UNIT – I

PLANNING FOR SEWERAGE SYSTEM

PART A1.What are the types of treatment processes?

1) Preliminary treatment 2) Primary treatment 3) Complete final treatment 4) Secondary treatment

Filters Aeration

Exudationponds&

Aerated lagoons

2. What are the various sources of wastewater generation? Industrial Wastes Domestic wastes Agricultural Wastes

3.List out the types of anaerobic bio logical units? Anaerobic lagoons Septic tank Inhofe tank

4.What is means by screening? Screening is the very first operation carried out at a sewage treatment

plant and consists of passing the sewage through different types of screens so as to trap and remove the floating matter such as process of cloth, paper, wood, cork, hair, fiber etc.

5. What is the purpose of providing screen? The main idea of providing screens is to protect the pumps and other

equipments from the possible damages due to the floating matter of the sewage.

It should be used for removing the floating matters.

6. What are the types of screen?Classification based on size of the opening

Coarse screens Medium screens Fine screens

Based on shape

Rectangular for coarse and medium screens

Disc or Drum for fore screen

7. Define bar screen?Rectangular shaped coarse and medium screens are made of steel bars fixed

parallel to one another at desired spacing on a rectangular frame and are called bar screen.

8. What is meat by movable screen? Movable screens are stationary during their operating periods. But they

can be lifted up bodily and removed from their partitions for the purpose of cleaning.

A common movable bar medium screen is a 3 – sided cage with a bottom of perforated plates. It is mainly used in deep pits ahead of pumps.

9. Define Communicators?Communicators or shredders are the patented devices, which break

the larger sewage solids to about 6 mm in size. When the sewage is screened through them such devices are used only in developed countries like USA.

10. What is meant by Screening?The material separated by screens is called the screenings. It

contains 85 to 90% of mixture and other floating matter. It may also contain some organic load which may putrefy, bad smells and nuisance.

11. What are the methods adopted for disposal of screenings?Burning

Burial Dumping

Burning of the screenings is done in the incinerators

Burial: The process is technically called composting

Another method of disposing of the screening is by dumping them in low lying areas (away from the residential areas) or in large bodies of water such as sea.

12. Define Grit Chamber?Grit chambers, also called or grit channels or grit basins, are intended to

remove the inorganic particles (specific graving about 2.65) such as sand, graver, grit, egg, shells, bones etc of size 2 mm or larger to prevent damages to the pumps and to prevent their accumulation in sludge digesters.

13. Define unit process?Methods of treatment in which the application of physical forces predominate

are known as unit operations while methods of treatment in which chemical or biological activities are involved are known as unit process.

14. What are the types of unit operations & processes?

Physical unit operations Chemical unit process Biological unit process

15.Give any two advantages of unit operations/ process? 1. It gives better understanding of the process as inherent in the treatment and of the

capabilities of these processes in attaining the objectives. 2. It helps in the development of mathematical and physical models of treatment

mechanisms and the consequent design of treatment plants.

16.Define phase transfer? Most waste water treatment process bring about changes on concentration of

a specific substances by moving the substance either into or unit of the waste water it self. This is called phase transfer.

17. Define definition time?The definition time (t) of a settling tank may be defined as the average

theoretical time required for the sewage to flow through the tank. Otherwise known as definition period or retention period

18. What is meant by principle of sedimentation?The turbulence is retarded by offering storage to sewage these

impurities tend to settle down at the bottom of the tank offering such storage. This is the principle of sedimentation.

19. Define the term “Sedimentation Burin”?The burin in which the flow of sewage is retarded is called the settling tank or

the sedimentation Tank or the sedimentation Burin.

20. Define the term “Detention Period”?

The theoretical average time for which the water is detained is called the detention period.

PART B

1. What are the different sources of wastewater generation?

ANS:Sources Of Wastewater Generation

The following are the major sources of waste water• Domestic (household wastes): Domestic waste includes wastewater from toilet, shower, cooking,

washing ,cleaning and laundry• Commercial/service: Commercial waste includes waste from schools, hospitals, restaurants,

offices, hotels, small businesses, shops, colleges• Industrial: Industrial wastewater includes wastewater from processing wastewater,

cooling water.Non-point sources:

a. Infiltration from groundwater into sewersb. Inflow from storm water surcharge into sewer manholesc. Combined storm/sewer overflows, older sewer systems where storm sewers

discharge into sanitary sewers(Chicago, San Francisco, etc.)d. Runoff from streets: sand and petroleum and tire residues (infiltration, not a

direct discharge).

2. Explain the Factors Influencing Sanitary Sewage Flow.ANS

Factors Influencing Sanitary Sewage Flow.

The quantity of sanitary sewage depends on Rate of water supply Population growth Type of area served Infiltration of ground water

Rate Of water Supply The quantity of wastewater produced from a community would depend on the rate

of water supply per capita per day. The quantity of wastewater entering the sewers would be less than the quantity of

sewers supplied because water will be lost in domestic consumption. Private source of water supply and infiltration of sub soil in sewers increases the

flow rate . The minimum capacity of the sewers per day should be 150 litres per capita /day.

Population GrowthSewers are generally designed considering the increase in population.The Population Forecast may be done by

Arithmetical Increase Method Geometrical Increase Method

Incremental Increase Method Deceased Rate Of Growth Method Graphical Extension Method Graphical Comnparision Method Zoning Method Ratio and Correlation Method Growth Composition Analysis Method

Design Period of a Sewer system is usually 30 years with the period of pumping plants for 10 years.

Type Of Area Served The quantity of wastewater produced depends upon the the serving area

(Residential, Commercial or Industrial) Wastewater from residential area directly depends on the rate of water supply,

if there are no infiltration and private supply. Wastewater from industrial area directly depends on the type and processing

in the Industry.\ Wastewater from Commercial area depends on the process of work in those

area.Infiltration Of Sub-Soil water

Ground water may infiltrate into sewers through leaky joints. Exfiltration also occurs where water flows from sewers to ground. Infiltration increases the quantity of water in sewers where as exfiltration

reduces the water quantity. Infiltration is most important in sewer designs since it increases the load on

the treatment works.

3. How will you estimate the storm run off?ANS

• Runoff – depends on various factors like,1. Type of Precipitation2. Intensity3. Duration of rainfall4. Rainfall distribution5. Soil moisture deficiency6. Direction of the prevailing storm7. Climate conditions

8. Shape, Size & type of Catchment basin• No proper formulas found to define the runoff ratetill now and all factors are interdependent.• So, a rational method was adopted commonly forcalculating (approx) the runoff of that area. But now urban storm drainage were analyzed through“Digital computer Simulations”.

