FUTURE GATE ACADEMY EE

Introduction 2) Determination of water demand for

The engineering discipline the future population
which deals with the study of design 3) Identification of source for the
and construction of structures, design demand
equipment and systems related to 4) Quality analysis of source water
protected water supply and quality sample
sanitary systems to ensure
environmental safety & public health 5) Treatment of water based on its
is called as Environmental quality parameters
Engineering. 6) Distribution of water towards the
 Before 1968, Environmental public by designing pipe
Engineering was called as networking.
Sanitary Engineering in US (or)  Water supply project is considered
Public Health Engineering in for an optimum (average) design
Great Britain, India etc., After period of 30 years, even though
1968, it was named as different components of the project
Environmental Engineering. may carry variable design periods.
 Environmental engineering deals  Design Period: The period in the
with control of effect of humans on future for which estimate (both
environment and vice versa. population and water quantity
 It is studied under three aspects prediction) is to be made is called
that are responsible for safety of Design Period.
environment. They are  Sanitary Engineering:
Water Supply Engineering The branch of environmental
Sanitary (or) Waste Water engineering that deals with the study
Engineering of design and construction of
structures related to the collection,
Pollution Aspects and its conveyance, treatment and disposal of
Control waste water collected from various
 Water Supply Engineering: sources is termed as Sanitary
The branch of environmental engineering.
engineering that deals with the study It is considered as the effect of
of design and construction of humans on environment.
structures related to the collection,  Pollution Aspects & its control:
conveyance, treatment and
distribution of potable water towards The various pollution aspects
the public is termed as Water supply (other than water pollution) and their
engineering. control are important to ensure public
health and environmental safety. It
The water supply scheme involves the includes the study of
following steps:
 Air Pollution and its control
1) Estimation of future population
 Noise Pollution
(Population Forecasting)
 Solid Waste Management

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FUTURE GATE ACADEMY EE

1. Population Forecasting
The probabilistic estimation of future  Applicable for old and settled cities
population of a town or city for the which have stabilized
future period based on previous  This method assumes that the
population data, using mathematical increase of population per decade
and graphical methods is called as is constant
Population Forecasting.
Pn = P0 + nC
 Population Density: No. of
persons occupied per unit area Where, n → No. of decades

 There are 2 reasons for change of C → Arithmetic Rate Constant

population : Ex 2: From the past population data
o Increase due to Births, given below. Calculate the population
Migrations and Annexations. in the year 2020, according to
arithmetic regression method?
o Decrease due to Deaths and
Migrations. Year 1970 1980 1990 2000
MATHEMATICAL METHODS
Pop. 8000 12000 17000 22500
A. Annual Rate of Increase Method:
C. Geometric Progression Method:
The rate of increase of population
per year is considered to be constant  This is uniform percentage growth
and then population for the given method.
design period is predicted as  This method is based on
Pn = P0 (1+i)n assumption that a constant
percentage of growth occur for
Pn → Population at the end of design
equal periods of time.
period (or) n years
 The geometric average percentage
P → Latest known population
of growth of last few decades is
1
Pn determined, and then the
i → Annual rate constant = n -1
P0 population forecasting is done
based on the geometric rate
Ex 1: The population of a certain town constant.
was 40,000 in the year 1950 and  This is best suitable for young and
50,000 in 1960. Determine its rapidly growing cities.
population in the year 1970?
 The method of population
1 50,000 forecasting accepted by census
Solution: i= 10 - 1 = 0.02
40,000 department as best suitable for
Indian population growth
P1970 = 50,000 (1+0.02)10 = 60,950 conditions is Geometric Increase
B. Arithmetic Increase Method (or) Method.
Arithmetic Regression Method: r 
Pn = P0 [1+ ]n
 The rate of change of population 100
with time is a constant value. r = Geometric Rate Constant
= [r1 x r2 x r3 . . . . . rn]1/n

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FUTURE GATE ACADEMY EE

Ex 3: From the past population data Ex 5: Calculate the population in the

given below. Calculate the population year 2030, according to decremental
in the year 2040, according to rate method?
geometric increase method?
Year 1980 1990 2000 2010
Year 1970 1980 1990 2000
Pop. 56000 68000 77000 83500
Pop. 48000 59000 67000 72500
2. Simple Graphical Method:
D. Incremental Increase Method: The method of population
The average increase in the forecasting in which a graph is
population is determined plotted using the population
according to arithmetical method against decades from the available
and then added to the average of data and the curve is smoothly
the net incremental increase once extended to forecast the future
for each future decade. population is called as Simple

Pn = P0 + n x +n
n  1 y Graphical Method.
2 3. Comparative Graphical Method:

y = Incremental Rate Constant The cities having the conditions

and characteristics to the city
Ex 4: From the past population data
whose future population is to be
given below. Calculate the population
estimated are first selected. It is
in the year 2030, according to
assumed that the present city will
incremental increase method?
develop like the selected cities
Year 1970 1980 1990 2000 have developed in the past. Then
the development curve for the
Pop. 36000 45000 58000 67500 similar cities was plotted by taking
population against decades and
E. Decremental Rate (or) Decreasing
then from the curve, the present
Rate Method:
cities population forecast was
 It was observed that growth of obtained.
population has a certain limit. The
The most precise and reliable
early growth may be at increasing
results are obtained by this
rate and the later growth may be
method.
different (at decreasing rate).
5. Logistic Curve Method:
 This method is suited particularly
for old cities that shows a plot of  Mathematical Determination
S-curve and indicates that was given by P.F. Vertulist.
saturation limit is attained.  The population at any time ‘t’

Pn = P0 [1+
r 0 d  ][1+  r 0  2d ] was given by,

100 100  PS  P   PS  P 0 
loge  P   log   = - K.Ps.t
r  ]……[1+  r ]
e
0  3d 0  nd    P0 
[1+
100 100
Ps
Pt =
1  m log e1 (n  t )

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FUTURE GATE ACADEMY EE

Class Room Objectives population growth follows

arithmetic growth at a rate of 7500
1. As compared to geometrical per decade and continues.
increase method, arithmetic Determine the population at the
increase method of population end of year 2035?
forecasting gives:
7. The populations of a city in the year
(a) Higher Value 2000 was 8,00,000. Thereafter, it
(b) Lesser Value (c) Equal Value increases in an arithmetic growth
(d) May vary depending on at a rate of 10,000 per decade and
population values also shows an incremental
2. The suitable method for increase of 2,000 per decade.
forecasting population for a young Determine the population at the
and a rapidly developing city is: end of year 2030?
(a) Arithmetic mean method 8. The population of India in the year
(b) Geometric mean method 2000 was 1 billion and it increased
(c) Comparative graphical method in an exponential manner at a rate
(d) none of these of 3.8% per year till 2020 and then
the growth rate is decreased to
3. The growth of population can be
2.4% per year and is continued at
represented by a ______ curve
the same level. What will be the
(a) Straight line (b) Semi-log population in the year 2060?
(c) Logistic (d) Logarithmic
9. In two periods each of 30 years, a
4. Given the population at a place is city has grown from 20000 to
75,000 in 2000. The average 140000 and then to 260000.
increase is population is 10,000 Determine the predicted
per decade till 2020. There after population after next two periods?
the population is increased in an
10. The population of a city in the
exponential manner at a rate of 2%
year 2010 was 250000 and then
per decade and maintained at the
for 20 years, the population
level. What will be the population
increases exponentially at a rate of
at the year 2050?
20percent per decade. Thereafter,
(a) 95,000 (b) 1,00,000 the population growth rate
(c) 1,25,000 (d) 1,00,815 increases constantly at a rate of
15000 per decade for 15 years.
5. The populations of a town in three
Estimate the population for 2045.
consecutive decades are: 1.6 lakh,
2.4 lakh and 2.84 lakh 11. Estimate the population for the
respectively. The population of year 2030 by the census method
this town in the fourth consecutive for the following data:
decade, according to geometric
Year 1970 1980 1990 2000
rate method, would be?
6. The populations of a city in the year Pop. 37500 46000 58000 69500
2010 was 20,00,000. Thereafter, it
12. The most suitable population
increases in an exponential
forecasting method in India
manner at a rate of 12% per
according to census department is
decade till 20 years and then the
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 FUTURE GATE ACADEMY EE

Previous Questions Reason (R) : The future population

1. The population of a city at previous depends on the trade and expansion
consecutive census year was of the city, discovery of mineral
400000; 558500; 776000 and deposits, power generation etc. (2000)
1098500. The anticipated (a) both A and R are true and R is the
population at the next census to correct explanation of A
the nearest 5,000 would be (b) both A and R are true but R is not
(Gate 91) a correct explanation of A
(c) A is true but R is false
2. The population figures in a growing (d) A is false but R is true
town are as follows : (IES 02)
6. Which one of the following
Year 1970 1980 1990 2000 Acts/Rules has a provision for “No
right to appear”? (IES 2004)
Pop 40000 46000 53000 58000
(a) Environment (Protection) Act, 1986
The predicted population in 2010 (b)The Hazardous Waste (Management
by Arithmetic Regression method is and Handling) Rules, 1989
(c) Manufacture, Storage and Import
(a) 62,000 (b) 63,000
of Hazardous Chemicals Rules,1989
(c) 64,000 (d) 65,000
(d) Environment Rules, 1992
3. On which of the following factors,
7. Which among the following brings
does the population growth in a town
about the Hazardous Waste
normally depend (IES - 2003)
Management and Handling Rules in
1. Birth and death rates India ? (IES 2007)
2. Migrations 3. Probabilistic growth
(a) Central Pollution Control Board
4. Logistic growth
(b) Ministry of Environment & Forests
Select the correct answer using the (c) Ministry of Urban Development
codes given below: (d) Ministry of Rural Development
(a) 1 and 4 (b) 1 and 2 8. When was the water (Prevention
(c) 1, 2 and 3 (d) 2 and 3 and Control of Pollution) Act enacted
4. Population levels over 5 decades of by the Indian Parliament ? (IES 1996)
a small town are given below (a) 1970 (b) 1974
(c) 1980 (d) 1985
1960 1970 1980 1990 2000
9. The population of new city is 18.5
250000 280000 340000 490000 490000 million and is growing at 20%
annually. How many years would it
The population of the town in the year
take to double at the growth rate?
2020, estimated by Arithmetic
increase method will be (IES 2014) (a) 3-4 yrs (b) 4-5 yrs
(a) 5,10,000 (b) 5,90,000 (c) 5-6 yrs (d) 6-7 yrs (Gate 2014)
(c) 6,10,000 (d) 6,90,000
Key for Previous Questions
5. Assertion (A) : The future
population is predicted on the basis of 1. 15.4 lakhs 2. c 3. b 4. C
knowledge of the city and its 5. a 6. a 7. b 8. b 9. a
environment.

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FUTURE GATE ACADEMY EE

2. Water Demand
The total quantity of the water (c) Watering of gardens, public
to be supplied every day for a city parks, and public fountains
throughout its design period to fulfill (d) Sanitation etc., is named as
various purposes of the city (or) town public demand.
is termed as Design Demand.
4) Compensate Losses Demand:
WD = Population x Percapita Demand Same portion of water is always
Types of Water demand worked due to

Type of LIG Need HIG %

(a) Defective pipe joints
Demand Need (b) Cracks in pipe line
(c) Faulty valves and Fittings
Domestic 135 lpcd 200 lpcd 55% (d) Opening of taps by consumers
Industrial 70 lpcd 70 lpcd 25% (e) Damaged public taps
(f) Unauthorized connection
Public 10 lpcd 10 lpcd 5%
5) Fire Demand: The amount of
Losses & water consumed for extinguishing
55 lpcd 55 lpcd 15 %
thefts the fire when a fire outbreak
TOTAL 270 lpcd 335 lpcd 100 occurs is named as Fire Demand.
Empirical Formulae:
1) Domestic Water Demand: The
quantity of water required in the 1) NBFU (National Board of Fire
houses for various purposes like Underwriters Formula):
drinking, bathing, cooking, Q = 4640 p (1-0.01 p )
washing etc., is called as Domestic
Q = Qty. of water required in lt/min
Water Demand (DWD).
P = Population in thousands
 The consumption of water by
domestic animals also included in 2) Freeman’s Formula:
this demand. P 
Q = 1135.5  10  and F = 2.8 Q
2) Commercial and Industrial  10 
Demand: The water demand in F = No. of simultaneous fire streams
commercial centers and buildings
3) Kuiching’s Formula: Q=3182 p
like office building, ware houses,
stores, hotels, hospitals, schools, 4) Bustons Formula: Q = 5663 p
factories, industries, railway & bus 5) Rational Formula:
stations, temples, theatres etc., is
If fire may occur frequently, it may be
named as Commercial & Industrial
determined from the following
Demand.
expression:
 The consumption will vary greatly
with the character of city 4360 T 0.275
Q= lt/min.
(t  12)0.757
3) Public Demand: The quantity of
water required for public utility t → Duration of fire (in min)  min.
purposes minimum 30 min fire is considered

(a) Washing & sprilinking of roads T → Period of recurrence occurrence

(b) Cleaning of sewers of fire in years

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FUTURE GATE ACADEMY EE

FLUCTUATIONS IN DEMAND: PRACTICE QUESTIONS

The average daily per capita
consumption is obtained by: 1. If the population of a city is 1 lakh,
average water consumption is 250
Quantity of water supplied during the year lpcd, then the filters and lift
No. of days  No. of population pumps will be designed for
The per capita consumption (or) (a) 45 MLD (b) 50 MLD
water demand varies from year to
(c) 25 MLD (d) 67.5 MLD
year, day to day, from season to
season and also hour to hour. 2. If the population of a city is 1 lakh,
 These variations are expressed as average water consumption is 250
percentage of annual average daily lpcd, the fire demand is 61 MLD
consumption. then the capacity of the
distribution system should be
a) Max. Seasonal Consumption:
130% of average annual daily rate (a) 510 MLD (b) 67.5 MLD
of demand (c) 106 MLD (d) none
b) Max. monthly Consumption : 3. As per IS : 1172 – 1993 in the
140% of average annual daily rate design of a water supply scheme
of demand for a town with full flushing
c) Max. Daily Consumption : 180% of system minimum water supply for
average annual daily rate of domestic water demand is
demand (a) 135 lpcd (b) 200 lpcd
d) Max. Hourly Consumption : 150% (c) 270 lpcd (d) 335 lpcd
of average for the day.
6. The per capita per day demand of
e) Max. Hourly consumption on Max. water is taken as an average value
 Max. Daily Demand  over a period of
Day = 2.7 x  
 24 (a) 24 hours (b) one month
f) Coincident Drat: (c) one year (d) > 1 year
Max. Daily Demand + Fire Demand 7. If the average water consumption
g) Total Draft (or) Capacity of a of a 300 lpcd and its population is
Storage Reservoir: 4,00,000 maximum hourly draft of
the maximum day and maximum
Max. Daily Demand + Fire Demand
daily draft will be
(or) Max Hourly Demand,
(a) 120 MLD and 216 MLD
Whichever is maximum that is
considered. (b) 1216 MLD and 324 MLD
Design Period: (c) 324 MLD and 216 MLD
The period of years in future for (d) none of these
which a provision is made in the 8. If the average daily water
water supply scheme is termed as consumption of a city is 24,000
Design period. m3, the peak hourly demand (of
the maximum day) will be:
The optimum design period is 30
(a) 1000 cum/hr (b) 1500 cum/hr
years for a water supply project.
(c) 1800 cum/hr (d) 2700 cum/hr

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FUTURE GATE ACADEMY EE

9. If the population of a city is 2 lakh, capacity, assuming an arithmetic

and average water consumption is rate of population growth, will be
200 lpcd, then the capacity of the
(GATE-CE-2004,2012)
pipe mains, carrying raw water
from source will be: (a) 5.5 years (b) 8.6 years
(a) 108 MLD (b) 72 MLD (c) 15.0 years (d) 16.5 years
(c) 60 MLD (d) 40 MLD 3. Consider the following statements :
10. Water supply project is designed The daily per capita consumption
for a city with a population of 1 of water apparently increases with
lakh and an area of 1000m2
constructed with 40% ordinary 1. Higher standard of living of people
and 60% Wooden constructions. If 2. Availability of sewerage in the city
the percapita water demand is 200
lpcd. Determine the capacity of 3. Metered water supply
storage reservoir? 4. Wholesome and potable quality of
(a) 36000.212m3 (b) 36212m3 water
(c) 36212lt (d) Data Inadequate Which of these statements are
correct? (IES 1999)
11. Determine the capacity of pumps
to lift water for a population of 1 (a) 1, 2 and 3 (b) 2, 3 and 4
lakh. With average, water
consumption of 250 lpcd? (c) 1, 3 and 4 (d) 1, 2 and 4

12. Determine the capacity of a 4. The present population of a

storage reservoir for a city of a community is 28,000 with an
population of 20 lakhs with an average water demand of 150
average water demand of 200 Lpcd. The existing water treatment
lpcd?
plant has a design capacity of
6000 m3/d. It is expected that the
PREVIOUS QUESTIONS
population will increase to 48000
1. Water distribution systems are during the next 20 years. What is
sized to meet the (GATE 1998) the number of years from now
a) Maximum hourly demand when the plant will reach its
b) Average hourly demand design capacity assuming an
c) Maximum daily demand and arithmetic rate of population
fire demand growth ? (IES-2006)
d) Average daily demand and fire (a) 8.6 years (b) 12.0 years
demand
(c) 15.0 years (d) 16.5 years
2. The present population of a
5. For water supply to a medium
community is 28,000 with an
town, what is the daily
average water consumption 4200
variation factor?
m3/d. The existing water treatment
plant has a design capacity of Key for Previous Questions
6000 m3/d. It is expected that the
1. c 2. c 3. c 4. b 5. 1.5
population will increase to 44,000
during the next 20 years. The
number of years from now when
the plant will reach its design

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FUTURE GATE ACADEMY EE

3. Sources of Water
The sources of water are classified as Infiltration Wells: A series of circular
Surface Sources and Ground Water shallow concrete wells constructed
Sources. along the river bank to entrap seepage
Surface Sources: losses through the river are called as
Infiltration wells.
(i) Streams (ii) Lakes (iii) Ponds
Jack Well: All the infiltration wells
(iv) Rivers (v) Reservoirs (vi) Oceans
discharge the collected water through
Ground Sources: a conduit into a large well named as
(i) Infiltration Galleries (ii) Springs Jack Well.
(iii) Infiltration Wells (iv) Wells
PRACTICE QUESTIONS
INFILTRATION GALLERIES:
1. A 600m3/day of water is
The horizontal rectangular tunnel
discharged from an infiltration
constructed below the river bed to
gallery at 6m depth from
entrap losses through infiltration from
subsurface water table. Find the
the river is considered as Infiltration
length of the gallery if the
Gallery.
drawdown in the gallery in
Ground water (or) River water after pumping does not exceed 4m. The
infiltration travels towards lakes, radius of influence may be
rivers or streams. assumed to be 100 m and ‘k’ as
 This water which is traveling can 100m/day?
be intercepted by digging a trench 2. Find the discharge in an
or by constructing a tunnel with infiltration gallery of length 50m
holes on sides at right angles to the running through the bed of a river
direction of flow of underground whose permeability in 0.07 m/mi.
water. If the Max. Drawdown allowed in
 Infiltration galleries allow water the well is 2m & thickness of
from both sides and one side as aquifer is 5m, the radius of
desired. influence maybe assumed as
100m. Workout the population
 Standard Yield: 1.5 x 104 L/day/m
served by the gallery if per capita
length of Infiltration Gallery.
consumption is 120L.
Dimensions: Length → 10 – 100m
3. The differentiating factor for
Width → 1m Height → 2m shallow and deep wells is
[H 2  h 2 ] a) Discharge b) Diameter
Q = KL
2R c) Depth d) Both a & b

