final year Project (civil)

INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT

BILHAUR, KANPUR NAGAR -209202

PROJECT REPORT OF THE FINAL YEAR -2013

PROJECT TITLE:-

“PRELIMINARY DESIGN OF WATER TREATMENT

PLANT”

Submitted by

AKHILESH PRATAP VERMA ABHISHEK TIWARI

ANKIT KUMAR JITENDRA KUMAR

AKHILESH SINGH ABHAY PRATAP YADAV

ASHAD AGNIHOTRI

In partial fulfillment for the award of the degree

Of

BACHELOR OF TECHNOLOGYIN

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CIVIL ENGINEERING

INDUS INSTITUTE OF TECHNOLOGY AND MANAGEMENT, BARAULIBILHAUR , KANPUR

SESSION 2014-2015

“DEPARTMENT OF CIVIL ENGINEERING”

PROJECT REPORT

UNDER THE ABLE GUIDANCE OF

MR.NITIN AGRAWAL MR.K. K. THAKUR

( H.O.D. ) (Lecturer) IITM,Kanpur IITM,Kanpur

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INDEX

S.no. Title Page No.

1 Acknowledgement 4

2 Certificate 6

3 Objective 8

4

5

Introduction

Survey

10

13

6 Structural Elements 15

7 Design Specification 19

8 Design 25

9 Drawing 37

10 Estimation 39

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11 Total cost of project 52

ACKNOWLEDGEMENT

We want to express our heartiest thanks to our most respected and honoured head of department of civil engineering, Mr. NitinAgrawal for his valuable teachings and the suggestions along with guideline which really help me in completing this project.

We want to thank our Project Head Mr.Krishna Kant Thakur for his dedication and the valuable time which he has spent with us during the completion of project. We are really thankful to entire faculty of theDepartment of Civil Engineering for the help and support we have got from them in any form. And lastly we want to thanks to all the members of group for all the cooperation and help we have got from them.

Thanking You!

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CERTIFICATEIt is certified that the project report entitled “ PRELIMINARY DESIGN OF WATER TREATMENT PLANT” is submitted by “, AKHILESH PRATAP VERMA,ABHISHEK TIWARI,ANKIT KUMAR,AKHILESH SINGH,ABHAY PRATAP YADAV,JITENDRA KUMAR,ASHAD AGNIHOTRI” student of Indus Institute of Technology And Management , Kanpur in practical fulfillment for degree in “Civil Engineering from the Board of Technical Education U.P.” during the academic session 2014- 2015.

The external examiner has checked and taken the oral viva – voice on the project report.

My blessing are always with him for his bright future.

MR. NITIN AGRAWAL MR.K. K.THAKUR

( H.O.D. ) ( Lecturer)

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OBJECTS OBJECTIVE

To prepare a project of Preliminary Design Of “ WATER TREATMENT PLANT”at Kanpur.

The objectives of the project are:-

Carrying out a complete analysis and designing of the main structural elements of an Water Treatment Plant including slabs, columns, and beams.

The structure should be able to accommodate all the machineries as well as all the required equipments needed in the water Treatment.

Use structural software (AutoCAD) to make the plan. Use IS codes. Getting real life experience with engineering practices. Use of all the necessary equipments needed for the survey of

the site. To study the various elements of the structure in detail. To estimate the cost of material as well as cost of labour along

with other indirect included cost incurred in the construction of civil structures.

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“INTRODUCTION”INTRODUCTION

The primary needs for any living being are the water,air,food,sheltered, for which water has the greatest importance water is required for various purposes such as drinking, cooking, bathing, washing, watering of lawns and garden or growing of crops for fire fighting, for heating and air-conditioning systems, etc. In the ancient times human required water for drinking, bathing, cooking etc. but with the advancement of civilization the utility of water enormously increased and now such a stage has come that without well-organized public water supply scheme it is impossible to run the present civic file and develop the towns.

In ancient times every individual or family was responsible to arrange for their water supplies. There was no collective effort by the whole community for it. But as the community developed it became essential to have public water supply soon the inhabitants realized that their local sources of water supply such as shallow wells, springs, cisterns etc are inadequate to meet the demand of the town, they started to collect the water from distant large sources and conveyed it to the town through aqueducts, canals etc. when the concentration of town increases, it becomes very difficult to locate wells. In addition to these sources of water, having good quality was less readily available. These all situations led to the development of public water supply schemes. Water treatment involves science, engineering, business, and art. The treatment may include mechanical, physical, biological, and chemical methods. As with any technology, science is the foundation, and engineering makes sure that the technology works as designed. The appearance and application of water is an art.

