Reed Bed System

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CHAPTER 1 INTRODUCTION 1.1 REED A tall, slender-leaved plant of the grass family, which grows in water or on marshy ground. 1.2 REED BED Reed beds are aquatic plant based systems which allow bacteria, fungi and algae to digest the sewage and clean the water. Reed bed is one of the natural and cheap methods of treating domestic, industrial and agricultural liquid wastes. Reed bed is considered as an effective and reliable secondary and tertiary treatment method where land area is not a major constraint. Generally reed bed is installed with a drain pipe in a bed of pieces of lime stones and filled up with pebbles and graded sand. In this sandy body, reed plants generally with hollow root which bring oxygen into the filter bed are planted. These systems use wetland plants, soils and their associated microorganisms to remove contaminants from wastewater. Plants provide an environment for microbes to live, they oxygenate the wastewater, 1

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Reed Bed System

Transcript of Reed Bed System

CHAPTER 1

INTRODUCTION

1.1 REED

A tall, slender-leaved plant of the grass family, which grows in water or on

marshy ground.

1.2 REED BED

Reed beds are aquatic plant based systems which allow bacteria, fungi

and algae to digest the sewage and clean the water.

Reed bed is one of the natural and cheap methods of treating domestic,

industrial and agricultural liquid wastes. Reed bed is considered as an effective

and reliable secondary and tertiary treatment method where land area is not a

major constraint.

Generally reed bed is installed with a drain pipe in a bed of pieces of lime

stones and filled up with pebbles and graded sand. In this sandy body, reed

plants generally with hollow root which bring oxygen into the filter bed are

planted. These systems use wetland plants, soils and their associated

microorganisms to remove contaminants from wastewater. Plants provide an

environment for microbes to live, they oxygenate the wastewater, providing

nutrients for the microbes to survive, they stabilize the soil and they also partake

in the reduction of nutrients.

Reed bed treatment system utilizes the active treatment capabilities of soil to

biologically treat effluents such as sewage, industrial wastewater, run-off and

leachates.

There are two types of flow in reed bed 

Vertical flow

Horizontal flow

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CHAPTER 2

LITERATURE REVIEW

2.1 TREATMENT OF TOXICITY SLUDGE BY REED BED

Jiangang Li, Yubo Cui carried out a project based on “Biological Toxicity of

Sewage Sludge Stabilized by Reed Bed” ,College of Environment and

Resources, Dalian Nationalities University, Dalian, China.

With the expanding scale of urban wastewater treatment, the resulting

excess sludge quantity is also growing. Excess sludge treatment and disposal

has become an important part of the sewage treatment. Sludge itself is rich in

essential nutrients of plant growth such as nitrogen and phosphorus, but it often

also contains harmful substances such as heavy metals. Tests are conducted in

this research by comparing the effects on the absorption and transformation of

toxic substances between traditional sludge drying bed and reed bed.

2.2 ENERGY EFFICIENT SLUDGE TREATMENT BY REED BED

Lakeville carried out a project based on “Energy Efficient Sludge Treatment

with REED BED Technology Demonstration Project”, NY, Water and Sewer

Authority, Livingston.

Reed bed technology involves the application of domestic wastewater

sludge to beds that have been planted with a specialized species of reeds Sludge

applied to reed beds is turned into a compost-like material that can be used as a

soil conditioner. Reed beds act to dewater and reduce the organic content of the

sludge, reduce the metals concentrations of the sludge, and stabilize the sludge

for subsequent disposal. This is the result of the following: first, the reed root

system provides oxygen to the sludge, which increases the activity and

population of microorganisms that mineralize the sludge; second, the growth of

the plants makes use of the nutrients, minerals, and water in the sludge.

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2.3 TREATMENT OF WASTEWATER USING REED BED

Dr.Rangarajan, Dr.Sakunthala, carried out a project based on “Construction

and Performance evaluation of reed bed wastewater treatment unit” is done

by Department of Civil Engineering, Vel Tech Multi Tech Engineering College,

Chennai 600062, Tamil Nadu.

Reed Bed System is the artificial wastewater treatment system consisting

of shallow ponds or channels which have been planted with aquatic plants, and

which rely upon natural, biological, physical and chemical processes to treat

wastewater. It typically has impervious clay or synthetic layer and engineered

structures to control the flow direction, water level and liquid retention time.

These plants can be used to treat variety of wastewaters including urban run-off,

municipal, industrial, agricultural and acid mine drainage. Reed beds have

positive characteristics of a natural wetland and can also be controlled to

eliminate the negative aspects of natural wetlands.

2.4 INDUSTRIAL & MUNICIPAL WASTEWATER TREATMENT

Mehali J. Mehta, carried out a project based on “Treatment of Municipal and

Industrial wastewater by REED BED Technology” Assistant Professor,

Department of Civil Engineering, Sarvajanik College of Engineering and

Technology, Gujarat, India.

Reed bed system for wastewater treatment has been proven to be effective and

sustainable alternative for conventional wastewater treatment technologies. Use

of macrophytes to treat wastewater is also categorized in this method. This new

approach is based on natural processes for the removal of different aquatic

macrophytes such as floating, submerged and emergent. Macrophytes are

assumed to be the main biological components of wetlands. These techniques

are reported to be cost effective compared to other methods.

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Various contaminants like total suspended solids, dissolved solids,

electrical conductivity, hardness, biochemical oxygen demand, chemical oxygen

demand, dissolved oxygen, nitrogen, phosphorous, heavy metals, and other

contaminants have been minimized using aquatic microphytes.

2.5 HOSPITAL WASTEWATER TREATMENT USING REEDBED

Badejo. A, carried out a project based on “Tertiary Hospital Wastewater

Treatment using Reed Bed Technology” Civil Engineering Department,

College of Engineering, University of Agriculture, Abeokuta, Nigeria.

Tertiary hospital wastewater in Nigeria constitutes a risk to public health

due to inadequate treatment. Reed bed technology using locally available

macrophytes holds tremendous potentials for biological wastewater treatment.

removal of organic and inorganic pollutants from tertiary hospital was therefore

investigated. Characteristics of wastewater such as pH, Nitrates (NO3-),

Phosphates (PO43-) and Ammonia (NH3) contents, Suspended Solids (SS),

Biochemical Oxygen Demand (BOD), Dissolved Oxygen (DO).