Springs: The natural outflow of 4. The method of treatment employed

underground water that held in the for Lake water is …………………..
soil and a pervious rock emerges out 5. The source of water for the well
as a stream of current above the water is
surface are called springs. There are 3
a) Water Table b) Aquifer
types of springs which include:
Surface, Gravity & Artesian Springs. c) Infiltration d) Both a & b

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FUTURE GATE ACADEMY EE

CONVEYANCE STRUCTURES (OR) V = Velocity of flow,

CONDUITS FOR WATER SUPPLY given by Hazen Williams equation
 The pipes and pumps which have V = 0.85CH (R)0.63 (S)0.54
been designed to convey the volume
of water from source towards water Conduits are also designed using
treatment plant are considered as Darcy’s weishbach equation
conveyance structures. hf = flv2 / 2gd (or)
 Water is generally conveyed using hf = flQ2 / 12.1d5 (or)
pipes and pumps (pressure hf = 4flv2 / 2gd
conduits) associated with Intake
The capacity of the pump required to
structures.
supply the water is given by the
INTAKES following equation:
The structure placed near surface BHP = γwQH/0.746 np nm
water source to permit the withdrawal
of water from the source and then PRACTICE QUESTIONS
discharge into a pump or conduit
though which it will flow into the 1. A town of design population of 4
water works system is called as lakh with an average water demand of
INTAKES (or) INTAKE WORKS. 150 lpcd wants to obtain water by
gravity from a reservoir of 13km away
 Intakes consist of opening, grating from it. Design the pipe diameter, if
or strainer through which the raw the minimum and maximum water
water from river, canal or reservoir levels in the reservoir are 106.550 and
enters and is carried to a sump well 102.250m. The elevation of water level
by means of conduits. Water from in the storage tank of the city is
the sump work is pumped through 96.5m. Assume f = 0.036
the raising mains to the treatment
plant. 2. A town with a population of 2 lakh
Intakes consists of supplied water @ 250lpcd gets its
water supply form a source which is
1) Conduit with protective works 6km away from it. Design the
2) Screens at the open ends diameter of the water main required if
3) Gates & valves to regulate the flow the elevation difference between
source and supply is 8m and frictional
PIPES & PUMPS:
factor is 0.002.
 The pump is used to drive water
from source to city or treatment 3. A storage reservoir situated 6km
plant when source is at lower away from the city carrying 3 lakh
depths compared to treatment population @ 200lpcd. The loss of
plant head from source to city does not
exceed 20m and pumping done 12
 The design of pressure conduits
hours only. Determine the size of
follow Darcy’s equation
supply main using Hazen William
Q = AV formulae assuming its coefficient as
Where, A= Area of the conduit=  /4 d2 130?

d = Diameter of the conduit

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FUTURE GATE ACADEMY EE

4. Find the discharge in an infiltration (b) Sluice valve: To control flow of

gallery of length 50m running water through pipelines
through the bed of a river whose (c) Air valve: To release the
permeability is 0.07 m/min. If the accumulated air
maximum drawdown allowed in the (d) Scour valve: To remove silt in a
well is 2m and thickness of the pipeline
aquifer is 5m. Assume radius of 5. Match List-I (Equation/Method)
influence as 100m. with List-II (Application) and select
5. For the above discharge, calculate the correct answer using the codes
the population equivalent if the per given below the lists : (IES-2007)
capita water demand is 120 lpcd? List-I
A. Manning’s Equation
PREVIOUS QUESTIONS
B. Dancy-Weisbach
1. A straight 100m long raw water C. Hardy Cross Method
gravity main is to carry water from
D. Rational Method
an intake structure to the jack well
of a water treatment plant. The List-II
required flow through this water 1. Frictional head loss estimation in
main is 0.21 m3/s. Allowable pipe flow
velocity through the main is 0.75
2. Sanitary sewer design
m/s. Assume frictional factor =
0.01, g = 9.81 m/s2. The 3. Storm sewer design
minimum gradient (in cm/100m 4. Water distribution system design
length) to be given to this gravity
Codes :
main so that the required amount
of water flows without any A B C D
difficulty is ____ (GATE 2014) (a) 2 1 4 3
2. Two reservoirs at different levels (b) 1 4 3 2
are connected by two parallel pipes
(c) 2 1 3 4
of diameter ‘2d’ and ‘d’. The ratio
of the flows in the two pipes (larger (d) 1 4 2 3
: smaller) is (IES-CE-1995) 6. Sonoscope is used for which one of
(a) 2:1 (b) 2 : 1 the following? (IES-CE-2008)

(c) 4 : 1 (d) 4 2 : 1 (a) Checking the accuracy of meters

3. A commonly used hand pump is (b) Regulating the fire hydrants
(a) Centrifugal pump (c) As a replacement of venturimeter
(b) Reciprocating pump for discharge measurement
(c) Rotary pump (d) Detection of leakage in
(d) Axial flow pump underground water mains
4. Which one of the following pairs is
not correctly matched? (IES1996) Key for Previous Questions
(a) Check valve: To check water flow 1.4.4 cm 2.D 3.B 4.A 5.A 6.D
in all directions
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FUTURE GATE ACADEMY EE

4. QUALITY ANALYSIS OF WATER

The laboratory procedures which  These impurities include Sand,
will analyze the type of impurities Silt, Clay, Bacteria, Algae and
present in water and provide the Fungi.
information regarding the methods of  Effects caused: Turbidity, Odour,
treatment to be carried out to convert taste, Colour and Acidity etc.,
the water into human consumable
 Treatment: Sedimentation
potable water is called as Quality
Analysis of Water. Colloidal Impurities:

 There are three methods of  Finely divide dispersed particles in

quality analysis for water, which water which are electrically
include Physical, Chemical and charged and remain in continuous
Biological Quality Analysis. motion are considered as Colloidal
Impurities.
 These methods of quality analysis
follow the protocols provided in  Mostly colour of the water is due to
the IS 3025-1983 (1991). colloidal impurities. Hence their
quantity is determined by colour
 The water which is fit for human
tests.
consumption and has both
characteristics of wholesome water  Size: 1  to 1 m 
and palatable water is termed as  Effects caused: Colour, Alkalinity
Potable water and is prepared and Acidity etc.,
according to the specifications
 Treatment: Coagulation
given in IS 10500.
Dissolved Impurities:
IMPURITIES IN WATER
 The impurities that are dissolved
 The impurities present in water
in water when it moves the rocks,
have been classified as Suspended,
soil etc., which include solids,
Dissolved and Colloidal on the
liquids and gases dissolved in
basis of size.
water are considered as Dissolved
 The impurities present in water Impurities.
have been classified as Organic
 Size: 10-7 mm to 10-9 mm in dia.
and Inorganic based on their
chemical nature.  The impurities include Dissolved
Salts, Gases (CO2, O2, Hydrogen
Suspended Solids:
Sulphate, Methane), Metals (Ar &
 The impurities that are solid Pb) and Bacteria.
particles of large size dispersed in
 Effects caused: Turbidity, Odour,
water in submerged state and in
Hardness, Alkalinity, Corrosion,
suspension during the flow are
Taste and cause diseases due to
considered as Suspended Solids.
harmful bacterial presence.
 Size: 10-1 mm to 10-3 mm in dia.
 Treatment:
Filtration & Disinfection
Page 12
FUTURE GATE ACADEMY EE

Physical Quality Analysis  Iron in water cause Red colour,

The quality analysis which gives Manganese (or) Fungi in water
information regarding the physical cause Brown colour and Algae (or)
appearance of water and reveals about vegetable matter can cause Green
visible impurities is considered as Colour.
Physical quality analysis. 3) Taste and Odour: The taste and
1) Temperature: Temperature of odour is due to the presence of dead
water determines the other properties or living micro organisms, dissolved
like density, viscosity, vapour organic or inorganic matter, dissolved
pressure & surface tension. gases (H2S, Methane, CO2) or O2
combined with organic matter,
 Desirable temperature: 4.4 to 100C
mineral substances like NaCl,
 Optimum temperature: 10 to 150C carbonates & sulphates of other
 Temperature of water reveals substances.
about the information regarding  Odour of water is measured by an
rates of chemical, biochemical & instrument, namely Osmoscope
biological activity due to bacteria. and gives Permissible odour
2) Colour: Allowable Limit is 5 to 20 number (PO number) and it is
TCU on Platinum Cobalt scale. graduated with values pO values
from 0 to 5.
 Colour is due to presence of
organic matter in colloidal  PO No. 0 : No perceptible odour
condition, minerals and dissolved  PO No. 5 : Extremely strong odour
organic & inorganic impurities.
 Odour in water is unobjectionable.
 The colour measurement scale is
 H2S cause Rotten Egg Smell
Burgess Scale.
 NH3 cause Pungent Smell
 It is measured by comparing the
colour of water sample with other  CO2 cause Foul Smell
standard Nesslar tubes (Glass  Odour of water is expressed by
tubes). Threshold Odour Number (TON)
 The colour produced by 1mg of and Taste of water is expressed by
Platinum Cobalt in 1L of distilled Flavin Threshold Number (FTN)
water has been fixed as unit of A B
TON (or) FTN =
colour as 1 TCU (True colour unit), A
this scale is termed as Platinum A = Vol. of sample
Cobalt Scale. B = Vol. of distilled water
 Colour of water is tested by TON (or) FTN is Dilution ratio.
instrument called TINTOMETER.
 Permissible limit: 1 to 3.
 Discharges from industries like
 Odour changes with temperature
tannery, textile and paper impact
and is tested normally at 200C to
colour due to presence of lignin,
250C
tannin etc., will cause color.

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FUTURE GATE ACADEMY EE

Ex 1: 20ml of raw water sample  Based on Light Absorption

containing certain odour is diluted phenomena, turbidity measuring
with 180ml of distilled water until it instruments include Jackson’s
loses its odour. Estimate TON? Turbidity Rod, Jackson’s, Baylis
and Hellige Turbid meters.
20  180
Solution: TON: = 10  These instruments measure
20
turbidity using Silica Scale: 1mg of
Ex 2: 25ml of raw water sample finely divided silica (Fullers earth)
containing certain taste is diluted with dissolved in 1L of distilled water.
distilled water until it loses its taste
and the final volume is 175ml.  Jackson’s Turbidity Rod:
Estimate TON of water? Similar to the measuring scale,
175 graduated aluminium (or) steel rod
Solution: TON: =9
25 with the values of turbidity from 0
to 50 JTU and a platinum pointer
4) Specific Conductivity: The total (needle) at the bottom end.
amount of dissolved salts present in Turbidity is measured based on the
water can be easily estimated by visibility of pointer.
measuring the specific conductivity of
 Jackson’s Turbid meter:
water.
It is used to measure high turbidity
 This was determined by portable
values of more than 25 JTU. This
ionic water tester (or) Diionic water
instrument contains a metal stand
tester
with source of light at bottom and a
 Specific conductivity is expressed graduated container at top.
in micro-mhos per cm at 250C. Turbidity is measured based on the
 Mho is unit of conductivity. visibility of light.
 Dissolved salts conc. (mg/lt)  Baylis Turbid meter:
= Specific Conductivity x 0.65 Measure even low turbidity values
 Permissible limit: conductivity of (0 – 2 ppm) using a metal box with
0 – 2500 micro-mhos per cm. 2 sample holders into which a light
source of 250 volts is transmitted.
5) Tubidity: A measure of the
The measurement is based on the
resistance of water to the passage of
passage of light and visibility of
light through it due to the presence of
colour matching.
Suspended and Dissolved Solids is
called Turbidity. It is also called as  Based on Light Scattering
phenomena, turbidity measuring
Opaqueness (or) Darkness.
instruments include Nephelometer
 Ground water generally less turbid and Turbido meters. These
than surface water. instruments measure turbidity
 Permissible limit: up to 10 ppm. using Formazin polymer (Hexa
methylene Tetramine) as a standard
 Turbidity is measured based on
reference solution. These
Light Absorption and Light
instruments express turbidity in
Scattering phenomenon.
NTU (Nephlometric Turbidity units)
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FUTURE GATE ACADEMY EE

 Turbid meter: where as only non volatile solids will

The apparatus in which the remain in the container.
measurement of turbidity is based W 2 W 3 W 3 W1
on the intensity of light as it passes VS = NVS =
V V
straight through the water sample
is called Turbid meter Ex 3: 100CC of water sample is
 Nephelometer: passed through a filter paper whose
initial wt is found to be 2.486 gm.
The apparatus in which the
After oven drying, the final weight of
measurement of turbidity is based
the filtrate and filter paper is
on the intensity of light scattered at
measured an 2.498 gm. Find the
right angle to the incident light is
suspended solids concentration?
called Nephelometer.
Solution:
Measure turbidity values of 0 – 1
ppm in terms NTU. 2.498  2.486
SS = X 106 = 120 mg/L
100
II. CHEMICAL QUALITY
ANALYSIS
2) Hardness: The inability of water
It reveals sanitary quality of water
for the formation of lather or foam (a
1) Solids: The residue left after white mass of bubbles) of the soap
evaporation of water is termed as which is caused due to the presence
Solids which are estimated by of Ca & Mg in the water is called
Gravimetry. Hardness.
 Total Solids: The amount of solids Measurement of Hardness:
that are present in suspension and
Versenate Method: Water is titrated
in dissolved form are called Total
against EDTA solution using
Solids.
Erichrome Black T (EBT) as indicator.
W 2 W1 While titrating the colour changes
TS =
V from wine red to blue.
 Suspended solids are estimated by
Soap Test: Standard soap solution
filtering the water sample through a
was added in the water and was
Whatmann filter paper No. 44 of 1
vigorously shacked to see the
pore diameter, drying and weighing.
formation of lather for 5 min. The
W 2 W1 hardness of water was calculated on
SS =
V the basis of soap solution added and
 Dissolved solids: Filtered water is the lather factor.
evaporated and the residue is Permissible limits:
weighed.
For Drinking purpose: 75 to 115 ppm
W 2 W1
TDS = For Industrial needs: < 75 ppm
V
If the residue of total solids is burned  If hardness < 75 ppm, it is soft
in a muffle furnace @ 700 to 10000c,  If hardness > 200 ppm, it is hard
the volatile solids will decompose

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FUTURE GATE ACADEMY EE

Total Hardness (mg/L as CaCo3) 4) Increased Soap Consumption which

leads to increased laundry expenses.
50 50
= [Ca+2] x + [Mg+2] x
20 12 Ex 5: If Total hardness is 250mg/L as
Ca+2 and Mg+2 concentrations must CaCO3 and Total Alkalinity is 200
be taken in mg/L. mg/L as CaCO3. Estimate CH and
NCH?

Ex 4: A water sample analysis reveals Solution: TH > TA:

the concentration of Ca+2 and Mg+2 So CH = TA = 200 mg/L
ions is 80mg/L and 60mg/L NCH = TH-TA = 50 mg/L
respectively. Determine Hardness?
Solution: Hardness is also expressed in various
scales and degrees of hardness.
50 50
TH = 80 x + 60 x = 450 mg/L One British Degree of Hardness
20 12
(Clark’s Scale) = 14.25 ppm CaCO3
Types of Hardness:
One French Degree of Hardness
1) Carbonate (or) Temporary Hardness
= 10 ppm CaCO3
2) Non Carbonate (or) Permanent
One American Degree of Hardness
Hardness
= 17.15 ppm CaCO3
TH = CH + NCH
3) pH – Value of water: The
 Carbonate hardness is caused by
negative logarithm of hydrogen ion
HCO3 and CO3 of Ca & Mg and can
concentration is called as pH value. It
be removed to some extent by
is a measure of acidity and alkalinity.
simple boiling or removed fully by
addition of lime.  Depending on the nature of
dissolved salts and minerals, the
 Non Carbonate hardness caused
water found in natural sources may
by sulphates, chlorides and
be acidic or alkaline.
nitrates of Ca and Mg & can be
removed by water softening  The acidity (or) alkalinity is
methods. expressed in terms of equivalent
weight of CaCO3 and usually
 Measurement of CH & NCH:
measured in ppm.
CH and NCH are measured by the
 pH = - log10 [H+]
relationship between TH and TA.
 pH : 0 – 6.5 is Acidic
a) If TH > TA, then
 pH : 6.5 – 7.5 is Neutral
CH = TA and NCH = TH – TA
 pH : 7.5 – 14 is Alkaline
b) If TH < TA, then
 Permissible Limit: 6.6 to 8.5
CH = TH and NCH = 0.
Measurement of pH
 Effects of Hardness:
1) Electrometric Method using pH
1) Bitter taste 2) Corrosion
meter: A pH meter consists of two
3) Scaling Problems electrodes: a glass electrode and a
Page 16
FUTURE GATE ACADEMY EE

reference electrode (calomel electrode). pOH = 14 – 12 = 2

Glass Electrode is sensitive to the
[OH-] = 10-2 mol/L
concentration of hydronium ions in
solution. Calomel electrode supplies a Ex 7: 40ml of 0.01 N H2SO4 solution
constant potential (E° = +0.24 V is consumed while titrating 80ml of
versus the standard hydrogen water sample. Find alkalinity in water
electrode) as determined by the half- sample as mg/L of CaCO3?
reaction Sol: Alkalinity = 40 x 0.01 x 50 x 103
Hg2Cl2 + 2 e– ---- 2 Hg + 2 Cl– 80
= 1000 mg/L of CaCO3
Calomel is the trivial name for the
compound Hg2Cl2. When both the Acidity
reference and glass electrodes are
It is the ability of water to neutralize
contained in a single unit, it is
the base. It is of 2 types
referred to as a combination
electrode. The potential of the glass A) Mineral Acidity: Strong acid of pH
electrode is proportional to the in the range of 0 to 4.2.
logarithm of the ratio of [H3O+] inside B) Carbonate Acidity: Weak acid of pH
and outside the electrode. The pH in the range of 4.2 to 8.5 and caused
meter measures the total potential by CO2
across the two electrodes and displays Effects of Acidity: Corrosion and
this measurement on a scale Tuberculation of pipelines.
calibrated in pH units.
Alkalinity
2) Titrimetric Method indicator &
It is the ability of water to neutralize
Titrant: pH is measured using
the acid. It is of 3 types
indicators like Methyl Orange and
Phenolphthalein that results colour A) Bicarbonate Alkalinity: Weak base
reactions with acid or base followed by of pH in the range of 4.2 to 8.5 and
titration using respective titrant. caused by HCO3-.
Based on volume of titrant consumed, B) Carbonate Alkalinity: Moderate
using the following formula, acidity or base of pH in the range of 8.5 to 10
alkalinity in mg/L of CaCO3 is and caused by CO3-2.
determined.
C) Hydroxyl Alkalinity: Strong base of
Acidity (or) Alkalinity = pH in the range of 10 to 14 and
VT x NT x 50 x 103 caused by OH-.

