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BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M

ENVIORNMENTAL ENGINEERING-II

UNIT – I

Planning for Sewarage Systems

PART – A

1. What are the types of treatment processes?

1) Preliminary treatment

2) Primary treatment

3) Complete final treatment

4) Secondary treatment

Exudation ponds

Filters Aeration &

tangles Aerated lagoons

2. What are the various sources of wastewater generation?

I. Industrial Wastes

II. Domestic wastes

III.Agricultural Wastes

3. List out the types of anaerobic bio logical units?

1) Anaerobic lagoons

2) Septic tank

3) 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

1) Coarse screens

2) Medium screens

3) Fore 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?

Comminutes 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, lacing bad smells and nuisance.

11. What are the methods adopted for disposal ofscreenings?

1) Burning

2) Burial

3) 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 damager

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?

1) Physical unit operations

2) Chemical unit process

3) 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. Define the term Displacement efficiency?

The ratio of the “Flowing through period” to the “detention period” is

called the displacement efficiency.

19. 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.

20. 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. 21. Define the

term “Detention Period”?


The theoretical average time for which the water is detained is called

the detention period.

22. Give any two advantage of chemical coagulationprocess in sewage

treatment?

i) Sedimentation aided with coagulation produces better efficient with lesser BOD and suspended solids, as compared to plain sedimentation.

ii) Coagulated settling tank requires less space than that

required by an ordinary plain settling tank.

23. What are the Demerits of coagulation in sewagetreatment?

i) Cost of chemicals is added to the cost of sedimentation, with out

much use, and thereby making the treatment costlier. ii) The

process of coagulation requires skilled supervision and handling of

chemicals.

24. What are the types of sedimentation tank?

Based on flow

1. Vertical flow tank

2. Horizontal flow tank

3. Radial flow tank

According to use

1. Primary

2. Secondary

3. inter mediate

25. What are the chemical used for precipitation ofsediment?

1. Alum

2. Ferrous sulphate

3. Ferric sulphate

4. Ferric chlorides

5. Sodium alluminate

6. Sulphuric acid

7. lime

8. copperas

26. What are the factors that affect the precipitations?

1. Kind of chemical

2. Quality of chemical

3. character and concentration of sewage

4. Ph values of sewage

5. time of mixing and flowlations


6. Temperature

7. Violence of agitation

27. What are the merits of chemical precipitation?

i) More rapid and through clarification

ii) Removal of higher percentage of suspended solids.

iii) Simplicity of operation

iv) Small size tank is enough

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 pathogence bacteria. Thus treatment, which is normally carried out

by chlorination

Shows diagrammatic sketches of some standard types of sewage

treatment plants 2. Describe the types of screens with need sketch.

Screening is the very first operation carried out at a sewage treatment

plant, so as to trap and remove the floating matter, such as pieces of cloth,

paper, wood cork, hair, fiber, kitchen refuge, fecal solids etc present in

sewage.

Thus, the main idea of providing screens it to protect the pumps and

other equipments from the possible damages due to the floating matter of the

sewage.

Types of screens depending upon the size of the openings and screens

may be classified as

1) Coarse screen

2) Medium screens

3) Fine screens Coarse Screen:

It is also known as racks and the spacing between the bars (i.e.,

opening size) is about 50 mm or more. These screens do help in removing

large floating objects from sewage. The material separated by coarse

screens, usually consists of rags, wood, paper etc.


Medium Screen:

The spacing between bars a about 6 to 40 mm. These screens will

ordinarily collect 30 to 90 lit of material per million liter of sewage. The

screenings usually contain some quantity of organic material may be

dispersed of by incineration of burial.

Rectangular Shaped coarse and medium screens are now – a – days

widely used at sewage treatment plants.

Now – a – days these screens are generally kept inclined at about 30

to 60 to the direction of flow, so as to increase the opening area and to reduce

the flow velocity and there making the screening more effective.

Fine Screens:

Have perforations of 1.5mm to 3mm in size. The installation of there

screens prove very effective and they remove as much as 20% of the

suspended solids from sewage. These screens, however, get clogged very

often, and need frequent cleaning. They are, therefore, used only for treating

the industrial waste waters or for treating those municipal waste wasters,

which are associated with heavy amounts of industrial wastewaters.

These screens will considerably reduce the load on further treatment units.

Brasses of Bronze plates or wire mesh are generally used for

constructing fine screens. The metal used to should be resistant to rest and

corrosion.

Find screens may be disc or drum type, and are operated continuously

by electric motor (Figure S.K.G Fig no 277)

3. Estimate the screen requirement for a plant treating a peak flow of 60 million liters/day of sewage.

Solution:

Peak flow = 60 ML/day

60 101000 6 cu.mday

0.694m3 sec

Assuming that the velocity through the screen (at peak flow) is not

allowed to exceed 0.8 m/sec we have the net area of screen openings reg


0.694 0.87m2

0.8

Using Rectangular steel bars in the screen, hawing 1 cm width and

placed at 5 m clear spacing The gross area of the screen required

0.876 1.04cm2

Assuming that the screen bars are placed at 60 to the horizontal

The gross area of the screen needed

1.04 1.04 2 1.2m2

Hence a coarse screen of 1.2m2 area is required while deigning the

screen, we have also to design its

cleaning.

Frequency:

The cleaning frequency is governed by the head loss through the

screen. The more the screen opening are clogged, more will be the head loss

through the screen generally not more than half the screen clogging is allowed

to know whether the screen has been clogged and needs cleaning.

The hL through the cleaning screen and half cleaned screen, can be

completed as follows Velocity through the screen = 0.8 m/sec

Velocity above the screen = 0.8 5

6 msec

= 0.67 m/sec

Head less through the screen = 0.7929(V2 – u2)

= 0.7929 (0.82 – 0.672)

= 0.0134 ray 0.013 m

When the screen openings get half clogged then the velocity through the

screen

V = 0.8 x 2 = 1.6 m/sec

Head loss =0.0729(1.62 – 0.672)

= 0.1538 say 0.15m

This shows that when the screens are totally clean, the head loss is

negligible is about 1.3cm only where as, the hL shoots up to about 15 cm at

half the clogging. The screens should therefore be cleaned frequently as to

keep the head loss within the allowable range.

3 3 2


4. A grit chamber is designed to remove particles with a diameter of

0.2mm, specific gravity 2.65. Settling velocity for there particles has

been found to range from 0.016 to 0.22 m/sec, depending on then shape

factor, A flow through velocity of 0.3 m/s. Will be maintained by

proportioning weir determine the channel dimensions for a maximum

waste water flow of 10,000 cum/day.

Solution:

Let us provide a rectangular channel section, since a proportional flow

wear is provided for controlling velocity of flow

Now,

Horizontal velocity of flow = Vn = 0.3 m/s

Setting velocity is b/w 0.016 to 0.022 m/s and hence let is be 0.020 m/s

Now

Velocity cross section =Vn A

= 0.116m3 sec

0.116 0.3A

A 0.116 0.385m2

0.3

Assuming a depth of 1m, we have the width (B) of the begin

1 B 0.385 B 0.385m

say0.4m

Now settling velocity Vs = 0.02 m/s

Detention time = Depth of the burin / settling velocity

10.02 50sec

Length of the tank = Vh x detention tome

= 0.3 x 50

= 15m

Hence, use a rectangular tank with dimensions

Length (L) = 15m

Width (B) = 0.4m

Depth (b) = 1.0m

5. Design a suitable grit chamber cum dexterities for a sewage treatment

plant getting a dry weather flow from a separate sewerage system @ 400

l/s. Assume the flow velocity through the tank as 0.2 m/s and detention

period of 2 minutes the maximum flow may be assumed to be three times

of dry weather flow.

Solution:


The length of the tank

= Velocity X Detention time

= 0.2 X (2 X 60) = 24 m

Now assuming that each detritus tank is designed for passing D.W.F, we have

The discharge passing through each tank

= 400 l/s = 0.4 m3 / sec

Cross-sectional Area required Discharge

Velocity

0.4 0.2

= 2 m2

Assuming the water depth in the tank to be 1.2 m, we have the width of the

tank

Area of X section Depth

2

1.2

= 1.67 m

Say 1.7 m

Hence, use a nitrites tank with 24 m X 1-7 m X 1.2 m size At the top, a

free-board of 0.3 m may be provided, and at the bottom a dead space depth

of 0.45 m for collection of detritus may be provided.

Thus, the overall depth of the tank = 1.2 + 0.3 + 0.45

= 1.95 m

The tank will be 1.7 wide up to 1.5 m depth, and then the sides will

slope down to form an elongated through of 24 m length and 0.8 m width at

the bottom with rounded corners, as shown in figure.

Detritus tanks:

Detritus tanks are nothing but grit chambers designed to flow with a

smaller flow velocity (of about 0.09 m/s) and longer detention periods (about

3 to 4 mini) so as to separate out not only the larger grit etc., but also to

separate out the very fine sand particles etc.

6. Design a grit chamber for a horizontal velocity of 25 cm/sec and a flow

which ranges from a minimum of 25000 cu-m/day to a maximum of

1,00,000 cu-m /day. Average flow is 62500 cu-m/days. Using

Bernoulli’s theorem


Total energy at U/s point in channel

= Total energy at critical point in control section

E1 Ec 1.5 V9C2

but E1 = D

D = 1.55 VC2 9

Using the value of Dmax, as equal to 1.16 at maximum discharge we have

Vc at Qmax 1.16X9.811.55

= 2.71 m/s

Also yc at Qmax V at QC2 max 2.71 2 0.74m g 9.81

The discharge through the control rection is

Q = (W. yc), VC

Where W is the throat width 8 W1 yc

is the flow area of the throat

W = Q/yc . Vc

W Qmax

y V bothatQc c max

0.289

0.74X2.71

= 0.144 m say 0.15 m Let us use throat

width 2 = 0.15 m For other flow conditions:

Using the above used two formulas we have yc VgC2

W . yc . Vc = Q

Vc = Q/w . yc


yc QW.yg c 2

g. yc . yc2w2 = Q2

yc2 g.wQ22

yc 3 gWQ22

Knowing Q and w, we can find yc at different discharges

D 1.55 VC2 g

D = 1.55 yc

Knowing y, at different discharges, we can find D at different discharges

Then finally we have for a parabolic section

2/3 X B X D = A

B 1.5A D

knowing A at various discharges, 7. Describe the skimming tanks with

neat sketch?

Skimming tanks are sometimes employed for removing oil and greeve

from the sewage and placed before the sedimentation tanks. They are,

therefore used where sewage contains too much of grease or oils which

include fats, wakes, soaps, fatty acids.

If such greasy and oily matter is not removed from the sewage before

if enters further treatment units, it may form unsightly and odourous scums on

the surface of the settling tanks or interfere with the activated sludge treatment

process.

These oil and greasy materials may be removed in a skimming tank, in

which air is blown by an aerating device through the bottom the rising air tens

to coagulate and longeal. (solidify) the grease and cause it to rise to the

surface (being pushed in separate compartments) from where it is removed.

The typical details of a skimming tank as shown


It consists of a long trough shaped structures divided into two or three

lateral compartments by means of vertical baffle walls (having slots in them)

for a short distance below the sewage surface as shown. The baffle walls help

in pushing the rising coagulated greasy material into side compartments

(called stilling compartments) the rise of oils and grease is brought above by

blowing compressed air into the sewage from diffusers placed at the bottom

of the tank.

The collected greasy materials are removed (i.e. skimmed off) either by

hand or by some mechanical equipment. It may then be disposed of either by

burning or burial.

A detention period of about 3 to 5 min is usually sufficient, and t he

amount of compressed air required is about 300 to 6000 m3 per million lit of sewate.

Surface Area = 0.00622 9

Vr g Rate of flow of sewage in m3/day

Vr Minimum rising velocity of greasy material to be

removed in m/min - 0.25 m/minute in most cases.

The efficiency of a skimming tank can be increased considerably (3 to 4 times) by pausing chlorine gas /2 mg / lit of sewage) along with the compressed air.

Unit-2

SEWER DESIGN

1. What are the Demerits of chemical precipitation?

1. High cost of chemicals

2. Large quantity of sludge which offers difficulty of its removal

3. Skilled attendance

4. Putrescible efficient

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. Write the expression for finding out the settling velocity?