In terms of business, RGF Environmental, Water Energy Technologies, Aquasana Store, Vitech, Recalyx Industrial SDN BHD and PACE Chemicals ltd are some of many companies that offer various processes for water treatment. Millipore, a Fisher Scientific partner, offers many lines of products to produce

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ultrapure water, using a combination of active charcoal membranes, and reverse osmosis filter. Internet sites of these companies offer useful information regarding water.

An environmental scientist or consultant matches the service provider, modify if necessary, with the requirement.

Natural Water includes some discussion on hard and soft water. Softening hard water for boiler, cooler, and domestic application is discussed therein. These treatments prepare water so that it is suitable for the applications.

Water Biology deals with water and biology. Drinking water is part of making water suitable for living. Thus, this link gives some considerations to drinking water problems.

There are many different industry types, and waters from various sources are usually treated before and after their applications. Pre-application treatment and water treatment offer a special opportunity or challenge. Only a general consideration will be given to some industrial processes.

General municipal and domestic water treatment converts used water (waste) into environmentally acceptable water or even drinking water. Every urban centre requires such a facility.

SURVEY

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A Survey is an inspection of an area where work is proposed, to gather information for a design or an estimate to complete the initial tasks required for an outdoor activity. It can determine a precise location, access, best orientation for the site and the location of obstacles. The type of site survey and the best practices required depend on the nature of the project. Examples of projects requiring a preliminary site survey include urban construction,specialized construction (such as the location for a telescope) and wireless network design.

The most important thing under the water supply schemes is the selection of source of water , which should be reliable and have minimum impurities. After the selection of source water, the next step is to construct intake works in order to collect the water and carry to the treatment plants, where this water will be treated. Type of treatment processes directly depend upon the impurities in water at the source and the quality of water required by the consumers. When the water is treated, it is stored in clear water reservoir , from where it is distributed to the consumers. The distribution system will also depend on the elevation of clear water reservoir and the elevavtion of distribution area.

STRUCTURAL ELEMENTSThe structural elements are those elements which form the supporting skeleton frame work of the supply scheme. It include the following things :-

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(i) Cascade type aerator(ii) Alum dose sand wash water tank(iii) Combined coagulation cum sedimentation

tank(iv) Chlorinator(v) Clear water reservoir(vi) Main distribution tank

(i) Cascade type aerator -since the raw water does not contain too much colour and odour, only nominal aeration is proposed by constructing an artistic cascade type aerator .This unit will help in main ting the wanted oxygen levels in water, remove dissolved iron and ,manganese, remove CO2 and H2S gases as well as the colour and tastes caused by volatile oils etc. released by algae and other micro-organisms. The raw water reaching the plant will be pumped into the the aerator tank through a 450mm dia pipe and will outflow through 4no. 250mm dia pipes.

(ii) Alum dossers and wash water tank - Alumdossers is the appurtenance through which measured quantity of alum is added to the water supply system before of water. The requisite alum dose shall be mixed in two polyethylene tanks of 3000L each, which shall be interconnected. The raw water shall first be added in

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one tank where the alum dose of shallbemixed manually. The alum mixed solution shall flow into the other tank, where a float control will be produced to ensure constant depth of wall

(iii) Combined coagulation cum sedimentation tank -It shall be constructed to allow formation of flocks and settlementof particles.

(iv) Rapid gravity filter -Filter units shall be constructed to filter the sediment water, as usual,with provision of wash water tank.

(v) Chlorinator- Considering the remote area and difficulty in transporting and storing the chlorine gas cylinders, it has been decided to use bleaching powder for disinfection by providing gravity typeof chlorinator.

(vi) Clear water reservoir- It shall be constructed to store treated water , from where it will be lifted by 2nd stage pump house to deliver it to main distribution reservoir. This pumping is to bed for 16 hrs.

(vii) Main distribution tank- The main distribution reservoir will obtain pumped water from 2nd stage pumped house and

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deliver it by gravity to service reservoirs of towns and villages for 18 hours per day.

DESIGN SPECIFICATION(1) Design of pre-sedimentation tank - We shall design the tank

to remove particles upto 0.1 mm size, and adopt the following general parameters.