Vs Total Alkalinity (mg/L as CaCo3)

3) Colorimetric Method : pH Paper 50 50
= [CO3-2] x + [HCO3-] x
30 61
Ex 6: If pH of water sample measured
using pH meter is 12, estimate OH- 50
ion concentration in mol/L? TA (mg/L as CaCo3) = [OH-] x
17
Solution: pH = 12 Permissible Limit: Up to 250mg/L

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FUTURE GATE ACADEMY EE

Effects of Alkalinity: Bitter taste to Total Kjeldahl N2 (TKN)

water, Incrustation & Sediment = Ammonia N2 + Organic N2
deposits in pipelines, difficulty in
5. Chlorides: NaCl(Sodium Chloride)
chlorination.
is the main substance of chloride
Ex 8: If the concentration of CO3-2 and water. The natural water near mines
HCO3- ions in water is found to be & sea water have dissolved Nacl.
90mg/L and 122 mg/L respectively.  Presence of chlorides may be due
Estimate Total Alkalinity? to mixing of saline water and
Solution: sewage in water.
50 50  Permissible Limit: Up to 250mg/L
TA = 90 x + 122 x = 250 mg/L  Excess of Chlorides is dangerous
30 61
and unfit for consumption.
4. Nitrogen Content: The presence of
 Estimation: Titration of water with
N2 in water indicates the presence of
silver nitrate and potassium
organic matter in water.
chromate. Presence of reddish
N2 exist in 4 forms: Ammonia brown color during titration
Nitrogen, Organic Nitrogen, Nitrites indicates presence of chlorides.
and Nitrates.
6. Chlorine: The dissolved chlorine is
A) Ammonia: N2 due to fresh pollution present in water due to the
in water. Ammonia represents zero disinfection with chlorine. It is
stage of decomposition and un- present as residual form.
decomposed organic matter in water.
Estimation: Starch Iodide Test (or)
Permissible limit: 0.15ppm.
orthotolodine Test
B) Albuminoid/Organic Nitrogen: N2
a) Starch Iodide Test:
at first stage of decomposition with a
permissible limit of 0.3ppm. Addition of starch solution and
potassium iodide with water and
C) Nitrites: Partial Oxidation of
titration using 0.01N Sodium
Organic N2 results Nitrites in water
thiosulphate, formation of blue color
which is very dangerous, considered
and then to colorless during titration
as half decomposed organic matter.
indicates presence of chlorine. The
Permissible limit: Nil
amount of chlorine can be determined
D) Nitrates: Completely oxidized by simple titration equation:
organic matter is converted into in-
Chlorine in mg/l = 0.355 x VT
organic and harmless nitrates.
Permissible limit: < 45ppm 7. Iron and Manganese: These are
generally found in ground water.
 Excess of nitrates in water
consumed by pregnant woman Permissible limit: Fe < 0.3ppm and
affects the new born child, causes Mn < 0.05ppm
the disease named as
Water is brownish red in colour due to
“Methemoglobenimea” (Blue Baby
presence of Fe and also causes
Disease).
discolouration of clothes.

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FUTURE GATE ACADEMY EE

Mn leads to corrosion of pipes, growth 13. Fluorides: Estimated using

of microbes and causes taste and colorimetry by zirconium spades
odour to water. solution.
8. Lead and Arsenic: Poisonous and  F < 1ppm causes formation of
dangerous for public health. dental cavities in teeth.
Permissible limit: Pb < 0.1ppm and  F > 1.5ppm causes Flourosis
Ar < 0.05ppm (mottling and yellow stains
formation on teeth) and
9. Dissolved Gases: It has been found
deformation of bones.
that water contain various dissolved
gases.  Permissible limit: Between 1ppm &
1.5ppm
a) Oxygen: Surface water contain,
large amount of dissolved O2 III. BIOLOGICAL QUALITY
because they absorb it from ANALYSIS
atmosphere. Algae and micro The identification of microbial flora
plants are responsible for and determination of its concentration
dissolved O2 & its presence is is important to estimate the quality of
necessary to keep water fresh water.
and sparkling. But more The microorganisms are classified as
quantity causes corrosion of pipe Bacteria, Algae, Fungi, Virus and
materials. Protozoa.
b) CO2: If water comes across Ca Bacteriological Analysis:
and Mg salts, the CO2 acts on
 Bacteria are single celled
them and converts into
organisms which are present
Bicarbonates & causes hardness.
everywhere (omni present).
 Water + Lime Solution → Milk
 These are microscopic in size
White Colour indicates
varying from 1 to 4 microns.
presence of CO2.
 For, water quality analysis
 CO2 presence in water
bacteria are classified as
indicates biological activity
Pathogenic and Non pathogenic
and causes corrosion in
in nature. Identification of
pipelines.
pathogenic bacteria which are
c) H2S: Impart bad taste and odour. harmful and capable of causing
10. Biochemical oxygen demand: diseases is important for water
BOD is Not allowable in drinking quality analysis.
water. Permissible limit: Nil  E.coli is used as an indicator
11. Copper: Affects lungs and organism for detection of
respiratory organs. pathogenic bacteria in water.
Permissible limit: 1 to 3ppm 1. Total count of Bacteria: The total
12. Sulphates: Cause Diarrhea. number of bacteria present in 1 ml
Permissible limit: < 250mg/L of water is counted.

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FUTURE GATE ACADEMY EE

 The 1 ml sample of water is sterilized membrane. All the bacteria

diluted in 99 ml of sterilized retained on the membrane. The
water. Then 1 ml of diluted membrane put in contact with culture
water is mixed with 10 ml of medium (M- Endo medium) in
Agar Gel culture medium. incubator for 24 hrs at 200C.
 This was kept in incubator at Membrane is taken out and colonies
200C for 48 hrs and then, the of bacteria are counted
sample will be taken out of microscopically.
incubator and colonies of Colony counts in the range of 20 – 80
bacteria are counted by means are valid count for Membrane Filter
of microscope. Technique.
 Then the product of number of 4. Most Probable Number Method
colonies and dilution factor give (MPN Test):
total number of bacteria per ml
The number which represents the
of undiluted water sample.
most probable count of bacterial
2. Bacteria Coil (B-Coli) Test (or) density in water is MPN. Laws of
E.Coli Test: It includes statistical probability are applied for
Presumptive Test and Confirmative E.coli test results and are expressed
Test. It is also known as multiple in terms of MPN. Different dilutions
tube fermentation test. [10 ml, 1 ml and 0.1 ml] of samples of
a) Presumptive Test: Definite volume water are placed in lactose broth and
of diluted water sample is taken in then incubated at 200C for 24 hrs.
fermentation tubes containing lactose Release of CO2 or Acid indicates
both as culture medium is kept in presence of B.coli.
incubator at 200C for 24 – 48 hrs. If Using statistics and standard charts
some gas is produced in the (provided by Mc. Cardy) MPN is
fermentation tube, it indicates the determined.
presence of B.coli, if not absence of
Allowable MPN: 1 per 100ml.
B.coli.
5) E.coli Index: The reciprocal of
b) Confirmative Test: Some sample
least positive dilution of confirmative
from presumptive test is taken and
test of MPN expressed as no. of
placed in another fermentation tube
bacteria per 100 ml is called as E.coli
with brilliant green lactose bile as
Index.
culture medium. It is kept again in
incubator at 200C for 48 hrs. If there Allowable Limit: 1 per 100ml.
is formation of gas in the tube, it Water Borne Diseases:
confirm the presence of B.coli and the
Diseases that spread through the
water is unsafe for use.
contaminated water by pathogenic
3) Membrance Filter Technique: A microbes are termed as Water Borne
simple and new method for B.coli. A Diseases.
sample of water is passed through
special bacteriological filter containing

Page 20
FUTURE GATE ACADEMY EE

Few important among them are as a) they are pathogenic causing

follows: intestinal diseases
b) their presence indicates viral
Bacterial Diseases: contamination of water
Cholera Vibrio cholera c) they are used as indicator
Typhoid Salmonella typhi organisms for probable presence
of pathogens
Paratyphoid Salmonella paratyphi
d) they represent unique indicator
Dysentery Schigella dysenteriae organism for sewage pollution
Diarrohea Excess E.coli / MPN 3. Breakpoint chlorination of water
Viral Diseases: involves addition of chlorine in an
amount sufficient to (GATE 95)
Polio Polio myelitis
(a) react with ammonia and readily
Jaundice Hepatitis B oxidizable organic matter
Gastroenteritis Gastric influenza (b) kill giardia cysts
Protozoan Diseases: (c) react with inorganic matter
Amoebiasis Entamoeba (d) reduce bacterial growth in
histolyitca filters
Giardiasis Giardia lamblia 4. The most important water quality
parameter for domestic use of
Entrozal Diseases:
water is (GATE - 1996)
Diseases caused due to physical
(a) CH (b) NCH
contact with contaminated water.
(c) Coliform group of organisms
Ex: Hay fever, Yellow fever, Gunea
worm and Tape worm infections etc., (d) Chlorides
5. MPN index is measure of one of the
PREVIOUS QUESTIONS following: (GATE 1997)
1. The pH of water admitted into a (a) Coli form bacteria (b) BOD5
treatment plant was 6.0 in the
(c) DO content (d) Hardness
morning. Consequent to inflow of
raw water from a different source, 6. Excessive fluoride in drinking
it changed to 8.0 in the next 24 water causes (GATE 1998)
hours. Assuming linear variation (a) Alzheimer’s disease
in time of the hydrogen ion
(b) Mottling of teeth and
concentration, the time mean pH
brittlement of bones
value of the water over this 24
hour period is …… (GATE1991) (c) Methemoglobinemia

2. Bacteriological examination of (d) Skin cancer

drinking water for Coli forms is
performed because (GATE1993)

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FUTURE GATE ACADEMY EE

7. Temporary hardness in water is (a) Turbidity followed by

caused by the presence of (1999) disinfection
(a) Bicarbonates of Ca and Mg (b) Fluorides and Hardness
(b) Sulphates of Ca and Mg (c) Iron, followed by disinfection
(c) Chlorides of Ca and Mg (d) Both (b) and (c)
(d) Nitrates of Ca and Mg 13. Results of a water sample
8. Blue baby disease in children is analysis are as follows: (2003)
caused by the presence of excess Cation Conc. (mg/L) Eq. Weight
(a) Chlorides (b) Nitrates
Na+ 40 23
(c) Fluoride (d) Lead
Mg+2 10 12.2
9. The minimum dissolved oxygen
content (ppm) in a river necessary Ca+2 55 20
for the survival of aquatic life is K+ 2 39
(a) 0 (b) 2 (c) 4 (d) 8 Hardness of the water sample in mg/L
10. The Ca2+
concentration and as CaCO3 is
Mg2+ concentration of a water (a) 44.8 (b) 89.5 (c) 179 (d) 358
sample are 160mg/lit and 40
mg/lit as their ions respectively.
14. Hardness of water is directly
measured by titration with
The total hardness of this water
ethylene-di-amine-tetracetic acid
sample in terms of CaCo2 in mg/lit
(EDTA) using (GATE 2004)
is a approximately equal to (2001)
(a) eriochrome black T indicator
(a) 120 (b) 200 (c) 267 (d) 567
(b) ferroin indicator
11. Aeration of water is to remove
(c) methyl orange indicator
(a) Suspended impurities
(d) phenolphthalein indicator
(b) Color (c) Dissolved Salts
(d) Dissolved gases
15. The organism, which exhibits
very nearly the characteristics of
12. The results of analysis of water an ideal pathogenic indicator is
sample are given below (2003)
(a) Entamoeba histolytica
Turbidity 5 mg/l
(b) Escherichia coli
pH 7.4
(c) Salmonella typhi
Fluorides 2.5 mg/l (d) Vibrio comma
Total Hardness 300 mg/l 16. A standard multiple-tube
Iron 3.0 mg/l fermentation test was conducted
on a sample of water from a
MPN 50 per 100 ml
surface stream. The results of the
From the data given above, it can be analysis of the confirmed test are
inferred that water needs removal of given below. (GATE-2004)

Page 22
FUTURE GATE ACADEMY EE

No. of No. of R. Bacterial concentration

Sample
positive negative S. Coagulant dose
size (ml)
results results
Group – II
1.0 4 1
0.1 3 2 1. BOD 2. MPN
0.01 1 4 3. Jar test 4. Turbidity
MPN Index and 95% confidence Codes: P Q R S
limits for combination of positive (a) 2 1 4 3
(b) 4 1 2 3
Results when five tubes used per (c) 2 4 1 3
dilutions (10 ml, 0.1 ml, 0.1 ml) (d) 4 2 1 3
21. The presence of hardness in
Combina MPN 95%
tion of Index per confidence excess of permissible limit causes
Positive 100 ml limit
(a) Cardio Vascular problems
4-2-1 26 12 65 (b) Skin discolouration
4-3-1 33 15 77 (c) Calcium deficiency
Using the above MPN Index table, the (d) Increased laundry expenses
most probable number (MPN) of the 22. The alkalinity and the hardness
sample is of a water sample are 250 mg/L
(a) 26 (b) 33 (c) 260 (d) 330 and 350 mg/L as CaCo3,
17. Most of the turbidity meters work respectively. The water has (2007)
on the scattering principles. The (a) 350 mg/L CH and Zero NCH
turbidity value so obtained is (b) 250 mg/L CH and zero NCH
expressed in (GATE-2004) (c) 250 mg/L CH & 350 mg/L NCH
(a) CFU (b) FTU (c) JTU (d) NTU (d) 250 mg/L CH & 100 mg/L NCH

18. Total Kjeldahl nitrogen is a 23. Match the estimation method

measure of (GATE-2005 & 15) (Group-I) with the corresponding
indicator (Group-II). (GATE-2008)
(a) total organic nitrogen
(b) total organic and ammonia N2 Group – I
(c) total ammonia nitrogen P. Azide modified Winkler method
(d) total inorganic & ammonia N2 for dissolved oxygen
19. If tomato juice is having a pH of Q. Dichromate method for COD
4.1, the hydrogen ion conc. In R. EDTA Titrimetric method for
mol/L will be (GATE-2005) Hardness
S. Mohr or Argentometric method
(a) 10.94 x 10-5 (b) 9.94 x 10-5
for Chlorides
(c) 8.94 x 10-5 (d) 7.94 x 10-5
Group – II
20. Match the property with
corresponding test. (GATE-2005) 1. Eriochrome Black T
2. Ferrion
Group – I
3. Potassium chromate
P. Suspended solids concentration
4. Starch
Q. Metabolism of biodegradable
organics
Page 23
FUTURE GATE ACADEMY EE

(a) P – 3, Q – 2, R – 1, S – 4 Group – I Group – II

(b) P – 4, Q – 2, R – 1, S – 3 P. Alkalinity 1. N/35.5 AgNO3
(c) P – 4, Q – 1, R – 2, S – 3 Q. Hardness 2. N/40 Na2S2O3
(d) P – 4, Q – 2, R – 3, S – 1 R. Chloride 3. N/50 H2SO4
24. A wastewater sample contains S. DO 4. N/50 EDTA
10-5.6
mmol/L of OH- ions at 250C. Codes: P Q R S
The pH of this sample is (2008) (a) 1 2 3 4
(a) 8.6 (b) 8.4 (c) 5.6 (d) 5.4 (b) 3 4 1 2
(c) 2 1 4 3
Common data for Questions 25 &26
(d) 4 3 2 1
Ion Conc. (mg/L) Atomic Wt.
29. Some of the nontoxic metals
Ca2+ 100 40
normally found in natural water are
Mg2+ 6 24
(a) arsenic, lead and mercury
Na+ 15 23
(b) calcium, sodium and silver
HCO3 250 H=1, C=12,
O=16 (c) cadmium, chromium & copper
SO42- 45 S=32, O=16 (d) iron, manganese & magnesium
Cl 39 Cl=35.3 30. For a sample of water with
ionic composition shown in the
25. Total hardiness (mg/L as
figure below,
CaCO3) present in the above water
sample is (GATE-2010)
(a) 205 (b) 250 (c) 275 (d) 308
26. Carbonate hardness (mg/L as
CaCO3) present in the above water
sample is (GATE-2010)
the carbonate and non-carbonate
(a) 205 (b) 250 (c) 275 (d) 289
hardness (in mg/L as CaCO3),
27. Anaerobically treated effluent respectively are: (2014)
has MPN of total coli form as
(a) 200 and 50 (b) 175 and 75
106/100mL. After chlorination, the
MPN value declines to 102/100mL. (c) 75 and 175 (d) 50 and 200
The percent removal (%R) and log 31. A ground water sample was
removal (log R) of total coli form found to contain 500 mg/L total
MPN is (GATE-2011) dissolved solids [TDS]. TDS [in % ]
(a) %R = 99.90; log R = 4 present in the sample is ___ (2015)
(b) %R = 99.90; log R = 2 32. Electrical conductivity (EC) of
(c) %R = 99.99; log R = 4 water and total dissolved solids
(d) %R = 99.99; log R = 2 (TDS) are interrelated. The value of
28. The correct match of water EC will be (IES-1995)
quality parameters in Group–I with (a) decrease with increase in TDS
titrants in Group–II is: (2013) (b) increase with increase in TDS

Page 24
FUTURE GATE ACADEMY EE

(c) decrease initially and then 38. A sample of ground water at a pH

increase with increase in TDS of 7.0 contains 122 mg/L of
(d) increase initially and then bicarbonates. What is the alkalinity of
decrease with increase in TDS this water (in terms of CaCO3)? 2005
33. Which one of the following (a) 120 mg/L (b) 60 mg/L
would contain water with the (c) 100 mg/L (d) 200 mg/L
maximum amount of turbidity? 39. Match List-I (Equipment) with
(a) Lakes (b) Oceans List-II (Parameter) and select the
correct answer (IES-2006)
(c) Rivers (d) Wells
List-I List-II
34. Zero hardness of water is A. Trintometer 1. Temperature
achieved by (IES -1997) B. Nephelometer 2. Colour
(a) using lime soda process C. Imhoff cone 3. Turbidity
(b) excess lime treatment D. Muffle furnace 4. Setleable solids
(c) ion exchange method
5. Volatile solids
(d) using excess alum dosage
Codes : A B C D
35. Match List-I with List-II and (a) 4 3 1 5
select the correct answer (1997) (b) 2 5 4 3
List-I List-II (c) 4 5 1 3
A. Hardness 1. 0.1 mg/L (d) 2 3 4 5
B. Nitrate conc. 2. 0.5 mg/L 40. What is the most common cause
C. Iron conc. 3. 200 mg/L of acidity in water? (IES-2006)
D. Fluoride conc. 4. 45 mg/L (a) Carbon monoxide
Codes : A B C D (b) Nitrogen (c) Hydrogen
(a) 3 4 2 1
(d) Carbon dioxide
(b) 3 4 1 2
(c) 4 3 2 1 41. If total hardness and alkalinity of
(d) 4 3 1 2 a water sample are 200 mg/L as
CaCO3 and 260 mg/L as CaCO3
36. Which of the following are the respectively, what are the values of
characteristic(s) of E.coli carbonate hardness and
1. Bacillus 2.Gram-negative non-carbonate hardness? (2006)
3. Ferments Lactose 4.Spore forming (a) 200 mg/L and zero
Select the correct answer using the (b) Zero and 60 mg/L
codes given below :
(c) Zero and 200 mg/L
(a) 1 alone (b) 1, 2 and 4
(c) 1, 2 and 3 (d) 2, 3 and 4 (d) 60 mg/L and zero
42. Hardness to water is caused by
37. What is the equivalent calcium
the presence of calcium (Ca2+) and
carbonate concentration of 110
magnesium (Mg2+) ions. Which are
mg/L of CaCI2?
the least soluble forms of calcium
(a) 50 mg/L (b) 58.5 mg/L and magnesium at normal water
(c) 100 mg/L (d) 117 mg/L temperature? (IES-2006)