Vs Ss 1 d2


18

For particle < 0.1 mm

Where

Vs Settling velocity in m/s

Unit t of water in leg / m3

Absolute or dynamic velocity on kg sec / m2

d die of particle in m

4. Write the hazen formula for determining the velocity?

Vs 418 Ss 1 d 3T100 70

Where

Vs Velocity in m/s d dia of particle in m T Temperature

in C

(For particles between 0.1 mm and 1mm)

5. What is do you mean by transitional setting zone?

Grit particles however, generally lie between 0.1mm and 1 mm, and

hence undergo settling which lies in between streamline settling and

turbulent settling. This settling zone is called the transitional settling zone

6. What are the users of Baffle?

1) Baffler are required to prevent the movement of organic matter

and its escape along with the efficient

2) Distribute the sewage uniformly through the cross section of

the tank.

3) It is used to avoid short circuiting

7. Write the equation for finding out the critical scour velocity?

VH 3 to 4.5 gd Ss -1

VH critical scour velocity

8. What are the classifications of biological process?

a) Aerobic processes


b) Anaerobic processes

c) Aerobic – anaerobic processes

9. List out the aerobic processes?

1. Activated sludge processes

2. Trickling filters

3. Aerobic stabilization pond

4. Aerated lagoon

10. List out the anaerobic process?

1. Anaerobic sludge digestion,

2. Anaerobic contact processes

3. Anaerobic filters

4. Anaerobic lagoons or ponds

11. What are the sources of waste water?

1. Domestic waste water (i.e sewage)

2. Agricultural return waste water

3. Industrial waste water

12. What are the methods involved in the treatment of waste

water?Mainly classified into

1. Conventional treatment methods

2. Advanced waste waster treatment

Conventional treatment methods

i. Preliminary

processes

ii. Primary treatment iii.

Secondary treatment

Advanced waste water treatment

i. Tertiary treatment

13. What are the functions involved in the chemical unit processes

1. Chemical precipitation

2. Gas transfer

3. Adsorption

4. Disinfection

5. Combustion

6. loss exchange

7. Electro dialysis


14. What do you understand by waste water treatment?

The waste water treatment or sewage treatment is a broad term that

applies to any process/operation or combination of processes and

operations that can reduce the objectionable properties of water carried

waste and render it less dangerous with the following.

1. Removal of suspended and floatable material

2. Treatment of biodegraslable organics

3. Elimination of pallogenic organisms

15. What is the detention periods range for sedimentation?

The detention periods range 45 to 120 min

16. Draw a general layout for sewage treatment process?

Raw sewage Treatment sewage

1 Screening

2 Sedimentation

3 Oxidation

4 Disinfection

17. Give example for single & Double storied sedimentation tank?

Single stored tanks septic tank

Double stored tank Inhofe tank

18. What is the detention period for detritus tank?

For detritus tanks, the detention period is 3 -5 minutes.

PART-B

1. What do you understand by unit operations and processes? What is

its importance in water and waste water treatment? Elaborate various

types of unit operations used for waste water treatment.

The waste water treatment is a bread term that applies to any

operation / process or combinations of operations and processes that can

reduce the objectionable properties of water-carried waste that render it less

dangerous and repulsive to man. Waste water treatment is a combination

of physical chemical, biological processes

Methods of treatment in which the application of physical forces

predominate are known as unit operations while methods of treatment in

1 2 3 4


which chemical or biological activities are involved are known as unit

processes.

Following are the important operations which constitute the physical,

chemical, biological unit operations/processes, employed in water and

waste water treatment.

1. Gas transfer : Aeration

2. Ion transfer

a. Chemical coagulation

b. chemical precipitation

c. Ion exchange

d. Adsortpion

3. Solute stabilization.

a. cj;promatopm b. Liming

c. Re carbonation

d. Super-chlorination

4. Solid transfer

a. Straining

b. Sedimentation

c. Floatation

d. Futration.

5. Nutrient or molecular transfer

6. Interglacial contact.

7. Miscellaneous operations

a. Disinfection

b. Copper sulphating

c. Fluoridation

d. Thermal desalination.

8. Solid concentration and stabilization

a. Thickening

b. Centrifuging

c. Chemical conditioning

d. Elutriation

e. Biological floatation

f. Vacuum filtration

g. Air drying

h. Heat drying

i. Sludge digestion

j. Incineration

k. Wet combustion

The unit operation approach in water anal waste water treatment has the

following advantages


1. It gives better understanding of the processes

inherent in the treatment and of the capabilities of

there processes in a chaining the objectives.

2. It helps in the development of mathematical and

physical models of treatment mechanisms and the

consequent design of treatment plants.

3. It helps in the coordination of effective treatment

procedure to attain desired plant performance and

efficient quality.

2. What do you understand by physical unit operations? Write a

noteon application of various physical unit operation employed in

waste water treatment.

The physical unit operation in which application of physical forces

predominate consist of the following

1. Screening

2. Communication

3. Flow equalization

4. Mixing

5. Foliation

6. sedimentation

7. Flotation

8. Estuation

9. Vacuum filtration

10.Micro screening

11.Air drying

Because the physical unit operations were deprival originally from

observations of the physical work, they were the first treatment methods to

be used. They form the basis of most process flow sheets in table gives the

applications of physical unit operating on waste water treatment

Operation Application

1. Screening Removal of coarse and settle able solids by surface spraining

2. Communication Girding of coarse solids

3. Flow equalization Equalization of flow and

mass loadings of BOD and

suspended solids.

4. Mixing Mixing of chemicals and

gases with waste water and

maintaining solids in

suspension.


5. Flocculation Promotion of the aggregation of small particle into larges particles

6. Sedimentation Removal of settle able

solids and thickening of

sludge

7. Floatation Removal of finally divided

suspended solids and

particulars with pensioner

close to that of water also

thickness biological sludge.

8. Filtration Removal of fine residual suspended solids removing after biological or chemical treatment

9. Micro screening Same of filtration, also

removal of algal from

stabilization pond efficient.

3. What is meant by chemical unit processes? Enlist the

applicationsof various chemical unit processes employed in waste

water treatment.

Chemical unit processes are those in which removal of continents

are brought about by chemical activity in the field of waste water treatment,

chemical unit operations are usually used in conjunction with physical unit

operation and biological unit processes. The following chemical unit

processes are commonly used for waste water treatment.

1. Chemical precipitation

2. Gas transfer

3. Adsorption

4. Disinfection

5. Combustion

6. Ton exchange

7. Electro dialysis

It should be clearly noted that chemical unit processes are additive

processes whereas physical unit operations and biological unit processes

are subtractive processes. This is an inherent disadvantage of chemical

unit operations because there is usually is net increase in the dissolved

constitution in the waste water because of the chemical unit processes. In

most of the caser, we add something else. Another disadvantage of

chemical unit processes is that they are all intensive in local operating cost.

Table gives the summary of the applications of chemical unit processes in

waste waster treatment.

Process Application


1. Chemical Removal of phosphoresces precipitation

and enhancement of

suspended solids removal in

primary sedimentation.

2. Gas transfer Addition and removal of gases

3. Adsorption Removal of organics

4. Disinfection Disinfection of disease – causing

organism.

5. De chlorination Removal of total combined chlorine

residuals

6. Misillaneous Achievement of specific

objectives in waste water treatment

4. Design a detritus tank for a DWF of 350 ips in a separate sewage

system. Make suitable assumptions wherever required.

Solution:

Let is assume the following

Detention time: 3m; Flow velocity: 0.2 m/s

Maxi flow : 3 times DWF

Hence Q max : 3 x 350 lit/s

Let us provide 3 tanks attached and running parallel to each other. Hence

design discharge for each tank is Q = 350 l/s = 0.35 m3/s

Cross – section area required

Qv 0.350.2 1.75m2

Let as provide a water depth of 1.2m, in the rectangular portion

Width of tank =

Areadepth 1.751.2 1.458m

Provide a width of 1.5m

Also, length of tank = Velocity x detention time

= 0.2 (3 x 60) = 36m

Making a provision of 6m for inlet and outlet arrangement, the total

length of tank = 42m. Thus each unit of the distribute tank will be 1.5m width


and 42m length. Provide of free board of 0.3m. Also, provide a bottom

depth of 1.5m for the accumulation of detritus and this depth be tapered of

an angle of 45 as shown in figure.

5. Design a detritus tank if the dry weather flow of a separate system

of sewage scheme is 130 l/s. Assume

(i) The maximum flow to be three time to average

(ii) The definition period as 45 seconds

(iii) The velocity as 30 cm/s The length of the tank =

45 x 30 = 1350cm

If the depth is 90cm taking actual velocity as 22.5 cm/s

Width of tank for average flow

64cm 65m ray

The detritus tank is there 13.5 m x 0.65 x 0.90 m. At the top, a free

board of 30 cm and at the bottom, 45cm. Storage of grit etc should be

provided. The tank may be 65 cm wide up to the depth of 120 cm and then


the sides will slope down to form an elongated through of 13.5m length and

90 cm width at the bottom with rounded corners. The total depth of the tank

= 90 + 30 + 45 = 165 cm

For maximum flow three tanks of above dimensions will be required

but normally one will be in use

6. 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 110 litday

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 5.5m2

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

Total screenings= 50,000 110 82.5lit

if the velocity in the screen chamber is 45 cm/s C/s

area of screen chamber

4220cm2

Detritus tank:


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

to 50,000 110 44000lit maximum quantity that flows through the

tank = 190 l/s

If the limiting velocity is 30 cm/s

c/s area = 6334cm2

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

42755lit

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 = 85cm, say

Width = 80cm, say

7. Design of circular settling tank unit for a primary treatment of

sewage at 12 million lit per day. Assume suitable values of detention

period (Presuming that trickling filters are to follow the sedimentation

tank) and surface loading.

Solution:


Assuming the normal detention period for such cases as 2 hr and surface loading as 40,000 lit / m2 /day

The quantity of sewage to be treated per 2 hours

12M.lit 224

1M.lit

1000m3

Capacity of tank = 1000 m3

Now, surface loading

Q Q

surface area of tank d2

40,000 12 10 d26 where

d is the dice of the tank

d2 4 12 1040,000 6

d 300 4

19.55m say 19.6m

Now, effective depth of tank

Capacity

Area of X - section

1000

4 19.6 2

3.2m say


Hence, use a settling tank with 19.6 m dia and 3.2m water depth with

free board of 0.3 m extra depth.

8. Design a suitable rectangular sedimentation tank 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 9.6

2M.lit

24

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

Capacity of the tank

length of the tank

22.2m2

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

The width of the tank required


Area of X - section

Depth

7.4m

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 36m x 7.4 m x 3.5 m can be used.

Unit-3

PRIMARY TREATMENT OF SEWAGE

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.

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

1) Acid fermentation

2) Acid repression

3) Alkaline fermentation

2. Define the term ripened sludge?

This digested sludge (geHong from Alkaline fermentation stage)is collected at the bottom of the digestion tank and is also called repented sludge.

3. What are the factors effecting sludge digestion 1)Temp eruterce

Thermopolis

Meropholic

2. Pit value

3. Seeding wotu digested sludge

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

4. What are functions of aeration in ASP?

1) oxygenation of the mixed log wor

2) Flocculation of the colloid in sewage influent

3) Suspension of activated sludge


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

i) Diffused air aeration air aeration ii)

Mechanical aeration iii) Combined diff used air and

Mechanical aeration

6. What are the patterns of mechanical aeration?

i) Haworth paddle or Sheffield aeration system ii) Hartley

paddle or bir Mangham Bio flocure lation system iii) Simplex

aeration system iv) Link belt aeration system

v) Kessner Brush aeration system

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

(1) Conventional process

(2) Tapered aeration process

(3) Step aeration process

(4) Contact slabolisection process

(5) Completely mixed process

(6) Modified aeration

(7) Extended aeration

8. What are the advantage of stabilization ponds or cagoins

(1) Lower initial lost than required for a mechanical plant.

(2) Tower operation costs

(3) Regulation of efficient discharge possible their provoelving control of

pollection during critical times of the year.

9. What are the disadvantage of tagoons?

(1) Requires extensive land area. Hence the method can be used only on

rural area.

(2) If used in urban areas, expansion of town and new developments may

encroach on the lagoon site.

