assumption that the populations increase at a constant rate i.e.., the rate of change of population with time (df/dt) is constant. Thus dp/df =k DP=>kdt. Integrating between first and last available census we get, Pe-Pf =k(te tf) Where the suffix ‘e and f” represent the last and the first census

respectively. K represent average increase per census .This method gives comparatively lower results. ii). GEOMETRICAL INCREASE METHOD: In this method, a graph is plotted from the available dates between time and population. The curve is then smoothly extended unto the desired gear. This

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method however gives very approximate results, as the extensions of the curve are done the intelligence of the designer II.PER CAPITA DEMAND It is the annual average amount of daily water required by one person a nd includes the domestic use, Industrial and commercial use, public use, wastes, thefts etc.., It may be therefore expressed as Liters per captia per day (LPCD). Total yearly water required of the city =365 x DESIGN POPULATION x PER CAPTIA DEMAND Per capita supply varies as per region and the people living standard in the region. For the rural area per capita should s hould be selected as per the t he following criteria. 1) Where house connection are not contemplated and supply through hand pumps or central stand post, per capita supply shall not be less than 40 lpcd. 2) Where the house connections are contemplated c ontemplated the supply shall be at least 70 lpcd. 3) A minimum of 135-lpcd areas preferably 150 for all residence provided with full flushing systems with modern sanitary.

III. SELECTION OF SOURCE The following order of priority may be given in the choice of suitable source among the various sources available.

a) Water that required no treatment except disinfections to meet water quality standard and would be supplied by gravity. b) Water that requires no treatment except disinfections to meet water quality standard but would require pumping. c) Water that requires simple treatment like chlorination.

IV.QUALITY REQUIREMENT The water required for domestic uses, particularly the water required for drinking must b e colorless, odorless, and tasteless. It should be free from turbidity and excessive of toxic. The maximum permissible limits for physical, chemical and bacteriological standards should satisfy the drinking water standard as given by I SI in the table shown. These limits in table are ideal and safe for public supplies, therefore in village they may not be followed rigidly due to the shortage of funds for treatment or due to

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lack of development .However, bacterial quality should no where be relaxed as to avoid serious water borne disease.

4. WATER QUALITY STANDARDS FOR DRINKING WATER

Type of

Type of

Permissible E

characteristics

impurity

limits

Physical

Absolute Max.Limits

1.Turbidity

5

10

On silica scale.

2. Colour

10

20

On cobalt scale.

3. Taste and

01

03

odor

Chemical

Remarks

Threshold no.

1. pH value

6.6

8.0

2. Hardness

75mg/ltr

110mg/ltr

Expressed as CaCo3 Equilatew/s.

3. Total solids

500mg/ltr

As per WHO

1500 mg/lit

international STDs. -do-

4. Mg & Na

500mg/ltr

1000mg/ltr

5. Chlorides.

200mg/ltr

400mg/ltr

6. Calcium.

75mg/ltr

200mg/ltr

7. Iron. 8. Fluorides.

Biological &

Coli form

micro-organic

Bacteria

-do-do-do-

1.0mg/ltr -

-do-

1.5mg/ltr Nil

1Coliform colony per

-

100m/n of B coli is ltd.to 0.3 mg/ltr1per 100me

Radiological

Emitters

Nil

-

-

After a complete treatment of water, it becomes necessary to distribute it to a number of houses and public places by means o f distribution network. The distribution system consistof pipes of various sizes, valves, water pumps, distribution reservoirs, hydrants, stand posts

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etc.., The pipes carry water to each and every house,roads,streets,while are controlled by valves. Mainly in village water flow will be designed as gravitational so the pumping is not required.

5. DESIGN PERIOD

The water supply schemes include huge and costly structure such as dams, reservoir, treatment plants ,penstock pipes etc.., which cannot be replaced are increase in their capacities easily and conveniently .For example the water mines laid underground including distribution system cannot be replaced or added without digging the road and disturbing the traffic .In order to avoid this future complication of expansion the various component of water supply scheme are purposely made larger so as to satisfy the community needs for a reasonable number of years to come. This future period or number of years far which a provision is made designing the capacities of various components of water supply scheme is known as DESIGN PERIOD. Such a scheme which is designed for a design period of say “X year “is supposed to satisfactorily serve the community need up to the end of ‘x years’ .The design period should neither be too long nor too short .The design period cannot c annot exceed the useful life of the components structure . Here the design period for water supply scheme is taken as 20 year.