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General parameters-1. Overflow rate = 20 to 80 m3/m2/day2. Minimum side water depth = 2.5m3. Detention time = 0.5 to 3 hrs4. Side slopes =10% from sides

5. Longitudinal slope =1% in rectangular tank 6. Settling velocity = 0.1mm

Hydraulic Design

Water required by the year 2030 =27.00 MLD =27.0*1000/24=1125m3/hr

pumping is for 16 hrs, thus discharge = 1125*24/16=1687.5m3/hr Water loss in desludging = 2% Design average flow = 1687.5*100/(100-2)=1721.9388m3/hr Assume detention period = 1.5 hrs Effective storage of sedimentation tank= 1727.9388*1.5=2582.91m3 Assuming effective depth = 3.0 m Area of tank required =2582.91/3.0=860.97 m2 Assume L:B = 3:1 3B*B = 860.97 B= 16.941m L= 3*16.941=50.823 Provide a tank of size 50.823m*16.941m*3.5m

(2) Design of raw water tank- A raw water storage tank of 8.0 hrs detention period is proposed.

Water requirement for the year 2030= 27.0 MLDTaking 9.00 hrs detention period capacity of tank =(8.00*27.0)/24= 9000 m3Provide water depth = 3.0 m

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Plan area of tank required 9000/3 = 3000 m2Assume a ratio of L:B =2.5 : 12.5B*B= 3000B=35.00ML= 87.00 MProvide a tank of size 87m*35m*3.5m

(3) Design of cascade aerator- water requirement for the year 2030 = 27.00 MLD Average water requirement Q = 27*1000000/(24*3600*1000)m3/s Q = .3125m^3/s for broad crest Q=1.65B*H^1.5 Taking width B= 4 mH=13cm

(4) Design of sedimentation tank- Overflow rate = 15 to 30 m3/d/m2 Minimum side water depth = 2.5 m Detention period for coagulated water = 2 to 4 hrs Weir loading = 300 m3/d/m2 Side slopes = 10 % Longitudinal slopes = 1% Settling velocity = 0.02 m

Water required by the year 2030 = 27.0 MLD = 1125m3/hr Water lost in desludging = 2%

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Design average flow = 1125 * 100/(100 – 2)= 1147.95 m3/hrAssume a total detention period of 2.6 hrs Effective storage of sedimentation tank = 1147.95 * 2.6 = 2984.69m3 Assume effective depth = 3.0 m Area of the tank required A = 2984.69/3 = 994.89 m2 Assume L/B = 3 :1 3B*B = 994.89 m2 B = 18.2 m L = 54.6 mProvide a tank of size 54.6m * 18.2m* 3.0m

(5) Design of rapid gravity filter- A-required flow of filtered water = 27.00 MLDB-quantity of back water used= 3% of filter outputC-time lost during back washing = 30 minD-design rate of filtration =5.4 m3/m2/hrE-length width ratio = 1.25F-size of perforation=9mmFilter water required =27.00 MLDFiltered water required per hour = 27.00*1000000/(24*10000)= 1125 m3/hrDesign flow for filter after accounting for backwash water3% and washing time 0.5hr =1125*(1+0.03)*24/(23.5)= 1183.4 m3/hrPlan area of filter required = 1183.4/5.4 = 219.14 m2

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Generally provide two filter units each of 112m2Provide 2 units, each 12.5 m * 9m size

(6) Designof chlorinator- Two polythene tanks,each of 3000L capacity, shall be installed to mix clear water with the requisite amount of bleaching powder, which water will flow by gravity into clear water tank to have contact period of more than ½ hr, as to cause disinfection.

Normal dose of chlorine to be taken = 0.3 ppm for a contact period of 30 minAverage daily demand of water of the year 2030 = 27.0 MLDChlorine required per day =0.3*27.00*10^6/(10^6)kg= 8.1 kgsince chlorine content in bleaching powder is 30% it means that 30 kg of chlorine in 100 kg of bleaching powder.Bleaching powder required per day = (8.1 *100)/(30)= 27 kgAnnual consumption of bleaching powder = 27* 365 = 9855kg