10.What do you understand by facultative ponds?

(1) A facultative panel combine the features of the acrobite and

anacrobic ponds.

(2) Constructed of intermediate depta (1, to 1.5m)

(3) A facilitative bond consists of three

(i) A aerobic Zone -Top

(ii) Facilative zone

(iii) Anacrobic zone --bottom

11.What are remedial measurement for rising bludge problem?

I) Increasing the return sludge age

II) Increasing the speed of the sludge scroper mechanism, where

possible

III) Decreasing the mech cell residence come by increasing the

sludge write rate


12.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 tance aeration

13.What are the advantage of increment 8 and filters?

(i) 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

(ii) The filter work under acrobic conditions, and hence there is no

trouble of odour, files and inserts

(iii) The operation is very simple, requiring no mechanical equipment

except for dosing

14.What are the disadvantages of intermittent sand filters?

i) 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

ii) They requires large area and large quantity of sand due to which their

construction is very lostly.

15.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 toght take of masonry walls and of rectangular shape. The depth of filtering media is kept b/w 1 to 1.8m

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

1. filling

2. Contact

3. Emptying

4. Oxidation

17.What are the advantage of contact of beds?

i) Contact beds can work under small heads. ii) Contact beds can be operated without exposing the sewage efficient to view. iii) There is no nuisance of filter flows iv) The problem of odour is

much less as compared to trill long filters.

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

i) Rate of loading is mech less in comparison to trilling filters. ii) Large areas of land is required for their installation iii) intermittent operation requires continceoces attendance iv) The cost of contact beds is mech more as compared to trick long

filters


19.What do you mean by tracking filters?

Tricking filters, also as percolating filters or sprinkling filters or sprinkling filters are similar to contact beds in construction, but their operation is confinceous 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? The

purpose of under drainage system is two fold

(i) to carry away the liquid efficient and sloughed biological solids.

(ii) To distribute air through the bed

21.What are the merits of conventional trickling filter?

1) 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

2) It has good dependability to produce good efficient under very

widely varying whether and other conditions

3) The working of truckling filter is simple and sheep and does not

require any skilled supervision

22.What are the demerits of conventional trickling filters?

1) The loss of head through the filter system is high their making the

automatic dosing through siphonic doing tanke necessary

2) The cost of construction of the filter is high

3) They require large area in comparison to their biological treatment

processes.

23.What is the necessary of Recirculation in T.F?

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 efficient is recycled through the filter.

PART-B

1. The decoyn flow of sewage is 3.8 mile l p day and the BOD of the raw

sewage is 300 mg/l Design a single stage Bio filter to produce an effluent

having a BOD of 45 mg/l or Less.

Total BOD present in raw sewage per day

= 3.8 X 300 kg = 1/40 kg

Assuming that 35% of this BOD is removed in the primary

sedimentation tank, we have the total daily BOD approed to the filter

= 0.65 X 1140

= 741 kg


Now the total daily BOD present on the effluent (permissible maximum)

= 3.8 x 45 kg

= 171 kg

Total daily BOD to be remove by the filter =

741 – 171 = 570 kg

Efficiency of the filter 100 76.92%

Assuming an organic loading of say 10,000

Kg/ha-m / day [ie between 9,000 to 14,000], we have volume of filter

required Total daily BOD removed

Organic loading

ha m

= 0.057 ha –m

= 570 m3 Now

using equation who have

100

1 0.0044 y

V F.

where y Total daily BOD approved to filter in leg

= 570 kg

V volume of filter = 0.0518 ha – m =

76.92 % (worked out earlier)

76.92 100


76.92 100

Thus gives R

= 1.47 (solving by trial)

I

Hence the recirculation ratio used will be 1.47 say 1.5

2. Explain Ecken folder trickling filter equation. Determine the BOD of

the effluent from a loco rate trickling filter that has a diameter of 35 m

and a depth of 1.5 m, if the hydraulic loading is 1900 m3/day and the

influent BOD5 is 150 mg/l. Assume the rate consistant as 1.89 d-1 and

= 0.67

(Engg services, 1994)

Ecken folder has developed an equation for measuring the performance of twinkling filters, on the basis of rate of waste removal. His final equation which helps to compute the BOD removed by the filter, is given as

yt e KD2

yo QC

Where yo --> BOD5 of the influent in terong the filters in mg/l

ytBOD5 of the efficient getting out of the filter in mg/l

KRate constant /day

D Depth of filter in m

2

2

2

570 0.0044 1

0.057

1 0.44 1

100 0.44 1 1.3

76.92

0.44 0.3

0.44 2.15

0.3

1

1 0.1

1 2.15

1 0.1

F

F

F

F

F

R I F

R I

R I

R I


QL Hydraulic loading rate per unit area of filter in m3/ day – m2

=Q/A

The values given in the question are yt =

BOD5 of the efficient =? yo = BOD5 of the

influences = 150 mg/l

= Depth of filter = 1.5m k =

Rate constant per day

= 1.89 d-1

=0.67

QL = Hydraulic loading rate in m3 /d-m2

1900m d3 /

= Area of filter

19002 1.976m d m3 / . 2

/ 4 (35) Substituting

the above

Yt 1.89 1.5

[0]

150 (1.976)0.67

( )e 1.769

1.7961 e

1

6.027

yt 150

24.89mg l/

6.027

Say 25 mg/L

Hence the BOD5 of the filter efficient = 25mg/L

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 between 0.6

m and 1.5 m water levels. Assuming an evaporation and seepage loss of

2.5 mm of water per day.

(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/day

BOD loading in kg/ha/day

kg d ha/ /

= 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/day

BOD loading on kg/ha/day

=33.3 log/ha/day

(ii) To calculate the number of days of storage between WL 0.6 m and

1.5 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/day

The sewage volume, which percolates and evaporates daily = 2.5 mm depth

2.5 1 mX surface area of tanks

10 100

90,000

= 225 m3

Not effective daily in flow of sewage

= (900 – 225) m3

= 675 m3/day

Winter storage available as days

vol. of storage in m3

= 3

Daily net sewage inflow on m / day


days

= 120 days.

4. Design the dimension of a septic tank for small colony of 150 persons

provided with an assured water supply from the municipal head works

at a rate of 120 lit/p/day assume any data, you may Reed. The quantity of

water supplied

= per capital rate X population =

120 X 150 lit / day = 18,000 l/d.

Assuming the definition time to be 24 hrs, we have the quantity of sewage

produced during the definition period (ie the capacity of the tank)

= 14,400 x 24/24

= 14,400 lit

Now assuming the rate of deposited sludge as 30 lit/capital/year; and

also assuming the period of cleaning as 1 year, we have

The volume of sludge deposited = 30 x 150 x 1

= 4,500 lit

Now assuming the rate of deposited sludge as 30 lit/capita/year;

and also assuming the period of cleaning as 1 year we have The volume

of sludge deposited = 30 x 150 x 1

= 4,500

Total required capacity of the tank

= capacity for sewage + capacity for sludge

= 14,400 + 4,500

= 18,900 lit = 18.9 m3

Assuming 1.5 m as the depth of the tank, we have the surface area of the tank

= m2 = 12.6 m2

If the ratio of the length to width is kept as 3:1, we have 3.B2 = 12.6

B 4.2 = 2.05 m say

= 2.10 m

Provide width = 2.1 m & provide length of the tank = 6 m

12.6

3


Area of C/s provided = 6 X 2%

= 12.6 m2 (same as required)

Thus, the dimensions of the specific tank will be

6m X 2.1 m X (1.5 + 0.3) m over all depth

(0.3 m used as free board)

Hence the use a tank of size 6m X 2.1 m X 1.8 m

5. Design a septic tank for the following data Number of people = 100

Sewage / 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 have

The quantity of sludge deposited = 30 x 100 x 1

= 3,000 lit

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

= 15,000 lit

= 15 m3

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

15 = 10 m2

1.5

Using L : B as 4 : 1 (given)

4 B2 = 10

B 2.5 1.5m

c = 4 X 1.5 = 6 m

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


6. In previous problem, what would be the size of its soak well if the

effluent from the septic tank is to be discharged in it.

Design of soak-well :

The soak-well or seak pit can be designed by assuming the per collation

capacity of the filtering media, say as 1250 lit per m3 per day.

Sewage out flow = 12000 l/d

Per collation rate = 1250 l/m3/d

Volume (of filtering media) required for the

seak – well 12000 l/d3

9.6m3 1250 l/m / d

If the depth of the seak well is taken as say 2 m, then Area of soakwell

required

9.6 4.8m2

2

Dia of soak-well required = 4.8 4

11

= 2.47 m

7. Estimate the size of a septic tank [ C l/w = 2.25) liquid depth 2m with

300 mm free board). Desludging intervals in years, and the total trench

area (m2) of the perlocation field, for a small colony of 300 people.

Assume water supply of 100 lit/cap/d, waste water flow of 80% of water

consumption, sludge production of 0.04 m3/cap/year and the refention

time of 3 days at start up. Deluding is done when the tank is one-third

full of sludge A per collation test indicated an allowable hydraulic

loading of 100 per square metre per day (Gate – 1995)

(i) Size of tank

(ii) Do sludging interval in years

(iii) Total trench area of percolation fields in m2 Given : L/B = 2.25

Dw = 2 m

Free-board = 0.3 m

Population = 300

Water supply = 100 L/c/day

Waste flow = 80% of water supplied

Sludge production = 0.04 m3/e/year


Retention time = 3 days

Using the above data, we have

Water supply to the colony=100L/c/dX300 persons

= 30,000 L/d

Sewage produced in 3 days (ie during retention period)

= 3 X 24,000 L/d

= 72,000 L = 72 m3

De sludging is done when the tank is filled upto 1/3rd of the capacity.

C) Hence, sludge volume collected is c/3

Capacity (c) = Max sewage volume retained

+ Sludge volume retained

C = 72 m3 + C/3

2/3 C = 72 m3

C = 72 X 3/2 = 108 m3

Hence the capacity of the tank = 108 m3

But C = C X BXDw = 2.25 B X B X 2 = 108 m3

2 108

B 24

4.5

B = 4.9 m

C = 2.25 X 4.9 = 11.10 m say

(i) Hence, thus tank size =11.1m X 4.9 m X (2 +0.3m) depth sludge

volume removed in de sludging =c/3=36 m2

m3

Sludge produced per year 0.04 300persons

capital year

= 12 m3/year

36 m3 of sludge will there fore be produced in

= 1/12 X 36 year

= 3 years

(ii) Hence, desludging interval = 3 years

Hydraulic loading of perolation trench

= 100 L/m2/day


out flowing sewage per 3 days = 72 m3 out flowing

sewage per 1 day = 24 m3 = 24,000 L/d (iii) Trench

Area required 24,000 L/d

2 100 L/m / d

= 240 m2

8. Describe the advantage and Disadvantages of septic tank?

Advantages 1. Septic tanks can be easily constructed and do not require any

skilled supervision during construction. More ever, there is no

maintenance problem (except periodical cleaning) as there is no

moving part in it.

2. Their cost is reasonable compared to the advantages and

sanitation they offer on rural or urban areas, where no sewage

system has been load.

3. An excellently functioning septic tank can considerably reduce the

suspended solids and BOD from sewage.

4. The sludge volume to be disposed of is quite less, as compared to

that in a normal Iedomentation tank. The quantity is reduced due

to digestion taking place in the tank itself. The reduction in volume

is about 60% and reduction in weight is about 30%

5. The effluent from the septic tank can be disposed of an land in a

soak-pit or a cost pool, without mach trouble.

6. They are best suited for irolated rural, areas, and for isolated

hospitals, buildings etc. Disadvantage

1. If the tank is not properly functioning, which happens many is

times then the effluents will be very foul, dark and even worser

than the influent.

2. They require too large sizes for serving many people.

3. Leakage of gases from the top cover of septic tank may cause bad

smells and environmental pollution.

4. Periodical cleaning, removal and disposal of sludge remains a

tedious problem.

5. The working of a septic tank is centre dictable and non-uniform.

UNIT – 4

SECONDARY TREATMENT OF SEWAGE

PART –A

1. What is the bawe deference between actovtvated sludge processes

and sludge processes and

srucking folter

Truck long filter Activated sludge

process


The bacterial film

coasting the grains of

the filter medice is

stationary

the bacterial film which

is kept moving is the

constant a gitation

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 idour nuisance iv) Capital cost is

less

3. What are the disadvantages of the activated study I plant?

i) High cost of operation, tooth 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 ertro phication?