Rational formula (Used for area more than 500 hectares)

Q = 1

36K . PC . A

Qp = Peak rate of runoff in cumecs,K = Coefficient of runoff,

A = Catchment area, in hectares,

Pc = Critical rainfall intensity, in cm/hr.Critical rainfall intensity = Rainfall intensity during the rainfall duration which is equal to the time ofconcentration.Time of concentration = Ti + Tf

Ti = (0.885L3

H )0.385

Tf = Lengt h Of T he DrainVelocity∈t h e Drain

¿¿

Intensity duration curve,

P = a

T+b

PC = P = Rain intensity in cm/hr T = Time in minutes, a & b = Constants

For T varying between 5 to 20 minutes, a = 75 & b = 10,

P = 75

T+10

For T varying between 5 to 20 minutes,a = 100 & b = 20,

P = 100

T+20

For localities where rainfall is frequent,

P = 343

T+18

For localities with rainfall frequency of 10 years,

P = 38

√T

For localities with rainfall frequency of 1 years,

P = 15

T0.62

Kuichling’s formula

P = 267

T+20

Storm frequency = 10 years

P = 305

T+20

Storm frequency = 15 years.

4. Explain the estimation of Sanitary sewage flow.

ANSSewage

Sewage is a water-carried waste, in solution or suspension, that is intended to be removed from a community. Also known as wastewater, it is more than 99%

water and is characterized by volume or rate of flow, physical condition, chemical constituents and the bacteriological organisms that it contains.

Sewer SystemsA Sewer system is a network of pipes used to convey

storm runoff and / or wastewater in an area. The sewage discharge which has to pass through a sewer must be estimated as correctly possible other wise it may lead to

– inadequate flow condition– overflow condition–wasteful investment.

Qty of Sewage = Domestic + Industrial

Domestic waste includes wastewater from toilet, shower, cooking, washing ,cleaning and laundry.

Industrial wastewater includes wastewater from processing wastewater, cooling water.

Net qty of sewage produced = Estimated wastewater generation rate + Unaccounted private water supplies + Infiltration – water losses – water not entering the sewage.

Wastewater generation rate is given by

Qmax = 5Q avg

P0.2

Qmax = (1+14

4+P0.5 )Qavg

5. A sewer has a catchment area of 50 hectares. Estimate the storm water flow corresponding to a rainfall of 40 mm during a time of concentration of 3o mins. Assume that the impervious area is equal to 55% of total catchment area. Use lloyed davis formulaeSoln:

Wastewater flow = r

6 tc

P cumecs

Where r = total rainfall in mm, During the time of concentration = 40 mmtc = time of concentration = 30 minsP = impervious area in hectares

= 0.55 x 50 = 27.5 hectares.

Wastewater flow = 40

6 x30x 27.5 m3/sec

= 6.1 cumecs= 6100 litres/sec

UNIT – II SEWER DESIGN

PART B

1. What are the Demerits of chemical precipitation?1. High cost of chemicals2. Large quantity of sludge which offers difficulty of its removal3. Skilled attendance4. Putrescible effluent

2. What do you mean by chemical precipitation?When certain chemicals are added to, sewage they produce a precipitate

known as flow which in insoluble or slightly soluble in water. The flow attracts small particles to form large size and thus size goes on increasing during the process of settlement

3. What are the users of Baffle?1) Baffler are required to prevent the movement of organic

matter and its escape along with the efficient2) Distribute the sewage uniformly through the cross section of the

tank.3) It is used to avoid short circuiting

4. What are the classifications of biological process? a) Aerobic processes

b)Anaerobic processes c) Aerobic – anaerobic processes

5. List out the aerobic processes?1. Activated sludge processes2. Trickling filters3. Aerobic stabilization pond4. Aerated lagoon

6. List out the anaerobic process?1. Anaerobic sludge digestion,2. Anaerobic contact processes3. Anaerobic filters4. Anaerobic lagoons or ponds

7. What are the sources of waste water?1. Domestic waste water (i.e sewage)2. Agricultural return waste water3. Industrial waste water

8. What are the methods involved in the treatment of waste water?Mainly classified into

1. Conventional treatment methods2. Advanced waste waster treatment

Conventional treatment methods i)Preliminaryprocesses ii.Primary treatmentiii.Secondary treatment

Advanced waste water treatment i. Tertiary treatment

9. What are the functions involved in the chemical unit processes1. Chemical precipitation2. Gas transfer3. Adsorption4. Disinfection5. Combustion6. loss exchange7. Electro dialysis

10. Give any two advantage of chemical coagulation process in sewage treatment?Sedimentation aided with coagulation produces better efficient with lesser

BOD and suspended solids, as compared to plain sedimentation. Coagulated settling tank requires less space than that required by an ordinary plain settling tank.

11.What are the Demerits of coagulation in sewage treatment? Cost of chemicals is added to the cost of sedimentation, with out

much use, and thereby making the treatment costlier. The process of coagulation requires skilled supervision and

handling of chemicals.

12. What are the types of sedimentation tank?Based on flow

Vertical flow tank Horizontal flow tank Radial flow tank

According to useo Primary

o Secondary

13.What are the chemical used for precipitation of sediment?o Alum

o Ferrous sulphate

o Ferric sulphate

o Ferric chlorides

o Sodium alluminate

o Sulphuric acid

o lime

o copperas

14.What are the factors that affect the precipitations?o Kind of chemical

o Quality of chemical

o Character and concentration of sewage

o pH values of sewage

o time of mixing and flowlations

o Temperature

o Violence of agitation

15. What are the merits of chemical precipitation?o More rapid and through clarification

o Removal of higher percentage of suspended solids.

o Simplicity of operation

o Small size tank is enough

16. Define the term Displacement efficiency?The ratio of the “Flowing through period” to the “detention

period” is called the displacement efficiency

17. Define the term “Detention Period”?The theoretical average time for which the water is detained is called

the detention period.

18.What are remedial measurement for rising sludge problem?o Increasing the return sludge age

Increasing the speed of the sludge scroper mechanism, where possible Decreasing the mech cell residence come by increasing the sludge write rate

19.What is meant by sludge bulking?

Sludge with poor setting characteristics is termed bulking sludge. It results on poor influent due to thee presence of excessive suspended solids and also in rapid loss of MISS from aeration tank.

20.Define the term “Raw sludge”?

The sludge, which is deposited in a primary sedimentation tank, is called Raw sludge. Raw sludge contains highly puterscible organic matter, and is thus, very objectionable.

PART B

1. Explain various types of pipe joint in transporting water.Ans: Different Types of .Joints:-

a)Socket and Spigot jointb)Flanged. Jointc) Couplingd)Flexible jointc)Collar jointd)Screwed Socket jointe)Expansion joint

Socket and Spigot Joint: The cast iron pipes which are to be joined by the socket and spigot joint. These are made in such a way that one of their end is enlarged and the other end is

normal. The enlarged end is called " socket'. The normal end is spigot. The spigot is fitted

into the socket. A few strands of jute are wrapped around the spigot before inserting in into the

socket and then more jute is packed into the joint. The remaining space between the socket and the spigot is finally filled with molten

lead.

Flanged Joint.- Flanged joints are used for pumping stations and some other locations where it may

be necessary to occasionally disjoint the pipe.

Cast iron pipe lengths to be joined by this joint are cast in such a way as to have flanges at both ends.

Two flanges are brought together keeping rubber washer in between them. so as to make them water They are fixed by means of nuts and bolts.Expansion Joint.

Expansion joints are provided at suitable intervals in the pipelines. so as to counteract the thermal stresses produced due to temperature variations.

