Design of Water Treatment Units for Kumarakom Panchayath, Kerala

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Design of Water Treatment Units for Kumarakom Panchayath, Kerala

Haritha M1, Rajalakshmi R S2

1 Department of Civil Engineering, Rajiv Gandhi Institute of Technology, Kottayam, Kerala

2 Department of Civil Engineering, Gurudeva Institute of Science and Technology, Kottayam, Kerala

AbstractIn Kumarakom Panchayath (a local government in Kerala state, India), it is observed that the main causes of deterioration in water quality are due to the discharge of domestic wastes, municipal wastes and terrestrial runoff from seepage sites. So, it is necessary to design a water treatment plant for kumarakom Panchayath. A layout of the proposed plant was made consisting of various elements like Screens, Cascade aerators, Flash mixer, Clariflocculator, Rapid gravity filter and a Chlorinator. In this paper, we report the design of a flash mixer unit, rapid sand filter and a chlorinator for the plant. Initially, a physic-chemical analysis was done by collecting data regarding the quality of water in Kumarakom Panchayath.The parameters analyzed are pH, turbidity hardness, alkalinity, acidity, sulphates, chlorides, residual chlorine, nitrates, iron, dissolved oxygen, biochemical oxygen demand (BOD), chemical oxygen demand (COD), most probable number (MPN). Thereafter, a population forecasting was done based on survey details collected with which we could find the capacity of the treatment plant. So, based on the quality of water and capacity of the plant, we made a layout of the treatment plant. Finally, the design of the components was done.

KeywordsTreatment plant, Flash mixer, Rapid sand filter

  1. INTRODUCTION

    Potable water is the fundamental need of man to sustain life. Present study aims to design a water treatment plant for Kumarakom panchayath based on characteristics of water and a schematic treatment plant layout was made using the analyzed data that would provide water to the individual residential lots, neighbourhood center, and utility tracts within the project area.[1] Thus, our work aims to make sure abundant purified water for all needs in the locality.

  2. METHODOLOGY

    1. Population Forecasting

      Initially the population of the Panchayath was forecasted so as to get an idea about how the design of the treatment plant should be done. Geometric Mean Method of Population Forecasting was done here. The population data for the year 1991, 2001, 2011 and 2016 was collected. From these data using the Geometric Mean Method, the Population for the next decade was forecasted.[2-4].From the forecasted population, assuming a per capita water requirement as 120lpcd the peak water demand was calculated. Thus it was found out that the water treatment plant is designed with a capacity of 10MLD for 10 years. The TABLE I shows geometric method for population forecasting. The detailed calculation works involved is as follows.

      TABLE I. GEOMETRIC MEAN METHOD FOR POPULATION FORECASTING

      Year

      Population

      Increase in population in each decade

      Percentage increase in population i.e., growth rate(ri)

      1991

      22500

      500

      4300

      300

      2.22

      18.69

      1.09

      2001

      23000

      2011

      27300

      2016

      27600

      Geometric mean of the growth rate is given by the following formula (1),

      = (1 × 2 × 3) (1) Where, t = 2.5 and r = 2.5 (2.22×18.69×1.09) = 16.81

      Assume design period of plant be 10 years. Therefore n = design period in decade = 1

      = 0 (1 + /100) (2)

      Where,

      P0 = last known population in the census = 27600 r = geometric mean of growth rate = 16.81

      n = 1

      We obtain, 1 = 32240

      Assuming per capita water requirement as 120 lpcd, Total water demand = 32240 × 120

      =3868800m3/day

      = 3.86 MLD

      Assuming a peak factor of 2.5,

      Peak water demand = 3.86 × 2.5

      = 9.6 MLD

      10MLD

      So the water treatment plant is designed with a capacity of 10MLD for 10 years

    2. Analysis of Water Quality Parameters

      The samples taken from the panchayath were tested in the laboratory. Table II shows the values of water quality parameters obtained after the analysis of the samples.