2.6 WASTEWATER TREATMENT USING WETLANDS

Jan Vymazal, carried out a project based on “Constructed Wetlands for

Wastewater Treatment Department of Landscape Ecology” Faculty of

Environmental Sciences, Czech University of Life Sciences, Czech Republic.

The constructed wetlands have evolved into a reliable wastewater

treatment technology for various types of wastewater. The classification of

constructed wetlands is based on: the vegetation type (emergent, submerged,

floating leaved, free-floating); hydrology (free water surface and subsurface

flow); and subsurface flow wetlands can be further classified according to the

flow direction (vertical or horizontal). In order to achieve better treatment

performance, namely for nitrogen, various types of constructed wetlands could

be combined into hybrid systems.

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CHAPTER 3

METHODOLOGY

3.1 REED BEDS WITH EMERGENT MACROPHYTES

Various emergent macrophytes species can be used in constructing

wetlands, including cattails, bulrushes, reeds, rushes. Constructed wetland for

wastewater treatment with emergent macrophytes can be constructed with

different design. In general these can be categorized into two major groups

according to their flow pattern: free water surface systems (FWS Wetlands) and

system with subsurface flow (SSF wetlands); subsurface flow wetland further

categorized into horizontal subsurface flow systems (HSSF or HF Wetlands)

and vertical subsurface flow systems (VSSF Wetlands).

3.2 DOWN FLOW OR VERTICAL FLOW (VF)

This design requires dosing of the bed’s surface using a network of pipes

using either a pumping or a siphon system. The idea is to flood the surface of

the reed bed a number of times per day. As the water flows down through the

bed, it draws air in, creating the right bacterial environment. VF reed beds are

very effective in removal of BOD, ammonia and some heavy metals and take up

less area for similar treatment compared to SSHF. The efficiency of SSHF and

VF reed beds may be improved by adding certain chemicals to the water during

the treatment. This dosing technique can be used for COD or phosphorous

removal in industrial process water, for example. Water can be treated

progressively through multiple reed bed stages and some or all of the above

systems can be incorporated into a complete treatment system.

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FIG 3.2 VERTICAL FLOW

3.3 MACROPHYTES & IT’S ROLE

The three types of macrophytes are emergent, free- floating and

submerged which have been discussed above. Macrophytes play a major role in

reed beds, influencing biological, chemical and physical treatment processes.

The most important function of macrophytes in reed beds has been categorized

by as physical and metabolic. Physical effects include: Filtration of suspended

material, protection against erosion by reducing turbulence and flow velocities

stabilization of sediments and providing the surface area for micro-organisms.

Metabolic functions of macrophytes include nutrient uptake and O2 release from

roots into the rhizosphere. Macrophytes have adapted to anaerobic conditions

by developing internal air spaces which transport O2to the root zone. Research

differs on the potential for macrophytes to release O2 from roots to the

surrounding thus providing aerobic conditions for plant nitrification to occur. A

study by concluded that internal O2 movement not only supplied to buried plant

tissues but also leaked O2 into the rhizosphere. Macrophytes can also provide

habitat for flora and fauna and increase aesthetic appeal. Research differs on the

significance of plant uptake in nutrient removal with nutrient loading is an

important part in the proportion of nutrient removal by plant uptake.

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3.4 LAYERS OF REED BED

3.4.1 CHARCOAL

Charcoal filters are used in the purification process of many liquids Charcoal

filters are used in air conditioning units and exhaust fans to rid air of unwanted

odors. The process is known as leaching Removal of materials by dissolving

them away from solids. Charcoal filters come in different forms such as solid

carbon, impregnated foam materials, powder & cloth, which is shown in the fig

3.4.1

FIG 3.4.1 CHARCOAL

3.4.2 COARSE & FINE GRAVEL

Coarse and fine layers of gravel filters are provided in the reed bed system.

Different sizes of gravel are provided for the effective filtration. Which removes

the minute suspended particles for the waste water because they form permeable

layers, Due to tiny spaces water can pass slowly. A slower flow of water

through the system produces clean water. Which is shown in the fig 3.4.2a and

3.4.2b.

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FIG 3.4.2a FINE GRAVEL

FIG 3.4.2b COARSE GRAVEL

3.4.3 LIME

Lime are used in Industrial waste water treatment to adjust pH and alkalinity

in coagulation, flocculation and biological treatment process. It is used to soften

process or boiler feed water, precipitate metals and non-metals and to adjust pH

with membrane treatment and to stabilize biosolids by killing pathogens, viruses

and reducing vector attraction to produce high quality agricultural fertilizer. It is

used to nutrient control of phosphorous, as lime precipitates phosphorous to

very low levels without biological treatment. Which is shown in fig 3.4.3.

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FIG 3.4.3 LIME

3.4.4 SAND

Sand filters can produce very high quality water free from pathogens, taste

odour without the need for chemical aids. It traps the suspended materials and it

reducing the numbers of bacteria and removing most of solids. Sand filters

become clogged with floc after a period in use and then they are back washed or

pressure washed to remove the floc. In these filters the sand traps residual

suspended material and bacteria and provides a physical matrix for bacterial

decomposition of nitrogenous material, including ammonia and nitrates, into

nitrogen gas. Which is shown in the fig 3.4.4.

FIG 3.4.4 SAND

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3.4.5 REED

A tall, slender-leaved plant of the grass family, which grows in water or on

marshy ground. Wet plants transfer atmospheric oxygen down through their

roots in order to survive in water logged conditions. This creates both aerobic

and anaerobic soil conditions. It creates a channel for the waste water to pass

through, and the roots introduce oxygen into the body of soil and provides an

environment for the bacteria to survive, These organisms are necessary to break

down many types of compound, in particular the oxidation of ammonia to

nitrate. Finally the plants themselves take up a certain amount of nutrients from

the waste water, which acts as a Natural Fertilizer. Which is shown in the fig

3.4.5.

FIG 3.4.5 REED

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3.5 MATERIALS USED

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FIG 3.5 MATERIALS REQUIRED

3.6 STEP BY STEP CREATION OF MODEL

STEP 1

FIG 3.6a FILLING WITH CHARCOAL

Placing and levelling of charcoal in the fifth (or) final layer for the effective filtration, for the height of about 100 mm.