The socket end is cast flanged -and the spigot end is plain. The socket end is connected rigidly to an annular ring which can slide freely over the spigot end.

While making this joint, a Small space (equal to L ± T.) is kept between the face of the spigot and the inner face of the socket, and the spigot is filled up by means of a rubber gasket file flanges are tightened by means of nuts and bolts'Flexible joint:

These joints are used wherelarge scale flexibilities are required. The pipes to be provided with such a joint are cast with special types of ends. The socket is spherical and tile. Spigot though plain is having a bead at the end. A retainer ring is placed over the bed which keeps the special rubber gasket. A split cast iron gland ring is then placed over it. They are then tightened by means of bolts and nuts.

Dresser Coupling (Mechanical Joint): This type of joints is used when it is required to join the plain ends of cast iron

pipes. A special type of metallic collar is then fitted and tightened over the abutting ends,

thus forming a mechanical joint. One of the most commonly used types of mechanical joint is dresser coupling. In this joint, an iron ring and a gasket are slipped over each of the abutting ends of

the pipes, and iron sleeve is inserted between the gaskets. The iron rings are then tightened by means of bolts.

2. Discuss the key features of testing of pipeline

Ans: The pipe line is tested from section to section, .Thus at a time Only one particular

section lying between two sluice values is taken up for testing, The downstream sluice value is closed. and water is admitted into the pipe through

the upstream valve.The air valves will be properly operated during filling upthe pipes

The upstream valve, through which water was admitted, is closed, so as to completely isolate the pipe section from the rest of the pipe,

Pressure gauges are then fitted along the length of the pipe section at suitable intervals on the crown, through holes left for this purpose.

The pipe section is then connected to the delivery side of a pump through a small by-pass valve, and the pump is started, so as to develop pressure in the pipe. The operation is continued till the pressure inside the pipe reaches the designed value, .which can he read from the pressure gauge fixed on the pipe.

The by-pass valve is closed and the pumping is discontinued. The pipe kept under pressure for 24.hrs, and inspected for possible defects, leakages

at the joints etc., Thepipe finally emptied through drain values and the observed defects are rectified,

so as to make the line fit for use. The pipe is again tested by repeating the same procedure.

3. Determine the size of a circular sewer for a discharge of 600lps running half-full. Assume I = 0.0001 and n =0.015.

SOLN:

dD

=0.5

= 12 (1−cos

∝2 )

Or cos ∝2=0

∝2=90°

∝=180 °

Areaof Cr oss sectionw h ile running partially full

= π D2

4 [ ∝360 °

− sin∝2 π ]

Now using this equation,

𝝅 = D2

4 [ π∝360°

− sin∝2 ]

= D2

4 [ π 180 °360 °

− sin180 °2 ]

= D2

4 [ π2−0] = π D2

8

Wetted Perimeter, while running partially full

= p = 𝝅D ∝

360°

Using this formula

p = 𝝅D ∝

360°

= 𝝅D 180°360°

= πD2

r = ap=¿

π D2

8.

1πD2

=D4

Using Manning’s formula,

q = 1n

a r2 /3 √S

0.6 = 1

0.015 π D2

8 ( D

4 )2 /3

√0.0001

D8 /3=0.6∗0.015∗8∗2.52∗100

π

D = 1.93 m.

4. Design a sewer to serve a population of 36,000; the daily per capita water supply allowance being 135 litres, of which 80 percent find its way into the sewer. The slope available for the sewer to be laid is 1 in 625 and the sewer should be designed to carry four timed the dry weather flow when running full. What would be the velocity of flow in the sewer when runningfull?Assume n = 0.012 in manning’s formula

Soln:

Population = 36000Per capita water supply = 135 lit/person/dayAverage water supplied daily = 36000 x 135 lit/dayAverage water supplied in cumecs

= 36000∗135

1000∗24∗60∗60

= 0.0562 cumecs. Average Sewage Discharge = 80 % of water supplied = 0.8 x 0.0562 cumecs =0.045 cumecs.Maximum discharge for which sewer should be designed running full = 4 x 0.045 cumecs = 0.18 cumecs

Using Manning’s formula, we have

Q = 1N

A R2 /3 √S

0.18 = 10.012 ( π

4D2)( D

4

2 /3) 1√625

0.18∗0.012∗4∗2.52∗52π

= D8 /3

=0.173

D = 0.1733 /8

= 0.31 mDiameter of sewer pipe = 0.31 m

Velocity of flow when running full

V = QA

= 0.18π4

(0.31 )2

V = 2.39 m/sec.

5. Write about one pipe system and two pipe system of plumbing

Two Pipe System:This system is the best system of plumbing used widely across the

world.It has two sets of pipes

1. For draining night soil2. For draining sullage

The first set of pipes carrying night soil is called soil pipesThe second sets of pipes carrying sullage from baths, etc are called sullage or

waste pipes.The soil fixtures such as latrines and urinals are connected through branch

pipes to vertical soil pipes.The sludge fixtures such as baths, sinks, wash basins are connected through

branch pipes to vertical waste pipes.Both the soil and waste pipes are ventilated separately by providing separate

vent pipes or antisiphonage pipes.This requires four pipes and is costly.

One Pipe system:Instead of using two separate pipes, only one main vertical pipe is used

to carry sullage and night soil.The main pipe is ventilated in itself by providing cowl at its top.An additional vent pipe is also provided.This system has two pipes instead of four pipes.

UNIT – III

PRIMARY TREATMENT OF SEWAGE

PART A

1.Define humus tank?The efficient of the filter is therefore, passed through a sedimentation take

called Humus tank otherwise called secondary clarifier or secondary setting take.

2.What are the distinct stages in the sludge digestion processes?

Acid fermentation Acid repression Alkaline fermentation

3.Define the term ripened sludge?This digested sludge is collected at the bottom of the digestion tank

and is also called repented sludge. 4. What are the factors effecting sludge digestion?

o Temperature

o Pit value

o Seeding out digested sludge

o Mixing and stirring of the raw sludge with digested sludge.

5. What are functions of aeration in ASP?

Oxygenation of the mixed log Flocculation of the colloid in sewage influent Suspension of activated sludge

6.What are the methods employed for the purpose of certain in ASP?

Diffused air aeration air aeration Mechanical aeration Combined diff used air and Mechanical aeration

7.What are the patterns of mechanical aeration?

Haworth paddle or Sheffield aeration system Hartley paddle or Bio flocculation system Simplex aeration system Link belt aeration system Kessner Brush aeration system

8.List out the important aeration processes in the ASP?

Conventional process Tapered aeration process Step aeration process Contact stabilization process Completely mixed process Modified aeration Extended aeration

9.What are the advantage of stabilization ponds

Lower initial lost than required for a mechanical plant. Tower operation costs

10.What are the disadvantage of lagoons?

Requires extensive land area. Hence the method can be used only on rural area. If used in urban areas, expansion of town and new developments may

encroach on the lagoon site.

11.What do you understand by facultative ponds?A facultative ponds combine the features of the aerobic and anaerobic

ponds.

o A facilitative pond consists of three

o Aerobic Zone

o Faculative zone

o Anaerobic zone

12.What are remedial measurement for rising sludge problem?

o Increasing the return sludge age

o Increasing the speed of the sludge scroper mechanism, where possible

o Decreasing the mech cell residence by increasing the sludge rate

13.What is meant by sludge bulking?