(7) Design of clear water reservoir- Treated water stored in clear water reservoir = 8.00 MLDSince pumping from clear water reservoir distribution tank is also 16 hrs, detention period of 8 hrs is provided in clear water reservoir to ensure 24 hrs supply capacity of clear water reservoir. Capacity of clear water reservoir = 27.00*10^6*10^-3 * 8.0/(24) = 9000m^3 Assume water depth as 3.0 m Plan area required =9000/3=3000 m^2 Assume a ratio of L:B of 2.5B*B=3000

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B=35m L= 87mThus , provide a tank of size 87m * 35m *3.0m

(8) Design of main distribution tank- water requirement for the year 2030=27.00 MLDSince the water will come in the tank in 16 hrs pumping from clear water reservoir and it will go to the service reservoirs in 18 hrs, a storage of 2 hrs is required on the tank.Capacity of tank required = 27.00*3*10^3/(24)=3375m3Provide depth of water = 3.0 mPlan area of tank required =3375/3=1125m2Assume L:B = 2.5 :12.5B*B=1125m^2 , B=21m, L=53m, ..provide a covered tank of size 53m*21m*3.5m as per detail

DESIGN1. Design of base slab -

Let us assuming that –

Angle of repose of soil = 30*

Bearing capacity of soil = 160 KN/m2

Unit weight of soil = 16.8 KN/m2

Height of wall above the ground level

H = height of wall above the plinth + plinth height

=

from ground level = 3+ 0.45 = 3.45 m

Over all depth of roof slab D = 150 mm

Live load on roof slab = 2000N/m2

= 2 KN/m2

2- Depth of base slab–

By Ramekin’sformula,

D = p/y × (1- sinα)^2 ÷(1-sinα)^2

Here,

D = 160/16.8 ×( 1- sin 30) ^2 ÷ ( 1- sin30) ^2

=

D = 1.06 m = 1.10 m (say)

Hence adopt depth of foundation = 1.10 m

3- Load calculation –

Self weight of wall per metre = L× B×H

= 1×0.3×3.45×19.2

= 17.28 KN/m

Width of slab = 2×t + 30

= 2×30 + 30 = 90 cm= 0.9 m

4- Design of roof slab –

Size of room = 4.0 m × 3.5 m (clear inside dimension) Thickness of wall supporting slab = 300 mm = 0.3 00 m

Live load on roof slab = 2000KN/m2

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The slab is simply supported on all four sides with corners not held down.

Using M15 grade of concrete and mild steel Fe415.

Design constants –

For M15 grade of concrete and Fe 415 grade reinforcement then,

fck = 5 N /mm2

Fst = 140 N/mm2, m =19

Neutral axis factor, k = x÷ d = (m ×fck×d)÷ (mfck + fst)

k = (19×5) ÷(19×5 + 140) = 0.404

Hence k = 0.404

Lever armfactor, j = (1- k)÷3

j = (1- 0.404)÷ 3 = 0.065

j = 0.065

Coefficient of moment of resistance, R = ( fck× j × k )÷2

R = (5×0.865×0.404)÷2

=

R = 0.874

Let,

Over all depth of slab = 150 mm

Assuming,10mm dia. Main bars and 15 mm clear cover

Hence,

Effective depth of slab d = overall depth – clear cover – 0.5 × dia. of main bars

= 150 – 0.5 ×10

= 130 mm

Effective depth, d = 130 mm

Length of room = width of room

L = 8 , B = 3.5

Hence, L/B = 4/3.5 = 1.1228 < 2

Therefore slab will be designed as two way slab and effective span shall be smaller of the following –

1. (a) Centre to centre bearing = 4000 + 300/2 = 4150 mm

= 4.150 m

=

(b) Clear span + effective depth = 4000 + 130 = 4130 mm

= 4.130 m

2. (a) Centre to centre bearing = 3500 + 300/2 = 3650 mm = 3.650m(b) Clear span + effective depth = 3500 + 130 = 3630 mm = 3.630 m