Page 25
FUTURE GATE ACADEMY EE

(a) CaCI2 and MgCO3 C. Odour 3. JTU

(b) Ca(HCO3)2 and MgCI2 D. Colour 4. MPN
Codes : A B C D
(c) Ca(OH)2 and Mg(HCO3)2
(a) 2 1 4 3
(d) CaCO3 and Mg(OH)2
(b) 3 1 4 2
43. Match List-I (Pathogen) with List-II (c) 2 4 1 3
(Epidemic) and select the correct (d) 3 4 1 2
answer (IES-2007)
47. Assertion (A) : Fluorides should
List-I List-II always be present in drinking
A. Bacteria 1. Gastroenteritis water upto a value 1.5 mg/l
B. Virus 2. Cholera
Reason (R) : Such a water helps
C. Protozoa 3. Worms
clean the teeth well
D. Helminth 4. Polio
(a) both A and R are true and R
Codes : A B C D
is the correct explanation of A
(a) 2 4 1 3
(b) both A and R are true but R
(b) 3 1 4 2
is not a correct explanation of A
(c) 2 1 4 3
(c) A is true but R is false
(d) 3 4 1 2
(d) A is false but R is true
44. Match List-I (Type of impurity) with
48. The maximum safe permissible
List-II (Harm caused) and select
limit of sulphates in domestic
the correct answer. (IES-2009)
water supply is (IES-2012)
List-I List-II
(a) 100 mg/L (b) 200mg/L
A. Excess nitrates 1. Brackish H2O
B. Excess fluorides 2. Goiter (c) 500 mg/L (d) 600mg/L
C. Lack of iodides 3. Fragile bones 49. 1 TCU is equivalent to the colour
D. Excess chlorides 4. Blue babies produced by (GATE-05)
Codes : A B C D (a) 1 mg/l of chloroplatinate ion
(a) 4 2 3 1 (b) 1 mg/l of platinum ion
(b) 1 2 3 4 (c) 1 mg/l of platinum in the
(c) 4 3 2 1 form of chloroplatinate ion
(d) 1 3 2 4 (d)1mg/l organo - chloroplatinate
45. The concentration of OH- ion in a
water sample is measured as 17
mg/L at 250C. What is the pH of
the water sample? (IES-2009)
(a) 10 (b) 11 (c) 12 (d) 13
46. Match List-I (Parameters) with List-
II (Units) and select the correct
answer (IES-2009)
List-I List-II
A. Turbidity 1. TON
B. Pathogen 2. TCU

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FUTURE GATE ACADEMY EE

PRACTISE QUESTIONS 6. Find out the pH of the following

1. A sample of water is analysed for the mixture:
coli form group using three sample Solution Volume (ml) pH
portions: 10mL, 60mL and 600mL. A 500 6
Each of these portions is filtered
B 500 5
through five filter membranes using
the membrane filter technique. The 7. The results obtained from a sample
results of the colony counts are of water are as follows (in mg/L):
given in tabular column. What is the
Na = 20 Sr = 2 Mg = 11
no. of coliforms per 100ml of the
sample? K = 30 SO4 = 5 Cl = 40
Ca = 6 HCO3 = 72 OH = 10
10 mL 60 mL 600 mL
6 30 350 Calculate total hardness (in mg/L)?
7 32 340 (A) 48 (B) 50 (C) 52 (D) 62
5 33 360 8. A factory discharges 50m3/day of
8 31 370 waste having pH = 11. If the waste
contains KOH only, find out the
6 25 340
quantity of KOH in kg/day?
(A) 50 (B) 30 (C) 100 (D) 60 (A) 2.80 (B) 3.4 (C) 3.6 (D) 4.2
2. TH is 240mg/L and TA is 120mg/L, 9. From the complete analysis of
determine CH water sample, it was found that the
(A) 240mg/L (B)1200mg/L total hardness value is 119 mg L .
(C) 120mg/L (D) 40mg/L The analysis further shows that,
3. 20 ml of 0.02N NaOH solution is the concentration of all the two
consumed to find total acidity in principal cations causing hardness
testing 100ml of water sample. Find are numerically the same.
the total acidity in mg/L as CaCO3? Determine the concentrations of
principal cations present in water
(A) 200mg/L (B) 100mg/L
(in mg/L)?
(C) 400mg/L (D) 20mg/L
(A) 15 (B) 16 (C) 17 (D) 18
4. pH value of a water sample is 8.0.
10. Match the following:
Determine OH- ion concentration in
mg/L? Group – I Group – II
(A) 17 (B) 17×103 (A) Hardness (1) Sodium Thiosulphate
(C) 17×10-3 (D) 0.17 (B) Chlorides (2) Banium Ions
5. The concentration of carbonate ions (C) Chlorine (3) EBT
is 180mg/L and Biocarbonate ions is
(D) Sulphates (4) Potassium chromate
122 mg/L. Determine the Alkalinity of
water sample? (A) 4321 (B) 3412
(A) 100mg/L (B) 200mg/L
(C) 3421 (D) 1234
(C) 300mg/L (D) 400mg/L

Page 27
FUTURE GATE ACADEMY EE

11. The results of standard multiple contains KOH only, the quantity of
tube fermentation are given the KOH in kg/day
below table. Calculate the E.coli (a) 2.80 kg/day (b) 1.68 kg/day
Index value per L of sample based
(c) 1.24 kg/day (d)0.096 kg/day
on results
Dilution Result 19. The pH of 1000 mg/L of Ca(OH)2

10 + (a) 12.4 (b) 10.8 (c) 7.4 (d) 6.19

1 + 20. The pH of the following mixture
0.1 + Volume pH
0.01 +
Sol. A 400 ml 5
0.001 -
Sol. B 600 ml 4
.0001 -
(a) 5.04 (b) 5.40 (c) 5.61 (d) 6.2
(A) 100 (B) 10 (C)1000 (D) 1
21. A water sample with pH 9 had a
12. If NaCl concentration found in caustic alkalinity of 70 mg/l, total
sample is 2 × 10-3 moles/lt., alkalinity of 230 mg/l and total
Determine its concentration in hardness of 300 mg/l, all as
mg/L as CaCO3? CaCO3. Calculate the amounts of
(A) 117 (B) 100 (C)110 (D) 200 the various forms of alkalinity
present and the amount of non-
13. Dissolved impurities in water
carbonate hardness.
consists of
22. If the hydrogen concentration is
(A) Bacteria (B) Iron
3 x 10-2 mol/l, calculate the
(C) Silt (D) Algae hydroxyl ion concentration.
14. The turbidity in water which can 23. What would be the pH of a
be seen easily by naked eye is solution containing 1.70 x 10-8 g of
(A) 2 JTU (B) 3 JTU hydroxide per L?
(C) 4 JTU (D) 5 JTU 24. Find out the pH of the mixture
15. If H+ ion concentration is 10-9 which will be formed by mixing the
following two volumes:
mol/lit. What is pOH
Volume pH
(a) 9 (b) 8 (c) 6 (d) 5
16. Two samples of water A and B Solution A .. 100 ml 6
have pH values of 5.6 and 6.5 then Solution B .. 900 ml 5
the average pH value is 25. There are three samples X, Y and
(a) 6.05 (b) 5.85 (c)4.13 (d) 3.16 Z of water having pH values of 4.5,
5.5 and 6.5 respectively. Calculate
17. The pH and pOH values of freshly
how many times X is acidic than Z.
prepared distilled water is
(a) 5, 9 (b) 6, 8 (c) 7, 7 (d) 8, 6 26. The Ca2+ concentration and Mg2+
concentration of a water sample
18. A factory discharges 30 m3/day of are 160 mg/l and 40mg/l as their
waste having pH = 11. If the water
Page 28
FUTURE GATE ACADEMY EE

ions respectively. The total that the concentrations of all the

hardness of this water sample in three principal cations causing
terms of CaCO3 in mg/L is hardness are numerically the
approximately equal to same. If the value of C.H. is 58
(a) 120 (b) 200 (c) 267 (d) 567 mg/l, calculate the following:

27. If for diluting 25 ml of water (1) the value of N.C.H.;

sample 175 ml of taste free water (2) the concentrations of principal
is required to be added to make cations; and
the water sample to just loose its (3) the value of total alkalinity in
taste, then the flavor threshold mg/l.
number (FTN) will be
32. The analysis of water from a bore
(a) 6 (b) 7 (c) 8 (d) 9 shows the following results in
28. The product of H+ ions and OH- mg/l: Ca = 60, Mg = 48,
ions in a stronger acid is Na = 103.5, K = 19.5
(a) 0 (b) 1 (c) 10-7 (d) 10-14 HCO3 = 244, SO4= 220.8
29. The analysis of a sample of water Find out the total hardness,
shows the following results in carbonate hardness and non-
mg/l: Na = 20 Cl = 40 carbonate hardness.
K = 30 HCO3 = 67 Ca = 5 33. How many gms of calcium will be
SO4 = 5 Mg = 10 NO3 = 10 required to combine with 90 gms
The concentration of Strantium of carbonate to form calcium
(Sr) is equivalent to a hardness of carbonate
2.29 mg/l and the carbonate (a) 20 (b) 60 (c) 90 (d) 120
alkalinity in this water is zero. 34. A 12.5 mL sample of treated waste
Calculate the total hardness, water requires 187.5 mL of odour-
carbonate hardness and non- free distilled water to reduce the
carbonate hardness in mg/l as odour to a level that is just
CaCO3. perceptible. What is the Threshold
30. The results obtained from a Odour Number (TON) for the
sample of water are as follows in wastewater sample?
mg/l: Na = 20 Sr = 2 35. The analysis of water from a bore
K = 30 Cl = 72 Ca = 6 shows the following results in
HCO3 = 67 Mg = 11 mg/L : Ca = 60, HCO3 = 244,
Mg = 48, SO4 = 220.8, Na = 103.5,
Find out T.H., C.H and N.C.H in
Cl = 78.1, K = 19.5 and pH is 7.5.
mg/l as CaCO3.
The total hardness of water is
31. The total hardness value obtained
from the complete analysis of a (a) 596.83 mg/L (b) 486.91 mg/L
water sample is found to be 116 (c) 346.72 mg/L (d) 212 mg/L
mg/l. The analysis further shows

Page 29
FUTURE GATE ACADEMY EE

36. For the above data carbonate and 39. What is the equivalent CaCO3
non carbonate hardness is concentration of 4 x 10-3 mol/L of
(a) 100 & 146.72 mg/lit NaCl in mg/L

(b) 200 & 146.72 mg/lit (a) 100 b) 200 c) 500 d) 1000

(c) 100 & 169.78 mg/lit 39. What is the equivalent CaCO3
concentration of 220mg/L of
(d) 200 & 169.78 mg/lit
Calcium chloride in mg/L
37. The total hardness value obtained
(a) 100 b) 20 c) 200 d) 1000
from the complete analysis of a
water sample is found to be 116
mg/L. The analysis further shows
that the concentrations of all the
three principal cations causing
hardness are numerically same. If
the value of carbonate hardness is
58 mg/L. What is the value of
total alkalinity is mg/L.
(a) 58 (b) 64 (c) 71 (d) 103
38. A standard multiple tube
fermentation test was conducted
on a sample as water. The results
of the analysis for the confirmed
test are given below.

Sample Positive Negative

size (ml) results results
10 4 1
1 2 3
0.1 1 4
0.01 0 5
MPN index for combination of
positive results when 5 tubes used
per dilution (10ml, 1.0ml, 0.1ml)
Combination of MPN Index per
positives 100 ml
5-4-3 280
4-3-1 33
4-2-1 26
2-1-0 7

Using the above MPN index table,

what is the most probable number
(MPN) of the sample?
(a) 280 (b) 33 (c) 26 (d) 70
Page 30
FUTURE GATE ACADEMY EE

KEY – Previous Questions KEY – Practise Questions

1. 6.29 2.C 3.A 4.C 5.A

1. B 2.C 3.A 4.C 5.D
6. B 7.A 8.B 9.C 10.D
11.D 12.D 13.C 14.A 15.B 6. 5.25 7.D 8.A 9.D 10.B

16.D 17.B 18.B 19.D 20.B

11. A 12.B 13.B 14.D 15.D
21.D 22.D 23.B 24.D 25.C
26.A 27.C 28.B 29.D 30.B 16. B 17. C 18. B 19.A 20.C
21. CA: 140 mg/lt, BCA : 90 mg/lt,
31. 0.05 32.B 33.C 34.C
CH: 230 mg/lt, NCH: 70 mg/lt
35.B 36.C 37.C 38.C 39.D
40.D 41.A 42.D 43.A 44.C 22. 1/3 x 10-12 mol/lt
45.B 46.D 47.A 48.B 49.C
23. 5 24. 5.041 25. 100

26. D 27. C 28. D

29. TH : 55.77, CH: 54.92, NCH: 0.85

30. TH : 62.36, CH: 59.02, NCH: 3.34

31. CH & NCH: 58 mg/lt Cations : 15

32. TH : 346, CH: 200, NCH: 146

33. B 34. 16 35. C

36. B 37. A 38. 70

39. B 40. C

Page 31
FUTURE GATE ACADEMY EE

TREATMENT OF WATER
Treatment of water is essentially PREVIOUS QUESTIONS
removal process in which removal of
impurities is achieved by Unit 1. Consider the following unit
Operations (applying physical forces processes commonly used in water
like gravity) and Unit Processes treatment; rapid mixing (RM),
(removal by chemical (or) biological flocculation (F), primary
forces).
sedimentation (PS), secondary
Influent: The incoming raw water into sedimentation (SS), chlorination (C)
treatment unit with impurities. and rapid sand filtration (RSF).
Effluent: The outgoing treated water The order of these unit processes
from treatment plant. (first to last) in conventional water
treatment plants is (GATE-2011)
The method of water treatment
directly depends on the impurities (a)PS → RSF → F → RM → SS → C
present in water. (b)PS → F → RM → RSF → SS → C
(c) PS → F → SS → RSF → RM → C
Type of Impurity Treatment Method (d)PS → RM → F → SS → RSF → C

Floating matters Screening 2. The potable water is prepared from

as leaves, dead turbid surface water by adopting
animals etc., the following treatment sequence
(GATE -2014)
Suspended Plain (a) Turbid surface water →
impurities Sedimentation Coagulation → Flocculation →
Sedimentation → Filtration →
Fine Suspended Coagulation+ Disinfection → Storage & Supply
Matter+ Colloids Sedimentation (b) Turbid surface water →
Disinfection → Flocculation →
Sedimentation Filtration →
Microorganisms Filtration Coagulation → Storage & supply
& Dissolved (c) Turbid surface water →
Solids Filtration → Sedimentation →
Dissolved Gases, Aeration Disinfection → Flocculation →
Taste & Odour Coagulation → Storage & supply
(d) Turbid surface water →
Hardness Softening Sedimentation → Flocculation →
Coagulation → Disinfection
Pathogenic Disinfection Filtration → Storage & supply
Bacteria
3. Match List-I (Type of water source)
Salts, Minerals, Demineralization with List-II (Treatment to be given)
Metals etc., / Desalinization and select the correct answer
using the codes given below the
Fluorides Deflouridization lists : (IES-1995)
List-I
A. Surface water (river or canal)
Page 32
FUTURE GATE ACADEMY EE

B. Water from infiltration gallery 6. Which of the following statements

C. Lake/pond water are correct? (IES-2001)
D. Tube well water 1. Groundwater is generally free
from suspended and dissolved
List-II
impurities
1. Aeration, coagulation,
2. Suspended matters often
sedimentation and disinfection
contain pathogenic bacteria
2. Disinfection
3. Rain water is soft and tasteless
3. CuSO4
4. Lake water may contain
4. Coagulation, flocculation, microscopic organisms
sedimentation, filtration and
Select the correct answer using the
disinfection
codes given below :
Codes: A B C D
(a) 1, 2, 3 and 4 (b) 1 and 2
(a) 4 1 3 2
(c) 2, 3 and 4 (d) 1, 3 and 4
(b) 1 4 3 2
(c) 1 4 2 3 7. Consider the following impurities :
(d) 4 1 2 3 1. CO2 and H2S
2.Finely-divided suspended matter
4. A river is the source of water for
water supply to a town. Its water is 3. Disease causing bacteria
very turbid and polluted. The 4. Excess alkalinity
correct sequence of steps for The correct sequence of the
treating the river water would be removal of these impurities in a
(a) Pre-sedimentation  pre- water treatment plant is (IES2003)
chlorination  coagulation  (a) 1, 2, 3, 4 (b) 1, 4, 3, 2
sedimentation  filtration  post (c) 1, 4, 2, 3 (d) 4, 1, 3, 2
chlorination
8. Which of the following treatment
(b) Coagulation  sedimentation
processes are necessary for
 post-chlorination
removing suspended solids from
(c) Coagulation  filtration  water ? (IES 2003)
sedimentation  post- 1. Coagulation 2. Flocculation
chlorination 3. Sedimentation 4. Disinfection
(d) Sedimentation  post- Select the correct answer:
chlorination (a) 1 and 2 (b) 1, 2 and 3
5. Match List-I (Nature of the solids) (c) 2 and 4 (d) 1 and 4
with List-II (process for its
removal): (IES-1998) 9. Dissolved gases are removed by the
process of
List-I List-II
(a) Reverse osmosis
A. Dissolved solids 1. Sedimentation
(b) CuSO4 (c) Coagulation
B. Colloidal solids 2.Reverse osmosis
(d) Aeration
C. Volatile solids 3. Coagulation
D. Settleable solids 4. Digestion
KEY – Previous Questions
Codes :A B C D
(a) 2 3 4 1
1. D 2.A 3.A 4.A 5.A
(b) 3 2 4 1
(c) 2 3 1 4
6. D 7.C 8.B 9.D
(d) 3 2 1 4

Page 33
FUTURE GATE ACADEMY EE

5. SEDIMENTATION
The physical unit operation for the TYPES OF SETTLING TANKS:
removal of suspended coarser
 Grit chamber: For removal of
particles heavier than water by
sand, grits, etc.
virtue of their self weight and
density due to gravity is called as  Plain sedimentation tank: For
Sedimentation. removal of discrete suspended
solids and settleable solids.
 The terms sedimentation,
settling, and clarification are  Chemical precipitation tank:
used interchangeably. for removal of very fine
suspended particles by adding
 The unit sedimentation basin
coagulants,
may also be referred to as a
sedimentation tank, clarifier,  Septic tanks: For doing
settling basin, or settling tank. sedimentation and sludge
digestion together in
 Types of Settling: Based on the
households
solids concentration and the
tendency of particle interaction,  Secondary settling tanks:
there are four types of settling After activated sludge or
which may occur in water and trickling filter treatment
wastewater settling operations. systems.
They are as follows:
PLAIN SEDIMENTATION
Settling Purpose of Sedimentation  Any particle which does not
alter its size, shape & weight
Type – I Plain (or) Primary
Sedimentation for separation of while rising or settling in any
Discrete
Settling
suspended solids during water fluid is called Discrete Particle.
treatment and Removal of Sand
and Grit in waste water treatment.  The process of settling of
discrete particle in dilute
Type – II Secondary Sedimentation
carried out for separation of suspension by gravity is called
Flocculent
Settling
flocculent particle formed after Plain Sedimentation.
chemical coagulation during water
treatment. PRINCIPLE OF SEDIMENTATION

Type – III Settling that occurs in sludge Suspended particles exist in water
Hindered /
thickeners and at the bottom of due to turbulence and flow velocity
Secondary clarifiers in
Zone Biological Treatment processes. even though they have more
Settling
density than water. The process of
When the concentration of
Type – V particles is high enough to bring plain sedimentation controls
Compression the particles into physical contact turbulence and provides settling
Settling with each other, compression
settling will occur. Consolidation velocity of particle more than flow
of sediment at the bottom of the velocity of water to achieve settling
clarifier is extremely slow.
by virtue of their mass.