STEP 2

FIG 3.6b FILLING WITH COARSE AND FINE GRAVEL

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Placing the Coarse Gravel sieve size is about 20mm & height 100mm and Fine Gravel sieve size is about 4.76 mm & height 75mm as a fourth layer.

STEP 3

FIG 3.6c FILLING WITH LIME

Placing the lime as a third layer of 50 mm height for reducing the chemical impurities.

STEP 4

FIG 3.6d FILLING WITH SAND

Placing the sand as a second for the initial arrestment of suspended impurities and sand sieve size is about 600 µ ; Height 75 mm.

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STEP 5

FIG 3.6e PLANTING REEDS

Plantation of REEDS as a initial layer or first stage of height is about 75 mm height, Free Board is provided as 75mm and Total Height of the filtration model is 550mm.

FINAL OUTPUT MODEL

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FIG 3.6f OUTPUT MODEL

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Flow Water

Free Board

Reed and it’s soil (75 mm height)

Sand sieve size is about 600 µ , Height 75 mm

Lime 50 mm height

Fine Gravel sieve size is about 4.76 mm

& height 75mm

Coarse Gravel sieve size is about 20mm

& height 100mm

Charcoal height is about 100 mm

Treated Water

FIG 3.6g : REED BED WITH DIMENSIONS

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CHAPTER 4

RESULTS AND DISCUSSIONS

4.1 COLLECTION OF WASTE WATER SAMPLES

• Effluent WasteWater is collected from DAIRY INDUSTRY @

Sholinganallur.

• WasteWater is collected from SEWAGE TREATMENT PLANT @

Agni College Of Technology.

4.2 ANALYZING THE CHARACTERISTICS OF SAMPLES

4.2.1 TOTAL SUSPENDED SOLIDS

FIG. 4.2.1 TOTAL SUSPENDED SOLIDS

Total suspended solids (TSS) gives a measure of the turbidity of the

water. We cannot see pH or other kinds of water qualities, but we can observe

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TSS directly. Suspended solids cause the water to be milky or muddy looking

due to the light scattering from very small particles in the water. Sometimes it is

mixed with color, but colored waters can also be clear. Normally, we notice

suspended solids before we notice anything else. Polluted waters are commonly

turbid and improvement is usually marked by greater clarity. Of course, good

and useful waters may be turbid, and many clean rivers are never clear because

they contain fine suspended minerals that never settle.

To determine total suspended solids, weigh a piece of filter paper as

accurately as possible. Filter a one liter sample of water through the weight

filter paper. Allow the filter paper to dry completely. Placing a lamp above the

filter paper may help the drying process, but take care in not getting the filter

paper too hot.

Preparation of the glass fiber filter disk Insert the filter disk onto the base

and clamp on funnel. While vacuum is applied, wash the disk with three

successive 20mL volumes of distilled water. Remove all traces of water by

continuing to apply vacuum after water has passed through. Dry in a oven at

103-105 C for one hour in china dish. When needed, remove dish from the

oven, desiccate, and weigh in dish. Re-dry and re-weigh filter until weight

change is less than 4% of previous weight or 0.5 mg. Place the filter on the base

and clamp on funnel and apply vacuum. Stir sample continuously while sub-

sampling and quantitatively transfer the sample to the filter using a 100 mL

graduated cylinder. Remove all traces of water by continuing to apply vacuum

after sample has passed through. Carefully remove the filter from the base. Dry

at least one hour at 103-105 C. Cool in a desiccator and weigh.

Reweigh the filter paper. The change in weight is the weight of the total

suspended solids. TSS values are commonly expressed in ppm (mg solids per

liter of water). which is shown in the fig. 4.2.1.

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TABLE 1 : TOTAL SUSPENDED SOLIDS FOR DIARY SAMPLE

Treatment Before AfterSample Sample 1 Sample 2 Sample 1 Sample 2Empty

Weight (g)

35.210 31.170 36.170 33.210

Paper

Weight (g)

1.070 1.070 0.980 0.980

Original

Weight (g)

36.300 32.250 37.160 34.210

Sample Sol.

(ml)

100 100 100 100

Suspended

Solids (mg/l)

2000 1000 100 200

TABLE 2 : TOTAL SUSPENDED SOLIDS FOR STP SAMPLE

Treatment Before AfterSample Sample 1 Sample 2 Sample 1 Sample 2Empty

Weight (g)

33.800 36.490 34.662 34.664

Paper

Weight (g)

1.080 1.080 0.950 0.950

Original

Weight (g)

34.890 37.590 35.617 35.621

Sample Sol.

(ml)

100 100 100 100

Suspended

Solids (mg/l)

100 200 50 70

TOTAL SUSPENDED SOLIDS

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Total Suspended Solids = Final Weight (g )−Original Weight (g)

Sample Volume (ml)×106

TOTAL SUSPENDED SOLIDS FOR DIARY SAMPLE

BEFORE TREATMENT

Sample 1 = 36.30−(35.21−1.07)100

×106× 10

= 2000 mgl

Sample 2 = 32.25−(31.17−1.07)100

× 106 ×10

= 1000 mgl

AFTER TREATMENT

Sample 1 = 37.16−(36.17−0.98)100

×106

= 100 mgl

Sample 2 = 34.21−(33.21−0.98)100

×106

= 200 mgl

TOTAL SUSPENDED SOLIDS FOR STP SAMPLE

BEFORE TREATMENT

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Sample 1 = 34.89−(33.80−1.08)100

×106

= 100 mgl

Sample 2 = 37.59−(36.49−1.08)100

×106

= 200 mgl

AFTER TREATMENT

Sample 1 = 35.617−(34.662−0.95)100

× 106

= 50 mgl

Sample 2 = 35.621−(34.664−0.95)100

× 106

= 70 mgl

4.2.2 pH-RANGE

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FIG 4.2.2 pH - METER

pH is measured using pH meter, which comprises a detecting unit

consisting of glass electrode or reference electrode, usually a calomel electrode

connected by a KCI bridge to the pH sensitive glass electrode and an indicating

unit which indicates the pH corresponding to the electromotive force is then

detected. Before measurement, pH meter should be calibrated by using at least

two buffers. Also it is recommend to use hydrated silica gel for the glass

electrode, and the electrode must be soaked with water or in suitable buffer

followed by rinsing in water. Electrode tips should be cleaned after use of

wiping

with tissue paper to remove adhering substances. Potassium level in the calomel

electrode is maintained an the cap should be removed during the measurement.