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6. UNITS OF WATER TREATMENT PLANT Design of intake well and screen :

The intake well is a circular or rectangular well with circular ends located in a river bed, so that it i t always remains surrounded with water, even during low flood stage. The well is built in masonry or concrete, and is rai sed above the river HFL and covered at the top by wooden sleepers etc; as to make it approachable from the river bank through a foot bridge arrangement. Screens are generally provided in front of the intake works, so as to exclude the large sized particles. Screens are normally inclined at about 45° 60° to the horizontal, so as to increase the opening area to reduce the flow velocity. Pre-sedimentation Pre-sedimentation Tank : In order to remove the large sized and rapidly settleable solids to reduce turbidity, it is proposed to construct a pre-sedimentation tank and raw water storage.

Cascade type Aerator: Since the raw water does not contain too much color and odor, only nominal aeration is proposed. This unit will help in maintaining the wanted oxygen levels in water , remove dissolved iron , manganese, CO 2 and H2S gases as well as color and taste caused by volatile oils etc.

Combined Coagulation cum Sedimentation Tank: It shall be constructed to allow formation of flocs and settlement of particles. The fine colloidal particles present in water, gets attracted and absorbed in these flocs, forming the bigger sized flocculated particles.

Rapid Gravity Filter: Filter units shall be constructed to filter the sedimented water, as usual, with provision of wash water tank. Rapid Gravity Filter is the best and most economical and, therefore invariably used for treating public supplies, especially for large towns and city. The treated water is not so much safe as obtained from slow sand filter and need further treatment before they can be supplied to the public.

Chlorinator:

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Considering the remote area and difficulty in transporting and storing the chlorine gas cylinders, it has been decided decided to use bleaching powder for disinfection by providing gravity type of chlorinator. Bleaching powder mixed clear water will be fed by gravity into clear water tank where contact period shall be more than half hour, which is the acceptable limit.

7. UNITS OF SEWAGE TREATMENT PLANT Primary Sedimentation Tank

The Clarification of sewage by the process of ‘sedimentation’ can be affected by providing conditions under which the suspended material present in se wage can settle out. This is is brought about in specially designed tanks called Primary Sedimentation Tanks .

Trickling Filters

Trickling filters are now almost universally adopted for giving secondary treatment to sewage. These filters, also called as percolating filters or sprinkling filters, consist of tanks of coarser filtering media, over which the sewage is allowed to sprinkle or trickle down, by means of spray nozzles or rotary distributors. The percolating sewage is collected at the bottom of the tank through a well designed under-drainage under-drainage system. The purification of the sewage is brought about manly by the aerobic bacteria, which form a bacterial film around the particles of the filtering filtering media. The action due to to the mechanic al straining of the filter bed is much less. In order to ensure the large scale growth of the aerobic aerobic bacteria, sufficient quantity of oxygen is supplied by providing suitable ventilation facilities in the body of the filter; and also to some extent by the intermittent functioning of the filter .

Secondary Sedimentation Tank

A filter does remove only a very small percentage of the finely divided suspended organic matter present in sewage, while the majority of this organic matter undergoes a change of character, only due to the biological oxidation and nitrification taking place in the filter. Due to this aerobic oxidation and nitrification that takes place in the filter, the organic solids are converted into coagulated suspended mass, which is heavier and bulkier, and would thus settle down by gravity, if allowed to dos so in sedimentation sedimentation tanks. The effluent of the filter is,

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therefore, passed through a sedimentation tank, called Secondary settling tank or Humus Tank.

Sludge Digestion Tank

The sludge withdrawn from the sedimentation basins contains a lot of putrescible organic matter, and if disposed of without any treatment, the organic matter may decompose, producing foul gases and a lot of nuisance, pollution, and health hazards. In order to avoid avoid such pollutions, the sludge is, first of all, stabilized by decomposing the organic matter under controlled anaerobic conditions, and then disposed off suitably after drying on drying beds, etc. The process of stabilization stabilization is called the sludge digestion; and the tank where the process is carried out is called the sludge digestion tank.

Three distinct stages have been found to occur in the biological action involved in the natural process of sludge digestion tank. These stages are:

(i) (ii) (iii)

Acid fermentation; Acid regression; and Alkaline fermentation.