Sludge with poor setting characteristics is termed bulking sludge. It results on poor influent due to thee presence of excessive suspended solids and also in rapid loss of MISS from aeration tank.

14.What are the advantage of increment 8 and filters? The efficient from intermittent sand filter is of better quality. It is more clean and

more stable and hence does not need further treatment before disposal The filter work under aerobic conditions, and hence there is no trouble of odour,

files and inserts The operation is very simple, requiring no mechanical equipment except for dosing

15.What are the disadvantages of intermittent sand filters? The rate of filtration and hence that of load long is very small per unit surface area

of the filter hence they cannot be employed for medium size or bigger plants

They requires large area and large quantity of sand due to which their construction is very lostly.

16.What do you understand by contact beds?

Confact beds, also called confact filters, are similar to inter mitten sand filters in construction, except that th filtering media is very coarse, consisting of broken stones called ballart of 20 to 50mm gauge.

A contact bed is a water trough take of masonry walls and of rectangular shape. The depth of filtering media is kept b/w 1 to 1.8m

17.What are the operations involved in the contact beds?

Filling Contact Emptying Oxidation

18.What are the advantage of contact of beds?

Contact beds can work under small heads. Contact beds can be operated without exposing the sewage efficient to view.

There is no nuisance of filter flows The problem of odour is much less as compared to trill long filters.

19.What are the disadvantage of contact beds in T.F?

Rate of loading is mech less in comparison to trickling filters. Large areas of land is required for their installation Intermittent operation requires continous attendance The cost of contact beds is mech more as compared to

trickling filters

19.What do you mean by trickling filters?

Trickling filters, also as percolating filters or sprinkling filters are similar to contact beds in construction, but their operation is continous and they allow constant aeration In this system sewage is allowed to sprinkle or trickle over a bed of coarse, rough hard filter media and it is then collected through the under drainage system

20.What are the purpose of under drainage system?

o The purpose of under drainage system is two fold

o To carry away the liquid efficient and sloughed biological solids.

o To distribute air through the bed

PART B

1. Explain classification of Treatment processes?

Sewage before being disposed of either in river streams or on land has generally to be treated. So as to make it safe

Sewage can be treated in difference ways treatment process are often classified as

1) Preliminary treatment 2) Primary treatment 3) Secondary or (biological) treatment 4) Complete final treatment

Preliminary treatment: Preliminary treatment consists solely in separating the floating

materials (Like dead animals, tree branches, papers, pieces

of rags, wood etc) and also the heavy settle able inorganic solids.

It also helps in removing the oils and greases etc. From the sewage this treatment reduces the BOD of the waste water, by about 15 to 30%.

The process used are screening for removing floating papers, rags, clothes etc.

Grit chambers or detritus tanks: For removing grit and sand Slimming tanks: For removing oils and greases.

Primary Treatment

Primary treatment consists in removing large suspended organic solids. This is usually this is usually accomplished by sedimentation on settling basins.

The liquid effluent from primary treatment often contains a large amount of suspended organic material and has a high BOD about (60% of original).

The original solids which are separated out in the sedimentation tanks (in primary treatment) are often stabilized by an anaerobic decomposition in a digestion tanks or are incinerated.

Sometimes the preliminary as well as primary treatments are classified to gather under primary treatment.

Secondary treatment

Secondary treatment involves further treatment of the efficient, coming from the primary sedimentation tank. This is generally accomplished through biological decomposition of organic matter, which can be carried out either under aerobic or anaerobic conditions.

In these biological units, bacteria will decompose the fine organic matter, to produce cleaner effluent.

The treatment reactors, in which the organic matter is decomposed (oxidized) by aerobic bacteria are known as aerobic biological units; and may consists of

i. Filters (intermittent sand filters as well as trick long filters). ii. Aeration tanks with the feed of recycled activated sludge (i.e., the

sludge which I settled in secondary sedimentation tank, receiving effluents from the aeration tank)

iii. Exudation ponds and Aerated legions. Since the there aerobic units,

generally make use of primary settled sewage, they are early classified as secondary units.

The effluent from the secondary biological treatment will usually contain a little BOD (5 to 10% of the original). The organic solids sledge separated out in the primary as wells as in the secondary settling tank will be disposed of by stabilizing them under anaerobic process in a sludge digestion tank.

The final or advanced Treatment

Thus treatment is sometimes called tertiary treatment, and consists in removing the organic local left after the secondary treatment, and particularly to kill the pathogenic bacteria. Thus treatment, which is normally carried out by chlorination

Shows diagrammatic sketches of some standard types of sewage treatment plants

2. Design a suitable rectangular sedimentation provided with mechanical cleaning equipment for treating the sewage from a city, provided with an assured public water supply system, with a max

daily demand of 12 million lit/day. Assume suitable values of detention period and velocity of flow in the tank. Make any other assumptions, wherever needed.

Solution:

Assuming that 80% of water supplied to the city becomes sewage, we have the quantity of sewage required to be treated per day i.e (max daily)

= 0.8 x 12 million lit= 9.6 M. lit

Now assuming the detention period in the sewage sedimentation tank as 2 hrs, we have

Q=9.624

x 2M.lit

= 0.8M.lit

800cu.m

Now assuming that the flow velocity through the tank is maintained at 0.3 m/min, we have

The length of the tank required

= velocity of flow x detention period

= 0.3 x (2 x 60)

= 36 m

C/s area pf the tank required

= C apa c i t y o f th e tan k length of the tank

= 80036

=22.2 m2

Assuming the water depth in the tank (i.e effective depth of tank) as 3 m

The width of the tank required

= A re a o f X – se ct io n

Depth

= 22. 2 = 7.4m

3

Since the tank is provided with mechanical cleaning arrangement no extra space at bottom is required for sludge zone.

No, assuming a free – board of 0.5 m, we have

The overall depth of the tank = 3 + 0.5

= 3.5 m

In overall size of 36.5m x 8 m x 4 m can be used.

3.Stabilisation ponds for a town of 3000 population are provided to operate in

serves. The larger cell has area of 60,000 m2, and the smaller one 30,000 m2. The

average daily was to flow is 900 m3/d containing 200 kg of BOD (222 mg/e)

(i) For series operation, calculate the BOD loadings based on both the total pond area and the larger cell only.(ii) Estimate the number days of winter storage available between0.6 m and 1.5 m water levels. Assuming an evaporation and seepage loss of 2.5 mm of water per day.