For the accurate measurement of pH, the temperature of the buffer should be

maintained for the standardization of pH meter is same.

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pH – RANGE FOR DIARY SAMPLE

BEFORE TREATMENT

Sample 1 : 7.5

Sample 2 : 7.5

AFTER TREATMENT

Sample 1 : 11.5

Sample 2 : 11.3

pH – RANGE FOR STP SAMPLE

BEFORE TREATMENT

Sample 1 : 7.5

Sample 2 : 7.7

AFTER TREATMENT

Sample 1 : 11.2

Sample 2 : 11.3

4.2.3 TOTAL SOLIDS

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FIG 4.2.3 TOTAL SOLIDS

Total solids, TS, is a measure of all the suspended, colloidal, and

dissolved solids in a sample of water. This includes dissolved salts such as

sodium chloride, NaCl, and solid particles such as silt and plankton If the levels

of total solids are too high or too low, it can impact the health of the stream and

the organisms that live there. High levels of total solids will reduce the clarity of

the water. This decreases the amount of sunlight able to penetrate the water,

thereby decreasing the photosynthetic rate. Reduced clarity also makes the

water less aesthetically pleasing. While this may not be harmful directly, it is

certainly undesirable for many water uses. When the water is cloudy, sunlight

will warm it more efficiently. This occurs because the suspended particles in the

water absorb the sunlight which, in turn, warm the surrounding water. This

leads to other problems associated with increased temperature levels.

Prepare two 250 mL beakers for drying and sample evaporation.

Carefully clean two 250 mL beakers and place them in a drying oven at 100–

105C for at least one hour to dry. Use an analytical balance to measure the

mass of each beaker. Transfer the samples to the beakers. Swirl the samples to

attain uniformity of suspended particles. Using a 100 mL graduated cylinder,

carefully measure 200 mL of sample water into each beaker. place the beakers

into the oven and allow the water to evaporate overnight at a temperature of

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around 100-105C. Next day Measure the mass of the beakers and solids.

remove the beakers

from the oven and place them in a dessicator, if available, to cool. A dessicator

will keep the samples from absorbing any water from the air that would increase

their mass. If no dessicator is available, the beakers can be cooled on a table top.

Obtain the mass of the solids by subtracting the mass of the empty beaker from

the mass of the beaker with the solids.

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TABLE 3 : TOTAL SOLIDS FOR DIARY SAMPLE

Treatment Before AfterSample Sample 1 Sample 2 Sample 1 Sample 2Empty

Weight (g)

33.92 36.53 34.24 36.19

Original

Weight (g)

34.05 36.65 34.23 36.18

Sample Sol.

(ml)

25 25 25 25

Total Solids

(mg/l)

5200 4800 400 400

TABLE 4 : TOTAL SOLIDS FOR STP SAMPLE

Treatment Before AfterSample Sample 1 Sample 2 Sample 1 Sample 2Empty

Weight (g)

34.67 30.80 33.80 36.53

Original

Weight (g)

34.69 30.82 33.79 36.52

Sample Sol.

(ml)

25 25 25 25

Total Solids

(mg/l)

800 800 400 400

TOTAL SOLIDS

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TOTAL SOLIDS = Final Weight−Empty WeightVolume of Sample

×106

TOTAL SOLIDS FOR DIARY SAMPLE

BEFORE TREATMENT

Sample 1 = 34.05−33.9225

× 106

= 5200 mgl

Sample 2 = 36.65−36.5325

× 106

= 4800 mgl

AFTER TREATMENT

Sample 1 = 34.24−34.2325

×106

= 400 mgl

Sample 2 = 36.19−36.1825

× 106

= 400 mgl

TOTAL SOLIDS FOR STP SAMPLE

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BEFORE TREATMENT

Sample 1 = 34.69−34.6725

×106

= 800 mgl

Sample 2 = 30.80−30.8225

× 106

= 800 mgl

AFTER TREATMENT

Sample 1 = 33.80−33.7925

× 106

= 400 mgl

Sample 2 = 36.53−36.5225

× 106

= 400 mgl

4.2.4 TOTAL DISSOLVED SOLIDS

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Total dissolved solids (or TDS) is the measure of all organic and

inorganic substances dissolved in a given liquid, revealing the proportion of

different solids. There are a number of different uses for TDS: it can measure

pollution levels in lakes and rivers or mineral levels in drinking water, for

example, and also has agricultural applications in irrigation.

A clean, properly sterilized beaker that is free of dust or other particles

sample of water, poured into the beaker, Filter paper, An evaporating dish, A

stirring stick, A pipette large enough to collect a 50 ml sample, A scale.

Weigh the evaporating dish in milligrams (mg). Make sure that it is

completely dry and completely clean of any extraneous particulate matter. Stir

the water sample in the beaker with your stirring stick. Stir vigorously enough

to agitate the solution. This ensures that any particulate matter is more or less

evenly distributed throughout the sample. Collect 50 mL of the water in the

pipette. Make sure you're still stirring the water while collecting the sample —

don't let the solution settle before you pipette your smaller sample. If you find

this difficult to accomplish, you might ask a friend to pipette the sample while

you stir. Extract the filtrate. Put the 50 mL water sample from the pipette

through the filter paper three times to ensure all particulate matter has been

collected in the filter. Extract the filtrate. Put the 50 mL water sample from the

pipette through the filter paper three times to ensure all particulate matter has

been collected in the filter.