Sludge Drying Bed

Sludge drying beds are open beds of land, 45 to 60 cm deep, and consisting of about 30 to 45 cm thick graded layers of gravel or crushed stone varying in size from 15 cm at bottom to 1.25 cm at top, and overlain by 10 to 15 cm thick coarse sand layer. Open jointed under-drain pipes (15cm dia) @ 5 to 7 cm c/c spacing are laid below the gravel layer in valleys, at a longitudinal longitudinal slope of about 1 to 100. The beds are about 15 X 30 m in plan, and are surrounded by brick walls rising about 1 meter above the sand surface.

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8. DESIGN OF WATER TREATMENT PLANT

Calculation of water Demand

1. Minimum domestic water consumption consumption as per IS- 1172-1993 required required water = 200 lit/hd/day 2. Parks-3.5lit/ sq m/day 3. Schools & colleges (day)- 45lit/hd/day 4. For Matta’s – 45lit/hd/day – 45lit/hd/day

  

Assumed area of park -10,000 sq m No of Schools & Colleges -5no, and having 200 students for each No of matta’s – 15no, – 15no, Capacity of each 80members

Water required

Domestic - 200*25,000 =50,00,000lit/day  Park’s - 10,000*3.5 = 35,000lit/day  Schools & Colleges - 5*200*45 = 45,000lit/day – 15*45*80 = 54,000lit/day  For Matta’s – 15*45*80 Total = 51,34,000lit/day 

Maximum water Demand = 1.5*5134 = 7701Cum/day Discharge = 7701.0/(24*60*60) = 0.0891Cum/sec

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DESIGN OF RAISING MAIN Using Lea’s formula for most economical diameter

D=0.972Q to 1.22Q D=23.81cm to 29.8cm Hence provide 30 cm pipe

Velocity = Q/A = 0.0891/0.00706 = 1.26m/sec

DESIGN OF PUMPS

Calculations of total head:H=Hs + Hd + Hf Hs = Section head Hd = Delivery head

Hf = Head loss due to friction

Hs = 952.0-944.5=7.5m Hd = 961.0-952.0=9.0m = 4flv2/2gd Hf = = 0.025*50*1.262/2*9.81*0.3 = 0.337m Total Head H= 16.5+0.337= 16.837 Adding 0.05% as Minor losses = 16.837*1.05= 17.679m

Break Horse power:-

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BHP = wQH/ η*.735 = 9.81*0.0891*17.67/0.65*0.735 = 32.34 ~ 35Hp

Design of Primary Sedimentation Sedimentation tank:Discharge = 7701cumec/day Detention time = 3hours Depth of sedimentation tank = 2.5m Specific gravity = 2.65 Velocity Vs = 0.23mm/sec

Volume of water to be treated / hour = 7701/24 = 320.87cumec / hour

The Raw water to be treated for a detention period = 7701*3/24 = 962.62cumec

Circular sedimentation tank:-

Depth of sedimentation tank = 2.5m Surface Area = 962.62 2.5 = 385.05sq m Diameter of circular tank = 22.14m Effective settling zone = 22.04

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2 = 11.07 Provide 20% of Diameter of tank for Inlet zone = 4.43m

Surface flow rate = 7701 385.05 = 20m/day

Horizontal flow velocity = 20 86400

= 0.00023m/sec

Weir loading = 7701/2Π*13.285 7701/2Π*13.285 = 92.258cumec/m/day

Inlet velocity = Q = a*v 0.0891 = ∏ *22.142*v/4

V = 0.00023m/sec

Sludge collecting capacity, assume suspended solid particles in water 300mg/lit

Sludge collection = 7701*10 6*300/105

= 2310lit/day

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Inlet pipe Provide 0.15m dia

Q = a*v

0.0891 = ∏ *.0152*v/4

V = 5.04m/sec

V = √2gH

H = v2 /2g

H = 5.042/2*9.81

H = 1.29m

Design of raw storage tank Capacity of water to be stored on tank approximately one day demand of water.