Hence,effective span lx (shorter) = 3.630 mEffective sanely (longer) = 4.130 m

Effective span = 3.630 m

Load calculations –

1. (a) Due to self weight of 150 mm thick slab = 0.15 ×25000

= 3750 N/m2

(b) Weight of 100 mm thick lime concrete = 0.100 × 19200 = 1920 N/m2

=

2. Live load = 2000 N/m2

Hence , total load = 3750 + 1920 + 2000 = 7670 N = 7.670 KN

Total load = 7670 N

By Ramekin’s formula –(A) Weight on shorter span –

Wb=( L^4 ×W) ÷ ( L^4 + B^4 ) = (4.130^4) × 7670 ÷ (4.130^4 + 3.630^4) = 4803.38 N

(B) Weight on longer span, Wl = W – Wb = 7670 – 4803Wl = 2867 N

Bending moment –

(a) Maximumbending moment on shorter span,

Mb = (Wb× B^2) ÷ 8= 4803× (3.630)^2 ÷ 8 = 7911.08 N-m

Bending moment = 7911081 N-mm

=

(b) Maximum bending moment on longer span, Ml = ( Wl× L^2)/8 = 2867 × (4.130^4) / 8 = 6112.76 N-m

Bending moment = 6112760 N-mm

Thickness of slab –

d = ((moment ofresistance)÷(0.874× 1000))^0.5

d = ((7911081) ÷(0.874×1000)) ^0.5 = 95.13 mm

d = 100 mm (say) < 130 mm

Therefore overall depth of slab = 150mm and

Effective depth of slab = 130 mm

Area of reinforcement along shorter span,

= (Moment of resistance)÷ (fst× j × d)

= (7911081) ÷ (140 × 0.865 × 130)

= 502 mm^2

By using 10 mm of main steel bars –

Area of one bar,Ast = ( × d^2) ÷ 4π

= ( × 10^2) ÷ 4 = 78. 5 mm^2π

And spacing = ( 1000 × 78.5)÷ 502

= 156.37 mm = 150 mm (say)

=

Hence, provide 10 mm diameter bars @ 150 C/C,

Bend up alternate bar at L/7 = 3.630 / 7 = 0.518 mm = 520 mm from the centre of bars.

Area of reinforcement along the span (Ly) perpendicular to the above span –

Ast = (Moment of resistance) ÷ { j × (d – d‘ ) × st }

= (6112760) ÷ {0.865 × (130 – 10)×140}= 420 mm^2

Centre to centre spacing of 10 mm diameter bars = (78.5×1000) ÷ 420

= 187 mm

This should not be more than 3d or 300 mm, so 3 × 120 = 360 , or 300mm,

Therefore centre to centre spacing = 190 mm

Bend up alternate bars @ L /7 = (4.130) ÷ 7 = 590 mm

Actual area of provided,Ast= (1000 × 78.54) ÷ 300

= 261.08mm2

=

Check –

1. Shear force for shorter span – Vb = (W × B) ÷ 3 = (7670 × 3.630) ÷ 3 = 9280 N

2. Shear force for longer span – Vl = [L ÷ B] × W × B ÷[ 2 + ( L ÷ B ) ] = [4.130 ÷ 3.630] × 7670 × 3.630 ÷ [ 2 + (4.130 ÷ 3.630) ] = 10091.31 N

Hence, = V ÷ (b × d) τ

= 10091 ÷ (1000 × 130) = 0.077 N /mm^2

= 0.077 N/mm^2τThe permissible shear stress * for M15 grade concrete; P = τ(100Ast) ÷ (b × d) = (100 × 261.08) ÷ (1000×130) = 0.21%And for slab 150 mm over all depth, from table = k × *τ

= 1.30 × 0.21 = 0.273 N/m2

But 0.273 >0.07; hence safe.

Check for development length –

1. Considering , shorter span –M1 = (fst × Ast × x × j × d) ÷ 2 = 140 × 502 × 0.865 × 130 = 3951443 N-mmM1 = 3.95 × 10 ^6 N-mm

=

So,

Development length Ld ≤ [ 1.3 (M1/V) + Lo ]

Anchorage length, Lo = 12 or d (max.)Φ = 12 × 10 or 130 mm = 130 mm

Hence,

Lo = 130 mm will be taken.

Development length = ( × fst )÷ ( 4 *)Φ τ= (10 × 140) ÷ (4× 0.6) = 583 mm Ld = 583 mm

[ 1.3 (M1/V) + Lo ] =[ 1.3 ( 3.95×10^6) ÷ 9280 ] + 130

= 683 mm

Since

M1 / V + Lo ≥ Ld

58 ≤ 683Φ = 683/ 58 = 11.78 mm = 12mm (say)Φ

But dia. of main bar is 10 , so bars safe in developmentΦ .