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TOTAL DISSOLVED SOLIDS

Total Dissolved Solids = Total Solids – Total Suspended Solids

TOTAL DISSOLVED SOLIDS FOR DIARY SAMPLE

BEFORE TREATMENT

Sample 1 = 5200 - 2000

= 3200 mg/l

Sample 2 = 4800 - 1000

= 3800 mg/l

AFTER TREATMENT

Sample 1 = 400 - 100

= 300 mg/l

Sample 2 = 400 - 200

= 200 mg/l

TOTAL DISSOLVED SOLIDS FOR STP SAMPLE

BEFORE TREATMENT

Sample 1 = 800 - 100

= 700 mg/l

Sample 2 = 800 - 200

= 600 mg/l

AFTER TREATMENT

Sample 1 = 400 - 50

= 350 mg/l

Sample 2 = 400 - 70

= 330 mg/l

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4.2.5 CHEMICAL OXYGEN DEMAND (COD)

FIG 4.2.5 CHEMICAL – OXYGEN DEMAND (COD)

The Chemical Oxygen Demand (COD) method determines the quantity

of oxygen required to oxidize the organic matter in a waste sample, under

specific conditions of oxidizing agent, temperature, and time. The method can

be applied to domestic and industrial waste samples having an organic carbon

concentration greater than 50 mg/L. For lower concentrations of carbon such as

in surface water samples, the Low Level Modification should be used. When the

chloride concentration of the sample exceeds 2000 mg/L, the modification for

saline waters is required. Collect the samples in glass bottles, if possible. Use of

plastic containers is permissible if it is known that no organic contaminants are

present in the containers, Extreme care should be exercised to avoid inclusion of

organic materials in the distilled water used for reagent preparation or sample

dilution. Glassware used in the test should be conditioned by running blank

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procedures to eliminate traces of organic material. Volatile materials may be

lost when the sample temperature rises during the sulfuric acid addition step. To

minimize this loss the flask should be cooled during addition of the sulfuric acid

solution. Reflux apparatus: Glassware should consist of a 500 mL Erlenmeyer

flask or a 300 mL round bottom flask made of heat-resistant glass connected to

a 12 inch Allihn condenser by means of a ground glass joint. Any equivalent

reflex apparatus may be substituted provided that a ground-glass connection is

used between the flask and the condenser. Distilled water: Special precautions

should be taken to insure that distilled water used in this test be low in organic

matter. Standard potassium dichromate solution (0.250 N): Dissolve 12.259 g K

2 Cr 2 O 7 , primary standard grade, previously dried at 103°C for two hours, in

distilled water and dilute to 1000 mL. Sulfuric acid reagent: Conc. H 2 SO 4

containing 23.5 g silver sulfate, Ag 2 SO 4 , per 4.09 kg bottle. With continuous

stirring, the silver sulfate may be dissolved in about 30 minutes. Standard

ferrous ammonium sulfate (0.25 N): Dissolve 98.0 g of Fe (NH 4 ) 2 (SO 4 ) 2

6H 2 O in distilled water. Add 20 mL of conc. H 2 SO 4, cool and dilute to 1

liter. This solution must be standardized daily against standard K 2 Cr 2 O 7

solution. Standardization: To approximately 200 mL of distilled water add 25.0

mL of 0.25 N K 2 Cr 2 O 7 solution. Add 20 mL of H 2 SO 4 and cool. Titrate

with ferrous ammonium sulfate using 3 drops of ferroin indicator. The color

change is sharp, going from blue-green to reddish-brown. Mercuric sulfate:

Powdered HgSO 4Phenanthroline ferrous sulfate (ferroin) indicator solution:

Dissolve 1.48 g of 1-10 (ortho) phenanthroline monohydrate, together with 0.70

g of FeSO 4 7H 2 O in 100 mL of water. This indicator may be purchased

already prepared Silver sulfate: Powdered Ag 2 SO 4. Sulfuric acid (sp.gr. 1.84)

Concentrated H 2 SO 4. Place several boiling stones in the reflux flask,

followed by 50.0 mL of sample or an aliquot diluted to 50.0 mL and 1 g of

HgSO 4. Add 5.0 mL conc. H 2 SO 4; swirl until the mercuric sulfate has

dissolved. Place reflux flask in an ice bath and slowly add, with swirling, 25.0

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mL of 0.025 N K 2 Cr 2 O 7. Now add 70 mL of sulfuric acid-silver sulfate

solution to the cooled reflux flask, again using slow addition with swirling

motion. Caution: Care must be taken to assure that the contents of the flask are

well mixed. If not, superheating may result, and the mixture may be blown out

of the open end of the condenser. If volatile organics are present in the sample,

use an allihn condenser and add the sulfuric acid-silver sulfate solution through

the condenser, while cooling the flask, to reduce loss by volatilization. Apply

heat to the flask and reflux for 2 hours. For some waste waters, the 2-hour

reflux period is not necessary. The time required to give the maximum

oxidation for a wastewater of constant or known composition may be

determined and a shorter period of refluxing may be permissible. Allow the

flask to cool and wash down the condenser with about 25 mL of distilled water.

If a round bottom flask has been used, transfer the mixture to a 500 mL

Erlenmeyer flask, washing out the reflux flask 3 or 4 times with distilled water.

Dilute the acid solution to about 300 mL with distilled water and allow the

solution to cool to about room temperature. Add 8 to 10 drops of ferroin

indicator to the solution and titrate the excess dichromate with 0.25 N ferrous

ammonium sulfate solution to the end point. The color change will be sharp,

changing from a blue-green to a reddish blue. The above same procedure is

repeated for blank determination.

33

TABLE 5 : COD ANALYSIS FOR DIARY SAMPLE

Treatment Before AfterSample Sample 1 Sample 2 Sample 1 Sample 2Sample

Solution

(ml)

20 20 20 20

Blank Titre

Value (ml)

12.7 13.7 12.6 12.6

Sample

Titre Value

(ml)

4.8 6.1 9.7 9.6

COD (mg/l) 790 760 290 300

TABLE 6 : COD ANALYSIS FOR STP SAMPLE

Treatment Before AfterSample Sample 1 Sample 2 Sample 1 Sample 2

Sample

Solution

(ml)

20 20 20 20

Blank Titre

Value (ml)

12.7 13.7 12.7 12.7

Sample

Titre Value

(ml)