The excess growth of algue and other aquatic plants in a river stream

is called ertroplication

5. What do you mean by secondary treatment?

The efficient from the primary sedimentation take contains about 60 to 80% of the unstable organic matter originally present in sewage. Thus colloidal organic matter, which passer the primary clarifies, without setting there, has to be removed by further treatment. This is called secondary or biological treatment

6. What are the filters used in sewage treatment?

1) contact beds (very small plant)

2) Intermittent fillers (small plant)

3) Trickling filters (commonly used in modern days)

4) raiscellaneous type filters (under special

circumference)

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

D10 (effective size) 0.2+0 0.5 mm

D 60 (uniformly coefficient ) --> 2 to 5 D10

8. What are the types of track long filters?

1) Conventional track long filter or ordinary or standard rate or low rate

trick long filter

2) High rate filters or high rate trick long filter

9. What are the advantage of track long filters?

i) Rate of filter loading is high as such requiring lesser land areas and smaller quantities of filter media for their installations. ii) They are self- clearing iii) Mechanical wear and tear is small as they contain less – mechanical equipment.


iv) Moisture content of sludge obtained from driclling filters is high as 99% or

80.

10.What are the disadvantages of trick long filters?

i) the head loss through these filters is high, making automatic during of

the filters necessary

ii) cost of construction is high iii) There filters cannot treat ratio sewage and primary sedimentation is a must.

11.Differentiate between low rate & high rate

Low rate

IF

High Rate

T.F

1. Hydraulic loading varies between 20 to 44 ML/ hec/ day

2. Dept of filter media rateje b/w1.6 to 2.4m

Varies form 110 to 330

M.C/hec/day

Varies b/w 1.2 b/w 1.2

to

1.8m

12.Draw a layout of single stage of recirculation processes.

13.Define the term recirculation ratio?

The ratio R

of the volume of sewage recirculated (R) to

I

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

14.Write the formula for recirculation factor?

1 R

F I

1 0.1RI 2

Where

F Recirculation factor

R Volume of sewage recirculated

I Volume of raw sewage

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

long filter?

(%) 100

1 0.0044 Y

V F.

Where

Y 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

V F.

uUnit organic loading on filter all symbols are already given

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

T.F?

final efficiency = '= 100

0.0044 Y '

1+

1- V F' '

Where

Y’ total BOD in efficient from first stage in kg /day

V’ Volume of second stage filter in ha-m

F’ Recirculation factor for the second stage filter ’ Final

efficiency

18.What are the types of high late Filters?

1) Bio filters

2) A ccelo filters

3) Aero filters

19.Draw a neat stitch of single complete treatment

(bio filters)?

20.What are the special types of filters?

1) Durban filter

2) Magnetic filters

3) Rapid sand filters

21.What do you mean by magnetic filters?

In this type of filter, a layer of crashed magnetic ore of Iron is provided in about 80mm, thickness, and is supported on a non-magnetic metal wire screen sewage is filtered through the magnetic layer which removes the impurities purely by mechanical starching action.

PART –B

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

clarofver to a standard rate trick long 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/dag.


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 4.5 10 6

3 Im/day

10

7,20,000 Im /day

Volume of the filter media required

Total B.O.B

organic loading

7,20,000 Im/day3

160 Im/ m /day

m3

=4,500 m3

Surface Area required for the filter

Total flow

Hydralic loading

4.5 10 6 c/d

2

2000 l/m d 4.5 10 6

2000 = 2.25X 103 m2 = 2250m2

depth of the bed required =

2m

Efficiency of the filter is given by

100

1 0.0044 u

u organic loading in kg / ha-m/day

= 100

1+0.0044 1600

100 100 85.03%

1 0.1761.176


2. Distinguish between standard rate filters and high rate filters.

S.No Characteristics Conventional or High rate filers

standard rate

filters 1. Dept of filter Veries between Varies between media 1.6 to

2.4m 1.2 to 1.8m

2. Size of filter 25 to 75mm 25 to 60mm media

3. Land required More land area Less land area is reqd as the is regd as the filter loading is filter loading is

less more

4. Cost of It is more for It is less of operation treating

equal treating equal quantity of quantity of

sewage. sewage

5. Method of Continuous Continous

operation application less application, flexible requiring more flexible less skilled and more skill supervision full operation is required

6. Type of The efficient is The efficient is efficient highly

nitrified nitrified cepto produced and

stabilized nitrite stage only with BOD in and is

thus less efficient 20 stable and ppm or so. hence it is of

slightly inferior quality B.O.D in efficient 30

ppm or so.

7. During interval It

generally

varies between

3 to 10 minutes

the

sewage is

generally

not approved

continuously

but is applied

at intervals.

It is not

more then

15

seconds

and the

sewage is

thus

applied

continually

8. Filter

loading

values

1.

Hydraulic

loading

Varies

between 20 to

44 M.L per

hectare per

day.

Various

between

6000 to

18,000 kg

of BOd5

per hect

metre of

filter

media per

day.

9. Recirculation

system

Not provided

generally

Always

provided

for

increasing


hydraulic

loading

10. Quality of

secondary

sludge

produced

Black highly

oxidized

with slight

fine

particles

Bown, not

fully

oxidized

with

fine

particular

3. A town having a population of 30,000 persons is produced the

following sewages:

(i) Domestic sewage @ 120 L.P.C.d having 200 mg/l of BOD

(ii) Industrial sewage @ 3,00,000 L.p.c.d having 800 mg/l of 30D

(iii) Design high rate single stage trick long filters for treating the

above sewage. Assuming that the primary sedimentation

removes 35% of BOD. Allow all organic loading of 10,000

kg/ha.m/day (excluding recirculated sewage). The recirculation

ratio is 1.0; and the surface loading should not exceed 170

mc/ha/day (including recirculated sewage). Also determine the

efficiency of the filter and the BOD of the efficient

Solution:

Quantity of domestic sewage produced perday

= 12X30,000 lit/day

= 3.6 M.L / day

BOD for domestic sewage = 200 mg/l Total BOD of

domestic sewage produced perday

= 720 kg/ day

Quantity of industrial sewage produced peer day

=3,00,000 lit

BOD of Industrial sewage = 800 mg/l Total

BOD of industrial sewage 3,00,000 8006

10 240kg

Total BOD of domestic as well as Industrial sewage Perday = 720+240

= 960 kg/day

Out of this BOD, 35% is already removed in primary

classifier

BOD to be removed by filter unit

=960X (0.65)

=624 Leg\day Volume of filter media

required

Total BOD removed

Organic loading


ha m

624m3

Now the total volume of sewage flowing

=3.6 X106 + 3,00,000 lit/day =3.9X106

lit/day =3.9 MC/day.

A recirculation ratio of /means that the volume of recirculated sewage ®

= Original volume

= 3.9 M.L day

Total volume is , original recirculated

= 2X3.9 M.L / day

= 7.8 M.L /day Filter area

required

Total flow volume

Surface loading

7.8 ml/d

170 ml/ha.d

hect

104m2

= 458.8 m2

Dia of the filter tank required

458.8 4 24.17m//

Hence, use, say, 24m dia tank with area as = 452.16m2

Depth of filter media required

Volume of filter media

Surface area

m3/ m2 1.38m//

Efficiency of this filter is given

= 100

1+0.0044 y/V.F

where y--> total organic lading ie total BOd applied to filter in kg/day

Vvolume of filter in ha – no

0.624ha m


fRecirculation factor as given by

1 R I/ Here R =I

1 0.1R I/

= 1+1 2 2/( 1)2 2 1.65

(1+0.1) 1.21

100 100 100 74.6%

1 0.004/1.28 1 0.34 1.34

BOD of the efficient left

100 74.6 624kg day/

100

624

158.49

Total volume of efficient = 3.9 M/day

BOD concentration in the efficient

158.49 10 6

6 mg l/

3.9 10

40.6mg l/

4. Determine the size of a high rate trick long filters for the following data.

i) Sewage flow = 4.5 mld ii) Recirculation ratio = 1.5 iii) BOD of row sewage =250 mg/l iv) BOD removal in primary tank = 30% v) Final efficient BOD desired = 30 mg/l Solution

Quantity of savage 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 mlX 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 kg

BOD concentration desired in final efficient = 30 mg/l

Total BOD left in the efficient perday

= 4.5 X 30kg = 135kg

BOD removed by the filter =787.5 -135


=652.5 kg BOD removed

100 Total BOD

Efficiency of the filter = 100

82.85%

Now using equation

100

1 0.0044 y V F/ . =82.85%

y= total BOD in kg = 787.15kg

1 R I/

(1 0.1R I/ )2

Here R/I = 1.5(given)

F = 1+1.5

F= [1+0.1 1.5] 2

2.5 2 2.5 1.89//

(1.15) 1.322

82.85 1000

1 0.0044 787.5

V 1.89

1 0.0044 416.6 0.2 V 0.0044

45.45 416.6

2066.10 V

V 0.2 hect m

= 2000 m3

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

m2

1333.3m2

surface area required =Dia of the circular filter required

1333.3 4/

=

=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.


5. Determine the size if a high rate trickling filter for the following data.

Flow = 4.5 m/d

Recirculation ratio = 1.4

BOD of raw sewage = 250 mg/l

BOD removed in primary clarifier = 251

Final efficient BOD derived = 50mg/l Calculate also the size of the

standard rate tricklong 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 efficient = 50m/l BOD allowed to

go into the efficient

= 50 mg/l X 4.5 nql

= 225 kg

BOD removed by the filter perday

= 843.75 – 225 =618.75kg BOD removed 100

Efficiency of the filter = Total BOD entering

100

Now efficiency of the filter is given by

100

1 0.0044 y V F/ .

yTotal BOD applied to the filter per day in kg

= 843.75 kg

FRecirculation factor

1 R I/

(1 0.1R I/ )2 R 1.4


1 1.4 2 2.4 2 1.85//

V

V 104m3 665.4m3

using 1.5m depth of the filter, we have

Area required = 413.6m2

Dia of the filter tanle required

413.6 4 23.8m

for an equivalent standard rate filter F =1

V

ha m104mm2 V= ha m.

104m3 1231m3

using depth of filter as 1.5m we have

surface area required = 1231 820.8m3 //

1.5

Dia of the filter tank required

820.8 4 32.3m

1

07

3 84

3.0.0

04

84

3.

1

00.0

04

1

.7

384

3.8

284

3.6

8

0.11.

4)

(

11

07

3 84

3.0.0

04 1

.4

5

1

00.0

04

1

.7

34

5

0

.8

20.0

044

56

8


6. 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

= 3.79 MC X 240 mg/l

= 909.6 kg

Now, filter volume required

Total BOD in raw sequence inlog

Given BOD loading rate of 11,086 kg/ha-m

ha m = 0.082 ha-m

Now assuming that 35% of BOD is removed in primary clarified we have

The amount of BOD approved to the filter

= 0.65X 909.6 kg

= 591.24 kg

Now using equation, we have

100

1 0.0044 Y

V F.

Where ytotal BOD applied to the filter on kg

= 591.24 kg

V Vol of the filter in ha-m

= 0.082 ha-m

F 1 R I/

(1 0.1 /1)R 2 HereR I/ 1

F 1 1

2 2/1.21 1.65

(1 0.1)

100

1 0.0044 591.24

0.082 1.65

100 100 77.45%

1 0.2911.291

The amount of BOD left in the efficient

= 591.24 ( 1-0.7745) kg


= 133.32 kg

BOD concentration in the efficient

Total BOD

sewage volume

133.32 10 6

6 mg L/

3.79 10

35.18mg L/

7. It is proposed to use a two stage plant instated of the single stage

plant in previous problem (6). The total volume of filter medium

remains the sama as war in one filter is 0.082 ha-m and each filter

is to contain half of thus material, and each filter is to contain

half of this material and the recirculation ratio is to be for each

filter Determined the BOD of

the plant of fluent Solution:

For each filter F = 1.65

For the first stage filter, the efficiency is given by

100

1 0.0044 y V F/ .