      TABLE II WATER QUALITY PARAMETERS

      Sl. No

      Parameters

      Units

      Observed Value

      Acceptable Limit

      1

      Turbidity

      NTU

      3

      5

      2

      pH

      8

      6.5-8.5

      3

      Acidity

      mg/l

      10

      4

      Alkalinity

      mg/l

      51

      200

      5

      Hardness

      mg/l

      417

      300

      6

      Sulphates

      mg/l

      6.2

      200

      7

      Chlorides

      mg/l

      96

      250

      8

      Residual chlorine

      mg/l

      Nil

      0.2

      9

      Iron

      mg/l

      Trace

      0.3

      10

      Nitrates

      mg/l

      Trace

      45

      11

      DO

      mg/l

      4.6

      4

      12

      BOD

      mg/l

      5.93

      30

      13

      COD

      mg/l

      12

      250

      15

      MPN

      count

      >1100

      Sl. No

      Parameters

      Units

      Observed Value

      Acceptable Limit

      1

      Turbidity

      NTU

      3

      5

      2

      pH

      8

      6.5-8.5

      3

      Acidity

      mg/l

      10

      4

      Alkalinity

      mg/l

      51

      200

      5

      Hardness

      mg/l

      417

      300

      6

      Sulphates

      mg/l

      6.2

      200

      7

      Chlorides

      mg/l

      96

      250

      8

      Residual chlorine

      mg/l

      Nil

      0.2

      9

      Iron

      mg/l

      Trace

      0.3

      10

      Nitrates

      mg/l

      Trace

      45

      11

      DO

      mg/l

      4.6

      4

      12

      BOD

      mg/l

      5.93

      30

      13

      COD

      mg/l

      12

      250

      15/p>

      MPN

      count

      >1100

      2)

      Aerator

      Aeration is done using cascade aerator. It is one of the oldest and most common type of aerator. It consists of a series of steps that the water flows over it similar to a flowing stream. In all cascade aerators, aeration is accomplished in the splash zones. Splash zones are created by placing blocks across the incline. Cascade aerators can be used to oxidize iron and to partially reduce dissolved gases. The Fig.2 shows the aerator.

      3) Flash Mixer Fig.2. Aerator

      After screening out debris and testing raw water, chemicals

    3. Layout of Proposed Treatment Plant

    A layout of the treatment plant was made consisting of various units such as screens, aerator, flash mixer, clari- flocculator, rapid sand filter and a chlorinator. The Fig.2 shows the components of Treatment Plant

    Fig.2. Components of Treatment Plant

    that encourage coagulation are added to the water stream. The mixture is agitated quickly and thoroughly in a process called flash mixing. [5-7].The chemicals introduced into the water stream will attract any very fine particles, such as silt, that will not readily settle or filter out and make them clump together. These larger, heavier formations are called floc, which are much easier to remove from the water. The Fig.3 shows the flash mixer.

    1) Screening

    Screening is the first operation at any water and waste water treatment works. This process essentially involves the removal of large non-biodegradable and floating solids that frequently enter a water works, such as rags, papers, plastics, tins, containers and wood.

    Efficient removal of these constituents will protect the downstream plant and equipment from any possible damage, unnecessary wear & tear, pipe blockages and the accumulation of unwanted material that will interfere with the required water treatment processes.

    Waste water screening is generally classified into either coarse screening or fine screening and they may be cleaned manually or mechanically.

    .

    Fig.3. Flash mixer

    1. Clariflocculator

      The clariflocculation is a chemical and physical water treatment process and it mainly consists in the removal of suspended substances. Its application is in the water treatment of surface water, industrial and municipal wastewater treatment, filtration pre-treatment, Reverse Osmosis, both in the municipal and industrial sectors.[8- 10]

      The clari-flocculation process is a traditional and consolidated treatment. The sequence of operations sedimentation-precipitation-coagulation-flocculation is the most common technique used in the world for the production of potable water. The process requires low costs for high volumes of purified water. It is a reliable process suitable for automatic control. The Fig.4 shows the Clariflocculator.