Page 34
FUTURE GATE ACADEMY EE

SETTLING VELOCITY: C) Hazen’s Formula : As 

depends upon temperature (T), Hazen
The uniform velocity with which
proposed a new equation for Settling
the discrete particles settle down
Velocity
through a quiescent fluid is called
as Settling Velocity. Settling  3T  70 
Vs = 418 (S – 1) d2 
velocity depends upon flow velocity  100 
and viscosity of water, size, shape
and specific gravity of particle. Where Vs = Settling Velocity in
mm/sec
d = Diameter of particle (mm)
T = Temperature (0C)
2) For “d” between 0.1mm and
1mm (more than 0.1mm & less
than 1mm):
Determination of Settling For particle diameter is greater
Velocity: It is based on Diameter than 0.1 mm, it is difficult to
of the particle maintain laminar flow conditions.
1) For d < 0.1mm (for Re<1):  3T  70 
Vs = 418 [S – 1]d 
Laminar conditions occurs and  100 
stoke’s law is valid
3) For d > 0.1mm (for Re>1000):
(when CD = 24/Re) Turbulent conditions occurs and
1 g 2 stoke’s law is not valid
A) Vs = . d (ρp – ρw)
18  (when CD = 0.4)
1 g Vs = 1.8 g (S  1)d
B) Vs = . d2 (S – 1)
18 
SEDIMENTATION TANK:
Where Vs = Settling Velocity
in m/sec g = 9.81m/s The structure which performs the
settling operation is considered as
ρp = Density of Particle
Sedimentation tank (or) Clarifier.
ρw = Density of Water
Based on mode of operation,
S = Specific Gravity settling chambers are classified
into 2 types.
d = Dia of particle (m)
1) Batch (or) Quiescent (or) Draw
μ = Dynamic viscosity
and Fill Type Settling Tank:
of water (N-s/m2)
The settling chamber which
 = Kinematic viscosity
functions intermittently and in
(m2/s)
which water is completely brought
to rest for duration of settling (30-
Page 35
FUTURE GATE ACADEMY EE

60 hrs, generally 24 hrs) is circular tanks with radial or

considered as Quiescent Settling spiral flow.
Chamber.
 After each round of
sedimentation, emptying of
tank and cleaning of sediment
is to be performed.
 It is rectangular in plan
 It needs more detention period,
labor and supervision.

Settling chamber consists of 4

zones:
A) Inlet Zone: Region in which the
flow is uniformly distributed
over the cross section such that
the flow through settling zone
follows horizontal path.
B) Settling Zone: Settling occurs
under quiescent conditions
C) Sludge Zone: For collection of
2) Continuous flow type tank:
sludge below settling zone.
The settling chamber in which D) settled material collects and is
water travels from inlet to outlet moved towards sludge hoppers
continuously and particles are for withdrawal. It is assumed
settled by gravity is considered as that once a particle reaches the
Continuous flow type settling tank. sludge zone it is effectively
This is the regular method of removed from the flow.
sedimentation used. E) Outlet Zone: Clarified effluent is
Generally 3 types of Continuous collected and discharge through
flow type sedimentation tanks outlet weir. Outlet weirs or
are available based on shape: submerged orifices shall be
Rectangular, Circular and designed to maintain velocities
Square. suitable for settling in the basin
and to minimize short-
Long narrow rectangular tanks circuiting.
with horizontal flow are
generally preferred to the

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FUTURE GATE ACADEMY EE

DESIGN OF CONTINUOUS FLOW Q

 Vo = → Rectangular Tank
SEDIMENTATION TANK: LB
The continuous flow type 4Q
 Vo = → for Circular tank
sedimentation tank is designed for d 2
Laminar flow conditions in such a
 If Vs > Vo : particles are
way that that the time taken by the
entrapped
particle to settle down is lesser
than the time taken by water to  If Vs < Vo : particles are escape
travel from inlet to outlet. It is from settling chamber
achieved by providing sufficient  So, sedimentation efficiency,
length of travel to reduce flow Vsi
velocity of water for efficient  100
V0
settling of particle.
2. Detention Time (Dt):
The suspended particle in water
experiences drag force caused by The average time taken by the
flow velocity of water in horizontal water to travel from inlet to
direction and also vertically outlet of settling chamber i.e.,
downward gravitational force due theoretical average time spent
to settling velocity. As a result, the by water in sedimentation tank
particle travels in parabolic is termed as Hydraulic
(inclined) path and settles down. Detention Time.
DESIGN CONCEPTS: Detention Time (DT) =
Volume of Settling Tank
1. Over Flow Rate [Vo] :
Q
 The volume of water loaded per
unit surface area of settling V LBH
 Rect. tank → Dt = =
tank in unit time duration is Q Q
called as over flow rate.
V
 For circular tank → Dt =
 It is also considered as Surface Q
loading rate (or) Surface over
flow rate (or) Hydraulic Loading d 2 (0.011d  0.785H )
=
Rate (or) Overflow Velocity. Q

 PST : V0 = 500 – 750 lit/hr-m2  For PST: Dt → 4 – 8 hrs

(or) 12 – 18 m3/day-m2  For SST: Dt → 2 – 4 hrs
 SST : V0 = 1000 – 1250 lit/hr-  Flow through period: The actual
m2 (or) 24 – 30 m3/day-m2 time taken by the water to
Flow Rate (m3 / s) travel from inlet to outlet of
 Overflow Rate  settling camber. It is always
settling surface area (m 2 )
less than detention period due
to short circuiting.

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FUTURE GATE ACADEMY EE

 Displacement Efficiency  Total amount of flow from tank

= Maximum Daily Demand with
Flow through period
= ×100 in 24 hrs.
Detention Time
 Volume of Settling Tank: V =
 It ranges between 25 – 50%
Q * Dt
3. Flow through velocity (or)
 Volume of Circular Tank: V =
Horizontal flow velocity (VH):
D2 [0.011D + 0.785h]
The speed with which water
 Particle removal is independent
travels from inlet to outlet of
of depth but dependent on
sedimentation tank is
surface area of tank.
considered as Horizontal flow
velocity. H L
 If  → 100% removal.
L
Vs VH
A) VH =
Dt  If detention time is considered
Q
B) VH = L H

BH VH Vs
VH = 0.15 – 0.9 m/min →
H .VH
Normal Range Then, Vs =
L
Best VH : 0.3 m/min
 Overall efficiency of
H sedimentation,
If Vs is given Dt =
Vs
ηoverall = Σ Pi.ηi
4. Weir Loading Rate [WLR]:
Where, Pi → % of ith
The volume of water flow particle removal
through unit length of the weir
ηi → Removal efficiency of ith
in unit duration is called as
particle
Weir loading rate.
 It is 60 – 70%
Q Q Q
WLR = (or) (or)
Lw d B  Concentration of particles
removed = Σ Ci.ηi
5. General Design Standards:
 Total quantity of suspended
Length of tank = VH * Dt solids settled = Q * Cs
Length is always 3 – 4 times of
width of the tank. (Max.4 times)
Depth of the tank = 3.5 – 5m
 Sludge zone depth = 0.8 – 1.2m

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FUTURE GATE ACADEMY EE

PREVIOUS QUESTIONS litres of water per day (4MLD). The

average temperature of water is
1. For a flow of 5.7 MLD (million 200C. The dynamic viscosity of
litres per day) and a detention time water is 1.002 x 10-3 N.s/m2 at
of 2 hours, the surface area of a 200C. Density of water is 998.2
rectangular sedimentation tank to kg/m3. Average specific gravity of
remove all particles have settling particles is 2.65. (GATE-CE-2007)
velocity of 0.33 mm/s is (1997)
5. What is the surface overflow rate
(a) 20m2 (b) 100m2 in the sedimentation tank?
(c) 200m2 (d) 400m2 (a) 20m3/m2/day
2. A town has an existing horizontal (b) 40m3/m2/day
flow sedimentation tank with an
overflow rate of 17 m3/day/m3, (c) 67m3/m2/day
and it is desirable to remove
(d) 133 m3/m2/day
particles that have settling velocity
of 0.1 mm/sec. Assuming the tank 6. What is the minimum diameter of
is an ideal sedimentation tank, the particle which can be removed
thepercentage of particles removal with 100% efficiency in the above
is approximately equal to (2001) sedimentation tank?
(a) 30% (b) 50% (a) 11.8 x 10-3 mm
(c) 70% (d) 90% (b) 16.0 x 10-3 mm
3. For a water treatment plant having (c) 50 x 10-3 mm
a flow rate of 432m3/hr, what is
the required plan area of a Type I (d) 160 x 10-3 mm
settling tank to remove 90% of the 7. A horizontal flow primary clarifier
particles having a setting velocity treats wastewater in which 10%,
of 0.12 cm/sec is (2002) 60% and 30% of particles have
(a) 120m2 (b) 111m2 settling velocities of 0.1 mm/s,
0.2mm/s, and 1.0mm/s
(c) 90m2 (d) 100m2 respectively. What would be the
total percentage of particles
4. An ideal horizontal flow settling
removed if clarifier operates at a
basin is 3 m deep having surface
Surface Overflow Rate (SOR) of
are 900 m2. Water flows at the rate
43.2m3/d/m2? (GATE-CE-2009)
of 8000 m3/d, at water
temperature 200C (  = 10-3 (a) 43% (b) 56%
kg/m-s and  = 1000 kg/m3). (c) 86% (d) 100%
Assuming Stoke’s law to be valid,
te proportion (percentage) of 8. A suspension of sand like particles
spherical sand particles (0.01 mm in water with particles of diameter
in diameter with specific gravity 0.10 mm and below is flowing into
2.65), that will be removed, is a settling tank at 0.10m3/s.
Assuming g = 9.81m/s2, specific
(a) 32.5 (b) 67 gravity of particles = 2.65, and
(c) 87.5 (d) 95.5 kinematic viscosity of water =
1.0105 x 10-2 cm2/s. The
Linked answer Questions 5 & 6 minimum surface area (in m2)
required for this settling tank to
A plain sedimentation tank with a
remove particles of size 0.06 mm
length of 20m, width of 10m, and a
and above with 100% efficiency is
depth of 3m is used in a water
_______ (GATE-CE-2014)
treatment plant to treat 4 million

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FUTURE GATE ACADEMY EE

9. Consider a primary sedimentation 3. It is the time taken for any

tank (PST) in a water treatment unit of water to pass through the
plant with Surface Overflow Rate settling basin
(SOR) of 40 m3/m2/d. The
4. It is usually more than the
diameter of the spherical particle
flowthrough period
which will have 90 percent
theoretical removal efficiency in Select the correct answer using the
this tank is ______ μm. Assume codes given below :
that settling velocity of the
particles in water is described by (a) 1, 2, 3 and 4 (b) 2, 3 and 4
Stokes’s Law.Given Density of (c) 1 and 3 (d) 4 alone
water = 1000 kg/m3; Density of
particle = 2650 kg/m3; g = 9.81 13. The raw water entering an ideal
m.s2; Kinematic Viscosity of water horizontal settling tank contains
(v) = 1.10 x 10-6 m2/s. (GATE2015) following two types of particles :

10. The drag force, FD, on a sphere Particle Settling Concentration

kept in a uniform flow field type velocity (m/h) (mg/L)
depends on the diameter of the
sphere, D; flow velocity. V: fluid I 3 200
density, ρ: and dynamic viscosity,
II 1 300
μ.Which of the following options
represents the non-dimensional When the surface overflow rate of
parameters which could be used to the settling tank is 3 m3/m2/h,
analyze this problem? the concentration of the particles
(GATE-CE-2015) in the settled water will be
FD  FD VD (a) 100 mg/L (b) 200 mg/L
(a) and (b) and
VD VD VD 2

(c) 300 mg/L (d) 400 mg/L
(c) FD VD (d) FD  14. If the specific gravity of a
and and
V 2 D 2  V 3 D 3 VD suspended particle is increased
from 2 to 3, the settling velocity
11. In a wastewater treatment plant,
will
primary sedimentation tank (PST)
designed at an overflow rate of (a) not change (b) get doubled
32.5 m3/day/m2 is 32.5 m long,
80m wide and liquid depth of (c) get increased by 1.5 times
2.25m. If the length of the weir is (d) get increased by 2.25 times
75m, the weir loading rate (in
m3/day/m) is ______ (GATE 2015) 15. Assertion (A) : The settling
velocity of a discrete particle will
12. Which of the following statement become five times when its
is/are true in relation to the term diameter doubles
‘detention period’ in a settling
tank? Reason (R) : The settling velocity
of a discrete particle is almost
1. It may be determined by proportional to the square of the
introducing a dye in the inlet and particle diameter.
timing its appearance at the
outlet (a) Both A and R are true and R is
the correct explanation of A
2. Greater the detention period,
greater the efficiency of removal (b) Both A and R are true but R is
of settleable matter NOT the correct explanation of A
(c) A is true but R is false

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FUTURE GATE ACADEMY EE

(d) A is false but R is true PRACTICE QUESTIONS

16. Which one of the following is a not 1. Water has to purify the water for a
a specific criterion for calculating town whose daily demand is 9 x
surface overflow rate in 106 litres/day. Design the suitable
sedimentation tank design ? sedimentation tank of the water
(a) Total quantity of water to be works fitted with mechanical
treated sludge remover. Assume the
velocity of flow in the
(b) Total surface area in the tank sedimentation tank as
(c) Total length of the tank 22 cm/minute and the detention
period as 8 hours.
(d) Total depth of the tank
2. Design a sedimentation for a water
17. Assertion (A) : A discrete particle works, which supplies 1.4 x 106
(of diameter d0) settling in a litre/day water to the town. The
circular sedimentation tank follows sedimentation period is 5 hours,
a parabolic path. the velocity of flow is 12
Reason (R) : The downward cm/minute, depth of water in the
settling velocity (V0) of the discrete tank is 4.0m. Assuming an
particle (of diameter d0) in a allowance for sludge is to be made
circular sedimentation tank does as 80 cm.
not change with time.
3. Design a circular sedimentation
18. What is the most important design tank fitted with mechanical sludge
parameter used in designing a remover for a water work which
continuous flow rectangular has to supply daily 4.2 million
sedimentation tank for removal of litres of water to the town. The
discrete particles ? (IES-CE-2009) detention period in the tank for
water is 4.5 hours, and the depth
(a) Length of the tank
of the water in the tank may be
(b) Surface overflow rate assumed as 3.3m.
(c) Depth of the tank 4. If a rectangular sedimentation
tank is treating 2.5 x 106
(d) Temperature of the water litres/day. The size of the tank is
19. What is the settling velocity of a 17.5 x 5.5 x 3.5m. If 80 p.p.m
discrete particle in a wide body of suspended solids are present in
water when the relevant Reynold’s the water, assuming the 75%
number is less than 0.5? The removal in the basin and the
diameter and specific gravity of the average specific gravity as 2.0,
particle are 2  10-3 cm and 2.65, determine the following:
respectively. Water temperature to i) Average flow through tank
200C. (Kinematic viscosity = 2  10-
2 cm2/sec.) (IES-CE-2014) ii) Detention time
(a) 0.018 cm/sec iii) Deposition of the solids in tank
(b) 0.025 cm/sec iv) Overflow rate
(c) 0.18 cm/sec 5. A water supply project has to
supply water for a population of
(d) 0.25 cm/sec 1,00,000 with an average demand
of 150/pcd and maximum demand
is more than 1.5 times the average
demand. Assuming Detention
time as 4 hours and velocity of
Page 41
FUTURE GATE ACADEMY EE

flow as 30 cm/min, design the 13. A settling tank is designed for a

dimensions of a sedimentation surface over flow rate of 30m3/day
tank. m2. Assuming specific gravity of
sedimentation Particles = 2.65,
6. A particle with a diameter of
density of water as 1000 kg/m3 &
0.1mm & Specific Gravity of 2
dynamic viscosity of water as
released in water at 300C. How
0.001 NS/m2 and stocks law is
long will it take travel 2m.
valid. The approximation
Kinematic Viscosity of water at
minimum size of particles which
300C = 8x10-3 stokes.
can be completely removed is
7. A particle with a diameter of
(a) 0.01mm (b) 0.02mm
0.5mm, specific gravity of 3
released in water with a (c) 0.03mm (d) 0.04mm
temperature of 25 C. 0 How fall
14. A town is required to treat 4.2
does the particle travel in 3 sec.
m3/min of raw H2O for a daily
8. A settling tank is designed for an domestic supply. Flocculating
overflow rate of 4000 litres per m2 particles are produced by chemical
per hour. What percentage of capsulation. A column analysis
particles of diameter (a) 0.05 mm given that an over flow rate of 0.2
(b) 0.02 mm, will be removed in mm/sec will produce satisfactory
this tank at 100C? particle removal in settling tank at
a depth of 3.5m. The required
9. Find the diameter of the particles
surface area is
with specific gravity of 1.2 removed
in a tank having a surface area of (a) 210 (b) 350
250 m2 and treating 8 million litres
(c) 1728 (d) 3500
of water per day. Assume
temperature = 260C. 15. Determine settling velocity of a
spherical particle with a dia. of
10. A sedimentation tank is to be
10μ and specific gravity of 2.3 in
designed to treat 1,00,000 m3 of
water at 250C. Take dynamic
water per day. If settling velocity
viscosity of water at 250C as 0.89 x
of the particles to be removed is 20
10-3 N-S/m2.
m / day, then the area required is
16. Two particles are released in H2O
(a) 1000 m2 (b) 3000 m2
at same time. Particle A has
(c) 5000 m2 (d) 6000 m2 0.4mm dia & particle B has
0.9mm. What is the ratio of
11. The particle having a size of 0.025
settling velocity of A to B assuming
mm with a specific gravity of 2.65.
equal densities?
If the kinematic Viscosity of water
is 0.01 cm2/sec, the settling 17. The gravity of water supplied to a
velocity of particle as per strokes city is 1 lakh m3/day and settling
less is velocity of particle is 20 m/day.
Determine surface area of tank?
(a) 0.056 cm/sec (b) 0.048 cm/sec
18. Detention time for a sedimentation
(c) 0.12 cm/sec (d)0.25 cm/sec
tank (continuous flow type) is
12. If the L, b & depth of water in a given for a tank, passing a
rectangular Sedimentation Tank of discharge = Q and having length =
discharge Q are L, b & d L, width = B and dept = H as
respectively, the settling velocity is
(a) BLH/Q (b) Q/BLH
(a) Q/L (b) Q/b
(c) Q/BL (d) none
(c) Q/Ld (d) Q/Lb