Quantity of water should be stored on tank = 7701Cum

Depth of water stored in a tank =4.5m

Provide circular tank. D = dia of the tank

D=

√4*A/∏

D=

√4*1711.32/∏

D = 46.67m

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Design of inlet well Providing an oblong shaped inlet well, with openings provided at 3 levels. The openings shall be fitted with bar screens made of 20mm dia steel bars of 50mm openings. Let the velocity through the bar screens be limited to 0.16m/s. The area of openings required at each level = Q/v = 0.133/.16 = 0.831sq m Let us provide 1m height of screen openings; then the clear length of the openings required = 0.831m

No. of openings required = 0.831/0.05 = 17no No. of bars=17

Length occupied by 20mm dia bars = 17*0.02 = 0.34m Total length of screen = 0.831+0.34 = 1.171 ≈ 1.2m

Design of pipe connecting intake well to jack well. Therefore,

Q=0.133cumec V=0.16met/sec

Area of pipe = Q/V = 0.133/0.16 = 0 .831m2

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Therefore, d = √4*0.831/∏ d = 1.028m Provide jack well dia = 10m.

Design of Cascade Aerator: Water requirement = 7.70 MLD

Q= 0.0891m/s Provide 3 steps, dia ratio 1:2:3. Depth of wall =0.75m

Diameter of weir proposed Top = 1m Middle = 2m Bottom = 3m

Sedimentation aided with coagulation tank: Average daily consumption = Volume of water treated per day =7701 m 3 Assume detention period of 4hrs. Quantity of water to be treated = 7701*4/24 = 1283.5m 3 .’. Capacity of tank = 1283.5 m 3 ≈ 1284m3 Assuming depth of water in tank = 4m , we have, Plan area = Breadth*L ength= capacity/depth = 1284/4 = 321m3 Take Breadth = 9m .’. length = 321/9 = 35.67m ≈ 36m Settling tank dimension = 36x9x4 Providing an extra depth of 0.5m at the starting tank ,I,e; a depth of 4.5 at the start and providing a slope of 1in30 along the length.

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We get, 4.5+37/30 = 5.7m, at the end.

Design of flocculation chamber:

In addition to 15m length of settling tank, the floc chamber at the entry has to be provided as half the depth of the sedimentation tank I.e. 2.5/2 =1.25 m

 1.5 m Detention time = 20 minutes

= 0.08912060 = 106.92cum Surface area

= 106.92/1.5 = 71.28m2

Provide rectangular chamber 1:2ratio Width of chamber = 6m Length of chamber = 12m Thus, floc chamber is of size 6.0 m

12m

1.5m.

Coagulation sedimentation plant: Coagulant added is Alum. Given pH=8

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The coagulant added : ALUM A LUM Alkalinity of water is 4.5mg/lit of CaCO3(nat) Quantity of water treated =7.701 MLD Quantity of filter ALUM required at 20 mg/lit = 20 x 7.701 x 10 6 mg/day =20 x 7.701 = 154 kg/day Thus the following chemical reaction will take place: Al2(SO4)3.18H2O+3Ca(HCO3)22AL(OH)3+3CaSO4+18H20+6CO2

 

CaCO3+H2O+CO2Ca(HCO3)2 CaCO3CaO+CO2

The molecular weight shall be: AL2(SO4)3.18H20={(2x27)+3x(32+(16x4))+18x((1x2)+16)}=666gm Ca(HCO3)2=40+(1+12+48)x2=162gm CaO=40+16=56gm It is clear that from the above equations 3x100parts of CaO 3 will produce the same alkalinity as that off 660mg of ALUM Quantity of CaCO3 required to produce the same alkalinity which is equal to 20mg/litre =>(3x100x20)/666=9.01mg/lit The natural alkalinity available in CaCO 3 =4.5mg/lit Additional alkaline required to be added to form the lime=40.5mg/lit of CaCO 3

Thus the above equation clearly states states that 40 parts of CaCO 3 produce by 56 parts of CaO. Therefore CaO required = (4.5 x 56)/100 = 2.52 mg/lit But Quick lime required =(2.52 x 100)/80 = 3.156 mg/lit The quantity of lime required for treatment treatment = 7.701 x 106 x 3.156 mg/day = 24.30 kg/day And, from above result,

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the quantity of alum =154 kg/day or, alum = 154/(24 x 3600) = 0.0018kg/sec. This coagulant is dispensed using a dry feeding device. These are in the form of a tank with a hopper bottom. Agitating plates are placed inside the tank so as to prevent the arching the coagulant. The coagulant is in the powdered form and is allowed to fall in the mixing basin. Its dose is regulated by the speed of a toothed wheel or a helical screw. The speed of the toothed wheel or the helical screw is controlled by connecting it to a venturi device installed in the raw water pipes bringing the water to the mixing basin.