  3. DESIGN UNITS

      1. FLASH MIXER

        Peak flow =416.6 m3/hr

        Assume detention time (Dt) =30sec

        [Range between 20-60sec]

        Fig.4. Clariflocculator

    1. Rapid Sand Filter

      The rapid sand filter or rapid gravity filter is a type of filter used in water purification and is commonly used in municipal drinking water facilities as part of a multiple-stage treatment system. Rapid sand filters use relatively coarse sand and other granular media to remove particles and impurities that have been trapped in a floc through the use of flocculation chemicalstypically alum. The unfiltered water flows through the filter medium under gravity or under pumped pressure and the floc material is trapped in the sand matrix. The Fig.5 shows a Rapid sand filter.

      Fig.5. Rapid sand filter

    2. Chlorination

    Water chlorination is the process of adding chlorine (Cl2) or hypochlorite to water.[11-15] This method is used to kill certain bacteria and other microbes in tap water as chlorine is highly toxic. In particular, chlorination is used to prevent the spread of waterborne diseases such as cholera, dysentery, etc. The Fig. 6 shows the chlorine containers in chlorine room.

    Volume of tank (V) = Flow x Dt ……………………………… (3)

    We obtain the volume as 3.5 m3

    But, Volume = Area x Depth of tank

    3.5 = × D2 x 1.5 D

    (Assume ratio of tank height to diameter =1.5:1) [Range is 1:1 to 3:1]

    D = 1.43 m 1.5 m

    So, provide tank diameter of 1.5 m Depth = 1.5 D = 1.5 x 1.5 = 2.25 m

    Provide a freeboard of 0.2 m Total depth = 2.25+ 0.2 = 2.45 m

    Power Requirements Power required (P) = G2v

    Assume velocity gradient (G) = 600/sec [permissible limit >

    300/sec]

    Absolute viscosity () = 0.798 x 10-3 NS/m2 at 30 C Volume to which power is supplied (V) = 3.5 m3

    P = G2v = 6002 x 0.798 x 10-3 x 3.5 = 1005.48 Watts

    1005 Watts

    Power per unit volume =

    =287.14 Watts/m3 Power per unit flow of water

    =

    = 2.41 Watts

    Assume alum dosage to be supplied is 15 mg/l

    Water to be treated = 10MLD =10 x106 l/day Hence total requirement of alum

    Fig.6. Chlorine containers stored in chlorine room

    =15 x 10 x 106 mg/day x 365

    = 5.475 x 1010 md/year

    = 54750 kg/year

    Step 2: Design of sand media

    Provide a media of 0.6mm size and uniformity coefficient of

    1.4 to a depth of 60cm

    = B x 29323 (check for depth of media)

    [Range:

    Depth, l =

    Uniformity coefficient – 1.2 to 1.6 Effective sizes (d) – 0.35 to 0.7mm Depth – 0.6 to 0.75 m]

    Fig.7. Design of flash mixer

    Assume height (h) as 2.5m

    1. RAPID SAND FILTER B = 4×10-4 (worst condition) Q = 2 x filtration rate

      Inflow to one Rapid sand filter =0.056m3/sec

      Step 1: Design of filter box

      Assume quantity of backwash water is 3% of total wash filtered

      Total water to be filtered = 204.134+ 204.134 x

      = 210.25 m3/ hr Actual filtration rate considering 30 min for backwash,

      =

      = 214.73 m3/hr Assume filtration rate = 5000 l/hr /m2

      Therefore, area of filter bed = =

      l = = 0.46 m = 46cm Given 60cm > 46cm.

      Therefore safe

      Step 3: Design of gravel bed

      We are providing four layers Top layer = 2mm Middle layer 1 =10mm

      Middle layer 2 = 20mm Bottom layer = 40mm Using the equation,

      l = 2.54 x k log d

      =42.9

      =43m2 (20-50 m2)

      Assume length to breadth ratio as 1.3:1 [Range is 1.2 to 1.33:1]

      where,

      l = depth and

      d = particle size

      =

      L = 1.3 B

      Area = length x breadth 43= L x B

      43 = 1.3 B x B

      1.3 B2 = 43

      Therefore, B = 5.75 m and L

      = 1.3 x B = 1.3 x 5.75

      = 7.47m

      Assume k = 12 (range is 10-14)

      ltop = 2.54 x 12 x log 2 = 9.17cm

      l middle1 = 2.54 x 12 x log 10 = 30cm l middle 2 = 2.54 x 12 x log 20 = 39cm l bottom = 2.54 x 12 x log 40 = 48cm Therefore ,