Page 42
FUTURE GATE ACADEMY EE

Common Data for Questions 19-22 velocity of the particle as per

Stokes law is
A sedimentation tank (6cm wise, 15m
long and 3m water depth) is treating 2 (a) 0.056 cm/sec
MLD of water.
(b) 0.048 cm/sec
19. The surface overflow rate is
(c) 0.12 cm/sec
(a) 8.58 bit/hr/m2
(d) 0.28 cm/sec
(b) 926 lit/hr/m2
26. A plain sedimentation tank has
(c) 1028 lit/hr/m2 dimensions 100  50 x 3 m and
receives flow 1,00,000 m3/d.
(d) None
Calculate the surface over flow rate
20. Detention time is in m3/m2/d and dia of the smallest
particle, which will be removed
(a) 4.26 hr (b) 3.84 hr 100%. Take density of particle =
(c) 4.8 hr (d) 3.24 hr 2.65g/cc,
 = 1.02  10-3cm2/sec.
21. If 70 mg/L is the concentration of
solids present in the turbid water, 27. An ideal horizontal flow settling
the amount of dry solids deposited basin is 3m deep having surface
in the tank per day with 70% tank are 900m2. Water flows at the rate
efficiency is of 8000m3/d, at water temperature
200C (  = 10-3 kg/m-s and
(a) 98 kg (b) 140 kg
 = 1000kg/m3). Assuming
(c) 70 kg (d) 49 kg Stoke’s law to be valid, the
22. If the water content of the sludge proportion (percentage) of
is 90%, then the total weight of the spherical sand particles (0.01 mm
sludge produced is in diameter with specific gravity
2.65), that will be removed, is
(a) 980 kg (b) 1400 kg
(a) 32.5 (b) 67
(c) 700 kg (d) 490 kg
(c) 87.5 (d) 95.5
Common Data for Questions 23 - 24
A rectangular sedimentation tank is
treating 1.8 MLD of raw water with a
detention period of 4 hours KEY – PREVIOUS QUESTIONS
23. The volume of the tank required is 1. C 2. B 3. C
(a) 180 m3 (b) 260 m3
4. C 5. A 6. B
(c) 300 m3 (d) 360 m3
24. If allowable overflow rate is 500 7. B 8. 0.02 mm 9. 22.58
lit/hr/m2 and L : B = 4 : 1, then
the length required for the tank is 10. C 11. 112.67 12. B

(a) 18.5 m (b) 24.5 m 13. B 14. B 15. D

(c) 27.5 cm (d) 36.5 m
16. D 17. B 18. B
25. A particle is having a size of 0.025
mm with a specific gravity of 2.65.
If the kinematic viscosity of water
is 0.01 cm2/sec, the settling

Page 43
FUTURE GATE ACADEMY EE

6. Coagulation
Coagulation aided with Sedimentation

The chemical unit process for the I) Coagulation (or) Rapid Mixing:
removal of fine suspended solids
and colloids by the addition of a The colloids contained in the
trivalent aqua metallic cation as water are negatively charged at
coagulant which interact with pH>pHiso and positively at pH <
charged particles (inter particle pHiso. These colloids are stable
bridging) resulting flocculent due to the repulsive forces
particle formation followed by its between the negative charges.
settling is considered as
These colloids are destabilized
Coagulation (or) Coagulation aided
with Sedimentation by positively charged ions
formed from the hydrolysis of
 Coagulation process is used to
coagulants and followed by
entrap fine suspended particles
their interaction.
which escape plain
sedimentation along with Destabilization of colloidal
charged colloidal particles. particles can be influenced by
the following mechanisms
 Colloidal particles are difficult (explained according to DLVO
to separate from water because Theory: D – Derdaght O – Over
they do not settle by gravity and deck L – Lander V – Veruey)
are so small that they pass
through the pores of filtration 1) Compression of Electronic
media. Double Layer

 Coagulation is a 3 step process 2) Adsorption & charge

which involves Neutralization
3) Entrapment in the precipitate
[Sweep coagulation]
4) Growth is colloidal particle →
Agglomeration / Flocculation
5) Settling of large sized colloidal
Lumps
II) Flocculation (Gentle Mixing):
Flocculation is the formation of
clumps or flocs of suspended
solids by agglomeration of smaller
suspended particles and also
because of interaction of colloids
and coagulants. Flocculation is
aided by mild agitation for a period

Page 44
FUTURE GATE ACADEMY EE

of 20 to 30 minutes, time required  Volume of Tank = Q * Dt

for maximum floc formation and  Depth = 1 – 1.5m
growth. The agitation should be  Surface Area = V /H
gentle, in order not to break flocs  Rapid mixing chamber is either
already formed. Gentle air square (or) circular in shape.
agitation has also been employed  Velocity Gradient(G) =
to promote floc growth. 700 – 1000 per sec
III) Type – II Settling (or) G= P
V
Secondary Sedimentation:
The process of settling of altered
CD AV 3
particle, i.e, flocculent particle G=
2V 
settling is called as Type –II
Settling. After the floc has formed Design of Gentle Mixing TankFor
and grown to its most effective
size, the waste passes to a
sedimentation chamber for solids
removal. Floc formation and
growth may be retarded or stopped
by surface-active chemicals such
as soaps and synthetic detergents.
Design of Clariflocculator:
The three chambered tank that
efficient floc formation, slow
perform the process of coagulation
mixing of water is carried out
aided with precipitation for
using either hydraulic (or)
separation of colloids is
mechanical mixing devices (drum
Clariflocculator. It consists of
or paddle rotators) .
 Rapid Mixing Chamber The design steps are as follows:
(Flash Mixer)
 Detention time = 20 – 30min
 Gentle Mixing Chamber
 Volume of Tank = Q * Dt
(Flocculator)
 Depth = 2.5 – 3.5m
 Sedimentation Chamber
 Surface Area = V/H
(SST)
 Rapid mixing chamber is either
Design of Rapid Mixing Tank: rectangular in shape.
 Velocity Gradient(G) =
Mechanical mixing chamber is 70 – 100 per sec
designed to solubilize the  Dimensionless number (Gt)
coagulant with water and to =104 - 105
achieve electronic double layer for effective floc formation
formation.
The design steps are as follows:
 Detention time = 1 – 3min

Page 45
FUTURE GATE ACADEMY EE

G t Type of Floc alginates. Depending on the

Higher Higher Breakage of Floc type of charge, when placed in
Higher Lower Smaller & Denser water, the polyelectrolytes are
Floc classified as anionic, cationic or
Lower Higher Larger & Lighter nonionic. These are available as
Floc emulsion or solution or
Lower Lower Only size powdered forms. Examples:
increases Polyacrylamide, Quaternized
Polyamines, Polyamines etc.,
Design of Secondary Settling
 Generally trivalent aqua
Tank (SST):
metallic cations of inorganic
SST performs type – II settling of
coagulants are commonly used.
flocculent particles settlement in
dilute suspension. The design The following are the most
parameters are similar to PST commonly used coagulants:
except the following:
a) Aluminum Sulphate - Alum
 Detention time = 2 – 4hrs
[Al2 (SO4)3.18H2O]:
 Over flow rate = 24 – 30
m3/d/m2 Universal Coagulant

COAGULANTS: The chemical  Alum is most widely used

components used for interaction chemical coagulant and
with colloids for floc formation available as grey solid in the
during coagulation are considered form of lumps containing about
as Coagulants. 17% aluminum sulphate.
Types of coagulants: Generally  Chemical reactions of Alum:
coagulants are classified as
 Al2(SO4)3.18H2O + 3Na2CO3 →
 Inorganic coagulants: These
2Al(OH)3 + 3Na2SO4 + 2CO2 +
include Alum, Lime, Sodium
18H2O
aluminate, Ferric sulphate,
Ferric chloride, Ferrous  Al2(SO4)3.18H2O + 3Ca(OH)2 →
sulphate, Poly aluminium 2Al(OH)3 + 3CaSO4 + 18H2O
chloride etc.,
 Al2(SO4)3.18H2O + 3Ca(HCO3)2
 Polyelectrolytes Polyelectrolytes
→ 2Al(OH)3 + 2CaSO4 + 6CO2 +
are water-soluble polymers
18H2O
carrying ionic charge along the
polymer chain and may be  Requirements:
divided into natural and o Alkalinity in water (achieved
synthetic polyelectrolytes. by addition of lime or soda)
 Important natural poly
electrolytes include polymers of o pH : 6.5 – 8.5
biological origin and those o Optimum Dosage: 14 mg/l
derived from starch products,
cellulose derivatives and
Page 46
FUTURE GATE ACADEMY EE

Advantages: chlorine to a solution of ferrous

sulphate in the ratio of 1 part
o Cheap, Easy to handle and apply
chlorine to 7.9 parts copperas.
o Produces effective floc
6FeSO4 + 3Cl2 →
o Also remove color, odour and improve 2Fe(SO4)3 + 2FeCl3
taste
It is very good coagulant and
o Produces less sludge than lime requires less amount of alkalinity
o Most effective between pH 6.5 and 7.5 in the water for floc formation.
The produced floc is tough and
Disadvantages: easily settles due to which only
o Impart permanent hardness small residue goes in the filters.
This coagulant removes colours
o Water becomes corrosive due to
very well.
release of CO2
d) Sodium Acuminate [Na2Al2O3]:
o Adds dissolved solids (salts) to water
This is an alkaline compound.
o Effective over a limited pH range The best grade it contains 55%
b) Ferric Coagulants: Generally of Al2O3, 34% of Na2O3, 4.5% of
ferric chloride (FeCl3), ferric Na2CO3, 6.3% of NaOH. This
sulphate [Fe2 (SO4)3] or the can be used for treatment very
mixture of both (or) ferrous easily in the water having no
sulphate (FeSO4.7H2O) is used alkalinity. It reacts very quickly
for coagulation purpose. and forms the precipitate of
Chemical reactions are as aluminum hydroxide. Its
follows: chemical equations are as
follows:
2FeCl3 + 3Ca(OH)2 →
2Fe(OH)3 + 3CaCl2 Na2Al2O3 + CaSO4 →
CaAl2O3 + Na2SO4
Fe2(SO4)3 + 3Ca(OH)2 →
2Fe(OH)3 + 3CaSO4 Na2Al2O3 + CaCl2 → CaAl2O3
+ 2NaCl
FeSO4.7H2O + Ca(OH)2 →
Fe(OH)2 + CaSO4 + 7H2O Na2Al2O3 + Ca(HCO3)2 →
CaAl2O3 + Na2CO3 + CO2
Ferrous Sulphate is considered
DOSAGE OF COAGULANTS:
as Copperas, react with
hydrated lime and produce floc. The optimum dose of coagulant
It is effective if pH>8.5. It is can be determined by Jar Test
cheaper than alum but does not apparatus.
react with colored water. The dose (or) concentration of
c) Chlorimated Copperas: It is coagulants added depends on
a mixture of ferric chloride and following factors:
ferric sulphate prepared by adding

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FUTURE GATE ACADEMY EE

1) Kind of coagulant 2) Turbidity of water

3) Color of water
4) PH value of water
5) Temperature of water
6) Mixing & Flocculation Time.
In case of more turbid water at lower temperature more quantity of
coagulants are required.

Turbidity Alkalinity Coagulant Predominant Mechanism

Dose

High Low Small Adsorption&charge

neutralization

High High Large Sweep Coagulation

Low High Medium Charge neutralization & Sweep

Coagulation

Low Low Small Cannot coagulated for any dose

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FUTURE GATE ACADEMY EE

PREVIOUS QUESTIONS (b) results in increase of pH of the

treated water
1. The calriflocculator is the unit in
(c) results in no change in pH of
which of the following things will the treated water
occur
(d) may cause and increase or
(a)Floc formation and its decrease of pH of the treated
subsequent removal by filtration water
(b)Floc formation and its Common Data for Questions 6 & 7
subsequent removal by
A water treatment plant treating
sedimentation
10mld of water requires 20mg/l of
(c)Floc formation and its filter Alum, Al2(SO4)3 18H2O. The
subsequent removal by water has 6 mg/l of alkalinity as
decantation
6. Total alkalinity requirement
(d) Removal of bacteria by filtration (106mg per day as CaCO3)
and chlorination (GATE-CE-1992) matching filter Alum, shall be
2. Design parameters for rapid (a) 180 (b) 120 (c) 90 (d) 60
mixing units are (GATE-CE-1997) 7. Quantity of Quick Lime required
(a) velocity gradient and the (106 mg per year as CaO) shall be
CaCO3 (Al = 26.97, S = 32, O = 16,
volume of mixing basin H = 1, Ca = 40 and C = 12).
(b) viscosity and velocity gradient (a) 2132 (b) 3000
(c) viscosity, velocity gradient and (c) 4132 (d) 6132
the volume of the mixing basin 8. The design parameter for
(d) detention time and viscosity of flocculation is given by a
dimensionless number Gt, where
water G is the velocity gradient and t is
3. Coagulation-flocculation with alum the detention time. Values of Gt
is performed (GATE-CE-1998) ranging from 104 to 105 are
commonly used, with t ranging
(a) immediately before chlorination from 10 to 30 mm. The most
preferred combination of G and t
(b) immediately after chlorination
to produce smaller and denser
(c) after rapid sand filtration flocs is
(d) before rapid sand filtration (a) large G values with short t
4. The following chemical is used for (b) large G values with long t
coagulation (GATE-CE-2000)
(c) small G values with short t
(a) Ammonium Chloride
(d) small G values with long t
(b) Aluminum Chloride
9. A circular primary clarifier
(c) Aluminum Sulphate processes an average flow of 5005
m3/d of municipal waste water.
(d) Copper Sulphate The overflow rate is 35 m3/m2/d.
5. Use of coagulants such as alum The diameter of clarifier shall be

(a) results in reduction of pH of (a) 10.5 m (b) 11.5 m

the treated water (c) 12.5 m (d) 13.5 m

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FUTURE GATE ACADEMY EE

10. A water treatment plant is required 14. For proper slow mixing in the
to process 28800m3/d of raw flocculator of a water treatment
water (density = 1000kg/m3, plant, the temporal mean velocity
gradient G needs to be of the order
kinematic viscosity = 10-6m2/s). of (IES-CE- 1998)
The rapid mixing tank imparts a
(a) 5 to 10 s-1 (b) 20 to 80 s-1
velocity gradient of 900s-1 to blend
35mg/I of alum with the flow for a (c) 100 - 200 s-1 (d) 250-350 s-1
detention time of 2minutes. The 15. Match List-I (Water treatment
power input (W) required for rapid units) with List-II (Detention time)
mixing is (GATE-CE-2008) and select the correct answer
using the codes given below the
(a) 32.4 (b) 36 lists : (IES-CE-1998)
(c) 324 (d) 32400 List-I List-II
11. A town is required to treat 4.2 A. Rapid mixing 1. 1.5 hours
m3/min of raw water for daily unit
domestic supply. Flocculating
B. Flocculator 2. 10 seconds
particles are to be produced by
chemical coagulation. A column C.Propeller 3. 30 seconds
analysis indicated that an overflow mixing unit
rate of 0.2 mm/s will produce
D. Sedimentation 4. 40 minutes
satisfactory particle removal in a tank
settling basin at a depth of 3.5 m.
The required surface area (in m2) Codes :
for settling is (GATE-CE-2012) A B C D
(a) 210 (b) 350 (a) 3 4 2 1
(c) 1728 (d) 21000 (b) 4 3 1 2
12. A surface water treatment plant (c) 4 3 2 1
operates round the clock with a (d) 3 4 1 2
flow rate of 35m3/min. The water
16. What is the predominating
temperature is 150C and jar
coagulation mechanism for raw
testing indicated an alum dosage water having high turbidity and
of 25mg/l with flocculation at a Gt high alkalinity ? (IES-CE-2007)
value of 4 x 104 producing optimal (a) ionic layer compression
results. The alum quantity
(b) Adsorption and charge
required for 30 days (in kg) of
neutralization
operation of the plant is ______
(c) Sweep coagulation
13. Assertion (A) : Tapered
flocculation is more efficient when (d) Inter particle bridging
compared to the conventional 17. The correct sequence of treatment
process of flocculation of typical turbid surface water is
Reason (R) : In tapered (a) Flocculation, coagulation,
flocculation, velocity gradient at sedimentation, filtration
the inlet is less than that at the
outlet of the flocculation unit

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FUTURE GATE ACADEMY EE

(b)Flocculation, coagulation, filtration, 3. The requirements of a city is 40 x

sedimentation 106 litres/day. The detention
period is one hour in the tank, and
(c) Coagulation, flocculation, filtration,
the flow velocity is 20 cm/sec.
sedimentation
Design baffle-wall sedimentation
(d)Coagulation, flocculation, tank. Any data not given may be
sedimentation, filtration suitably assumed.
18. In a water treatment, the optimum 4. At a water treatment plant, 12 ML
time of flocculation is usually of water is treated daily, using
given as alum dosage of 16 mg per lt. Find
30 minutes. In case the time of (a) Total gravity is alum used daily
flocculation is increased beyond (b) Amount of CO2 released.
this value, the flocs will
5. At a water treatment plant 12
(a) become heavy and settle down million lt of water is treated daily
in flocculation itself using FeSO4 & lime. If the dosage
of FeSO4 is 10mg per litre;
(b) Entrap air and will float in the determine total gravity of FeSO4 &
sedimentation tank lime required daily?
(c) break up and defeat the 6. Water works of a town treat 35 x
purpose of flocculation 106 lt/day. The water is treated by
(d) stick to the paddles coagulation – sedimentation.The
gravity of filter alum is consumed
at 20 mg/lt of water. If the
PRACTICE QUESTIONS alkalinity of raw water is
equivalent to 4.5 mg/l of CaCO3
1. Waterworks of a town treat 35 x determine Gravity of filter alum &
106 litres/day. The water is the guice lime (containing 80% of
treated by coagulation- CaO) reg. per month by the water
works.
sedimentation tanks. The quantity
of filter alum is consumed at 20 7. Determine gravity of guice lime
required/year to satisfy alkalinity
mg/litres of water. If the alkalinity
present in 10MLD of water. If
of the raw water is equivalent to required alkalinity is 9 mg/L as
4.5 mg/litre of CaCO3, determine CaCO3 & present alkalinity is
the quantity of filter alum and the 6mg/L.
quick lime (containing 80% of 8. Determine total alum required per
CaO) required per month by the year for treatment of 50MLD of
water works. Molecular weights H2O with alkalinity equivalent of
are given as [Ca = 40, C = 12, S = 5mg/L as CaCO3 & alum dosage
as 16 mg/L. Also determine
32, O = 16, Al = 27 and H = 1].
alkalinity deficiency. Total
2. Determine the quantity of coppers alkalinity requirement and
and the lime required per year to quantity of lime req/year for
treat 4 x 106 litre/day, if 11 mg of solving alkalinity deficiency.
coppers is consumed with lime at 9. For treatment of 12MLD of water
a coagulation basin. Molecular alum dose required is 14 ppm
weight of Fe = 55.85, S = 32, O = Determine Qty of alum reg/day
16, H = 1, Ca = 40. and also CO2 released / lt of H2O.

