Upload
truongtram
View
219
Download
0
Embed Size (px)
Citation preview
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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”?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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:
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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:
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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:
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
= 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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
=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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
= 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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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”?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
= 200 l/s
B.O.D. of sewage = Cs = 250 mg/l.
B.O.D. of river = CR = 6 mg/l.
B.O.D. of the diluted mixture
= 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
B.O.D. of the diluted mixture
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
=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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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:
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
(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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
= 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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
= 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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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?
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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.
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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).
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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:-
BHARATHIDASAN ENGG COLLEGE ENVIRONMENTAL ENGG 2 2M & 16M
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