Mixing the coagulant in a mixing basin with baffle walls. Design: Assume Velocity of flow = 0.3 m/sec Detention time = 30min. Volume of water to be treated per day = 7701m 3/day .’. Volume of water required to be treated treated in 30min = Capacity of tank = 7701*30/(24*60) = 160.43 m3 Length of flow = velocity*Detention period = 0.3*30*60 = 540m The required c/s between each baffle is assumed to be 0.5m. .’. Depth Depth of water in the basin = 0.29/0.5 = 0.58 ≈ 0.6m 0.6m The clear opening between end of each baffle is kept equal to 1.5 times the distance between baffles = 1.5*0.5 = 0.75m Assume clear with of clear compartment as 9m. Effective length of each channel = clear width of each compartment – compartment – (2*0.5 (2*0.5 of clear opening) = 9 – 9 – (0.5*2*0.75) (0.5*2*0.75) = 8.25m .’. No. of channels = Total flow length/eff. Flow length of each channel

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= 540/8.25 = 65.45 ≈ 66

.’. The total no. of channels =66 ; and 66/2 = 33 in each half. Clear length of tank excluding baffle walls and side walls = No. of channels x distance b/w baffles = 33*0.5 =16.5m Assume thickness of baffle = 7.5cm .’. Overall inside length of tank = 16.5 + ( (7.5/100) x no. of baffles) = 16.5 + ( (7.5/100) x (33-1)) = 18.9m ≈ 19m.

Sludge treatment: The sludge collected from the primary and secondary sedimentation tanks gets stored in a large storage tank. From here it is taken to the sludge drying beds. Once the water from the sludge drains into the channel after passing through the sand and gravel layers over a period of one week, the dry sludge is taken out manually and transported elsewhere. Design of sludge drying beds Quantity of sludge generated per day = 2310 litres/day = 2.310m3/day Height at the starting of the sludge bed = 2.5m .’. Cross sectional area of the bed = 4.5/2.5 4.5/2 .5 = 1.8m2 Assuming width to be 1.2m, we have, Length of the bed = 1.8/1.2 = 1.5m Providing a sand layer of depth 1.5m and a gravel layer of depth 1.5m, we have, Total height at the start of the sludge bed = 2.5+1.5+1.5 = 5.5m Providing a gradient of 1in10 along the width, we have, Depth at the end of 1.2m ,i.e, depth at the channel transporting the water from the sludge beds = (2.5+1.2/10)+1.5+1.5 = 2.62+1.5+1.5 = 5.62m

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Now there will be 3 such units on the either side of the channel.

Design of Rapid sand filters:

General provisions 1. Required flow of filtered water = 7.701 MLD 2. Quantity of back wash water used = 3% of filter output 3. Time lost during back washing = 30 minutes 4. Design rate of filtration = 500lit/sq.m/hr. 5 00lit/sq.m/hr. 5. Length to width ratio =1:2.5 6. Under drainrage system = central manifold.

Filter media Fine grain - (3.5mm-5mm) Coars grain - (3mm-6mm) Fine sand - (0.52mm effective size) Filter water required per hour = 320.87cum/hr Design flow for filter after accounting for backwash water (3%) & washing time (20min) = 320.87(1+0.03)24/23.5 = 337.53cum/hr Plan of filter required = 337.53/.5 = 675.06 sq m Generally the plan area of filter unit is kept as 10 sq m to 80 sq m Provide 1 unit if 80 sq m

Size of tank 12.65m

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Hence provide 8 filter unit Now, let a manifold and lateral system be provided below the filter bed, for receiving the filtered water and to allow back washing for cleaning the filter. This consists of central manifold pipe with laterals having perforations at the bottom. .’. Total area of perforations = 0.2% * filter area = (0.2/100)*(6.35*12.65) m2 = 0.16 m2 Assuming the area of each lateral = 2 times the area of perforations Total area of laterals = 2*Total area of perforations = 2*0.1.6 m2 = 0.32 m 2

Assuming the area of manifold to be about twice the area of laterals, The area of manifold = 2*0.32 = 0.64m 2 .’. Dia of manifold (d) is given by, π/4*d2 = 0.64 .’.