      Total depth = 48cm 50cm

      Depth of each layer

      Top layer

      Depth = 9.17cm 10 cm Average size = 2 mm

      Middle layer 1

      Depth = 30- 9.17 = 20 cm Average size = 10 mm

      Middle layer 2

      Depth = 39- 30 = 9 cm 10 cm Average size = 20 mm

      Bottom layer

      Depth = 48 – 39 = 9 cm 10 cm Average size = 40 mm

      x1 = 0.178

      x = 0.25 + x1 = 0.25 +0.178 = 0.428

      = 0.428

      Total area of lateral

      = 0.250 m2

      iii) Area of manifold = 1.5 to 2 times area of lateral

      = 1.6 x area of lateral

      = 1.6 x 0.250

      = 0.401m2

      Step 4: Design of under-drainage system Diameter of manifold

      1. Total area of perforation

        Assume size of perforation = 10 mm

        Assume = 0.0025

        (Range is 0.002- 0.003)

        Total area of perforation = 0.0025 x (L x B)

        = 0.0025 x (7.47 x 5.75)

        =0.1073 m2

      2. Assume = x

      For 5mm perforation, ratio is 0.25 For 12mm perforation, ratio is 0.5

      We need to find the ratio for 10mm perforation

      x d2 = area

      x d2 = 0.401

      Therefore diameter = 0.714m 71cm Provide 71cm piping

      Assuming a spacing of 15cm for the laterals,

      Total number of laterals = = = 49.8 50 Therefore number of laterals = 50

      Area of one lateral pipe

      = =

      = 5 x 10-3 m2

      x d2 = 5 x 10-3

      Therefore diameter = 0.079m 79mm

      Perforations

      x = 0.25 + x1

      =

      Fig.8. Calculation of X

      Given 10mm perforations,

      Total area of perforation = 0.1073 m2

      Area of one perforation = x d2 = x (10 x 10-3)2 = 7.85 x 10-5 m

      Total number of perforations = = 1366.87 1367

      Number of perforation in one lateral =

      = 27.34

      = 28

      Length of one lateral = =

      = 2.52m

      Spacing of perforation = = 0.09m

      (Spacing must be between 8-20cm) Hence OK

      71cm manifold is provided for a length of 7.47m.It is to be connected with 79mm laterals at a spacing of 15cm c/c .Each lateral should be provided with 10mm perforation at spacing of 9mm c/c providing one perforation at a cross section.

      So, in one trough ,Q = 0.429/3 = 1.36 x 0.4 x y 3/2 (assume b= 0.4)

      y = depth of trough = 0.410m 41cm

      Total depth of filter basin = 0.5m (freeboard) + 1.2m (water)

      + 0.6m (sand) + 0.5m (gravel) + 0.71m (manifold) =3.51m

      Fig.11. Design of wash water trough

      Outflow from rapid sand filter, reducing 3% of sludge

      = 210.25 x 210.25 = 204 m3/hr

    2. CHLORINATOR

    Inflow to disinfection = 204 x 2 = 408 m3/hr = 9.8MLD

    Dosage of chlorine required to have 0.3mg/l of residue = 2mg/l

    Assuming HOCl is applied to flash mixer

    Quantity of HOCl provided = 2mg/l x 9.8 x 106 l/day

    Fig.10. Design of under drainage system

    Step 5: Wash water trough

    Assume backwash water flow rate = 60cm/min = 60cm3/min/cm2

    Total quantity of water coming from the filter = Q

    Using the equation,

    Where,

    = 19.6 x 106 mg/day

    = 19.6 kg /day 20kg /day

    C 0.86 t p = 6.3

    Using the equation,

    Q= 1.376by3/2

    tp = contact time which is the detention time

    C= dosage of chlorine required to have 0.3mg/l of residue

    Q= 60cm/min = 0.6m/min = 0.6(7.47 x 5.75) m3/min

    = 25.77/60

    = 0.429 m3 /sec

    Align the wash water trough along the length of basin ,total of three numbers ,provided at a spacing of 1.5m c/c