SOLN:

(i) a) BOD Loading based on total pond area

Total pond area of both cells joined in serves

= 60,000 m2 + 30,000 m2

= 90,000 m2 = 9 hec

Total BOD per day = 200 kg/dayBOD loading in kg/ha/day

= 200

kg / d / ha

9

= 22.2 kg/ha/day(i) (b) BOD loading based on area of larger cell only

Area of larger cell = 60,000 m2 = 6 hac

BOD = 200 kg/dayBOD loading on kg/ha/day

= 200

9=33.3 log/ha/day

(ii) To calculate the number of days of storage between WL 0.6 m and1.5 m, we have depth available for storage= 1.5 – 0.6 = 0.9 m

Total area = 90,000 m2

Volume of stage available

= 90,000 x 0.9 = 81,000 m3

Daily in flow of sewage = 900 cum/dayThe sewage volume, which percolates and evaporates daily = 2.5 mm depth

= 2.510

x1

100m x surfacearea of tanks

= 2.5

1000x 90000

= 225 m3

Not effective daily in flow of sewage

= (900 – 225) m3

= 675 m3/day

Winter storage available as days

= Vol. Of Storage∈m3

Daily net sewage inflow onm3

day

= 81000

675 = 120 days.

4.Design a septic tank for the following data Number of people = 100Sewage / capital/day = 100 lit De – sluding period = 1 year Length = width = 4 : 1

Quantity of sewage produced per day = 12,000 lit/day

Assuming the defention period to be 24 hrs, we have the quantity of sewage produced during the defention period is the capacity of the tank.

= 12,000 x 24/24

= 12,000 lit

Now assuming the rate of sludge deposit as 30 lit/capita/year and with the given 1 year period of cleaning, we haveThe quantity of sludge deposited = 30x100 x1= 3,000 lit

Total required capacity of the tank = 12,000 + 3,000

= 15,000 lit

=15m3

Assuming the depth of the tank as 1.5 m, the c/s area of the tank

= 151.5

=10 m2

Using L : B as 4 : 1 (given)

4B2=10

B = √2.5=1.5 m

C = 4 x 1.5 = 6m

The dimensions of the tank will be 6 m X 1.5 m X (1.5 + 0.3 m) as overall depth with 0.3 m free board. Hence, use a tank of size 6 m X 1.5 m X 1.80 m

5. Design in a preliminary treatment unit the screen and the detritus tanks for 50,000 people. The dry weather flow is 110 lit / h / day. Assume the maximum flow as 3 times the DWF. Assume suitably the data not given

Screens Total Flow

= 50000 x 110 lit/day

= 50000

24 x60 x 60=63.65

ls

Maximum flow = 3 x 63.65 = 190.95 l/s ray 190 l/s

Using one screen with openings of 25mm, at the rate of 1160 cm2

per thousand people

Submerged area required = 50 x 1160 cm2

= 5.8 m2

Alternatively, the area of the rack may be @ 1.0 cm2 per 100 lit of DWF

i.e 5 0, 00 0

= 5.5m2

100How assuming that 15 lit screenings per ML of flow are separated.

Total screenings =

15106

x50000 x 110

= 82.5 lit

if the velocity in the screen chamber is 45 cm/sC/s area of screen chamber

= 19 0 x 100 0

30

= 6333.3 m2

De tritus ta nk :

Assuming the maximum capacity of tank as 0.8% of DWF,it is

equal to 0.8100

x50000 x 110=44000 lit

Maximun quantity that flows through the tank = 190 l/s

If the limiting velocity is 30 cm/s

c/s area = 19 0 x100 0 =

6334cm2

30

If the detention period is 45 sec

Length of the tank = 45 x 30 = 1350 cm

Providing 5 tanks of 13.5 length each

Total capacity of the tank

= 5 x 135 0x633 4

1000

= 42755 lit

Quantity of grit at the rate of 151/ML/day

= 82.5 lit/day

If the cleaning period is 2 weeks

Storage for 2 weeks = 82.5 x 14 = 1155lit

Total capacity of 5 tanks

= 42755 + 1155 = 43910 = 44000 lit say

Which is equal to maximum capacity required,

Depth of tank = 135 0 =

85cm, say

16

Width = 633 4 =

80cm, say

85

UNIT – IV

SECONDARY TREATMENT OF SEWAGE

PART A

1.What is the basic difference between activated sludge processes and trickling filter

Trickling Filter Activated Sludge ProcessThe bacterial film coating the grains of the filter is stationary.

The bacterial film which is kept moving is due to constant agitation.

2. Give any 4 advantage of activated sludge plaof? i) Lesser land area is reqd ii) The head loss on the plant is quite low iii) There is no fly or odour nuisance

iv) Capital cost is less

3. What are the disadvantages of the activated study plant? i) High cost of operation, take greater power consumption ii) A lot of machinery to be handled iii) The sudden change in the quantity and character of sewage may produce

adverse effects on the working of the process thus producing inferior efficient

4. Define the term eutrophication? The excess growth of algae and other aquatic plants in a river stream is called eutroplication.

5. What do you mean by secondary treatment? The effluent from the primary sedimentation tank

contains about 60 to 80% of the unstable organic matter originally present in sewage. Thus colloidal organic matter which passes the primary clarifier without settling there, has to be removed by further treatment. This is called secondary or biological treatment.

6. What are the special types of filters?

Durban filter Magnetic filters Rapid sand filters

7.What is the range of sand particle in the filtering média?

D10 (effective size) = 0.2+0 0.5 mmD60

D10

(uniformly coefficient ) --> 2 to 5

8. What are the types of trickling filters?

1) Conventional trickling filter or ordinary or standard rate or low rate trickling

filter 2) High rate filters or high rate trickling filter

9. What are the advantage of trickling filters?

Rate of filter loading is high as such requiring lesser land areas and smaller quantities of filter media for their installations.

They are self- clearing Mechanical wear and tear is small as they contain less – mechanical

equipment. Moisture content of sludge obtained from trickling filters is high as 99%

or 80.

10. What are the disadvantages of trickling filters?

the head loss through these filters is high, making automatic cleaning of the filters necessary

cost of construction is high There filters cannot treat raw sewage and primary sedimentation is a must.

11.Define the term Recirculation Ratio?

The ratio ( RI ) of the volume of sewage recirculated

(R) to the volume of raw sewage (I) is called recirculation ratio.

12. What is the formulae for recirculation factor?

F=1+ R

I

(1+0.1RI )

2

WhereF = Recirculation factor

R = Volume of sewage recirculatedI = Volume of Raw sludge

13. What do you understand by “Littoral zone”?The shallow water near the shore in which rooted plants grow, is called

the literal zone. The extent of the littoral zone depends on the slope of the lake bottom, and the depth of the euphosic zone.

14. What is meant by Behthic zone? Give exampleThe bottom sediments in a lake comprises what is called the benthic zone.

As the organisms living in the overlying water die, they settle down to the bottom, where they are decomposed by the organisms living in the bent hic zone. Bacteria are always present on this zone.

15.Write the formula for finding the efficiency of single high rate trick long filter?

η (% )= 100

1+0.0044√ YVF

WhereY = The total organic loading in kg/day applierd to the filter is the total bon

in kg. V = Filter volume in hec-m

F = Recirculation factor

16.Write the equation for unit organic loading?

U = Y VF

u = Unit organic loading on filter

17. Write the expression for finding out the final efficiency of two stage T.F?

𝛈 = 100

1+ 0 .00441−η √ Y '

V ' F '

WhereY’ =total BOD in efficient from first stage in kg /day

V’ = Volume of second stage filter in ha-mF’ =Recirculation factor for the second stage filter 𝛈’ = Final

efficiency

18.What are the merits of conventional trickling filter?

