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1. Environmental Engineering- II
By
Akash Padole
Department of Civil Engineering
Characterization and
Primary Treatment of Sewage

2. Contents
2.1 Need for Analysis, Characteristics of sewage:
Composition, Biochemical characteristics, aerobic
decomposition, anaerobic decomposition, Sampling
of sewage, Analysis of sewage.
2.2 Treatment processes:
Objective, methods of treatment, flow sheets
showing Preliminary, Primary, Secondary and Tertiary
treatment. Primary treatment: Screening, Grit
removal, Oil and Grease removal, settling tank.
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3. • Most of the organic matter present in water is
unstable and decomposes readily through chemical
as well as the biological processes.
• The organic matter which can be decomposed by
bacteria under biological action, is called
biodegradable organic matter.
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4. • Most of the organic matter present in water is
biodegradable and hence undergo biological
decomposition, which can be divided into:
i. Aerobic decomposition
ii. Anaerobic decomposition
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5. Aerobic decomposition
• Aerobic decomposition is caused by both aerobic
bacteria operating aerobically, in presence of air or
oxygen which is available in the waste water in the
dissolved form.
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6. Anaerobic decomposition
• If free dissolved oxygen is not available to the water
or sewage, then the anaerobic decomposition, called
putrefaction will occur.
• Anaerobic bacteria operating anaerobically, will then
flourish and convert the complex organic matter into
simpler organic compounds of nitrogen, carbon, and
Sulphur.
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7. • These anaerobic bacteria survive by extracting and
consuming like nitrate and sulphates.
• Gases like ammonia, hydrogen sulphide, methane
etc. are also evolved in this decomposition,
producing obnoxious (bad) odour.
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8. Facultative Micro organisms:
– There are certain types of micro organisms which
can act both presence or absence of oxygen, these
micro organisms are referred Facultative micro
organisms.
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9. Techniques of Sampling
❖Grab Sampling
• A single sample of water collected from sampling
spot at any instant
❖Composite Sampling
• Mixture of grab samples at same point for different
time
• Interval is 30min /60min/120minutes for 24 hrs.
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10. Characteristics of Sewage
• Physical Characteristics
• Chemical Characteristics
• Bacteriological Characteristics
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11. Physical Characteristics
• Turbidity
• Colour
• Odour
• Temperature
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12. Chemical Characteristics
• Tests conducted for determining the chemical
characteristics of sewage help in indicating: the stage
of sewage decomposition, its strength, and extent
and type of treatment required for making it safe to
the point of disposal.
• Chemical analysis is, therefore, carried out on
sewage in order to determine its chemical
characteristics.
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13. It includes tests for determining:
(i) Total Solids, Suspended Solids, and Settleable Solids
(ii) pH value
(iii) Chloride Content
(iv) Nitrogen Content
(v) Presence of Fats, Greases, and Oils
(vi) Dissolved Oxygen (D.O.)
(vii) Chemical Oxygen Demand (C.O.D.)
(viii) Bio-chemical Oxygen Demand (B.O.D.)
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14. Total Solids, Suspended Solids & Settleable
Solids
• Sewage normally contains very small amount of
solids in relation to the huge quantity of water
(99.9%).
• It only contains about 0.05 to 0.1 per cent (i.e. 500 to
1000 mg/l) of total solids.
• Solids present in sewage may be in any of the four
forms: suspended solids, dissolved solids, colloidal
solids, and settleable solids.
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15. • Suspended solids are those solids which remain
floating in sewage.
• Dissolved solids are those which remain dissolved in
sewage just as salt in water.
• Colloidal solids are finely divided solids remaining
either in solution or in suspension.
• Settleable solids are that portion of solid matter
which settles out, if sewage is allowed to remain
undisturbed for a period of 2 hours.
10-3 µm 1 µm
Dissolved Solids Colloidal Suspended Solids
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16. • The proportion of these different types of solids is
generally found to be as given below:
– It has been estimated that about 1000 kg of sewage
contains about 0.45 kg of total solids, out of which
0.225 kg is in dissolved, 0.112 kg is in suspension, and
0.112 kg is settleable.
– Further, the solids in sewage comprise of both the
organic as well as inorganic solids.
– The organic matter works out to be about 45 per cent
of the total solids, and the remaining about 55 per
cent is the inorganic matter.
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17. • The quantity of settleable
solids can be determined
easily with the help of a
specially designed conical
glass vessel called Imhoff
cone
• The capacity of the cone is
1 litre, and it is graduated
up to about 50 ml.