11.3 11.8 11.8 11.6

COD (mg/l) 140 190 90 110

34

COD ANALYSIS

COD = (Blank TitreValue−Sample Titre Value)Volume of Sample

× N × 8 ×1000

Normality of FAS = 0.25 N

COD ANALYSIS FOR DIARY SAMPLE

BEFORE TREATMENT

Sample 1 = (12.7−4.8)×0.25 ×8× 100020

= 790 mgl

Sample 2 = (13.7−6.1)× 0.25× 8×100020

= 760 mgl

AFTER TREATMENT

Sample 1 = (12.6−9.7)×0.25 ×8×100020

= 290 mgl

Sample 2 = (12.6−9.6)× 0.25 ×8×100020

= 300 mgl

35

COD ANALYSIS FOR STP SAMPLE

BEFORE TREATMENT

Sample 1 = (12.7−11.3)× 0.25 ×8 ×100020

=140 mgl

Sample 2 = (13.7−11.8)×0.25 ×8×100020

=140 mgl

AFTER TREATMENT

Sample 1 = (12.7−11.8)×0.25 ×8×100020

= 90 mgl

Sample 2 = (12.7−11.6)×0.25 ×8× 100020

= 110 mgl

36

4.2.6 BIO-CHEMICAL OXYGEN DEMAND

The test for biochemical oxygen demand (BOD) is a bioassay procedure

that measures the oxygen consumed by bacteria from the decomposition of

organic matter. The change in DO concentration is measured over a given

period of time in water samples at a specified temperature. Procedures used to

determine DO concentration are described in NFM 6.2. It is important to be

familiar with the correct procedures for determining DO concentrations before

making BOD measurements. BOD is measured in a laboratory environment,

generally at a local laboratory.

There are two stages of decomposition in the BOD test:

carbonaceous stage and

nitrogenous stage

FIG 4.2.6 BOD CURVE

The carbonaceous stage, or first stage, represents that portion of oxygen

demand involved in the conversion of organic carbon to carbon dioxide.

37

The nitrogenous stage, or second stage, represents a combined carbonaceous

plus nitrogeneous demand, when organic nitrogen, ammonia, and nitrite are

converted to nitrate. Nitrogenous oxygen demand generally begins after about 6

days. For some sewage, especially discharge from wastewater treatment plants

utilizing biological treatment processes, nitrification can occur in less than 5

days if ammonia, nitrite, and nitrifying bacteria are present. In this case, a

chemical compound that prevents nitrification should be added to the sample if

the intent is to measure only the carbonaceous demand. The results are reported

as carbonaceous BOD (CBOD), or as CBOD5 when a nitrification inhibitor is

used.

The standard oxidation (or incubation) test period for BOD is 5 days at 20

degrees Celsius (°C) (BOD5). The BOD5 value has been used and reported for

many applications, most commonly to indicate the effects of sewage and other

organic wastes on dissolved oxygen in surface waters. The 5-day value,

however, represents only a portion of the total biochemical oxygen demand.

Twenty days is considered, by convention, adequate time for a complete

biochemical oxidation of organic matter in a water sample, but a 20-day test

often is impractical when data are needed to address an immediate concern.

Samples can degrade significantly during extended storage. To minimize

sample degradation, and thus avoid negative bias in the measurement of BOD5,

analyze samples promptly or store chilled without freezing (maintain a

temperature from 1 to 4°C). Chilling the sample is not necessary if the analysis

begins within 2 hours of collection

If a sample is refrigerated prior to analysis, allow the sample to warm to

20°C before starting the test. A sample may be removed from an ice chest or

refrigerator during transit to allow it to warm to 20°C before analysis begins.

38

It is optimum to start the BOD5 analysis immediately after sample

collection to minimize changes in bacterial concentration. The maximum

holding time of a sample to be analyzed for BOD is 24 hours.

Determine the amount of sample to be analyzed; if available, use the

historical results of a previous test of BOD5 for a particular sampling site, and

refer to table.

Place a clean, calibrated thermometer into the constant temperature

chamber.

Turn on the constant temperature chamber to allow the controlled

temperature to stabilize at 20°C ±1°C.

Turn on the DO instrument, but not the stirring attachment. Some DO

instruments need to be turned on 30 to 60 minutes before calibration—check the

manufacturer’s instruction manual. Aerate dilution water before adding nutrient

solutions. After aeration, Add to dilution water and 1 mL each of the potassium

phosphate, magnesium sulfate, calcium chloride, and ferric chloride solutions

per 1 L of dilution water. Shake the container of dilution water for about 1

minute to dissolve the slurry and to saturate the water with oxygen. Place the

dilution water in the constant temperature chamber to maintain a temperature of

20°C until sample dilutions and analyses begin.

The initial and final (after 5 days ± 4 hours) DO tests of the dilution water is

determined and recorded simultaneously with each batch of environmental

samples.

Check the temperature of the air incubator or water bath using a laboratory

thermometer to ensure that the temperature has been maintained at 20° ± 1°C. A

minimum/maximum recording thermometer can be used to audit the

temperature during times when checks cannot be made.

39

Place the sample container in the constant-temperature chamber or water

bath to begin warming the sample to 20°C ± 1°C. While the sample is warming,

insert the air diffusion stone into the container and aerate the sample for about

15 minutes. After removing the air diffusion stone, allow several minutes for

excess air bubbles to dissipate. The initial DO of the BOD sample needs to be at

or slightly below saturation.

Measure the appropriate amounts of sample necessary for the analysis.

BOD5 dilutions should result in a DO residual of at least 1 mg/L and a DO

depletion of at least 2 mg/L after a 5-day incubation to produce the most reliable

results. Prepare the dilutions to obtain a DO uptake in this range using the

dilution water prepared earlier.

For each subsample, mix thoroughly by inverting 20 times.

Use a large-bore pipet for sample volumes less than 50 mL. Withdraw a

subsample that is representative of all the particle sizes present.

Use a graduated cylinder for sample volumes greater than or equal to 50 mL.

Dilute two additional samples to bracket the appropriate dilution by a factor

of two to three. Prepare at least three samples diluted according to volumes

specified.

Pour the sample from the pipet or graduated cylinder into a clean BOD

bottle.

Agitate the dilution water and fill the remaining portion of the BOD bottle

with dilution water.

Prepare three samples containing only dilution water. These samples serve

as blanks for quality control. If two of the three samples meet the blank-water

criterion, accept the data.

Calibrate the DO instrument in accordance with the procedures outlined in

NFM 6.2.