The efficient obtained from truckling filters is highly nitrified and stabilized. The efficient can there fore be disposed of in smaller quantity of deputation water

It has good dependability to produce good efficient under very widely varying whether and other conditions

The working of truckling filter is simple and sheep and does not require any skilled supervision

19. What are the demerits of conventional trickling filters? The loss of head through the filter system is high their making the

automatic dosing through siphonic doing tank necessary. The cost of construction of the filter is high . They require large area in comparison to their biological treatment

processes.

20.What is the necessary of Recirculation in Trickling Filter?Recirculation is necessary to provide uniform hydraulic

loading as well as to dilute the high strength waste waters.In constant to the

low rate filters, in high rate filters a part of settled or filter effluent is recycled through the filter.

PART B1. The sewage is flowing @ 4.5 million liters per day from a primary

clarifier to a standard rate trickling filter. The 5-day BOD of the influent is 160 mg/l. the value of the adopted organic loading is to be 160 gm / m3/ day and surface loading 2000l/m2/day. Determine the volume of the filter and its depth. Also calculate the efficiency of this filter unit.Solution

Total 5-day B.O.D present in sewage

= 160 x 4.5 x106

103

= 720000 gm/day

Volume of the filter media requiredTotal B . O. D

Organic Loading Rate

=720000

160gm /d

gm /m3/d=720000/160m3

=4500m3

Surface area required for the filter

= Total Flow

Hydraulic Loading Rate

= 4.5 x106c /d2000 l /m2 d

= 4.5 x106

2000

= 2250 m2

Depth of the bed req = 45002250

=2mEfficiency of filter is given by𝛈 =

100

1+0.0044√u

U ---organic Loading kg/ha-m/day

𝛈 = 100

1+0.0044√600

= 100

1+0.176

= 100

1.176 = 85.03%

2. Determine the size of a high rate trickling filters for the following data.

Sewage flow = 4.5 mld Recirculation ratio = 1.5 BOD of row sewage =250 mg/l BOD removal in primary tank = 30% Final efficient BOD desired = 30 mg/l

Solution Quantity of sewage flowing into the filter per day = 4.5 M.L /day BOD concentration in raw sewage

250 mg/l

Total BOD present in raw sewage=4.5 ml x 250 mg/l =1125 kg

BOD removed in primary tank = 30%BOD left in the sewage entering per day on the filter unit = 1125 X 0.7

= 787.5 kgBOD concentration desired in final effluent = 30 mg/l

Total BOD left in the effluent perday = 4.5 X 30kg = 135kg

BOD removed by the filter =787.5 -135 =652.5 kg

Efficiency of filter = B O D Removed

ToTal B O Dx100

= 652.5787.5

x100

= 82.85%Now using Equation,

η= 100

1+0.0044 √ YV F

Y = Total BOD in kg = 787.5 kg

F = 1+R/ I

(1+0.1 R /I )2

= RI=1.5 (given )

F= 1+1.5

[1+0.1+1.5 ]2

= 2.5

1.152=1.89

82.85= 100

1+0.0044 √ 787.5V x1.89

V = 2000 m3

Assuming the depth of the filter as 1.5m, we have

Area = 20001.5

m2

= 1333.3 m2

Dia req = √1333.3 x4π

= 41.2m.Hence, use a high rate tricking filter coth 41.2m dia, 1.5m deep filter media and wo the recirculation (single stage)ratio of 1.5.

3.Determine the size if a high rate trickling filter for the following data. Flow = 4.5 m/d Recirculation ratio = 1.4BOD of raw sewage = 250 mg/lBOD removed in primary clarifier = 251 Final effluent BOD derived = 50mg/lCalculate also the size of the standard rate trickling filter to accomplish the above requirement

Solution:-Total BOD present in raw sewage perday= 4.5Ml X 250 Mg/l =1125kg BOD removed in the primary clarifier =25% BOD inferring per day in the filter units =0.75 X 1125 kg =843.75 kg Permissible BOD concentration in the effluent = 50m/l

BOD allowed to go into the effluent

= 50 mg/l X 4.5 nql = 225 kg

BOD removed by the filter perday = 843.75 – 225 =618.75kg

Efficiency of filter = B O D Removed

ToTal B O Dx100

= 618.75843.75

x 100=73.33 %

Now using Equation,

η= 100

1+0.0044 √ YV F

Y = Total BOD in kg = 843.75 kg

=1+R/ I

(1+0.1 R /I )2

= RI=1.4 ( given )

F= 1+1.4

[1+0.1+1.4 ]2

= 2.4

1.142=1.85

73.33= 100

1+0.0044 √ 843.75V x1.85

V = 664.5 m3

Assuming the depth of the filter as 1.5m, we have

Surface area = 664.5

1.5m2

= 413.6 m2

Diameter req = √413.6 x4π

= 23.8m.For an equivalent standard rate filter, F =1

73.33= 100

1+0.0044 √ 843.75V x1

V = 1231 m3

Assuming the depth of the filter as 1.5m, we have

Surface area = 12311.5

m2

= 820.8 m2

Diameter req = √820.8 x4π

=32.3 m

4. A single stage filter is to treat a flow of 3.79 M.L.d of raw sewage BOD of 240 mg/l.It is to be designed for a loading of 11086 kg of BOD in raw sewage per hence fare metre, and the recirculation ratio is to be l. what will be strength of the efficient, according to the recommendation of the National Research Council of U.S.P Solution:-

Total BOD present in raw sewage = 3.79M.l X 240 mg/l

= 909.6 kgNow, filter volume required

Total BOD in raw sequence Given BOD loading rate of 11,086 kg/ha-m

= 909.611086

ha−m=0.082 ha−m

Now assuming that 35% of BOD is removed in primary clarifier we haveThe amount of BOD approved to the filter= 0.65X 909.6 kg = 591.24 kg Now using equation, we have

η= 100

1+0.0044 √ YV F

Where y= total BOD applied to the filter on kg = 591.24 kgV = Vol of the filter in ha-m =0.082 ha-m

F =1+R/ I

(1+0.1 R /I )2

RI=1

F= 1+1

1+ (0.1 )2

= 2

1.21=1.65

η= 100

1+0.0044 √ 591.240.082 x1.65

= 77.45%The amount of BOD left in the effluent

= 591.24 ( 1-0.7745) kg = 133.32 kg

BOD concentration in the effluent = Total BOD

Sewage Volume

= 133.32 x106

3.79 x 106

` = 35.18 mg/L

5.It is proposed to use a two stage plant instated of the single stage plant in previous problem (4). The total volume of filter medium remains the same as was in one filter is 0.082 ha-m and each filter is to contain half of this material and the recirculation ratio is to be for each filter. Determine the BOD of the plant effluent

Solution:For each filter F = 1.65

For the first stage filter, the efficiency is given by

η= 100

1+0.0044 √ YV F

Y = Total BOD applied to filter = 591.24 kg (from previous problem) V = Volume of filter = 0.082/2 =0.041 ha-m

η= 100

1+0.0044 √ 591.240.041 x1.65

= 1001.41

=70.92%

percentage of BOD removed in first stage filter,

= 70.92 %

Amount of BOD left in the effluent from that filter = 591.24 (1-0.7092)

= 171.9 kg

For the second stage filter, the efficiency is given by

η= 100

1+ 0.00441−η √ Y '

V F '

Y’ = 171.9 V’ = 0.041 ha.m F’ = 1.65𝛈 = 0.7092

η'= 100

1+ 0.00441−0.7092 √ 171.9

0.041 x 1.65

= 100

1.762=56.75 %

The Amount of BOD left in the effluent plant =

171.9( 100−56.75100 )

= 74.35 kg

BOD concentration in the effluent = Total BOD

Sewage Volume

= 74.35 x 106

3.79 x 106

` = 19.61 mg/L

UNIT V

DISPOSAL OF SEWAGE ON LAND

PART A

1. Define the term “Dilution Factor”?

The ratio of the quantity of the diluting water to that of the sewage is known as the Dilution Factor.