=

2. Considering longer span of slab –

M1 = (fst×Ast× y × z) ÷ 2

= (140 × 420 ×0.865 × 130) ÷ 2 = 3.3 × 10^6 N-m

Ld = 583 mm

1.3 M1 + Lo =1.3 (3.3× 10^6 ÷ 100.91) + 130

= 555 mmNow , 58 = 555 mm Φ Φ = 9.58 mm (say = 10 mm)Since used bars are also 10 mm , so , bars are safeΦ

DRAWING

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ESTIMATION

METHOD OF ESTIMATION

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Centre Line Method

Long Wall and Short

WallCrossing Method

Long wall short wall method

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Length of long wall

l1=c/c length + width of step

Length of long wall

l1=c/c length + width of step

Length of short walll2=c/c length -width of step

1 Raw water tank-

Description of work Measurements

quantity RemarkNo. L (m) B (m)

h/d

Base slabLong wall 2 50.67 0.90 0.75 68.40 m3 49.77+0.9 =

50.67 m

Short wall2 28.87 0.90 0.75 38.97 m3 29.77 – 0.9 =

28.87 m

Concrete work4 18.37 0.60 0.60 26.45 m3 17.77 +0.6 =

=

Long wall 18.39 m

Short wall3 14.71 0.60 0.60 15.88 m3 15.31 – 0.6 =

14.71 mLong wall 1 12.6 0.60 0.60 4.53 m3 12 + .6

=12.60mShort wall 2 7.05 0.60 0.60 5.07 m3 7.65 – 0.6 =

7.05mOther wall 1 18.54 0.6 0.60 6.67 m3

Total 166 m3

2 Cascade Aerator-

Description of work Measurements

quantity RemarkNo. L (m) B (m)

h/d

Base slabLong wall 2 50.67 0.90 0.30 27.36 m3

Short wall2 28.87 0.90 0.30 15.58 m3

=

Concrete Work

Long wall 4 18.37 0.60 0.15 6.61 m3

Short wall3 14.71 0.60 0.15 3.97 m3

Long wall 1 12.6 0.60 0.15 1.14 m3

Short wall 2 7.05 0.60 0.15 1.26 m3

Other wall 1 18.54 0.6 0.15 1.67 m3

Total 57.6 m3

3 Sedimentation Tank-

Description of work Measurements

quantity

Remark

No. L (m) B (m)

h/d

1ST STEP

=

Long wall 2 50.34 0.57 0.15 11.26 m3 49.77 +0.57 = 50.34 m

Short wall2 29.20 0.57 0.15 4.99 m3 29.77- 0.57 =

29.20m

Long wall 4 18.17 0.40 0.15 4.36 m3 17.76 +.4 =

18.17mShort wall 3 14.91 0.40 0.15 2.68 m3

Long wall 1 12.40 0.40 0.15 0.74 m3 15.31-0.4 =14.91m

Short wall 2 7.25 0.40 0.15 0.87 m3 12+.4 =12.4mOther wall 1 18.54 0.40 0.15 1.12 m3 7.65-0.4 =

7.25m2ND STEP

Long wall 2 50.23 0.46 0.30 13.86 m3 49.77 +0.46 = 50.23 m

Short wall2 29.31 0.46 0.30 8.10 m3 29.77- 0.46 =

29.31m

Long wall 4 18.09 0.30 0.07

51.62 m3 17.76 +.3 =

18.09mShort wall 3 15.01 0.30 0.07

51.01 m3 15.31-0.3

=15.01mLong wall 1 12.30 0.30 0.07

50.276 m3 12+.3 =12.3m

Short wall 2 7.35 0.30 0.075

0.331 m3 7.65-0.3= 7.35m

Other wall 1 18.54 0.30 0.075

0.41m3

3ND STEP

Long wall 2 50.12 0.35 0.40 14.94m3 49.77 +0.35 = 50.12 m

Short wall2 29.42 0.35 0.40 8.23 m3 29.77- 0.35 =

29.42m

Long wall 4 17.96 0.20 0.62

58.98 m3 17.76 +.2 =

17.96mShort wall 3 15.11 0.20 0.62

55.66 m3 15.31-0.2

=15.11mLong wall 1 12.20 0.20 0.62

51.525 m3 12+.2 =12.2m

=

Short wall 2 7.45 0.20 0.625

1.86 m3 7.65-0.2= 7.45m

Other wall 1 18.54 0.20 0.625

2.138m3

TOTAL 94.33m3

4 Rapid Gravity Filter-

Description of work Measurements

quantity RemarkNo. L (m) B (m)