Y Total BOD applied to filter

= 591.24 kg (from previous problem)

V = Volume of filter = 0.082/2

=0.041 ha-m

100

1 0.0044 591.24

0.041 1.65

100

= 70.92%

1.41

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

100


1 0.0044 y

1 vF '

y’ = 171.9 kg V’ = 0.041 ha.m

F’ = 1.65

= 0.7092

' 100

1 0.0044171.9

1 0.70920.041 1.65

100

=

1.762

= 56.75%

The amount of BOD left in the effluent from the plant

171.9 100 56.75

100

= 74.35 kg

BOD concentration in the effluent

Total BOD

sewage volume 74.35 10 6

3.79 10 6

19.61mg l/

UNIT – V

DISPOSAL OF SEWAGE AND

SLUDGE

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 sewagedisposal?


1. Dilution is disposal in water.

2. Effluent Irrigation or Broad Irrigation or Sewage forming is disposal on

land.

3. What are the conditions adopted for disposal bydilution?

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

and settlable solids.

2. When the dilution water has a high dissolved oxygen (D.O.) content.

3. 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?

1. Dilution and dispersion.

2. Sedimentation

3. Oxidation – reduction in sun-light.

4. Oxidation

5. Reduction

5. What are the factors affecting self purification ofpolluted 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. What is meant by photo synthesis?


The sun light has a bleaching and stabilizing effect of bacteria. It also helps

certain micro organisms to derive energy from it and convert themselves into

food for other forms of life, thus absorbing Co2 and releasing O2 by a process

known as Photo synthesis. 9. What do you mean by Oxidation?

The oxidation of the organic matter prevent in sewage effluents, will start

as soon as the sewage out falls into the river water containing dissolved

oxygen. The deficiency of oxygen so created will be filled up by the

atmospheric oxygen. The process of oxidation will continue till the organic

matter has been completely oxidized. This is the most important action

responsible for effecting self purification of rivers.

10. What do you understand by Reduction?

Reduction occurs due to hydrolysis of organic matter settled at the bottom

either chemically or biologically. An aerobic bacteria will help in splitting the

complex organic constituents of sewage into liquids and gases and thus

paving the way for their ultimate stabilization by oxidation.

11. 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.

12. 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.

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

B.O.D. of the diluted mixture

C = C Q C Qs. s R. R f

Q Qs R

Where

Cs B.O.D. of sewage

CR B.O.D. of river

Qs Sewage discharge

QR Discharge of the river

14. 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.

15. 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.

16. 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.

17. What do you understand by monocline? Giveexample.

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.

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

The most important biological zones are

(i) euphotic zone

(ii) Litt oral zone

(iii) benthic zone

19. What do you understand by “Euphotic Zone”?

The upper layer of lake water through which sunlight can penetrate is called

the euphotic 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.

20. Define the term secchi dosk? Draw a neat sketch

The depth of the exphotic zone can be approximated and measured by a

sample device called the secchi disk as shown in figure.

21. 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.

22. What is meant by Behthic zone? Give example

The 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.


23. Define the term “productivity of a lake”?

The productivity of a lake is defined as a measure of its ability to support a

food chain. Since the algae forms the base of this food chain, which is required

by the other forms of loving organisms to thrive. Its presence measures the

lake productivity. 24. What are the types of lakes?

Depending upon the increasing level of its productivity the lakes may be

classified as

1. Oligotrophic lakes

2. Mesotrophic lakes

3. Ecetrophic lakes

4. Senescent lakes

25. What are the requirements for disposal of Noglit soil?

1. It should be located away from the building, on the leeward side.

2. Its floor should be atleast 1.2 m above the general level.

3. Its floor should be of impervous material.

4. It should be well ventilated.

26. Give examples for soil waste?

1. garbage

2. ashes

3. rubbish

4. dust

27. What are the methods adopted for disposal of refuge?

Refuse or solid waste can be finally disposed of the following methods.

1. Controlled tipping

2. Filling of low lying areas. (Land filling)

3. Dumping into sea

4. Pulverization

5. Incineration

6. composting

28. What are the advantage of land filling methods ofdeposal?

1. It is simple and economical

2. No plant / equipment is required

3. There are no by products and hence there is no problem of the

disposal of the by-products.

4. Separation of varies materials of the refuge is not required.

29. What are the disadvantages of land fillingmethods of disposal?


1. Proper site may not be available near by

2. Wind direction map not be favourable.

3. Large band areas are required.

4. It may be difficult to get large quantities of covering material.

30. What do you understand by pulverization?

In this method, the dry refuse is pulverized into power form, without

changing its chemical form. The power can either be used as a poor quality

manure, or else be disposed of by land filling.

31. What are the factors considered during incineration?

1. The refuge charging should be carefully observed during incineration.

2. Each batch of refuge entering furnace should be well mixed.

3. Auxiliary burners are usually installed above the refuge to ignite it and

to establish the draft at the beginning of the cycle. This is all the more

necessary when the moisture content of air is high.

32. What are the advantages of incineration method of disposal?

1. This is most hygienic method, since it ensures complete destruction of

pathogens.

2. There is no odour trouble or dust nuisance.

3. The heat generated can be used for saving steam power.

4. Clinker produced can be used for road purposes.

33. What are the disadvantages of incineration ofmethod of disposal?

1. Large initial expenditure.

2. Improper operation results in air pollution problems and incomplete

reduction of the waste materials.

3. Disposal of the remaining residue is required.

4. High slacks needed for natural draft chimneys present safety problems.

34. 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.

35. What are the methods adopted for composting?

1. Composting by trenching.

2. Open window composting.


3. Mechanical composting.

36. What is meant by “humus”?

The refuge gets stabilized in about 4.5 months period, and gets changed

into a brown coloured odourless innocuous powdery form known as humus,

which has high manure value became of its nitrogen content. 37. Describe

the term open window composting.

In this method, a large proportion of mineral matter like dust, stone, broken

glass pieces etc. are first removed from the refuge. The refuge is then dumped

on the ground in the form of 0.6 to 1 m high, 6 m long and 1 to 2m wide piles

at about 60% moisture content. The pile is then covered with high Bool, cow

dung, cattle urine etc. through which the organisms or germs that are

necessary for fermentation are added after which compost is ready for use as

manure when an temperature falls considerably.

38. What do you understand by mechanicalcomposting?

The open window method of composting is very laborious and time

consuming process. Also it requires large area of land which may not be

available in big cities these difficulties are overcome by adopting mechanical

composting in which the process of stabilization is expedited by mechanical

devices of turning the compost.

39. What are the operations involved in themechanical composting?

The operations involved in a large scale composting plant as follows:-

1. Reception of refuse

2. Segregation

3. Shredding or pulverizing

4. Stabilization

5. Marketing the humus

40. Give three important methods of disposal ofsludge.

1. Sludge disposal into water.

2. Sludge disposal by application on land.

3. Sludge disposal by clogging.

4. Sludge disposal by composting.

41. What are methods adopted for sludge drying?

1. Drying the sludge on prepared sand beds.

2. Drying the sludge on centrifuges.

3. Drying the sludge by heat dryers

42. What is meant by house refuse?


This consists of vegetable and animal waste matters, ashes, cinders,

rubbish, debries from cleaning and demolition of structures.

43. What is meant by organic waste?

It includes dry animal and vegetable refuse, cow dung, excreta of birds,

tree leaves, sticks, plastic bottles, paper waste, rags. This waste is subject to

decay with time and evolve highly offensive odour and gases which are highly

detrimental to health.

44. What do you understand by inorganic waste?

This consists of non-combustible materials such as grit, dust, mud, metal

pieces, metal containers, broken glass and crockery, tiles waste building

material. It is not subjected to decay and is therefore not harmful to public

health.

45. 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 H2S, Co2, CH4. This phenomenon of soil getting

clogged is known as sewage sickness.

46. What are the types of preventive measure in adopted for sewage

sickness?

1. Primary treatment of sewage

2. Choice of land 3. Under-drainage of sool.

4. Giving rest to the land.

5. Rotation of crops

6. Applying shallow depths.

47. Define the term “Raw sludge”?

The sludge, whock is deposited in a primary sedimentation talk is called

Raw sludge. Raw sludge is colourous, contains highly puterscible organic

matter, and is thus, very very objectionable.

48. Define the term “secondary sludge”?

The sludge, whock is deposited in a secondary clarifier is called secondary

sludge. It is also put rescible, through a little less objectionable.


49. What are the unit operation / process on the sludge treatment?

Sludge treatment may include all or a combination of the following unit

operations and processes

1. Thickening or concentration

2. Digestion

3. Conditioning

4. Dewatering

5. Drying

6. inconconeration

50. What is the purpose of thickening?

The purpose of thickening is to reduce moisture content of the sludge, and

consequently to increase the solids concentration. 51. What are the types of

thickening?

1. Gravity thickening

2. Air Floatation.

3. Centro fugation.

52. What do you mean by “Digestion”?

The principle objectives of sludge digestion is to subject the organic matter

present in the settled sludge to anaerobic or aerobic decomposition so as to

make it innocuous and amenable. 53. What is meant by “conditioning”?

Conditioning improves the drainability of digested sludge. Prior conditioning

of sludge before application of dervatering methods renders it more amenable

to dervatering.

54. What are the purpose of derveefering?

The purpose of derveefering is to further reduce the volume of sludge and

thereby increase the solids concentration.

55. What are the characteristics of sources ofsludge?

1. Sludge from primary settling tanks 2. Chemical precipitation

3. Trickling filter huming.

4. Activated sludge

56. Write the formula for determing the volume of sludge?

Vsl Ws

PS Pw s1 s

Where


Vsl volume of sludge

Ws wt of dry solids

Ss1 sp. Gravity of sludge Ps percent solids expressed as a

decimal

Pw density of water

(103 kg/m3 at 5o C)

57. Define the term dissolved air floatation?

Air floatation units employ floatation of sludge by air under pressure or

vacuum the former process more commonly used is known as dissolved air

floatation or pressure type floatation.

58. What is meant by centrifugal thicknening?

Centrifuges are used both to thicken and to dewater sludge. Their

application in thicknening is normally limited to waste activated sludge.

Thickening by centrifugation involves the setting of sludge particles under

the influence of centrifugal forces.

59. What are the basic types of centrifugalthickening?

The three basic types of centrifuges currently available for sludge

thickening

1. Nozzle disc

2. Solids bowl

3. Basket centrifuges.

60. What are the types of digertors?Sludge dogestors can be of two types

1. Conventional or low rate digestor

2. High rate digester

61. What are the elements involved in the design ofdigester?

1. Number of unit

2. Tank shape and size

3. Water depth and F

4. Roofing

5. Mixing of digester contents

62. What do you mean by Aerobic digestion?


The function of aerobic digestion is to stabilize waste sludge by long term

aeration, thereby reducing the BOD and destroying volatile solids. Aerobic

digestion is generally defined as a process in which micro-organisms obtained

energy by endogenous or auto-oxidation of their cellular protoplasm.

63. What are the factors considered for design ofaerobic digestion?

The factors that should be considered on designing an aerobic digester

include

1. Defention time

2. loading criteria

3. oxygen requirement

64. What are the advantage of aerobic digestion?

1. Lower BOD concentration in digester supernatant

2. Production of odourless and easily dewaterable biologically stable

digested sludge.

3. Lower capital lost

65. What are the disadvantage of aerobic digestion?

1. Higher power costs generate higher operating costs comparable

with anaerobic digestion.

2. No methane gas is produced for recovery as a by-product.

66. What is meant by “Elutriation”?

Elutriation is literally a “washing” of the sludge. It is a unit operation in which

a solid or a solid-liquid mixture is intimately mixed with a liquid for the purpose

of transferring certain components to the liquid.

67. What are the purpose of dewatering?

1. Cost of trucking sludge to ultimate disposal site is reduced, because

of reduce sludge volume consequent to dewatering.

2. Eare in handling dewatering sludge.

68. What are the methods of “elutrication”?

1. Single stage electrication

2. Multi stage electrication

3. counter current washing

69. What are the advantages of “tow stagedigestion”?


1. Two stage digestion is an effective method of preventing any

tendency for the sludge to short-circuit.

2. Two stage digestion offers the freedom from large slum formations

in any other digestion tanks.

70. What is meant by lagoon?

The raw sludge is kept at rest in a large shallow open pond, called a lagoon.