2.What are the methods adopted for sewage disposal?

5) Dilution is disposal in water. 6) Effluent Irrigation or Broad Irrigation or Sewage forming is disposal

on land.

3.What are the conditions adopted for disposal by dilution?

7) When sewage is comparatively fresh (4 to hr old) and free from floating and settlable solids.

8) When the dilution water has a high dissolved oxygen (D.O.) content. 9) When the out fall sewer of the city or the treatment plant is situated

near some natural waters having large volumes.

4.What are the natural forces of purification?

o Dilution and dispersion.

o Sedimentation

o Oxidation – reduction in sun-light.

o Oxidation

o Reduction

5. What are the factors affecting self purification of polluted streams?

a) Temperature b) Turbulence c) Hydrography such as the velocity and surface expanse of the river stream. d) Adviable dissolved oxygen and the amount and type of organic matter. e) Rate of re aeration.

6. What are the types of self purification?

The self purification divided into four zones.

1. Zone of degradation. 2. Zone of active decomposition. 3. Zone of recovery 4. Zone of Cleaner water

7. What is meant by “Self purification phenomenon”?

When sewage is discharged into a natural body of water, the receiving water gets polluted due to waste products, present in sewage effluent. The natural forces of purification such as dilution, sedimentation, oxidation – reduction in sun light go on acting upon the pollution elements and bring back the water into its original condition. This automatic purification of polluted water, in due coarse is called the self purification phenomenon.

8.Define the term Re-oxygenation curve.

In order to counter – balance the consumption of D.O. due to de-oxygenation, atmosphere supplies oxygen to the water and the process is called re-oxygenation.

9. What is mean by “Oxygen sag curve”?

The amount of resultant oxygen deficit can be obtained by algebraically adding the de-oxygenation and re-oxygenation curves. The resultant curve so obtained is called the oxygen sag curve or the oxygen deficit curve.

10. Write the equation for find out the B.O.D. of the diluted water.

B.O.D. of the diluted mixture

C=C sQ s+CR QR

QW+QR

Where

Cs = B.O.D. of sewageCR= B.O.D. of river

Qs = Sewage discharge

QR = Discharge of the river

11. Define the term “limnology

A study of the lake systems is essential to understand the role of phosphorous in lake pollution. The study of lakes is called limnology.

12. What is meant by epilimnion zone?

The water of a lake gets stratified during summers and winters. Since such turbulence extends only to a limited depth from below the water surface, the top layers of water in the lake become well mixed and aerobic. This warmer, well mixed and aerobic depth of water is called epilomnion zone.

13. What is meant by hypolimnion zone?

The lower depth of water in the lake which remains cooler, poorly mixed and an aerobic, is called are hypolimnion zone.

14. What do you understand by monocline? Give example.

The water of a lake gets stratified during summers and winters. The change from epilimnion to hypolimnion can be experienced while swimming in a lake. When you swim in top layers horizontally you will feel the water warmer and if you dive deeper, you will find the water cooler. The change line will represent monocline.

15. What are the classification of biological zones in lakes?

The most important biological zones are

(i) eutrophic zone (ii)Littoral zone (iii) benthic zone

16.What do you understand by “Eutrophic Zone”?

The upper layer of lake water through which sunlight can penetrate is called the eutrophic zone. All plant growth occurs in this zone. In deep water, algae grow as the most important plants, whole rooted plants grow in shallow water near the shore.

17. Define the term secchi disk? Draw a neat sketchThe depth of the eutrophic zone can be approximated and measured by a

sample device called the secchi disk as shown in figure.

18.What is meant by “humus”? The refuge gets stabilized in about 4.5 months period, and gets changed into

a brown coloured,odourless innocous powdery form known as humus, which has high manure value became of its nitrogen content

19.What do you understand by composting? Composting is a method in which putrescible organic matter in the solid

waste / refuge is digested anaerobically and converted into humus and stable mineral compounds.

It is a hygienic method which converts the refuge into manure through the bacterial agencies.

20.Define the term “Sewage sickness”.When sewage is applied continuously on a piece of land, the soil

pores or voids may get filled up and clogged with sewage matter retrained in them. The tome taken for such a clogging will, of course depend upon the type of soil and the load present in sewage.

The organic matter will thus, of course, be mineralized, but with the evolution of four gases like H 2S, Co2, CH4. This phenomenon of soil getting clogged is known as sewage sickness.

PART-B1.Enumerate the two general methods adopted for sewage disposal and explaining the conditions favourable for their adoption

There are two general methods of disposing of the sewage effluents.

a. Dilution is disposal in water. b. Effluent Irrigation or Broad Irrigation or sewage farming is

disposal on land.

Disposal by dilution:-

Disposal by dilution is the process whereby the treated sewage or the effluent from the sewage treatment plant is discharged into a river stream, or a large body of water, such as a lake or sea. The discharged sewage in due course of time, is purified by what is known as self purification process of natural waters. The degree and amount of treatment given to raw sewage before disposing it of into the river stream in question, will definitely depend not only upon the quality of raw sewage but also upon the self purification capacity of the river stream and the intended use of its water.

Conditions favouring Disposal by dilution.

The dilution methods for disposing of the sewage can favourably be adopted under the following conditions.

1. When sewage is comparatively fresh (4 to 5 hr old) and free from floating and settleable solids. (or are easily removed by primary treatment)

2.When the diluting water (is the source of disposal) has a high dissolved oxygen (0-0) content.

3. Where diluting waters are not used for the purpose of navigation or water supply for at least some reasonable distance on the downstream from the point of sewage disposal.

4. Where the flow currents of the diluting waters are favourable, causing no deposition, nuisance or destruction of aquatic life.

5. When the out fall sewer of the city or the treatment plant is situated near some natural water having large volumes.

Disposal on land:-

Disposal of Sewage Effluents on land for Irrigation:

In this method, the sewage effluent (treated or diluted) is generally disposed of by applying it on land. The percolating water may either soon the water table or is collected below by a system of under drains. This method can then be used for irrigating crops.

This method, in addition to disposing of the sewage may help in increasing crop yields (by 33% or so) as the sewage generally contains a lot of fertilizing minerals and other elements.