40

After bringing the samples to saturation and preparing the dilutions, measure

the initial DO concentration (D1) of each sample and each dilution blank.

Carefully insert the self-stirring sensor into the BOD bottle, avoiding air

entrapment. Turn on the stirrer and allow 1 to 2 minutes for the DO and

temperature readings to stabilize. Record the bottle number, date, time, and D1

on a form similar to that shown in figure. Turn off the stirrer and remove the

sensor from the BOD bottle. Rinse the sensor and stirrer with deionized water

from a wash bottle. Discard rinse water into a waste container. Add glass beads

to the BOD bottle, if necessary, to displace the sample up to the neck of the

bottle so that inserting a glass stopper will displace all air, leaving no bubbles.

Carefully cap the BOD bottle with the ground-glass stopper. Tip the bottle to

one side and check for an air bubble.

• If an air bubble is present, add glass beads to the bottle until the bubble

is removed. Cap the bottle and check again for an air bubble. Repeat if

necessary.

• If no bubble is present in the sample, create a water seal by adding

distilled or deionized water to the top of the BOD bottle around the glass

stopper. Then place the overcap over the stopper on the BOD bottle to

minimize evaporation from the water seal.

Place the sealed BOD sample in the air incubator or water bath and incubate

the sample at 20°C ± 1°C for 5 days.

At the end of 5 days ± 4 hours, remove the BOD bottles from the incubator,

remove the overcap, pour off the water seal, remove the ground-glass stopper,

and measure the final DO concentration (D2).

• The DO uptake (DO0 days - DO5 days) in the dilution water should not

be greater than 0.2 mg/L and preferably not more than 0.1 mg/L.

41

Exceeding the 0.2-mg/L criterion could be grounds for rejecting results of

the BOD analysis of the environmental sample.

• Dilution water of poor quality will cause an oxygen demand and appear

as sample BOD. Improve purification or get the dilution water from

another source if DO uptake exceeds 0.2 mg/L (see section 7.0.5,

Troubleshooting).

Complete the field form by recording the date, time, and D2 for each

respective sample bottle.

The BOD5 test can be quite variable. Collect sufficient field and split

replicates (10 to 20 percent) to provide an estimate of method variability.

TABLE 7 : BOD ANALYSIS FOR DIARY SAMPLE

Treatment Before AfterBottle A B C A B C

Seed (ml) 10 15 20 20 25 50

Initial D.O.

(mg/l)

15.68 17.04 17.92 9.2 9.84 11.76

Final D.O.

(mg/l)

2.48 3.52 2.4 2.88 2.56 2.72

BOD (mg/l) 330 225 194 79 72.8 45.2

TABLE 8 : BOD ANALYSIS FOR STP SAMPLE

Treatment Before AfterBottle A B C A B C

Seed (ml) 10 20 25 100 150 200

42

Initial D.O.

(mg/l)

8.48 11.44 13.36 9.44 10.72 13.44

Final D.O.

(mg/l)

2.64 1.68 2.16 2.08 2.48 2.08

BOD (mg/l) 146 122 112 18.4 13.73 14.2

BOD ANALYSIS FOR DAIRY SAMPLE

BEFORE TREATMENT

DO Calculation For sample A

Initial D.O.

Volume of Na2 s2 o3 = 19.6 ml

Strength of Na2 s2 o3 = 0.01N

Volume of Sample = 100 ml

Strength of Sample = ?

Strength of the Sample = 19.6× 0.01100 = 1.96 ×10−3 N

D.O. = Strength of the Sample × Equivalent weight of Oxygen × 1000

= 1.96 ×10−3 × 8 ×1000

= 15.68 mg/l

5 Day D.O.

Volume of Na2 s2 o3 = 3.1 ml

Strength of Na2 s2 o3 = 0.01N

Volume of Sample = 100 ml

43

Strength of Sample = ?

Strength of the Sample = 3.1× 0.01100 = 3.1 ×10−4 N

D.O. = Strength of the Sample × Equivalent weight of Oxygen × 1000

= 3.1 ×10−4 × 8 ×1000

= 2.48 mg/l

B.O.D = ( Initial D . O.−Final D . O)Seed (ml)

× Volume of Iodine flask

B.O.D = (15.68−2.48)10

×250

= 330 mg/l

The above Calculation Procedure is repeated for Sample B and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 21.3 ml for First Day

Volume of Na2 s2 o3 = 4.4 ml for Fifth Day

The above Calculation Procedure is repeated for Sample C and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 22.4 ml for First Day

Volume of Na2 s2 o3 = 3.0 ml for Fifth Day

AFTER TREATMENT

The above Calculation Procedure is repeated for Sample A and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 11.5 ml for First Day

Volume of Na2 s2 o3 = 3.6 ml for Fifth Day

44

The above Calculation Procedure is repeated for Sample B and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 12.3 ml for First Day

Volume of Na2 s2 o3 = 3.2 ml for Fifth Day

The above Calculation Procedure is repeated for Sample C and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 14.7 ml for First Day

Volume of Na2 s2 o3 = 3.4 ml for Fifth Day

BOD ANALYSIS FOR STP SAMPLE

BEFORE TREATMENT

DO Calculation For sample A

Initial D.O.

Volume of Na2 s2 o3 = 10.6 ml

Strength of Na2 s2 o3 = 0.01N

Volume of Sample = 100 ml

Strength of Sample = ?

Strength of the Sample = 10.6× 0.01100 = 1.06 ×10−3 N

D.O. = Strength of the Sample × Equivalent weight of Oxygen × 1000

= 1.06 ×10−3 × 8 ×1000

= 8.48 mg/l

5 Day D.O.

Volume of Na2 s2 o3 = 3.3 ml

Strength of Na2 s2 o3 = 0.01N

45

Volume of Sample = 100 ml

Strength of Sample = ?