71. Define what is meant by sludge concentratorunit?

Since the sludge obtained in a sludge digestion plant contains too much of

moisture 98% to 99%) and is therefore very bulky, may sometimes be reduced

in its moisture content by first sending is to a sludge thickener unit also called

a sludge concentrator unit. 72. Define the term “high rate digestion”?

The process of sludge digestion using a sludge thickener before the

digestion tank, helps in reducing the capacity of the digestion tank which

further reduce their capacity and the rate of digestion is also made high. Such

a digestion which is used in modern large sized plants is called high rate

digestion.

73.Define the term “Sludge bulking”?

The settled sludge may contain more moisture and thus resulting in the

swelling of the sludge volume. This phenomenon is known as sludge bulking.

74. What are the factors assist on the development ofsludge bulking?

1) Presence of harmful industrial waster waters, especially those containing high carbohydrate content antiseptic or other such properties.

2) Accumulation of sludge at the bottom of the aerofuon

tanks.’

75. What are the disadvantages of sludge bulkong?

1. When sludge bulkong occurs, the sludge does not readily settle

down, and is, remains in suspension in secondary clarifier is even

seen in the efficient of the secondary clarifier.

2. When sludge bulkong occurs, naturally large volume of sludge will

have to be handled.

76. What are the remedial measure to adopted for controlling the

bulkong of sludge?

1. by heavily chlorinating the sewage

2. by increased aeration


3. by reducing the suspended solids in the sewage

77. Define the term MLSS?

The total microbial mass in the aeration system (M) is computed by

multiplying the ever. Concentration of solids in the mixed liquor of the aeration

tank called Mixed Liquor suspended solids (MLSS)

78. Explain the following term

1. SRT

2. MLSS

3. MCRT

4. F/M

1. SRT Solids Retention Time

2. MLSS Mixed liquor suspended solids

3. MCR Mean cell Residence time

4. F/M Food / Micro organisms

79. Define the term “sludge age”?

It is defined as the ratio between mars of suspended solids (MLSS) in the

system (M) to Mass of solids leaving the system / day. It is denoted as QC

Mass of suspended solidds (MLSS)

QC in the system(M)

Mass of solids leaving un system / day

80. What are the methods adopted for disposed ofwet digested sludge?

1. Disposed by Dumping into the sea

2. Disposed by Burial in the trenched 3. Incineration.

81. What do you understand by ‘dumping’?

Dumping in an abandoned mine clearly can be resorted to only for sludges

and solids that have been stabilized so that no decomposition or nuisance

conditions will result. Thus method can be safely adopted for digested sludge,

clean frit and incinerator residue. 82. What do you understand by “Vacuum

filtration”?

Vacuum titration is the most common mechanical method of dewatering. It

is used to dewater raw or digested sludges preparatory to heat treatment by

vacuum filtration because the warse solids are rendered fine during digestion.

83. What do you mean by “sludge drying beds?


This method of dewatering and drying the sludge is specially suitable for

those locations where temperature are higher, similar to the one prevailing in

our country. The method consists of applying the sludge on specially prepared

open beds of card. 84. Give Brief notes on sludge Lagooning ?

A Lagoon is a shallow earth basin into which untreated or digested sludge

is deposited. Untreated – sludge lagoons stabilize the organic solids by an

aerobic and aerobic decomposition, which may give rise to objectionable

odours. Hence the lagoons should be located away from the town.

85. What do you understand by “chemical conditioning”?

Chemical conditioning is the process of adding certain chemicals to enable

coalescence of sludge particle facilitating easy extraction of moisture.

Exp: Alum, ferric, chloride, lime.

86. Write the expression for determining the capacityof digester in

parabolic shape?

V Vf 32 Vf Vd T1 V Td 2

Where

v volume of digester vf volume of fresh sludge added per day vd volume of digested sludge withdrawn per day

T1 digestion time in days

T2 mooning storage in days

87. Write the expression for determining the capacity of digester in

linear?

V Vf 2Vd T1 V Td 2

88. What do you understand y temp of digestion?

Digestion of sludge in temperature dependent. Here, rate of digestion

increases with increase in temp upto a temp of 40 C the digestion is brought

about by a particular type of organisms.

While a totally different type of organisms establish in the digester at temp

higher than 45 C. The former range is called the mesospheric range and the

latter is known as Thermopolis range.

89. What are the factors depends upon the capacityof digester?


The capacity of digester, depends on

(i) Daily volume and moisture content of input sludge and

digested sludge

(ii) Temperature of digestion

(iii) Desired degree of destruction of volatile solids

(iv) Storage capacity of for digested sludge.

90. What do you understand by Daily volume andmoisture content of

sludge?

The volume of daily sludge verves depending upon degree of removal of

suspended solids in primary and final setting links, moisture content and sp.

Graving of sludge.

91. What are the purpose of sludge concentration?

1. To permit increased loadings to sludge digesters

2. To increase feed solids concentration of vacuum filters

3. To economize on transport costs as in ocean barging incase of raw

sludge.

UNIT – B

1. 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 town is to be discharged into a river stream. The

quantity of sewage produced per day is 8 million litres, and its BOD is

250 mg/l. If the discharge in the river is 200 l/s and if its BOD is 6 mg/l.

find out the B.O.D. of the diluted water.

Solution:-

Sewage discharge = Qs

8 10 6 l/s

24 60 60

92.59 l/s

Discharge of the river = QR


= 200 l/s

B.O.D. of sewage = Cs = 250 mg/l.

B.O.D. of river = CR = 6 mg/l.


= C = C Q C Qs. s R. R Q Qs R

250 92.59 6 200

92.59 200

83.21 mg/l

3. In the above problem, what should be the river discharge, if it is

desired to reduce the B.O.D. of diluted water to 20 mg/l.

Solution:-

Here C = 20 mg/l


C = CQ C Qs s 12 12 Q Qs R

20 250 92.59 6 QR

92.59 QR

20 (92.59 QR ) 250 92.59 6QR

20 92.59 20QR 250 92.59 6QR

QR = 1521 l/s

4. A city discharges 1500 lit per second of sewage into a stream whose

minimum rate of flow is 6000 lit/sec. The temperature of sewage as well

as water is 20 C. The 5 day B.O.D. at 20 C for sewage is 200 mg/l and

that of river water is 1 mg/l. The D.O. content of sewage is zero, and that

of the stream is 90% of the saturation D.O. if the minimum D.O. to be

maintained in the stream is 4.5 mg/l, find out the degree of sewage

treatment, required. Assume the de-

oxygenation coefficient as 0.3

Solution:-

From the table given at the end of the book, the value of saturation D.O. at

20 C is found out as 9.17 mg/l

D.O. Content of the stream

= 90% of the saturation D.O


9.17

8.25mg

l

D.O of mix of the start point is at t = 0

8.25 6000 0 1500

6.6mg

l

DO = initial DO deficit

= [saturation D.O at mix temp – D.O. of mix]

= 9.17 – 6.6 = 2.57 mg/l

(Assuming instances mixing]

Minimum DO to be maintained in the stream

= 4.5 mg/l Maximum permissible saturation deficit (i.e critical DO deficit)

DC = 0.17 – 4.5

=4.67 mg/l

Now, using equation the first stage BOD of mixture of sewage and stream (L)

is given by

D fL f 1 f 1 f 1 DLo c

Do 2.57mg

l

DL 4.67mg

l


f kR 0.3 3

kD 0.1

We get

4.67 3L 3 1 3 1 3 1 2.57L

L 2 3 1 5.14L

14.01

Solving by hit & travel we get the value L = 21.1 mg/l

New Using

Yt L 1 10 KD.t

Maximum permissible 5 day BOD of the mix (at 20 C)

Y5 21.1 1 10 0.1 5

Where K at20 D C 0.1

14.43mg

l

Now, using equation

C C QS S C GR R QS GR

C stands for concentrations of BOD

14.43=CS 1500 1 6000

1500 6000

Where Cs will represent the permissible BOD5 (at 20 C of course of the

discharged waste water)

Solving, we get

Cs = 68.16 mg/l

Degree of treatment regd (%)

Original BOD of sewage-Permissible BOD Original BOD


65.9%

5. A city discharges 100 cumecs of sewage into a river, which is fully

saturated with oxygen and flowing at the rate of 1500 cumecs during it’s

lean days with a velocity of 0.1 m/s. The 5 – days BOD of sewage at the

given temp is 280 mg/l. Find when and where the critical D.O deficit will

occur in the down stream portion of the river and what its amount is,

assume coefficient of purification of the stream (f) as

4.0, and coefficient of de – oxygenation (KD) as 0.1 Solution:

The initial D.O of river

= saturation D.O at the given temp

= 9.2 mg/l (say)

D.O of mix at t = 0 is at start

9.2 1500 0 100

assuming that D.O of sewage is nil

8.62mg

l

Initial D.O deficit of the stream

=D.O = 9.2 – 8.62 = 0.58 mg/l

Also, 5 – day BOD of the mixture of sewage and stream is given by

C C QSS Q CR R

QS QR

280 100 1500 0

17.5mg

l

5 day BOD of mix at the given temp

Y5 = 17.5 m/l

The ultimate BOD of the mix is L

0.68417.5 25.58mgl


Now using

DC.fL f 1 f 1 f 1 DoL

25.58 3 4 1 3 0.5825.58

Dc 4

DC 4.12mg

l

Now, from equation tc KD 1f 1 log10 f 1 f

1 DOL

tc 0.1 4 1 1 log10 4 1 3 0.5825.58

1 0.571 1.905days

0.3

Now, distance = Velocity of river x Travel lime

= 0.1 m/s x (1.905 x 24x 60x 60 sec)

= 16, 460 m

= 16.46kN

Hence the most critical deficit will occur after 1.905 days and at pt 16.46

km down stream of the pt of sewage disposal.

6. A Town with a population of 30,000 has to design a sewage treatment

plant to handle industrial as well as almost waste waters of the town. A

sanitary survey revealed the following:

Dairy waster of 3 million lit/day with BOD of BOD mg/l and sugar mill

waste of 2.4 million lit/d with BOD of 1500 mg/l are produced. In addition

domestic sewage is produced of the rate of 240 lit/ca/d. The per capita

BOD of domestic sewage being 72 gm/d. An overall expansion factor of


10 percent to be produced. The sewage effluents are to be discharged

to a river stream with a minimum dry weather flow of 4500 lot / sec and

a saturation dissolved oxygen content of 9 mg/s. It is necessary to

maintain a dissolved oxygen content of 4 mg/l in the stream. Determine

the degree of treatment regd to be given to the sewage. Assume suitable

values of coefficient of de-oxygenation and re – oxygenation.

Solution:

Per Capital BOD of the domestic sewage

= 72 gm/day

= 72 x 1000 mg/d

The per capita sewage produced

= 240 lit/d

BOD per lit of the domestic sewage

72 1000240 mgl

300mg/l

Amount of domestic waste water produced per day

= 30,000 x 240 lit

= 7.2 million lit

Net BOD of all waste waters (i.e domestic _+ industrial)

7.2 300 3 1100 2.4 1500

719mg

l

Total waste water discharge

Vol of waste waters entering /day No of sec in 1 day

3ML 2.4ML 7.2ML

1 24 60 60sec

l/s

145.8 l/s


Total waste water discharge with 10% expansion factor

= 1.1 x 145.8 l/s = 160 l/s

Initial DO of saturated stream water

= 9 mg/s (i.e saturation D.O as given)

DO of mixture of t = 0 is at start pt

D.O of river QR DO of sewage Qs QR QS

(Assuming that the bio of water wafers on N.o1)

=8.69

initial b.o defliot =Do=9-8-69 (assuming instaneous having =0.31mg/l.

Also, critical D.o deflect is allowable max D.o deflect =Dc = 9-4-0=5mg/l Now,

using equ.

L f1 f 1 f1 DoL

Dcf

where Dc5mg/l

Do 0.31mg/l

KD=0.1KR=0.3 f=3

(assumed values of mix temp)

t 2 3 1 2x0.31L

5x3

solving by not and trial l=26.65mg/l

max, permissible 5 day B.O.D of max at max

temp=y5=L[1-(1-0)-0.1x5] ten at max temp is assumed =0.1.