Imhoff cone
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18. • Sewage is allowed to stand in this Imhoff cone for a
period of two hours, and the quantity of solids
settled in the bottom of the cone can then be directly
read out.
• However, in order to obtain precise amount of
settleable solids, the liquid from the cone should be
decanted off, and the settleable solids collected at
the bottom of the cone should be dried and
weighed.
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19. Typical quantities of Solids present in
Domestic Sewage in mg/l
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20. Fats, Oils and Greases
• Fats, oils and greases are derived in sewage from the
discharge of animals and vegetable matter, or from
the garages, kitchens of hotels and restaurants, etc.
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21. • Such matter form scum on the top of the
sedimentation tanks, clogs the voids of the filter
media and affects the diffusion of oxygen.
• They thus interfere with the normal treatment
methods. Hence these detection and removal is
important.
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22. Dissolved Oxygen (D.O.)
• The determination of dissolved oxygen present in
sewage is very important, because: while discharging
the treated sewage into some river stream, it is
necessary to ensure at least 4 ppm of D.O. in it ; as
otherwise, fish are likely to be killed, creating
nuisance near the vicinity of disposal.
• To ensure this, D.O. tests are performed during
sewage disposal treatment processes.
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23. • The D.O. test performed on sewage before
treatment, helps in indicating the condition of
sewage.
• It is well known by now, that only very fresh sewage
contains some dissolved oxygen, which is soon
depleted by aerobic decomposition.
• Also, the dissolved oxygen in fresh sewage depends
upon temperature.
• If the temperature of sewage is more, the D.O.
content will be less.
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26. • The solubility of oxygen in sewage is 95% of that in
distilled water.
• The D.O. content of sewage is generally determined
by the Winkler’s method which is an oxidation-
reduction process carried out chemically to liberate
iodine in amount equivalent to the quantity of
dissolved oxygen originally present.
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27. Chemical Oxygen Demand (C.O.D.)
• The amount of oxygen required to carry out the
decomposition of both bio degradable and non-bio
degradable organic matter present in the system is
termed as CHEMICAL OXYGEN DEMAND
• It can be determined by adding Potassium
Dichromate and H₂SO4 acid in the water to be tested
and noting the amount the oxygen consumed for
decomposition of organic matter.
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28. • Potassium Dichromate is such a strong oxidizing
agent that in of some case it also carries the
oxidation of inorganic substance present in the
system. Hence result in higher value of COD then
actual.
• Thereby COD is also refers as Dichromate Demand.
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29. Theoretical Oxygen Demand
• If the exact formula and the concentration of organic
matter present in the system is known then the
amount of O2 required for its decomposition can be
computed theoretically and is being referred as
THEORETICAL OXYGEN DEMAND (ThOD)
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31. Total Organic Carbon (TOC)
• TOC is just another way to express the organic matter
present in the wastewater in terms of its carbon
content.
• Carbon (C) is the primary constituent of organic
matter hence chemical formula of every organic
compound will reflect the extent of carbon present in
that compound.
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33. Biochemical Oxygen Demand (BOD)
• The organic matter, in fact, is of two types
– biologically oxidized (i.e. oxidized by bacteria) and
is called biologically active or biologically
degradable
– that which cannot be oxidized biologically, and is
called biologically inactive.
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34. • While testing a waste water, we are mainly interested
in finding out the amount of biologically active
organic matter present in it; whereas the COD test
gives us the total of biologically active as well as
biologically inactive organic matter.
• Hence, further testing is carried out to determine the
biochemical oxygen demand (B.O.D.) of sewage,
which directly gives us the amount of biologically
active organic matter present in sewage.
• The amount of oxygen consumed in this process is
the BOD.
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35. • Biochemical Oxygen Demand (BOD) is defined as the
amount of oxygen required by the microorganisms
to carry out decomposition of Biodegradable
Organic Matter under aerobic conditions.
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36. • Polluted waters will continue to absorb oxygen for
many months, and it is not practically feasible to
determine this ultimate oxygen demand.
• Hence, the BOD of water during 5 days at 20°C is
generally taken as the standard demand, and is
about 68% of the total demand.
• A 10 day BOD is about 90% of the total.
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37. Why do we determine BOD?
• The BOD test is widely used to determine the
pollution strength of domestic and industrial wastes
in terms of the oxygen that they will require if
discharged into natural watercourses (it is the one of
the most important test in stream pollution control
activities).