d = 0.7m

Hence, use a 70 cm dia manifold pipe laid lengthwise along the centre of the fi lter bottom. Laterals running perpendicular to manifold ( widthwise) emanating from the manifold may be laid at a spacing of say 15cm. The number of laterals is then given as, = 12.65/0.15 = 84.33 say 85 On the either side of manifold. Hence use 170 laterals in all, in each unit. Now, length of each lateral = (width of filter/2) – filter/2) – (Dia (Dia of manifold/2) = (6.35/2) – (6.35/2) – (0.7/2) (0.7/2) = 2.82m Now, adopting 13mm dia perforations in the laterals, we have, Total area of perforations = 0.16 = 1600cm2 = x*(π/4)*(1.3)2 Where x = Total no. of perforations in all 170 laterals. .’. x = 1600*(π/4)*(1/1.3) = 966.64 say 967

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.’. No. of perf orations orations in each lateral = 967/170 = 5.68 say 6. .’. area of area of perforations per lateral = 6*((π/4)*(1.3) 6 *((π/4)*(1.3)2) cm2 = 7.96 cm2 . Now, area of each lateral = 2*Area of perforations per lateral = 2*7.96 =15.92cm2 .’. Dia of each lateral lateral = √(15.92*(π/4)) (15.92*(π/4)) = 3.53cm Hence, use 170 laterals each of 3.53cm dia, @ 15cm c/c, each having 4 perforations of 13mm size, with 70cm dia manifold.

Disinfection process: Chlorination: Assume the chlorine dosage rate to disinfect water be 3PPM. Therefore, chlorine requirement = Avg quantity of water  dosage =7.701106310-6 = 23.10 kg/day Generally chlorine is stored for 3 months Therefore chlorine for 3 month = 23.10 90 =2079kg Therefore no. of chlorine cylinder required = 12+1 stand by.

Design of the raising main: Using the empirical formula given Lea, dia of the pipe= 0.972 to1.22 Q Q = 7.701 MLD Q = 0.0891 D = 1.220.089

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= 36.4 cm

 40 cm Therefore provide 30cm dia pipe .’. area of pipe = 0.07m 2 .’. velocity = Q/A = 0.0 = 1.27m/s

Design of pump. To pump water from from the treatment plant plant to water storage tank (over (over head tank)

H = Hs + Hd + Hf Hs= 3m Hd = 100m Hf = 4flv2/2gd = 4*.0075*1250*1.272/(2*9.81*.3) = 10.27m Total head = 3 + 100 + 10.27 = 113.3m

Brake horse power of the pump (BHP) Assume efficiency of pump as 0.65 √WQH/η*0.735 BHP = √W = (9.81*0.0891*113.3)/(0.65*0.735)

BHP =207.28H.P. Assume the capacity of pump available in the market : 50H.P, 100H.P, 150H.P, 200H.P, 250H.P, 500H.P, 750H.P, 1000H.P, .’. BHP provide = 250H.P

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Design of clean water reservoir: Treated water is stored in clear water reservoir = 7.701 MLD Since the pumping from clear water reservoir to distribution tank is for 24 hrs, detention period of 24hr is provided in clear water reservoir to ensure 24 hr supply capacity of clear reservoir. Capacity of clear water reservoir = 7.701 24/24 =7.701 MLD = 7701cum Assuming effective depth = 4.50m Plan area required =7701/4.5 = 1711.3

Provide circular tank

Area = ∏ D2/4 1711.33 = ∏ D2/4 D = 46.6 .’. D = 47m

Design of overhead tank: Assuming following pattern of % draw off from daily supply 7.00am-8.00am

30% = 2.31MLD

8.00am-5.00pm

35% = 2.69MLD

5.00pm-6.30pm

30% = 2.31LD

6.30pm-7.00am

5% = 0.385MLD

Total storage req. = 7.7 MLD Providing 3 hrs of static storage Capacity = 7.710003/24

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= 962.5cum Assume depth = 4.5 Dia = 16.5 Therefore dia of overhead tank= 9m and depth =3m and distribution mains are designed for peak demand i.e. for 2.69 MLD and diameter diameter of main is 25cm. 25cm.