h/d

Long wall 2 50.12 0.35 ----- 35.08m2

Short wall2 29.42 0.35 ----- 20.59m2

Long wall 4 17.96 0.20 ----- 14.37m2

Short wall 3 15.11 0.20 ----- 9.06m2

Long wall 1 12.20 0.20 ----- 2.44m2

Short wall 2 7.45 0.20 ----- 2.98m2

Other wall 1 18.54 0.20 ----- 3.70m2

TOTAL 88.247m2

5 Chlorinator-

Description of work Measurements

quantity RemarkNo. L (m) B (m)

h/d

Long wall 2 50.0 0.23 3.0 69.00m3 49.77 +0.23 = 50.0 m

Short wall2 29.54 0.23 3.0 40.76 m3 29.77- 0.23 =

29.54m

=

Long wall 4 17.65 0.11

54.75 38.57 m3 17.76 +..115 =

17.65mShort wall 3 15.19 0.11

54.75 24.90 m3 15.310.115=1

5.19mLong wall 1 12.115 0.11

54.75 6.618 m3 12+.115

=12.115mShort wall 2 7.353 0.11

54.75 8.232 m3 7.65-0.115=

7.535mOther wall 1 18.54 0.11

54.75 10.12m3

TOTAL 198.217 m3

6 Clear water reservoir at WTP-

Description of work Measurements

quantity RemarkNo. L (m) B (m)

h/d

RCC WORK 1 18.626 15.83

0.10 29.485m3

L= 17.76+.30+.3+.23= 18.626B= 15+.23+.23+.23= 15.83

7 PLASTER WORK

=

Description of work Measurements

quantity RemarkNo. L (m) B (m)

h/d

Long wall 4 50.0 ------ 3.0 600m2

Short wall 2 29.54 ------ 3.0 354.48m2

Long wall 4 17.65 ------ 4.75 670.85m2

Short wall 3 15.19 ------ 4.75 433m2

Long wall 1 12.115 ------ 4.75 115.09m2

Short wall 2 7.353 ------ 4.75 139.39m2

Other wall 1 18.54 ------ 4.75 176.13m2

TOTAL 2488.97m2

8 TOTAL MASSONARY WORK

T. MASSONARY WORK 94.339 m3

=

Estimation(Material cost)

S. N0. Items quantity Rate (Rs.) Cost (Rs.)

1 Cement 19850 bags 350/bag 6947500

2 L.Sand 987 m3 1000/m3 987000

3 C.Sand 829 m3 2100/ m3 1740900

4 Bricks 1124100 5000/1000bricks 224820

5 Grit 524 2200/ m3 1152800

6 Steel 15123 Kg 45/ Kg 680535

Total material cost in Rs. Rs. 11733555/-

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(Labour cost)

S.No. Items Quantity Rate Cost (Rs.)

1 Base slab 166 m3 70/ m3 116202 P.C.C. 57.6 m3 320/ m3 184323 Flouring 1552.37 m2 65/m2 100904.054 Brick work 88.46 m3 530/ m3 46883.85 Staging for brick work 2488.97m2 16/m2 39823.526 Shuttering (slab) 378.22m2 130/m2 49169.697

Plastering 15mm 2488.97m2 35/m2 87113.95

12mm 2488.97m2 42/m2 104536.748 R.C.C. 29.485 510/ m3 15037.35

Total 473521.10

Total cost of project =

COST OF PROJECT

Total cost of project –

Total civil works = Rs. 985261.35/-

Adding –

(a)- 20% of civil work for necessary (electric, water,supply, sanitary fittings ) = Rs. 224381.27/-

(b). 0.5% of civil work for survey work = Rs. 5609.6/-

(c). 4% of civil work for internal road = Rs. 4487.68/-

(d). 2% of civil work for work charge establishment = Rs. 3244/-

(e). 3% of civil work for contingencies = Rs. 5365.76/-

(f). 10% of civil work for contractor profit = Rs. 102190.135/-

Net total cost = Rs. 13608387.79/-

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FIG- SHOWING VARIOUS UNIT OF WATER SUPPLY PROJECT

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FIG- SHOWING RAW WATER TANK

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FIG- SHOWING PRE-SEDIMENTATION TANK

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FIG- SHOWING RAW WATER TANK

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