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38. • This test is of prime importance in regulatory work
and in studies designed to evaluate the purification
capacity of receiving bodies of water.
• It is also useful in design of wastewater treatment
plant and also to measure the efficiency of some
treatment processes
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41. Derive the Equation of first stage BOD
• The sanitary engineers are more concerned with the
first stage demand, since the oxygen consumed in its
satisfaction is not recoverable.
• Hence, the term BOD is usually used to mean the
first stage BOD, i.e. the demand due to presence of
carbonaceous matter alone.
• “The rate at which BOD is satisfied at any time,
depends on temperature, and also on the amount
and nature of organic matter present in sewage at
that time.”
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47. De-Oxygenation Constant (KD)
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48. BOD Curve
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49. Limitation of BOD Test
• It measures only the biodegradable organic matter.
• Time duration of the test is very long i.e. 5 days, so if
quick results are needed it is not useful.
• Pretreatment is needed if the sample contains toxic
waste.
• Nitrifying bacteria can cause interferences and could
give higher results. To avoid them proper care must
be taken.
• It is essential, to have high concentration of active
bacteria present in the sample.
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50. BODu/COD ratio
• BODu/COD ratio indicates the quality of sewage with
regards to the presence of biodegradable and non-
biodegradable organic matter present in it.
• If the ratio of untreated sewage is greater than 0.92
to 1, it is considered to be treated biologically and
• If it is less than 0.5 it cannot be treated biologically
as it consists of toxic impurities.
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51. COD/TOC ratio
• The ratio also indicates the quality of sewage with
respect to the type of organic matter present in it.
• Its value varies between 0 to 5.33.
– Its 5.33 for methane,
– 0 for organic matter that are resistant to
dichromate oxidation.
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52. Population Equivalent
• Industrial wastewaters are generally compared with
per capita normal domestic wastewaters, so as to
rationally charge the industries for the pollution
caused by them.
• The strength of the industrial sewage is thus, worked
out as below:
[ Standard BOD5 of industrial Sewage]=
[ Std. BOD5 of domestic sewage per person per day] x
[ Population equivalent]
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53. Example:
• The average standard BOD5 of domestic sewage is
worked out to be about 0.08 kg/day/ person. Hence,
if the BOD5 of the sewage coming from an industries
is worked out to be 350 kg/day. PE=?
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54. Calculate the Population Equivalent of a city
• The average sewage from the city is 95 x 106 l/day, &
• The average 5 day BOD is 300 mg/l.
Example:
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55. Relative Stability
• Relative Stability of sewage effluent may be defined
as the ratio of oxygen available in the effluent ( as
DO, nitrite or nitrate) to the total oxygen required to
satisfy its first stage BOD demand.
• It is expressed as percentage of the total oxygen
required, and can be expressed by the equation.
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56. • The decolorization caused by enzymes produced by
anaerobic bacteria is an indication of the available
oxygen in oxidizing the unstable organic matter.
• The sooner the decolonization takes place, the earlier
the anaerobic condition develops, which means
lesser availability of oxygen.
• Hence if the decolorization takes place sooner (within
4 days), the effluent sample may be taken as relative
unstable.
• But samples which do no decolorize in 4 days can be
taken as relative stable, and thus can be discharged
into streams without any troubles.
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57. Example:
• If the period of incubation is 10 days at 200C in the
relative conductivity test on sewage, calculate the
percentage of relative stability.
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58. • A 2 percent solution of sewage sample is incubated
for 5 days at 20 degree C. The depletion of oxygen
was found to be 4 ppm. Determine BOD of sewage.
Example:
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59. Example:
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60. Solution:
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63. Example:
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64. Solution:
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65. Example:
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66. Solution:
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67. Example:
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68. Flow Chart
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69. Flow Chart
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70. Unit Operations and Unit Processes
• Unit Operations are the application of Physical
forces.
• In this, no other external forces are required.
• Example: Screening, aeration, sedimentation, etc.
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71. • Unit processes are the types of treatment in which
removal of contaminants is brought about by the
addition of chemicals or by microbial activities.
• Based on the type of agent used, Biological Unit
Process are
– Suspended Growth Process (Activated Sludge
Process, Oxidation Pond, Aerated Lagoon, etc.)
– Attached Growth Process (Trickling Filters,
Rotating Biological Contractors, Biofilters, etc.)
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72. Classification of Treatment Processes
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74. Preliminary Treatment
• Separating the floating materials (dead animals, tree
branches, papers, pieces of woods)
• Heavy settleable inorganic solids.