However, the sewage effluent before being used as irrigation water, must be made safe. In order to lay down the limiting standards for sewage effluents, and the degree of treatment required, it is necessary to study as to what happens when sewage is applied on to the land as irrigation water.

The pretreatment process may be adopted by larger cities which can afford to conduct treatment of sewage when sewage is diluted with water for disposal for irrigation, too large volumes of dilution water are generally not needed, so as not to require too large areas for disposal.

2.The sewage of a certain town contains 600 ppm of suspended matter. Assuming that 55% of this settled down in plain sedimentation tank, and the sludge collected has a water content of 95% calculate has a water content of 95% calculate its quantity per million litre, both in bulk and weight. Assume sp. Gravity 1.2

Solution:-

Suspended matter in sewage

= 600 ppm = 600 mg / l

For 1 million litre of sewage, we have the suspended matter.

¿ 600

106x106 kg

¿600 kg

Now, 55% of this matter is settled as sludge, and therefore quantity of sludge solids.

= 0.55 x 600

= 330 kg.

The sludge is having 95% m.c. which, means 5 kg of dry solids will made 100 kg of wet sludge.

5 kg of dry solid make = 100 kg of sludge

330 kg of dry solids make

¿ 1005

x330

¿6600 kg of sludge

Hence, the wt. of sludge formed per million litre of sewage = 6600 kg.

Volumeof sludge= Wt . of sludgeUnit Wt Of Sludge

(Unit wt of sludge=Specific Gravity x Unit wt of water

= 1.02 x 1000 = 1020 kg /m3)

¿ 6600 kg

1020 kg/m3

¿6.47 m3

Hence volume of sludge formed per million litre of sewage = 6.47 cu.m

3.Write short notes oni) Efficient irrigation and sewage farming.

ii)sewage sickness.

Efficient irrigation and sewage farming:

Both these terms are used as synonyms to each other, yet there is one basic difference b/w them.

This difference is that : in “efficient irrigation” (or broad irrigation ), the chief consideration is the successful disposal of sewage, while in sewage farming, the chief consideration in the successful growing of the crops.

o Hence in broad irrigation, the raw or settled. Sewage is discharged on

vacant land which is provided under neath with a system of properly laid under – drains.

o These under –drains basically consist of 15 to 20 cm river process tile

pipes with a spacing of 12 to 30m. o The effluent collected in these drains after getting filtered through the

pores is a generally small (as a large quantity gets evaporated) and well stabilized, and can be early disposed into some natural water courses, with out any further treatment.

o In case of sewage farming, however the tress is load upon the use of

sewage efficient for irrigation crops and increasing the fertility of the soil.

o The pre-treatment of sewage in removing the ingredients which may

prove harmful and toxic to the plant is there fore, necessary in this case.

Sewage sickness:

When sewage is applied continuously on a Piece of land, the soil pores or void may get filled up and clogged with sewage matter retained in them.

The time taken for such a clogging will, of course depend upon the type of pores and the load present in sewage.

But once these voids are clogged, free circulation for air will be prevented and anaerobic conditions will develop on the pores.

Due to those the aerobic decomposition of organic matter will stop, and anaerobic decomposition will start.

The organic matter will there, of course, be minor but with the evolution of foul gases like H2S, CO2, CH4. this phenomenon of soil getting clogged is known as sewage sickness of land.

4. A treated waste water is discharged at the rate of 1.5 m3/s into a river of minimum 710 to 5m3 /sec. The temperature of river flow and waste water flow may be assumed at 250C The BOD removal rate constant k is 0.12/d (base 10) . The BOD5 at 250C of the waste water is 200 mg/l and that of the river water upstream of the waste water out full is 1mg/l. the efficiency of waste water treatment is 80% Evaluate the following.

i) BOD5 at 250C if river water received un treated waste water.

ii) BOD 5 at 250C if river water recieves treated waste water.

Solution:

Discharge of waste water = QW=1.5m3/sDischarge or river QR =5m3/sTemperature T =250C

KD(250)=K1=0.12/d

CW-conc of BOD5 for untreated water=200mg/l

CR-conc. Of BOD5 for river water =1 mg/l(i) conc of BOD5 of the mixture if untreated waste water is discharged

into the river

C=CW QW +CR QR

QW+QR

¿ 200 x 1.5+1 x51.5+5

= 46.92 mg/l(ii)BOD5 of the treated waste water is given by cw=20% of the BOD5 of

untreated waste water

(li efficiency of waste water treatment a 80%) =20% x CW

=20% x200 mg/l =40mg/l

BOD5 of mixture if treated waste water is discharged into the river

C=CW QW +CR QR

QW+QR

¿ 40 x1.5+1 x51.5+5

= 10 mg/l

5.A sewage containing 200mg/l of suspended solids is passed through primary setting tanks, tricking filters, and secondary settling tanks, how much gas will

probably be produced in the digestion of sludge from on million litre of sewage?

Solution:

Total suspended solids in sewage =200mg/l

Assuming 90% removed of suspended solids in complete treatment, we have.

The suspended solids removed =90% x 200 mg/l=180mg/l.

Assuming volatile solids to be equal to 70% of suspended solids, we haveVolatile solids removed

=70%x180mg/l =126mg/lNow, assuming that the volatile solids (matter) is reduced by 65% in the sludge by digestion, we have

Volatile solids reduced

=65%x126mg/l=81.9mg/l

Volatile matter reduced per million litre of sewage produced per kg of volatile matter reduced, we have the gas produced per million litre of sewage

=0.9x81.9 cu.m. =73.71 cu m. =73710 litres.

6.What are the preventive measure of sewage sickness by the land disposal? Describe it.

In order to prevent the sewage sickness of a land, the following preventive measures may be adopted.

1. Primary treatment of sewage. 2. Choice of land 3. under drainage of soil 4. Giving rest to the land. 5. Rotation of crops 6. Applying shallow depths.

Primary treatment of sewage.

The sewage should be disposed of, only after primary treatment, such as screening, grit removal and sedimentation. This will help in removing settle able solids and reducing the B.O.D load by 30% or so. And as such, soil pores will not get clogged, quickly.

Choice of land:

The piece of land used for sewage disposal should normally be sandy or loamy, dayey lands should be avoided.

Under – drainage of soil:The cannel on which un sewage is being disposed of, can be better drained if a

system of under – drains (ie open joined proper ) is laid below, to collect the efficient; and those will also minimize the possibilities of sewage sickness.

Giving rest to the land:

The land which the sewage being used for disposal should be given rest, periodically by keeping some extra land as reserve and stand-by for diverting the

sewage during the period the first land is at rest more over, during the rest period, the land should be thoroughly planned so that it gets broken up and aerated.

Rotation of crops;

Sewage sickness can be reduced by planting different crops in rotation instead of growing single type of a crop. This will help in utilizing the fertilizing elements of sewage and help on aeration of soil.

Applying shallow depths:

The sewage should not be filled over the area in large depth, but it shocked be approved in this layers. Greater depth of sewage on a land does not allow the soil to receive the sewage satisfactory and ultimately results in it clogging.

Sewage –sick land can be improved and made useful by thoroughly plugging and treating the soil, and exposing it to the atmosphere.