Strength of the Sample = 3.3× 0.01100 = 3.3 ×10−4 N

D.O. = Strength of the Sample × Equivalent weight of Oxygen × 1000

= 3.3×10−4 × 8 ×1000

= 2.64 mg/l

B.O.D = ( Initial D . O.−Final D . O)Seed (ml)

× Volume of Iodine flask

B.O.D = (8.48−2.64)10

× 250

= 146 mg/l

The above Calculation Procedure is repeated for Sample B and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 14.3 ml for First Day

Volume of Na2 s2 o3 = 2.1 ml for Fifth Day

The above Calculation Procedure is repeated for Sample C and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 16.7 ml for First Day

Volume of Na2 s2 o3 = 2.7 ml for Fifth Day

AFTER TREATMENT

The above Calculation Procedure is repeated for Sample A and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 11.8 ml for First Day

Volume of Na2 s2 o3 = 2.6 ml for Fifth Day

46

The above Calculation Procedure is repeated for Sample B and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 13.4 ml for First Day

Volume of Na2 s2 o3 = 3.1 ml for Fifth Day

The above Calculation Procedure is repeated for Sample C and for

determining the Initial D.O. and Final D.O.

Volume of Na2 s2 o3 = 16.8 ml for First Day

Volume of Na2 s2 o3 = 2.6 ml for Fifth Day

4.3 TREATMENT BY USING REED BED SYSTEM

A reed bed is essentially a basin that is lined with an impermeable

membrane, filled with gravel and planted with macrophytes such as reeds and

rushes. Wastewater (black or grey) passes through the root zone of the reeds

where it undergoes treatment via physical, chemical and biological interactions

between the wastewater, plants, micro-organisms, gravel and atmosphere. Inlet

and outlet pipes are positioned below the gravel surface, so that the water

always remains below the surface, thus minimising the risk of human exposure

to the wastewater, mosquito breeding and unpleasant odours.

Raw wastewater from the house flows into a collection tank for primary

treatment to remove large solids, grease and oils. The partially clarified effluent

from the collection tank passes through an effluent filter to trap any large solids

that remain, and then flows into the reed bed. Once inside the reed bed, the

wastewater undergoes a complex series of natural treatment processes as it

moves laterally through the root zone from one end of the bed to the other. The

wetland plants leak small amounts of oxygen out through their roots, creating

47

small oxygenated sites within an otherwise anaerobic environment. This mix of

aerobic and anaerobic conditions creates an ideal environment for the growth of

micro-organisms on the surface of the gravel and plant roots. These micro-

organisms are largely responsible for the pollutant removal that occurs in a reed

bed, as they feed on and breakdown organic matter and nutrients, and compete

against pathogenic organisms. Earthworms have also been found to inhabit reed

beds, and assist with the breakdown of organic matter and solids.

Most of the pollutant removal processes in reed beds are time dependent.

Thus, the residence time (length of time water spends in the reed bed) is an

important design parameter. Reed beds are generally designed to detain the

wastewater for a period of 5 to 7 days in order to allow sufficient time for the

settling and filtering of suspended solids, breakdown of organic matter, binding

of some contaminants onto the gravel, and removal of nutrients by plants and

micro-organisms.

They are environmentally friendly using only natural sustainable

ecological processes. Gravity driven systems don’t require any energy input.

Maintenance requirements are low and can be carried out by anyone with the

modification of gardening skills and common sense.

They are highly effective when properly designed and can be used in

combination with ponds and wetlands to produce near river quality water.

Vertical flow reed-beds are more effective at nitrifying effluents, converting

ammonia into nitrates and nitrites, than most package of sewage system.

48

FIG 4.3 REED BED PROJECT MODEL

4.4 RATE OF FILTRATION

Rate of Filtration = Sample WaterTime

Rate Of Filtration time for 3 litres is 12 minutes.

Rate of filtration = 312× 60

= 4.167 × 10−3 litsec

= 4.167 ×10−3× 100060 ×60

= 1.157 ×10−3 m3

sec

49

Rate Of Filtration time for 1 litres is 4 minutes.

Rate of filtration = 14 × 60

= 4.167 × 10−3 litsec

= 1.157 ×10−3 m3

sec

1000 litres of wastewater can be treated in 2.7 days by using our REED BED Project.

4.5 EFFICIENCY OF REMOVAL

TABLE 9: EFFICIENCY OF REMOVAL

Efficiency of Removal (%)S.no. Treatment Dairy Waste STP

Sample 1 Sample 2 Sample 1 Sample 2

1.Total Suspended

Solids95 80 50 65

2. Total Solids 92.3 91.67 50 50

3. Total Dissolved 90.63 94.74 50 45

50

Solids

4.Chemical oxygen

demand63.29 60.52 35.71 42.11

5.Biochemical

Oxygen Demand76.77 67.64 87.32 88.74

BEFORE TREATMENT AFTER TREATMENT

FIG 4.5 STP SAMPLES

CHAPTER 5

CONCLUSION

The project gives us the idea about Biological Treatment for treating the

waste water using REED BED system. Plants provide an environment for

microbes to live, they oxygenate the wastewater, providing nutrients for the

microbes to survive, they stabilize the soil and they also partake in the reduction

of nutrients.

Reed bed treatment system utilizes the active treatment capabilities of soil

to biologically treat effluents such as sewage, industrial wastewater, run-off and

leachates.

51

This Project can be extended by Treating other Industrial wastewater and

comparing its efficiency for its future use.

REFERENCES

Badejo. A, “Tertiary Hospital Wastewater Treatment using Reed Bed

Technology” Civil Engineering Department, College of Engineering,

University of Agriculture, Abeokuta, Nigeria.

Jan Vymazal, “Constructed Wetlands for Wastewater Treatment

Department of Landscape Ecology” Faculty of Environmental Sciences,

Czech University of Life Sciences, Czech Republic.

52

Jiangang Li, Yubo Cui, “Biological Toxicity of Sewage Sludge

Stabilized by Reed Bed” College of Environment and Resources, Dalian

Nationalities University, Dalian, China.

Lakeville, “Energy Efficient Sludge Treatment with REED BED

Technology Demonstration Project” NY, Water and Sewer Authority,

Livingston.

Mehali J. Mehta, “Treatment of Municipal and Industrial wastewater

by REED BED Technology” Assistant Professor, Department of Civil

Engineering, Sarvajanik College of Engineering and Technology,

Gujarat, India.

Sean O'Hogain, Liam McCarton, “The Operation of Hybrid Reed Bed

and Willow Bed Combinations” Dublin Institute of Technology, Ireland.

53