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FUTURE GATE ACADEMY EE

10. Determine qty of FeSO4 used and 16. At a water treatment plant 12MLD
qty of lime req/day to treat 12MLD is treated daily using ferrous
of H2O with ferrous sulphate sulphate and lime. If the dosage of
treatment of dosage – 10mg/L.
ferrous sulphate is 10mg/L,
11. A water treatment unit has a determine the total required daily.
discharge of 3000m3/hr. The
Use the following chemical
water is sent into a coagulation
unit of determine time 20 mins & equation
Min.Vel. gradient of 40/sec. Fe SO4. 7H2O + Ca (OH)2
Assuming ‘μ’ as 1.0087 x 10-3,
Determine power. Also design the =Fe (OH)2 + CaSO4 + 7H2O
coagulation unit assumint L = 2B
17. Design a floculator for a design
& Depth as 0.4 times the breadth.
flow of 3000m3//hr. Assume
12. A rapid mixing unit which is Detention time of
square shaped has to treat 5MLD
20 minutes. G = 40 sec . Also
–1
of water with a detention time of 1
minute. Determine the velocity determine the power requirement
gradient required and side of if  = 1.0087  10-3 N. S/m2.
mixing basin if depth is assumed Length to width ratio 2 and depth
as 1mt and μ is 10-3 N-S/m2.
may be assumed as 0.4B.
13. The volume of rapid mixing unit is
2m3, mean vel. Gradient of 600 per
second determine the power
applied by assuming dynamic
viscosity as 10-3 N-sec/m2.
14. A coagulation sedimentation plant
clarifies 50 MLD. The dosage of KEY – PREVIOUS QUESTIONS
filter alum required is 16 mg/L. If
1. B 2. C 3. D
the raw water is having an
alkalinity equivalent to 5mg/L of 4. C 5. B 6. C
CaCO3, determine the quantities of
filter alum and quick lime 7. D 8. A 9. D
(containing 87% of CaO) required
10. D 11. B 12. 37800
per year for the plant. Given the
molecular weights as follows 13. C 14. B 15. A
15. Determine the quantity of alum
16. C 17. D 18. B
required in order to treat 12
million liters of water per day at a
treatment plant, where 14ppm of
alum does is required. Also
determine the amount of CO2 gas
which will be released per litre of
water treated. [Atomic weights A1
= 27, S = 32, O = 16, H = 1]

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FUTURE GATE ACADEMY EE

7. Filtration
The process of passing the water matter and because of the
through beds of sand or other presence of gelatinous coating
granular materials (filter media) is formed on the sand grains by
called Filtration.
previously deposited suspended
 Media retain particles & produce
matter attract the other fine
possible free water.
 Media: (a) Silica Sand (b) particles.
Anthracite Coal (c) Garnet 3. BIOLOGICAL METABOLISM:
 The impurities which escape Suspended impurities contain
through coagulation process and some portion of organic impurities
other dissolved impurities are like Algae, Bacteria, and Plankton
removed by Filteration Process. etc., which are food of various
types of Microorganism.
 For removing bacteria, colour,
The bacteria caught in the voids of
taste, odour, iron and Manganese
sand grains utilize organic
to produce clear & sparkling water
impurities present in water and
filters are used.
convert them into harmless
Theory of Filtration / Mechanism compounds by the complex
of Filtration: biochemical reactions.
The phenomenon of filtration has The harmless compounds so
formed are deposited at the
been explained on the basis of
surface of the sand in the form of a
following actions:
layer which contains a zoological
1. MECHANICAL STRAINING:
jelly in which the biological
The particles of suspended matter
activities are at their highest. This
that are of size larger than the size
layer is called “SCHMUTZDECKE”
of voids are arrested and removed
(Dirty Skin). This layer further
by the action of Mechanical
helps in absorbing and straining
straining.
out the impurities.
Small particles of suspended
4. ELECTROLYTIC ACTION: The
impurities move through the pures
sand particles of filter media are
in sand come in contact with sand
changes with electricity of some
surfaces and adhere causing
polarity.
further reduction tank and settled
So, suspended and dissolved
particles from a mat on the top of
particles of optical polarity come
sand bed which further arrests
into contact, they neutralize each
very fine suspended particles &
other and it results in changing
removes them from water.
the chemical characteristics of
2. FLOCCULATION, SEDIMENTATION
water. After long use, the electric
& ADSORPTION:
charge of filter sand is exhausted,
The interstices (voids) between the
which is renewed by wasting the
sand grains act as minute
filter bed.
sedimentation tanks [settling
basins] in which particles will
settle and adhere to the sides of
sand grains due to the physical
attraction between two particles of

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 FUTURE GATE ACADEMY EE

CLASSIFICATION OF FILTERS: Single Media, Gravity, Slow Filter Type

SLOW SAND FILTER RAPID GRAVITY FILTER PRESSURE FILTER
1) Developed by James Developed in USA during 1900  This is rapid gravity
Simpson in 1829 in – 1910 filter works under
England Move advantages than slow pressure.
2) Efficiency: sand filters  This is employed for
Bacterial Load: Less efficient for bacterial small water plants,
Remove about 98 – 99%. load. Remove 80 – 90% of individual colour (or)
Colour: Less efficient to Bacteria in water. industry
remove the colour Highly efficient in colour  Costlier
Turbidity: Removes 50 removal  Water is forced through
ppm filters at pressure
3) Cleaning of Filters: These Cleaning is done by back ranging 3 – 7 kg/cm2
are cleaned by scrapping washing system for 10 –  Rate of filtration 6000 –
and removal of top sand 15min. Troughs are kept to 15000 lt/hr/m2
layer of 1.5 to 3cm receive the back wash water.
thickness. Water requested for back
 Pretreatment not washing is 2 to 5% of total
required post amount of water filtered. Back
treatment is optimal wash water rate is 5 – 7 times
 Coagulated H2O should filtration rate.
not allow through this.
4) Size of Filter Bed: Varies
from 10m2 – 2000 m2 (or) 10m2 – 50m2
more.
5) Filter Media of Sand: Fine
sand grains Coarse Sand Grains
Effective Size: 0.25 to
0.35mm 0.45 to 0.70mm
Uniformity Coefficient: 3
to 5 1.3 to 1.7
Thickness: 90 to 110cm
reduced to not less than 60 to 75cm not reduced by
40cm by scrapping washing
6) Base Material of Gravel:
Base: 3 to 65 mm
Thickness: 30 to 75mm
7) Coagulation 2 to 50mm
8) Under Drainage System 45 to 60cm
Not required Essential
9) Rate of Filtration Provided to receive filtered
10) Amount of wash water Provide only to receive filtered water & also to supply water
11) Filter Box water for back washing of filter
3000 to 6000 lt/hr/m2
12) Period of Cleaning 100 to 200 lt/hr/m2 2 to 5% of filtered water
0.2 to 0.6% of filtered water Open water tight
Open water tight rectangular rectangular for concrete box
tank with masonry (or) 1 to 3 days
concrete box1 to 3 months

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FUTURE GATE ACADEMY EE

DESIGN OF FILTERS FLOW THROUGH EXPANDED BEDS:

For the design of sand filters and gravity Le 1  n
filters, the following design aspects are 
L 1  ne
considered. For a town/city having
Le = Expanded Bed thickness
certain population, average demand of the
L = Original Bed thickness
town is calculated.
Average Demand of the Town = Percapita Vb = Vs x ne4.5
Demand x Total Population n = Porosity of Original Bed
 The maximum water demand must be Vb = Vel. Of Backwash
always greater than some % (or) some ne = Porosity of expanded Bed
times than the average demand. Head Loss during Back Washing:
 The rate of filtration required in water hfb = L[1-n] [s-1] (or) Le [1-ne] [s-1]
per day is calculated by knowing the Efficiency of particle removal:
capacity of filtration unit per hour. 1 100 1 100
 Total surface area of filters required is
ln  ln
Z1 100  1 Z n 100   2
the ratio of maximum Daily demand to
Z1 = Initial Depth
that of rate of filtration/day.
Z2 = Increased Depth
Total Surface Area of Filters
η1 = Initial Efficiency
needed =
η2 = Increased Efficiency
Maximum Daily Demand
Rate of Filtration/Day SLOW SAND FILTERS

 Surface area of each unit = Construction: Water tight shallow tanks

Total Surface Area Depth → 2.5 to 4m
No.of Filter Units S.A. → 10 to 2000m2
No. of Rapid Graints Filters units Sand Bed Thickness → 60 – 90cm →
Filtering Media supposed by Gravel Bed
N = 1.22 Q given by Morrel & Gravel Bed Thickness → 30 – 60cm
Wallace Gravel → 3 to 4 layers each
Q → Qm of water in ML/Day → 15 – 20cm thickness
Minimum No = 2 Filters Coarse gravel is placed in bottom &
Filter Hydraulics: smallest one
Carmen – Kozney Equation: Head loss Effective Size → 0.25 – 0.35cm is in the
through a granular process medium top most layer.
fLVs2 (1  n)  The size of bottom layer gravel → 40
hf = …. For uniform size of – 60 mm
gd n 3  
 The size of Intermediate layer → 20
particles
– 40 mm & 6mm – 20mm if 2 layers
(1  n)
Where f = 150  1.75 Re = are used.
Re  Size of top most gravel layer →
Vs.d Vs.d . 3.6mm
. (or)
  FILTER TROUBLES IN GRAVITY TYPE
d = Grain Size, f = frictional factors, Re = FILTERS:
1. Formation of Mud Balls:
Reynold’s number
L = Depth of FB, n = Porosity of media Conglomerations (group of diff. things
 = Particle shape factor → 1 for sphere, gathered together) of the sand grain,
floc & other binding materials →
lessthan 1 for other wages
Vs = Filtration Velocity found must densly collected at or near
sand bed surface → 25 – 50mm dia

Page 55
FUTURE GATE ACADEMY EE

Reason: Due to insufficient washing PREVIOUS QUESTIONS

of sand bed.
Removal of Mud Balls: (1) Breaking 1. A small filter of 0.05 m depth removes
up with rakes (2) Washing by high 90% of particles present in water. If
the particles removal required is 99%,
velocity surface wash (3) Washing with what should be the depth of filter?
chemical sol. Of caustic soda, H2SO4 (a) 0.10 m (b) 0.50 m
etc, (4) Digging out with shavels (c) 0.75 m (d) 1.00 m
2. Cracking (or) clogging of filter bed:
2. A dual-media rapid sand filter plant is
Occurs due to shrinkage of filter bed to be constructed for treatment of 72
& pulling away sand from side walls million liters of water per day. A pilot
→ prevention measures as in (1) plant study indicated that a filtration
rate of 15m/h would be acceptable.
3. Air binding: Negative pressure
Allowing one unit out of service for
exceeds, water tends to release back washing, how many 5m x 8m
dissolved gases in the form of bubbles filter units will be required? Determine
which seriously affect working of filler the net production in million liters per
day of each filter unit if backwashing
→ Air Bubbly. is done at 36m/h for 20 minutes and
Reasons: (1) Move Negativeve head (2) the water is wasted for the first 10
Increase in temperature of H2O (3) minutes of each filter run.
Release of O2 by algae 3. The following characteristics pertain
Prevention: (1) Avoid negative head to the sand filters used in water
(2) Avoid warming up of the (3) industry.
I. Filtration rate is 1 to 4 m3/(m2
Control growth of algae (4) Avoid
day).
super separation of the with air. II. Typical duration of operation in
4. Sand Incrustation: Caused due to (a) one run is 24 to 72 hours.
deposition of the from influent water III. Operation cost is low.
Which of the above characteristic
(b) presence of CaCO3 sand grains pertain to slow sand filters?
enlarge and effective sand size is (a) I, II and III (b) I and II
changed. (c) II and III (d) I and III
Prevention: Carbonation of influent
Linked answer questions 4 & 5 :
→ dissolves both CaCO3, MnCO3 A city is going to install the rapid sand
deposited on sand grain filter after the sedimentation tanks. Use
5. Jetting & Sand : Due to back the following data. Design loading rate to
the filter. (GATE-CE-2005)
washing of filter → Minimized by
Design loading to the filter: 200 m 3/m2 d
surface wash Design flow rate : 0.5 m3/s
6. Sand Leakage: Down ward migration Surface area per filter box : 50m2
and escape of fitness → Minimized by 4. The surface area required for the rapid
proportionally sand & gravel layers. sand filter will be
(a) 210m2 (b) 215m2
Loss of Head: Difference in water (c) 216m 2 (d) 218m2
levels between filter tank & outlet
5. The number of filters required shall be
chamber due to resistance offered by
(a) 3 (b) 4 (c) 6 (d) 8
sand grains to flow.
 It is exceeds specified value, Common data for questions 6 & 7
cleaning of filter should be done. In a rapid sand filter, the time for
reaching particle break through (TB) is
defined as the time elapsed from start of
filter run to the time at which the
turbidity of the effluent from the filter is

Page 56
FUTURE GATE ACADEMY EE

greater than 2.5NTU. The time for

reaching terminal head loss (TH) is defined
as the time elapsed from the start of the PRACTICE QUESTIONS
filter run to the time when head loss 1. Design a set of 3 rapid gravity filters
across the filter is greater than 3m. for treating the water at a water
6. The effect of increasing the filter depth works, which has to supply the water
(while keeping all other conditions
to a town of population 1, 00,000.
same) on TB and TH is
(a) TB increases and TH decreases PCD of town is 270 lt/day. The rate of
(b) both TB and TH increase filtration of rapid gravity filters may be
(c) TB decreases and TH increases taken as 4500 lt/hr/m2.
(d) both TB and TH decreases
2. Design a rapid sand filter unit for
7. The effect of increasing the filter
loading rate (while keeping all other treating 5x106 lt/day supply for a
conditions sames) on TB and TH is town the filter are to use day and
(a) TB increases and TH decrease night. Take ROF 4,500 lt/m2/hr.
(b) both TB and TH increase 3. Design 5 slow sand filters beds from
(c) TB decreases and TH increase
the following data for the water works
(d) both TB and TH decrease
of a town of population 75,000
8. A town is required to treat 4.2 m 3/min Percapita Demand = 135
of raw water for daily domestic supply. lt/day/capita
Flocculating particles are to be Rate of Filtration/hr = 210 lt/m 2/hr
produced by chemical coagulation. A Assume maximum demand as 1.5
column analysis indicated that an
times the average demand out of 5
overflow rate of 0.2 mm/s will produce
satisfactory particle removal in a units. One is to be kept as stand by
settling basic at a depth of 3.5m. The and used while repairing other works?
required surface area (in m 2) for
settling is (GATE-CE-2012) 4. Find the diameter of pressure filter to
(a) 210 (b) 350 treat 1 MLD of water with rate of
(c) 1728 (d) 21000 filtration 10,000 lt/hr/m 2.
9. Uniformity coefficient of filter sand is
given by (IES-CE-1997) 5. a) A dual media rapid sand filter plant
D50 D50 is to be constructed for treatment of
(a) (b)
D5 D10 72 MLD of water.A pilot plant study
indicated that a filtration rate of
D60 D60
(c) (d) 15m3/hr/m2 would be acceptable.
D5 D10 Allowing one unit out of service for
10. Assertion (A) : In the case of dual back washing how many 5m x 8m
media filter, the rate of filtration is filter units will be required?
more than that of rapid sand filter
b) Determine the net production in
Reason (R) : The direction of flow is
MLD of each filter unit if back
from fine medium to coarse medium
washing is done at 36 m3/hr/m2 for
20 mm & the water is wasted for front
11. The effective size (ES) of sand and its
uniformity coefficient (UC) are the 10 min of each filter run?
usually specified parameters for sand
filters. In slow sand filters, as 6. A city is going to install the rapid
compared to rapid sand filters, sand. Filter after the sedimentation
(a) ES is less but UC is more Tanks.
(b) ES is more but UC is less Use the data: Design loading rate of
(c) Both ES and UC are more filter → 200 m3/m2.d surfaces
(d) both ES and UC are less Area per filter box → 50m2
Design flow rate → 0.5 m3/sec

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The surface area required for rapid

sand filter will be
KEY – PREVIOUS QUESTIONS
(a) 210 m2 (b) 215 m2
(c) 216 m2 (d) 218 m2 1. A 2. 13620 3. D

7. A city has to treat 24 MLD of turbid 4. C 5. C 6. A

water using Rapid sand filter with a
filtration rate of 5m3/hr/m3 7. D 8. B 9. B
(a) Determine the size of the Filter
bed if L : B : 2 : 1, only one unit of 10. C 11. A
filter is provided
(b) Determine the percentage of
filtered water required for
backwashing the filter, if the
rate of back wash is 6 times the rate
of filtration and duration of back
wash is 10min. Back washing is
done once in a day.

8. A filter plant has 6 filter units each

with a rated capacity of 2.5MLD. The
rapid sand filter loading rate is 125
l/m2/min. What percent of treated
water is used for back washing if each
filter is washed every 48 hours at 7.5
times the loading rate for 5 minutes
duration?

9. A bed of uniform sand, having particle

size 0.65 mm dia and specific gravity
2.66, porosity 0.42 and depth 65cm is
to be washed hydraulically. Compute
(a) back wash rate so that
expansion will be 50%
(b) head loss at this rate.
Take  = 1.3  10-2 cm2/sec
and assume CD = 24/R e and
Laminar flow condition

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8. Disinfection
The process of killing infective bacteria 3) Iodine & Bromine Pills: → Costly
from the water & making it safe to the → can’t be used for public water
user is called Disinfection. supply
 It does not mean total destruction of  Provides long lasting protection
living things in the medium treated against pathogens & reduced
because sterilization means total offensive taste & odours.
destruction.  Used in water supply for army
 The chemicals or substances which troops, private plants,
are used for killing bacteria → swimming pools.
Disinfectants  Quantity → should not exceed
 In case of emergencies, when water is 8 ppm → kill bacteria with
supplied to the army, if all treatments minimum constant period → 5
are not possible, the disinfection of min.
water is done to keep it safe. 4) Treatment with excess time:
 During disinfection, it is also Effectively kills bacteria, but can’t
important to add sufficient quantity of safeguard future population
disinfections to all bacteria during its  Enough lime is added to H2O to
conveyance from water works to raise its pH value to 9.5 (or) more.
consumers.  Recarbonation → The process of
 Chlorine → most ideal disinfectant. removal of excess lime before it is
Methods of Disinfection: supplied to public → it is
necessary.
1) Boiling of water: Most effective 5) Disinfection with Ozone: Most
method of disinfection → boiling 15 powerful disinfectant than chlorine
– 20 min. & also costly than chlorine.
 Can’t be used for huge quantity  Does not provide protection
of public water supplies. against recontamination & can’t
 Used for Domestic purposes in be stored.
emergencies, costly  Used for disinfection on a small
 Can’t take care of future scale → Swimming pool.
contamination.  Gaseous in forms → Faintly blue
2) UV Rays: These are invisible light in colour → Pungent Odour.
rays of 1000 – 4000 m wave length
 Unstable allotropic form of O2.
 Sun light is the best source of
 It is produced by passing a high
UV rays. Mercury vapours
tension electric current through a
enclosed in quartz bulb &
stream of air in a closed chamber.
passing current in it produce UV
3O2 → 2O3
rays artificially.
 It is highly unstable: O3 → O2 + O
 Very costly & effective method
Nascent Oxygen.
 Adopted for water supply
 The nascent O2 is very powerful
installation of institutions &
oxdising agent & it kills the
private buildings, treating small
bacteria as well as oxdises org.
quantity of water in hospitals,
matter present in water.
dispensaries for surgical use,
 Process: During disinfection, O3
swimming pools etc.,
gas is released into sterilizing
 Water thickness allowed to pass
chamber water enters into it
through UV rays is → 10cm (or)
through inlet & comes out through
less
outlet.