• Removes Oil and Grease.
• Reduces BOD by 15-30%
– Screening
– Grit Chambers or Detritus Tank
– Skimming Tanks
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75. Primary Treatment
• Removing Large Suspended Organic Solids.
• Removes 60-70% of organic solids and satisfy 30-40%
of BOD
• Organic solids which separated out in sedimentation
are often stabilized by anaerobic decomposition in
Digestion tank.
• Residue left for land fills and soil conditioner.
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77. Secondary Treatment
• Provided to remove OM present in waste water by the
action of micro-organisms. (Biological decomposition of
OM)
• This can be carried out either by Aerobic or Anaerobic
Condition.
• The effluent contain little BOD (5-10% of the original)
• Aerobic Biological units
– Activated Sludge Process
– Tricking filter
– Oxidation ponds
– Aerated lagoon
➢Anaerobic Biological units
– Anaerobic Lagoon
– Septic tank (uses raw sewage)
– Imhoff tank (uses raw
sewage)
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78. Final or Advanced Treatment
• Also known as Tertiary treatment.
• Used to kill the pathogenic bacteria (chlorination)
• Adopted when the outfall of sewage is near to the
water intake of some nearby town.
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80. Screening
• To trap and removing the floating materials such as
pieces of cloth, paper, wood, kitchen refuse, etc.
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83. • Comminutors are device used to break the large
sewage solids to 8mm which is based upon cutting
action to ease the process of digestion.
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85. • The head loss through clean flat bar screens is
calculated from the following formula:
where, h = head loss in m,
V = velocity through the screen in m/s,
v = velocity before the screen in m/s.
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86. Grit Chamber
✓ GRIT:
• Grit is the heavy mineral material found in raw
sewage and it contains sand, gravel, silt, broken
glasses, small fragments of metal and other small
inorganic solids.
• Subsiding velocity or specific gravity substantially
greater than those of the organic solids in water.
• Examples: Eggshells, bone chips, sand, gravel, silt etc.
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87. ✓ Grit Chamber:
• Grit Chamber is the second unit operation used in
primary treatment of wastewater and it is intended
to remove suspended inorganic particles such as
sandy and gritty matter from the wastewater.
• Limited to the municipal waste water.
• It is provided to protect moving mechanical
equipment from abrasion and abnormal wear; avoid
deposition in the pipelines, channel and conduits; and
to reduce frequency of digested cleaning
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90. • It is design in the form of long narrow channel which
may either rectangular or parabolic section.
• Velocity control sections:
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92. Detritus Tank:
• These are nothing but Rectangular Grit Chamber
designed to flow for smaller velocity (about 0.09m/s)
and longer detention period (3-4 mins).
• To separate out larger grit and Fine Sand Particles.
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93. Flow Equalization Chamber
• This unit is provided to overcome the operational
problem troubles caused due to variation in rate of
flow so as to increase the performance and reduce
the size and cost of downstream treatment.
– To make flow and concentration of impurities
uniform
– To avoid shock loading
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94. Skimming Tank
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95. Vacuator
• Grease can be removed from the sewage by vacuum
floatation method, by subjecting the aerated sewage
to a vacuum pressure of about 0-25 cm of mercury
for 10 to 15 minutes in a Vacuator.
• This causes the air bubbles to expand and move
upward through the sewage to the surface.
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96. • The rising bubbles lift the grease and the lighter
waste solids to the surface where they are removed
through skimming troughs.
• Heavier solids settle to the tank bottom, where they
are collected and carried away for sludge treatment
and disposal.
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97. Sedimentation
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99. Primary Sedimentation Tank
• To remove the Organic Suspended Solids.
• Sludge formed are mostly organic which is called as
Primary Sludge.
• Prior to disposal, volume need to be reduced and
reduction of volume occurs in Sludge Digester.
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100. Types of Sedimentation Tank
1. Quiescent Type Tank or Intermittent Settling Tank
2. Continuous Flow Type
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101. Quiescent type tank or Intermittent settling tank
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102. Continuous Flow Type
1. Rectangular Settling Tank:
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103. 2. Circular Settling Tank:
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108. Cleaning and Sludge Removal
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109. Example:
• Design a suitable rectangular sedimentation tank
(provided with mechanical cleaning equipment) for
treating the sewage for a city provided with a public
water supply system with a Maximum Demand of 12
million litres per day. Assume suitable value of
detention period and velocity of flow in the tank
make any assumption wherever needed
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