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9.Flow diagram of water treatment plant

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10. FLOW DIAGRAM OF WATER SUPPLY SCHEME

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11. DESIGN OF SEWAGE TREATMENT UNITS

Quantity of water supplied = 7.701MLD Assuming 80% of supplied water becomes sewage =0.87.701106 =6.16MLD Assuming detention period = 2hr Quantity of sewage to be treated in 2hr i.e. capacity of tank Q =6.162/24 = 0.513 MLD = 513.4cum Now, assuming flow velocity = 0.3m/min The length of tank req. = vel. Of flow  detention period = 0.3260 = 36m c/s area of the tank req.= capacity of tank/length of tank = 513.4/36 = 14.26m2 Assume effective depth = 3mc The width of tank = area of c/s /depth = 14.26/3 = 4.75m Assume free board = 0.5m Overall depth =4.75+0.5 = 5.25m

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Therefore dimensions of primary sedimentation tank are 36 m

4.75m

5.25m.

Design of trickling filter:

Assume BOD =150 ppm Total BOD present in sewage to be treated per day = 6.16 150 = 924kg Assume organic loading = 1500kg/ha-m/day Volume of filtering media req. = 924 10000/1500 = 6160 cum Assuming effective depth = 2m. We have The surface area of filter req. = 6160/2

= 3080sqm Using 15 m dia circular trickling filter units Area of each unit = 176.71 sq m No of units req = 3080/176.71 = 18 nos

Check for Hydraulic loading Assume Hydraulic loading = 25ML/ha/day Surface area required = total sewage to be treated per day/hydraulic loading per day = 6.1610000/25 = 2464sqm Surface area chosen 3080sqm >2464sqm, therefore safe.

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Design of sec. Sedimentation tank: Assume detention period = 1.5hr We have capacity of tank = 6.16 (1.5/24) = 0.385 capacity of tank = 385cum Volume of each conical tank=V= (1/3) (d2/4) depth Using 2 numbers of 5.0m depth of each tank, we have total volume required for all 2 Tanks = 385/5 = 77sq m Therefore dia d = 9.9m

 10m Hence use 2 tanks with Dia of 10m and height of 5.0m and 1 stand by.

Design of Sludge digestion tank: Assuming suspended solids in raw sewage = 300 ppm. Moisture content = 85% Therefore suspended solids in 6.16MLD of sewage flowing per day = 3006.16106/106 = 1848 kg Assuming that 65% solids are removed in primary settling tank, we have Wt of solids removed in primary settling tank = 0.65 1848 = 1201.2 kg/day Assuming that the fresh sludge has o m c of 95%, we have 5 kg of dry solids make 100 kg of wet sludge And 1201.2 kg of dry solid will make = (100/5) 1201.2 = 24024kg of wet sludge/day

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Assuming the sp.gr. Of wet sludge = 1.02 i.e. unit wt = 1020 kg/cum The volume of raw sludge produced per day = V 1 = 24024/1020 = 23.55 cum/day Now, the volume of digested sludge V 2 at 85% m c is given by V2= V1 [100-P1/100-P2] = 23.55[100-95/100-85] V2 = 7.85cum/day Now, assuming the digestion period as 30 days, we have the capacity of required Digestion tank = [v1-2(v1-v2)/3]t = [23.55-2(23.55-7.85)/3]30 = 392.5 cum Provide 5m depth cylindrical tank C/s area of tank = 392.5/5 = 78.5 sq m

 Dia of tank= d = 10 m

Disposal of digested sludge:

Volume of wet sludge from sewage = 23.55cum And let it spread in 20 cm layer on under drained bed then, Area of bed required = 23.55/0.2 = 117.75 sqm/day Under Indian tropical condition , the bed get dried about 10 days and hence taking 2 weeks as Avg drying time = 52/2 = 26 times in an year Therefore area of bed required per year = 117.75 365/26 = 1653.03 sq m

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Making 100% allowance for space for storage, repair, and resting of beds etc, we have The total required area of beds = 2*1653.03 = 3306.06sqm Using 1025 m beds we have no of beds required = 3306.06/(10*25) No of beds required required = 13.22 No of beds required  14beds Provide 14 beds with size as Area = 3306.06/14 = 236.15 sq m Using width = 10m

 Length = 236.15/10 =23.61m Provide 14 no of (10 23.61) m beds.

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12. FLOW DIAGRAM OF SEWAGE TREATMENT PLANT

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13. BIBLIOGRAHY

36

Environmental Engineering 1

S K GARG

Environmental Engineering 1

B C PUNMIA & ASHOK JAIN

Environmental Engineering II

S K GARG

Environmental Engineering II

B C PUNMIA & ASHOK JAIN


Batch A2


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Water Supply

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