=0.684L

=0.684x25.65 =17.54mg/l using equ

C C Qs s C QR R QS QR


17.54 C x160S 0x4500

160 4500

Cs maxi permissible B.O.D5 of waste waters. Cs = 510.99mg/l

Initial B.O.D of coty waste waters

=719mg/l

Degree of treatment required.

=

=28.93%

7. In the previous prob determine what should be as direction ration if

no treatment was required. And thus determine the over dk change for

such as condition.

Solution:

When no treatment is required. The value of max. permissible Bon5 of waste

water is Cs should be 719, QR can then be determined as

17.54 719x160 0xQR

160 QR

17.54 160 QR 719x160

160 QR 6559

QR 6399l/s(ray) Dilution ratio= 399.99say

40times.

Hence when the dilution ration is 40 and the minimum river dos charge is 6400

l/s, no treatment will be required.

8. A waste water efficient of 560 l/s wo/u a BOD =50mg/l Do=3.0mg/l and

tempof 230C enters a river where the flow is 28 m3/sec, and BOD =4.0mg/l

and temp of 170c k1 of the waste is 0.10/ day at 200C. The vel of water in

the river down stream is 0.18m/s and depart of 1.2m. determine the

following after moving of waste water in the low over water; i) combined

discharge ii) BOD iii) Do and iv) Temperature

Solution:-

Particulars of sewage particulars of River

Thrown

Qs=560 c/s QR=28 m3/sec

=0.56m3/sec


Concentration(cs) Concentrations (CR)

Bon=50mg/l BoD=4.0mg/l

Do=3.0mg/l Do=8.2mg/l

Temp=230C Temp=170C

K1 at 200=0.1 perday

i) combined discharge =9s+9R

=0.56+28

=28.56m3/sec Now, using equ

C C Qss C QR R

QS QR

(ii) BOD of max

4.9mg/l

(iii) Do of mix

3.0x0.56 8.2x28

0.56 28

8.098mg/l

(iv) temp of max

17.120C

9. 125 ciemecs of sewage of a city is discharged in a perennial river

which is fully saturated with oxygen and flows at a minimum rate of 1600

ciemecs with a minimum velocity of 0.12 m/s if the 5 day BOD of the

sewage is 300mg/l final out where the oritual do will occur in the river

assume.

i) the co efficient of purification of the river at

4.0 ii) The coefficient of Do as 0.11 iii) The ultimate BOD as 125% of the 5 day BOD of the mixture of sewage and river water. Solution:

Assume saturation D.O concentration. Of the given river Ds=9.2


The D.o of the river at the mixing Dt after disposal of

Sewage

8.53mg/l

Initial D.O. default (DO)=DS-D

=9.2-8.53

=0.67mg/l

Bon5 of the river at the mixing pt after disposal of sewage y5

21.74mg/l

The ultimate BoD of river (max) at moving

Pt(L) = 125%BOD

=125x21.74=27.17mg/l

Noco using equ

BOD5=L[1-(10)-kdx5]

21.74=27.17[1-(10)-kdx5]

0.8=[1-(10)-kdx5]

(10)-5KD=0.20

-5KDlog10=log0.20

KD=0.14

The coefficient of DO or BOD (KD) Is given in assumption NO. (ii)m to be

0.11 as against its value of 0.14 computed above on the basis of

assumption (iii) Even finally there is some in consistency in the given data,

and the examiner should have given only one of the two assumption is

either ii) or iii) which would have suffice purpose.

Under such difficult situation, 10e may solve the question by using both the

values of KD is 0.11 as well as 0.14. The KD value of 0.14 will, displace the

critical pt upstream and will thus provide more conservative design values.

Case (1) :- when KD=0.11

tc k (fD 1

1)log 1 f 1Do

L f

tc 0.11(41 1)log 1 4 1 27.170.67 x4


1.723

The distance along the river, where the critical

F.O. defilit will occur =S=velocity x time =0.12m/s x1.723x24x3600sec.

=17.86km sat 18jn

Hence, critical D.O. difficult will occur at is km down stream of the sewage

disposal pt Case(2) : when KD=0.14 tc 0.140.11

x1.723 1.354days

S 17.86x1.354

14.04km

1.723

Hence, critical D.O. difficult will occur at 14km downstream of sewage

disposal pt.

10. 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/s

Discharge or river =Qn5m3/s

Temperature =T=250C

KD(250)=K1=0.12/d

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

CR-conc. Of BOD5 for river water =1/mg/l

(i) conc of BOD5 of the maxture if un treated waste water is dischargd

into we river

C C Qw w C QR R Qw QR

46.92mg/l

(ii) BOD5 of the treated waste water is given by c+w=20% of the BOD5

of un treated 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

C1 CTW.QW C .QR R QW QR

10mg/l

11. In impervious prob(40) found out the ultimateBOD of the river

water after is receives treated waste water.

BOD5 of river water after is receives treated waste water

=10mg/l (as computed above prob (10))

ultimate BOD of this mixture

=yu=C=?

Using equ

Yt(day) = L[1-(10) –kp.t]

Y5=L[1-(10) -0.12x5]

10=L[1-(10)-0.6]

L=13.35 mg/l

12. A town having population of 40,000 disposessewage by label

treatment. It gets a per copier assured water supply from water works

at a rate of 130 d. assuming that the land used for sewage disposal

can absorb 80m3 of sewage per her per day, determine the lanel area

reqd and (t) cost at the rate of Rs./ 25,000 per thee make suitable

assumptions where needed.

Solution:

Population =40,000

Rate of water supply =130 c/d/per

Total water supplied per day

=40,000x130l =52,00,000lot =5,200cu.m.

Assuming that 80% of this water appears as sewage, we have

The quantity of sewage produced per day

=0.8x5200

=4160 cu.m


Area of land reqd for disposing sewage

52hect

providing 50% extra land for rest and rotation, we have the total land area

reqd.

=1.5x52 =78hect.

Cost of lanel involved = Rs25,000x78

=Rs. 19,50,000

13. A Town disposes sewage by lanel treatment. It has a sewage farm of

area 150 hect. The area included an extra provision of 50% for rest and

rotation the population of the town being 50,000 and rate of water suppy

140 lot/ capila / day. If 75% of the water is converted into sewage

determine the consuming capacity of the soil.

Solution:

Quantity of water produced per day

=50,000x140 lit/d =70,00,000 l/d

=7,000 cu. m/d.

Quantity of sewage produced

=0.75x7000

=5,250 cu. m/d.

Area of farm land provided

=150 hec worth 50% additional reserve

Area provided for immediate need

100hect

100 hect is capable of passing 5250 cum/d

consuming capacity of 800%

52.5cu.mhe/d

14. Write short notes on

i) Efficient irrigation and sewage farming.

ii) sewage sickness.

Efficient irrigation and sewage for morning:

Although, out wardly, both these terms are used as synonyms to each other,

yet there is one basic difference b/w the. 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.

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

pipes, load open founded at a spacing of 12 to 30m. The efficient collected in

these drains after getting filtered through the 5001 pores is a generally small

(as a large quantity gets evaporated) and well stabilized, and can be early

disposal into some natural water courses, with out any further treatment.

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. 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 once. 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 $001

and the load prevent in sewage. But when once these voids are clogged, free

circulation for air will be prevented and anaerobic conditions will develop

coottion the pores. Due to those the aerobic de composition of organic matte

will stop, and anaerobic decomposition will start. The organic matter will there,

of course, be miner lord but with the evolution of foul gases live H2S, CO2,

CH4. this phenomenon of soil getting clogged is known as sewage sickness

of land.

15. 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.

16. A sedimentation tank is treating 4.5 million lit of sewage per day

containing 275 ppm of suspended solids. The tank removes 50% of

suspended solids. Calculated the quantity of sludge produced per day

in bulk and lot if (a) moisture content of sludge is

98%

Solution:

Volume of sewage treated

= 4.5 M.Lit / day

Since suspended solids amount to 275 mg / 2, we have that wf of suspended

solids present in sewage

106 kg/d

1237.5 kg/d

Since 50% of solids are removed in sedimentation tank, we have the wt of

solids removed in sedimentation tank


1237.5

618.75 kg/d

(a) When moisture content of sludge I s98% then 2 kg of solids (dry sludge)

will make

= 100 kg of wt sludge

618.75kg of solids (dry sludge) will make

618.75

30937.5 kg30940 kg

Hence wet sludge or sludge produced per day

=30, 940 kg 30.94 ton

Assuming the specific gravity of wet sludge (sludge) as

1.02, we have unit wt. of sludge

1.02 1 t/m3

1.02t/m3

unit wtof water 1t/m3

volume of wet sludge produced per day

= wt =30.94=30.33m3 unit wt 1.02

Vol. of sludge (when its m.c. is 98%)

= 30.33 cu.m

17. In the same prob (1) Finding out the moisture content of sludge is

96%.

When moisture content is 96% then

4 kg of solids will make

= 100 kg of wet sludge

618.75 kg of solids will make (Refer previous prob)

618.75 kg of wet sludge

15468.75 kg of wet sludge

15,470 kg say of wet sludge

=15.47 tonnes of wet sludge


Hence, wt. of, sludge (when its m,c, is 96%)

= 15.47 tonnes if sp – gravity of sludge

is 1.02 then

Volume of sludge (when its m,c, is 96%)

m3

15.17m3

Hence, the vol. of sludge at 96% m.c.

= 15.17 cu.m

18. There is a sewage sludge with volume containing a certain moisture

content P1 (%) what will be the volume of this sludge if its moisture

content is reduced to P.(%)

Solution:

Let the given sewage contains soids = w kg. let its volume to v1 at a moisture

content of p1(%) and v at a moisture content of p(%).

At moisture content of P11 we have (100 – P1) kg of

solids will make w kg of solids will make

100.w kg of wet sludge 100 P1

or wt. of sludge produced

100.2 kg

100 p1

if r2 is the unit wt of sludge in kg / m3, then vol of sludge

produced 100.w . 1 m3

100 p1 rS

V1 100100 .wP r1 . 1S

At moisture content of P(%), similarly, we have vol. of sludge produced (v)

100.w . 1 m3


100 P rS r 100.w . 1

100 P rS

From equation

100 P v .v

100

Equating (1) & (2)

100 P v v1 1S 100 P v.v S

100 100

v v 1 1

0 01 0 0

P1

P

19. The moisture content of a sludge is reduced from 95 to 90% in a

sludge digestion tank. Find the percentage decrease in the volume of

sludge.

Solution:-

Using equation

v v1 100100 PP1 v v1

100100 9590

v1 5/10

v1

2

Thus, the volume at 90% moisture will be half of that at 95% moisture.

Hence the percentage decrease in moisture will be 50%.

w 1 1 S

100

100 P v.v

w S

....(1)

.... 2


20. Design a digestion tank for the primary sludge with the help of

following data:

(i) Average flow = 200 mcd

(ii) Total suspended solids in raw sewage =

300 mg /l

(iii) Moisture content of digested sludge

= 85%

Assume any other suitable data you require Solution:

Average sewage flow = 20 m.c.d

Total suspended solids = 300 mg / l

wt of suspended solids in 20 Mc of sewage flowing per day = 300 1020 106

6

= 6000 kg / 1 day

Assuming that 65% solids are removal on primary settling tanks, we have wt

of solids removed in the primary settling tank

65% 6000 kg/d

3900 kg/d

Assuming that the fresh sludge has a m,c, of 95% we have 5 kg of dry solids

will make = 100 kg of wet sludge and 3900 kg of dry solids will make

3900kg of wet sludge per day

Assuming sp – gravity of wet sludge as 1.02 i.e. unit wt = 1020

kg / m3, we have the volume of raw sludge produced / day

v1 780001020 m /d3

76.47m /d3

The volume of the digested sludge (V2) at 85% m.c. is given by the formula


v2 v1 100100 PP1 v2 v1

100100 9585

v2 v1 5/15 31

v1

76.47 m /d3

v2 25.49 m /d3

Now, assuming the digestion period as 30 days, we have capacity of the reqd

digestion tank, given by equation capacity v1 82

v1 v2 t

76.47 76.47 25.49 30

76.47 3250.98 30

1274.5 1275 m3

Now, providing 6.0m depth of the cylindrical digestion tank, we have c/s area

of the tank

= 212.5 m2

of tank212.5

14 16.45 16.5m

Hence provide cylindrical sludge digestion tank 6m deep 16.5 m , with

additional hoppered bottom of 1% slope for collection of digested sludge.