    Substituting value of C and tp in the above equation ,

    2 0.86 t p = 6.3

    Therefore tp = 3.47min Provide a safety factor of 2.5

    Therefore, tp = 3.47 x 2.5 = 8.675 min 9 min

    Volume = Discharge (Q) x Detention time(Dt)

    = 408m3/hr x 8.67

    = x 8.675

    = 58.99m3 60m3

  4. RESULTS AND DISCUSSIONS

    1. FLASH MIXER

      Provide a tank of 1.5 m diameter and 2.45 m depth which includes 0.2m freeboard for flash mixing of chemicals.

      Power required per unit volume of water = 2.41 Watts Alum dosage to be applied = 15 mg/L Therefore,

      Total alum requirement for an year = 54750 Kg / year

    2. RAPID SAND FILTER

      Filter box

      Inflow to one R.S.F = 0.056 m3/sec Total water to be filtered = 210.25 m3/hr

      Actual filtration rate considering 30min for backwash

      = 214.73 m3/hr

      Area of filter bed = 43 m2

      Length of filter bed =7.47m

      Breadth of filter bed = 5.75m

      Sand media

      Depth, l = 0.46m Gravel bed

      Total depth = 50cm

      1. Top layer

        Depth = 9.17cm 10 cm Average size = 2 mm

      2. Middle layer 1

        Depth = 30- 9.17

        = 20 cm Average size

        = 10 mm

      3. Middle layer 2

        Depth = 39- 30 = 9 cm 10 cm Average size = 20 mm

      4. Bottom layer

      Depth = 48 – 39 = 9 cm 10 cm Average size = 40 mm

      Under drainage system

      Total area of perforation = 0.1073m2

      Total area of laterals = 0.250m2 Area of manifold = 0.401m2 Diameter of manifold = 71cm

      Total number of laterals = 50

      Area of one lateral pipe = 5 x 10-3

      Diameter of lateral pipe = 79mm

      Area of one perforation = 7.85 x 10-5m2 Total number of perforation =1367

      Number of perforation in one lateral = 28 Length of one lateral = 2.52m Spacing of perforation = 9cm

      Wash water trough

      Total quantity of water coming from the filter

      = 0.429 m3/sec

      Quantity of water coming from one filter if three filters are provided = 0.143 m3/sec

      Depth of trough = 41cm Total depth of filter basin = 3.51m

      Outflow from filter reducing 3% of sludge

      =204 m3 /hr

    3. CHLORINATOR

    Inflow to disinfection = 9.8 MLD

    Dosage of chlorine required to have 0.3mg/l of residue

    = 2mg/l

    Quantity of HOCl provided = 20kg /day Detention time tp = 9 min Volume of tank = 60m3

  5. CONCLUSION

    The study concludes that lake water of study area was moderately polluted in respect to analyzed parameters. pH, total hardness, chloride and fluoride were found within permissible limit prescribed BIS. But the higher values of BOD and bacterial content in present study attributed lake water were not fit for drinking purpose. It is necessary to aware local villagers to safeguard the precious river and its surrounding. The concerned authorities should strictly monitor the quality of drinking water being supplied to the consumers to ensure public health. A majority of the people as well as the resorts here now buy drinking water. Those who can't afford to do that depend on the lake water and water supplied by the panchayat, which is not very regular. So a water treatment plant is designed to purify the lake water to potable water standards. The design of water treatment plant is done by conventional method of water treatment plant design by assuming some constant values involves processes that alter the chemical composition or natural "behavior" of water. A coarse strainer including a removable basket screen with 50 to 100 mesh, is positioned at the intake point of surface water to remove larger particulate matter and this water is fed to the raw water tank. Treatment steps include the addition of chemical coagulants (clotting agents), pH-adjusting reagent chemicals that react to form floc. Floc then settles using the force of gravity into settling tanks or is removed as the water percolates through gravity filters. The clarification process is designed to removes particles larger than 25 microns.

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