21. Raw waste water is entering a treatment plant and contains 250 mg /

l suspended solids. It 55% of these solids are removed in sedimentation.


(a) Find the volume of raw sludge produced permillion litre of waste

water. Assume that the sludge has a moisture content of 96% and

specific gravity of solids is 1.2.

(b) Find the unit weight of raw sludge

(c) If 45% of raw sludge is changed in liquid and gasin the digestion tank,

find the volume of digested sludge per million litre of waste water.

Assume that the moisture constant of the digested sludge is 90%.

Solution:

(a) Suspended solids in waster water = 250 mg / 2

Since 55% of these solids are removed in

sedimentation, we have

The solid removed in sedimentation as sludge

= 55% 250 mg / l

= 137.5 mg / l

If volume of waste water is 1 million litre, then solids removed as sludge

106 kg

137.5 kg

Sludge produced will, thus, have 137.5 kg solids, and rest will be water.

Now, since the moisture constant of sludge is 96%, we have 4 kg of solids will

produce 100 kg of wet sludge, by joining with 96 kg of water.

Water contained in 4 kg of solids

= 96 kg

Water contained in 137.5 kg of solids

137.5

3300 kg

Hence volume of sludge produced per million litre of water

Weight of solids wt. of water

unit wt. of solids unit wt. of water


= 1.2 1000137.5 10003300 cu.m

unit weight of solids

= sp. gravity of solids unit wt. of water

= 1.2 1000 kg/ m3

0.115 3.3

3.415 cu.m

Hence, vol. of sludge produced per million litre of waste water.

= 3.415 cu. M

(b) unit wt of raw sludge

wt. of solids + wt. of water volume of sludge

kg/m3

kg/m3

1007 kg/m3

© 45% of raw sludge is changed into liquid and gas, means that 45% of solids

are consumed (digested). wt. of dry solids left in the digested sludge

= ( 100 – 45) % of total solid

= 137.5 kg

= 75.625 kg

since digested sludge contains 90% m.c. we have The volume of digested

sludge

wt. of solids left in digested sludgeunit weight of solids unit

wt. waterwt. of meter

1.2 100075.625 75.6251000 10090 m3

0.063 0.681

0.744 cu. m


Hence, the volume of digested sludge per million litre of waster water =

0.744 cu.m

22. 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.

= 106 kg

= 600 kg.

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

solids.

= 0.55 600 = 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

330

6600 kg of sludge

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

Volume of sludge = wt. of sludge

unit of wt.

of sludge

10206600kg/mkg 3

[ unit wt. of sludge

= sp. Gravity unit wt. of water.


= 1.02 1000

= 1020 kg / m3 ]

= 6.47 m3

Hence, the vol. of sludge formed per million litre of sewage.

= 6.47 cu.m

23. Design a sludge digestion tank for 40,000 people. The sludge content

per capita per day is 0.068 kg. The moisture of the sludge is 94% The sp.

Gravity of the wet sludge is 1.02 and 3.5 percent of the digester vol. is

daily filled with the fresh sludge, which is mixed with the digested

sludge.

Solution:-

Dry sludge content produced by 40,000 persons

= 0.068 40, 000 kg

= 2, 720 kg / day

94% moisture content means that 6 kg of dry sludge will produce 100 kg of

wet sludge.

6 kg of dry sludge produces wet sludge = 100 kg

2720 kg of dry sludge produces wet sludge

6. 2720

45333 kg

45.3 t/day

Volume of wet sludge produced

wt. of sludge

unit.wt. of sludge

45.3m /day3

1.02

Unit wt. of sludge in 6 / m3 sp. gravity unit wt. of water i.e 1t

/ m 3

1.02 1


1.02 t/m3

44.4 m /day3

44.4 m3 of fresh sludge is added to the tank daily, to fill 3.5% of the digester

capacity.

capacity of digester

=volume of fresh sludge produced daily

=44.4 m3

or capacity of digester required

=

1268.9 cu.m

Providing 30% additional capacity for fluctuations, we have,

The required digester capacity,

= 1268.9 1.3

= 1650 cu. M (say)

Now, providing 6 m depth of the cylindrical digestion tank, we have

The cross sectional area of the tank

275 m2

Dia of tank = 257

/4

= 257

0.785

=18.7m.

Hence, provide a cylindrical sludge digestion tank, 6m deep and 18.7 in

diameter, with an additional hoppered bottom of 1:1 slope foe collection of

digested sludge.

24. 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 have

Volatile solids removed

=70%x180mg/l

=126mg/l

Now, 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.

25. A sewage containing 200 mg/l of suspended solids is passed through

primary settling tank. The solids from the primary settling tank are

digested to recover the gas. Find the likely volumes of mathene and

carbon di oxide produced in the digestion of the sludge from 10,000 m3

of sewage. Calculate the fuel value of the gas produced. State clearly the

assumption made.

Solution:

Total suspended solids in sewage =200mg/l

Assuming that 60% of suspended solids are removed in the primary settling

tank, we have the suspended solids removed as sludge. =60%x200 mg/l

=120mg/l

Now assuming that the volatile solids present are 70% of the suspended

solids, we have

The volatile solids removed

=70%x120mg/l

=84mg/l

Further, assuming that the volatile matter is reduced by


65% in sludge digestion, we have Volatile matter reduced =65%x84mg/l

=54.6mg/l. Hence, volatile matter reduced in 10,000 cu m of sewage

54.6 kg.

546kg

Now, assuming that 0.9 cu. M of gas is produced per kg of volatile matter

reduced, we have Total quantity of gas produced

=0.9x546 cu m. =491.4 cu. M.

Assuming that the produced gas contains 65% of methoane and 30% of carbon di oxide, we have Methane produced =0.645x491.4 cu m.

=319.41 cu. M.

carbondioxide produced = 0.30x491.4 cu m.

=147.42 cu m.

Now assuming that the methane in the sludge gas has a fuel value of 36,000

kg/m3, we have

The fuel value = 36,000x391.41kj = 11.50M.kj.

Now, assuming a boiler efficiency of 80% we have the amount of heat that

can be furnished by the boiler.

=80%x11.50mkj =9.2mkj

= 9.2 mkc2

4.18

=2.2 million kilo calorie.

26. (a) calculate the area of land required for drying the sludge from the digestion tank for 40,000 population, designed in qn. No. 8. (b) Also design the dimensions of beds.

Solutions:

(b)The volume of wet sludge from the sewage of 40,000 population was

worked out as 44.4m3/day

Let it be spread in 22.5 cm thick layer (ie between 20 to

30 cm thick layer) on under drained said beds, then

The area of beds required

44.40m2 = 225

197.3m /day.2

Under tropical Indian conditions, the beds get dried out in about to days and

hence taking 2 weeks as average drying time including wet days of rainy season, we can

utilize the same bed= 52

26 times in an year *. 2


Area of bed required per year

2770m (2 say)

Making 100% allowance for space for storage, repairs, and resting of beds,

etc, we have

The total area of beds required

=2x2770 m2

=5540 m2

=0.554 hectares Ans.

(b) Now, using 15x30 m sized beds, we have the No. of beds required

12.3

So let us use 14 nos. of beds, with size as:

Area

=395.7m2

using 15m width, length=

=26.4m.

Hence, use 14 beds, of size 15mx26.4m in plan. The beds should be provided

with under drains and side walls, with typical section and plan as shown in

figure.

27. Describe the mechanical methods of dewatering sludge?

Droed or dewatered by mechanical means, such as by vacuum filtration or by

high speed centrifuges.

In vacuum filtration process, the sludge is first mixed with a consequent such

as ferric chloride and then conveyed to a vacuum filter, consisting of a hollow

rotating drum, covered with a replaceable filter cloth. The drum rotator partly

submerging into the sludge. The vacuum created by a pump with in the drum

draws the moisture from the sludge through the cloth. The sludge cake which

is formed on the out side of the drum is removed by a scraper as we drum rotates.

High – speed centrifuges: are also used for drying of raw or digested

sludges, and are becoming more popular because of small area requirements.

There methods may remove about 50% moisture.


Vacuum filtration or centrifugation of raw sludge is often adopted in

situation where sludge is to be disposed of by incineration (i.e. burning). These

mechanical methods of drying are generally used when the available area is

less than there required for sludge drying beds, or where the clomater are too

cold or at places where rains are frequent as not to permit natural drying or as

a preliminary to heat drying for making fertilizer.

28. Determine the liquid volume before and after digestion and

percentage reduction for 600 kg (dry basis) of primary sludge having the following characteristics.

Primary Digested

Solids (%) 6 12

Volatile matter (%) 65 65

Specific gravity of

fixed solids

2.5 2.5

Specific gravity of

voluble solids

1.0 1.0

Solution:-

1) Computation of average spe. Gravity of all the solids in primary sludge

100 35 65 SS 2.5 1.0

From which SS = 1.266 (primary solids)

2) Computation of sp. Gravity of primary sludge

100 6 94 SSC 1.266 1 SS1

1.013

3) Computation of volume of primary sludge

VS1 S PwSCsS 1000 1.013 0.06 600

9.874 m3

4) Computation of % volatile mater after digestion fixedmatter in primary

sludge = 0.35 600 = 210 kg

volatile matter in primary sludge = 0.65 600 = 390 kg

Volatile matter after digestion = 0.35 390 (65% of 360 kg has been digested

on digestion).


Total matter after digestion = 210 + 0.35 390

Volatile matter = 2100.35 390 0.35 390 100

39.39

5) Computation of average spe. Gravity of all the solids indigested sludge

100 60.61 39.39

SS 2.5 1

1.571 digested sludge

6) Computation of sp. Gravity of digested sludge

100 12 88 SdsL 1.571 1

SdsL 1.046

7). Computation volume of digested sludge

VdsL SwdsLds .Ps

wds 200 0.35 390 336.5 kg

VdsL 1000 1.046 0.12 336.5 2.681m3

8) Reduction of sludge volume after digestion

% reduction 9.874 2.681

100 72.85%

9.874

29. Design a gravity thickens for thickening the combined primary and

activated sludge from a treatment plant for 200,000 population.

Solution:-

Let us assume / capita settle able solids in primary sludge as 54gm / day

and per capita settle able SS in activated sludge as 31gm / day making a total SS as 54 + 31 = 85 gm / day table 16.1


lot of combined sludge = 85 10 3 200,000

17000 kg/d

Recommended average surface loading = 40 kg / m2 / d

Hydraulic loading regd = 25 m3 / m2 /d

Let as assume sp. gravity of wet mixed sludge = 1.008 and solids in combined

sludge as 3%. The volume of wet sludge / d is given

VS1 S PwS1s s

562m /d3

Surface Area needed 425m2

Flow needed from giving hydraulic loading of 25 m3 / m2 / d = 25 425 = 10625

m3 / m2 /d

Balance of 10625 – 562 = 10063 m3 / d is made available by blending with

primary or secondary effluent. Let us provide a side water depth of 3m

sludge detention period = V/Q

24

54.5 hrs

This is more than 24 hrs and hence o.k.

Let us provide a circular sludge blanket type thicken

of tank 425 4

23.26m

Let as provide a 24m dia. Tank

Sludge blanket restricted to 1m is adopted. Expected solids in the thickened

sludge = 6%

30. Design a sludge drying bed for digested sludge from an activated

sludge plant serving 200,000 people. Solution:-


From cable 16.1, total solids remaining on digested sludge (combined

primary activated)

= 57 gm / cap / d

Daily solids = 200,000 57 10-3 = 11400 kg / d

Let us adopt a dry solids loading of 100kg 1m2 / y

Area of bed needed = 41610m2

Check for per capita Area = 0.208m2

(This is within the recommended range of 0.175 to 0.25)

Let us adopt 8m wide 30m long beds with single pt discharge and a bed

slope of 0.5%

No. of beds = 174

Assuming 2 months of rainy season in a year and sweets for drying and

one week for preparation and

12 2 4 10 repair of bed,

number of cycles / year =

4 Let us Assume 7%. Solids and

sp. Gravity of 1.025, the volume of digested sludge is given by

VSl .S .Pwsls s

1.59 m /d3

Depth of application of sludge =

0.139m

14 cm

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