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 Residual 2ml 0.1ppm → for 13 min Chlorination

& then 2-3 ppm after contact period.
(Universal Disinfection)
 Quantity of residual ozone →
 Capable of providing residual action
measured by → Orthotolidine Test.
 Effective for long periods
presence of light yellow colour
 It satisfied all requirements of an
presence of residual ozone.
ideal disinfectant.
 O3 also remove colour, task & odour
 It prevents future contamination
from water.
also.
 Ozonation method is used if
 Only disadvantage: Imparts bed taste
electricity is easily & cheaply
 It does not bring significant change
available.
in pH value of natural water
 It being unstable nothing remains in
 It is not effective if water is alkaline.
water till it reaches consumers.
 It makes water more tasty &
Action of Chlorine:
pleasant.
When chlorine is added to water
6) Potassium Permanganate: Used for
Cl2 + H2O ↔ HOCl + H+ + Cl- Hydrolysis:
disinfection of well water supply →
Both are reversible reactions
generally contaminated with low
H+ + OCl- ↔ HOCl : Ionisation
amount of bacteria.
 Remove 100% of organisms At pH < 5.5 → 100% HOCl & no OCl
causing cholera.
PH > 9.5 → 100% OCl & no HOCl
 Oxidises org. matter → Dose 1-2
mg/L with a contact period of 4 – PH = 7.5 → 50% HOCl & 50% OCl
6 hrs.
 Efficiency → 98%. HOCl (Hypochlorous Acid) & OCl-
 In rural areas, to bacteria of (Hypochlorite ions) accomplish
water, 0-10mg/Lt KMO4 is added disinfection of water.
& mixed with water of a well These two HOCl & OCl combinely called
frequently. The water in bucket as Free Available Chlorine.
are pink in colour after adding  HOCl is 80 – 100 times more
KMnO4. powerful than OCl-
7) Silver (or) Electro – Katadyn  The free Cl2 reacts with compounds
Process: like NH3, Proteins, AA & phenol that
 Removers algae & take care of may generally present in water to
future pollution also. form Chloramines & Chloro
 Silver Ions → Introduced into derivatives called combined chlorine.
water → By passing solid silver  Free Chlorine in 25 times more
electrodes tubes into water & powerful than combined chlorine.
then passing electricity through  Residual Chlorine = Chlorine Dosage
1.5V DC battery → It kills all – Chlorine Demand
bacteria.  Green & Stump after long
 Dose → 0.05 – 0.1 mg/L → contact experiments reported that the death
period → 15 – 180 min. of microbes results from a chemical
 It is important to remove surp. reaction of HOCl with the enzyme
Org. matter &H2S before using system of cell which is essential for
this method, if not the efficiency metabolic activities of living cells &
is greatly reduced. reduce the ability of org. to ferment
glucose.
Results death of organism.

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Chlorine acts in 2 ways:  When chlorine is added to water,

a) It can oxidize the sulphydryl group some of it is consumed in killing
(-SH groups) of triose-dehydrosenase bacteria & oxidizing organic
[an enzyme responsible for matter to form its compounds
fermentation of glucose] (chloramines) & the remaining →
b) Chlorine acts by penetrating the residual chlorine → indicated by
cell wall & attacking the enzyme. line Q.
This factor explains the fact that why  Fall of lines → indicates →
cysts & spcres have greater resistance oxidation of chlororganic
to disinfection as compared to compounds & chloramines →
bacteria. indicated by bad odour & taste in
 Chlorine Demand = Amount of water.
chlorine added to water – Amount of  Any further increase in chlorine
chlorine remaining at the end of dose beyond point B will appear as
contact period [free + comb chlorine] residual chlorine only
 Shown by line S.
Chloramines:  Applied ‘Cl’ in ppm
 NH3 + HOCl → NH2Cl + H2O  B → Greak point
 NH2Cl + HOCl → NHCl2 + H2O  The sudden decrease in res.
 NHCl + HOCl → NCl3 + H2O Chlorine is due to conv. of
chloramines → nitro compounds.
 Mano & Di → got disinfectant property DOSAGE: The dose of ‘Cl’ which leaves
but NCl3 → have negligible residual chlorine of about 0.2 mg/L at the
disinfection end of 10min. Contact period is selected
 If pH < 4.4 → only NCl3 exists which gives opt. dose of chlorine for the
 pH → 4.4 – 5.5 → only NHCl2 exists given water sample.
 pH → 5.5 – 8.4 → both mono &  Normal Dosage → 0.3 to 1.1 mg/L
dichloramines exists  ‘Cl’ dose increased during rainy
 pH → 8.4 → only mono chloramines season
exists  Residual Chlorine → 0.1 to 0.2 mg/L
Break Point Chlorination:  ‘Cl2’ dosage = Cl2 demand + Cl2
residual
 Super Chlorination: Administration of
excess dose → during epidencies of
water borne diseases
 2 – 3 ppm beyond break point → gives
strong odour & taste
 Added at the end of filtration →
Removed by dechlorination

Dechlorination: Partial (or) complete

M → Destruction of ‘Cl’ residual by reduction of ‘Cl’ in water by chemical (or)
reducing compounds (Fe, Mn, H2S) physical treatment.
N → Formation of Chloro org. compounds Chemical used for dechlorination: SO2,
& chloramines Sod. Thiosulphate, Sod. Biosulphate,
O → Destruction of chloro org. activated carbon, Pot. Permanganate,
compounds & chloramines Ammonia.
P → Formation of free Cl and presence of
chlororganic comb. not destroyed.

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PREVIOUS QUESTIONS 7. Chlorine gas (8mg/L as Cl2) was

added to a drinking water sample. If
1. Pathogens are usually removed by
the free chlorine residual and pH was
(a) Chemical precipitation measured to be 2 mg/L (as Cl2) and
(b) Sedimentation 7.5, respectively, what is the
concentration of residual OCI ions-
(c) Activated sludge process
water? Assume that the chlorine gas
(d) Chlorination
added to the water is completely
2. Among the following disinfectants of converted to HOCI and OCI-. Atomic
waste water, the one that is most Weight of Cl : 35.5. (GATE-CE-2011)
commonly used is (GATE-CE-1997) OCl- + H+  HOCl, K = 107.5
(a) Chlorine dioxide (b) Chlorine (a) 1.408 x 10-5 moles/L
(c) Ozone (d) UV-radiation (b) 2.817 x 10-5 moles/L
(c) 5.634 x 10-5 moles/L
3. The disinfection efficiency of chlorine
in water treatment (GATE-CE-2000) (d) 1.127 x 10-4 moles/L
(a) is not dependent on pH value
(b) is increase by increased pH value 8. 16 MLD of water is flowing through a
(c) remains constant at al pH value 2.5 km long pipe of diameter 45cm.
(d) is reduced by increased pH value The chlorine at the rate of 32kg/d is
applied at the entry of this pipe so
4. In disinfection, which of the following
forms of chlorine is most effective in that disinfected water is obtained at
killing the pathogenic bacteria the exit. There is a proposal to
(a) Cl (b) OCl increase the flow through this pipe to
(c) NH2Cl (d) HOCl 22 MLD from 16 MLD. Assume the
dilution coefficient, n = 1. The
5. Chlorine gas used for disinfection
minimum amount of chlorine (in kg
combines with water to form
per day) to be applied to achieve the
hypochlorous acid (HOCI). The HOCI
same degree of disinfection for the
ionizes to form hypochlorite (OCI) in a
enhanced flow is (GATE-CE-2014)
reversible reaction:
(a) 60.50 (b) 44.00
HOCI  H+ + OCl- (c) 38.00 (d) 23.27
(k = 2.7 x 10-8 at 200C),
the equilibrium of which is governed 9. An effluent at a flow rate of 2670m 3/d
by pH. The sum of HOCI is the more from a sewage treatment plant is to be
effective disinfectant. The 90% disinfected. The laboratory data of
fraction of HOCL in the free chlorine disinfection studies with a chlorine
residual is available at a pH value dosage of 15mg/l yield the model Nt =
(a) 4.8 (b) 6.6 (c) 7.5 (d) 9.4 Noe-0.145t where Nt = number of micro-
organisms surviving at time t (in
6. An aerobically treated effluent has min.) and No = number of micro-
MPN of total coliform as 106/100 mL. organisms present initially (at t = 0).
After chlorination, the MPN value The volume of disinfection unit (in m 3)
declines to 102/100mL. The percent required to achieve a 98% kill of
removal (%R) and log removal (log R) micro-organisms is________
of total coliform MPN is
(a) %R = 99.90; log R = 4
(b) %R = 00.90; log R = 2
(c) %R = 99.99; log R = 4
(d) %R = 99.90; log R = 2

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FUTURE GATE ACADEMY EE

PRACTICE QUESTIONS hypochlorous acid (HOCL). The HOCL

is ionizes the form hypochlorite (OCL)
1. The water works at a town of in a reversible reaction : HOCL  H+
population 25,000 has to meet its + OCL – (k = 2.7  10-8 at 200C), the
water demand @ 135 lpcd. If the equilibrium of which is governed by
pH. The sum of HOCL and OCL is
disinfection is to be done by bleaching
known as free chlorine residual and
power having 45% available chlorine, HOCL is the more effective
determine the quantity of bleaching disinfectant. The 90% fraction of
power required per year. The HOCL in the free chlorine residual is
required. dose of ‘Cl’ at water works is available at a pH value
0.3 ppm for disinfection?
(a) 4.8 (b) 6.6
(c) 7.5 (d) 9.4
2. Chlorine usage in the treatment of
25000 m3/day is 9kg/day. The
7. To obtain 99.7% kill of bacteria, ozone
residual chlorine after 10 minutes
is to be used in water with a residual
contact is 0.2 mg/l. Calculate the of 0.6 mg/L. The reaction constant
dosage in mg/l and the chlorine under these conditions is 3  10-2 per
demand of the water. second. Calculate the time of contact
required
3. A sample of raw water contains
200mg/l, alkalinity, 50 mg/l, 8. If 0.7 mg/L of chlorine is required for
hardness as CaCl2 and 75mg/l satisfactory disinfection of water at pH
= 7, what dosage will be necessary at
hardness as MgSO4. Compute the pH = 8. K = 2.7  10-8 @ 200C
quantities of lime and soda required to
treat 1M.I. of water. If slaked lime of
85% purity is available in place of
pure lime, what will be the required
quantity of slaked lime?
KEY – PREVIOUS QUESTIONS
4. It is required to supply water to a
population of 20,000 at a per capita 1. D 2. B 3. D
demand of 150 lit per day. Determine
how much bleaching powder is 4. D 5. B 6. C
required in kg on an average per day if
the chlorine dosage is 0.2ppm. 7. A 8. A 9. 15.02M3
Assume bleaching powder contains
30% of available chlorine.
(a) 0.6 (b) 0.18
(c) 2 (d) 0.8

5. Chlorine usage in the treatment of

20,000 cu m of water per day is 8 kg.
The residual after 10min. contact is
0.15 mg/l. The chlorine dosage and
demand of water in mg/l, are
respectively
(a) 2.5 & 0.25 (b) 2.5 & 2.35
(c) 0.4 & 0.25 (d) 0.4 & 0.15

6. Chlorine gas used for disinfection

combines with water to form

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FUTURE GATE ACADEMY EE

9. Miscellaneous Water Treatments

Water Softening: Removal of Hardness waste to treatment → Act.
Carbon granuals of 0.1 – 1mm
1. Lime Soda Process : size.
(a) Lime [Ca(OH)2 & soda ash [sodium  Acts in 2 ways (1) Filter Media
carbonate Na2CO3] used for (2) Fine powder Feed
removal of hardness
3. Use of CuSO4: Control growth of
(b) Lime → Quick lime
Algae, Bacteria & Weeds applied in
[CaO]/Hydrated lime [Ca(OH)2]
Distribution pipes / open
can be used
reservoirs
Recarbonation: Diffusal of CO2 gas
II. Fe & Mn: It available without any
through effluent
organic matter, removed by Aeration
Insoluble CaCO3 + Mg[OH]2 →
→ Coag → Sed → Filtration.
CaHCO3 + MgHCO3
Effects: (1) Develop Reddish /
Ion Exchange
Brownish strains in cloths when H2O
2. Zeolite Process: Zeolite → complex of used in laundry
Aluminium, Silica & Soda (2) Deposited in distributed pipes &
 Hard H2O is passed through bed of cause blocking of pipes, meters etc.
Ion exchanger, namely Zeolite (3) H2O reg in industries paper
making, photographic film,
 Ca & Mg are removed from H2O &
manufacturing, ice making must be
substituted by Sodium free from Fe, Mn etc.,
 Silicates of Na & Al →Zeolite Sand
→ Ion Exchange III. 1) FLUORIDATION: Raising Fluoride
content of H2O
3. Demineralisation / Deionisation:
NaF → adopted for public H2O
Similar to Zeolite process but Ca+2 &
supplies
Mg+2 are exchanged for H+ ions.
 Apply flouride after other
Ion Exchange → Carbonacedus
treatment & mixing can occur
Material / Resin
before H2O leaves treatment
 H2O similar to distilled H2O used
works.
for Industries
(2) Deflouridation: Reducing ‘Fl’ conc.
I. Removal of Colour Taste & Odour: in H2O (a) Nalgonda Technique (b)
Activated Carbon (c) Lime Soda
1. Aeration: H2O is contact with
Process
atmospheric air → promote
Nalgonda Technique:
exchange of gases between H2O &
H2O → Sod. Aluminate or lime →
air.
Bleaching Power → Filter Alum added
 Expell CO2 & H2S → which
to fluoride H2O in sequence
impact taste & odour
Nalgonda Tech: Coagulation →
 Oxidise Fe & Mn to certain Sedimentation → Disinfection →
extent Deflouridation
 Device for mixing of air with Aeration:
H2O → Eductor This is the process of bringing waters into
intimate contact with air with the object
2. Treatment by Activated Carbon:
of driving out objectionable dissolved
Available in Granular / powder
gases and oxidizing other soluble
form
compounds present in the ground waters
 Remove org. contaminates by
or in stagnant waters of pools and
Adsorption. Most commonly
reservoirs.
used adsorbent in H2O &
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Aeration is effected in many ways (i) by (Na) in the zeolite until Na becomes
causing the water to flow over weirs and exhausted. The sodium is then restored
waterfalls called Cascade aerators, (ii) by by regenerating the zeolite with a solution
dropping water through perforated plates, of common salt (NaCl) resulting in the
(iii) by forcing it through spray nozzles, reversal of ionic reactions viz, Ca and Mg
(iv) by filtering perforated trays, coke taking the place of Na and being washed
beds, and (v) through special devices out as chlorides while the Na remains in
which aspirate air by diffusion through the zeolite so that it is again ready to act
porous plates. The spray nozzle is the as a softening agent.
most effective aerator. Aeration is effective
in removing 75 percent of the odours. DESALINATION: Saline H2O → Fresh H2O
Removal of carbon dioxide is equally high. 1) Distillation: Most common Saline
H2O → Boiled → Evaporated
Taste and Odour Control: Tastes and Fresh H2O ← Condensation ← Vapours
odours in water supplies may be caused 2) Reverse Osmosis: H2O is forced
due to the presentee in water of any of the across a semi permeable membrane
following: by Mechanical force
Decaying organic matter resulting 3) Electro Dialysis: Electrical currents
from alage and other micro organisms, (ii) applied to remove salt ions using
industrial wastes such as phenols (iii) micro process membrane
chlorophenol compounds resulting from 4) Freezing: Saline H2O → Low Temp →
combination of residual chlorine with Ice crystals → Melt → Fresh H2O
phenol (iv) dissolved gases like H2S, CO (v) 5) Demineralisation: Ion exchanger
excess amount of cholorine. 6) Solar Evaporation : Solar Energy
Methods for the control of tastes
and odours are (a) copper sulphate
PREVIOUS QUESTIONS
treatment, (b) Ammonia cholorine process
(c) use of activated carbon (d) Use of
1. The absorbent most commonly used
cholorine dioxide
in water and waste treatment is
Water softening Processes: there are three
(a) Sand of grain size from 0.1 to 2
general methods used for water softening:
mm
(i) Lime Process (ii) Lime and soda
(b) Activated carbon granules of size
ash process and (iii) base
0.1 to 2 mm
Exchange Process
(c) Ordinary wood shavings of find
Lime Processes reduces only the
size
carbonate hardness. The principle
(d) Coal – tar
involved is to neutralize the CO2 with milk
of lime i.e., Ca (OH)2, forming normal
2. Aeration of water is done to remove
carbonates which precipitate out when
(a) Suspended Impurities
present in excess and are removed by
(b) Color
settlement and filtration. The process is
(c) Dissolved Salts
also known as the Clark Process.
(d) Dissolve Gases
Chemical reactions involved are
CO2 + Ca (OH)2 = CaCO3 + H2O
3. Zero hardness of water is achieved by
(a) lime soda process
Base Exchange Process: In this process,
(b) excess lime treatment
hard water is passed through a bed of
(c) ion exchange treatment
zeolite sand (complex silicates of
(d) excess alum and lime treatment
aluminium and sodium) whereby it
exchanges its Ca and Mg for the sodium

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4. Match the following:

Group I (Type of water impurity)
P. Hardness
Q. Brackish water from sea
R. Residual MPN from filters KEY – PREVIOUS QUESTIONS
S. Turbidity
Group II (Method of treatment) 1. B 2. D 3. C
1. Reverse Osmosis
4. D 5. A
2. Chlorination
3. Zeolite Treatement
4. Coagulation and Flocculation
5. Coagulation, Flocculation, and
Filtration
Codes:
P Q R S

(a) 1 2 4 5

(b) 3 2 2 4
(c) 2 1 3 5

(d) 3 1 2 5

5. Match Group I (Terminology) with

Group – II (Definition / Brief
Description) for wastewater treatment
systems
Group I
P. Primary treatment
Q. Secondary treatment
R. Unit operation
S. Unit process

Group II
1. Contaminant removal by physical
forces
2. Involving biological and/or
chemical reaction
3. Conversion of soluble organic
matter to business
4. Removal of solid materials from
incoming waste water

Codes:
(a) P – 4, Q – 3, R – 1, S – 2
(b) P – 4, Q – 3, R – 2, S – 1
(c) P – 3, Q – 4, R – 2, S – 1
(d) P – 1, Q – 2, R – 3, S – 4

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10. Distribution System

Pipes – mains, sub mains, branches and  Advantage is that during fire
laterals accidents quantities of water at high
Valves – For controlling flow pressures be pumped to put off the
Hydrants – For releasing water during fire fire.
breakouts
2.3 Combined or dual systems:
Meters – For measuring discharges
“Pumping storage system” or “Direct –
Pumps – For lifting water
indirect system this
Service reservoir – For storing the treated
system, the treated water is pumping
water & stabilizing pressures.
constant rate into an elevated
Importance: It involves 40 to 70% of the
total cost of W.S. scheme. reservoir as directly into distribution
system.
1. Recommendations: Thus it is a combination of gravity
Pressure for single storey buildings is system direct pumping system. This
7m for two storey buildings is 12m of is most adopted system in water
water for three storey buildings is 17m supply schemes obvious advantages
of water such as Pumps operated with uniform
The fire hydrant’s pressure should not speed at their capacities.
be less than 1 Kg/sq. cm. 1. It is a reliable system as there is
Minimum pipe size: 100 mm some reserve water in elevated
tank augment the supply during
2. Systems of Distribution:
peak demand.
1) Gravity System
2. In case of fire accidents large
2) Direct Pumping System
quantity water can be drawn.
3) Combined System or Dual System
3. It is an economic and efficient
system.
2.1 Gravity System:
Water is distributed by gravity only, to 3. Lay out of Distribution Systems:
the consumers points. It is suitable 1. Dead end or tree system
for situations where the source of 2. Grid iron system or Reticulat
water is located at a sufficiently higher Interlaced system.
level than the town. This system is 3. Circle or Ring system
economical and reliable since no 4. Radial system
pumping is involved at any stage. It
needs a lake or storage reservoir as a 3.1 Dead end system:
source of supply located at a sufficient
higher level.

2.2 Direct Pumping System: Treated

water is directly pumped into the
distribution pipes by means of high
lift pumps without storing anywhere.
 Pumps should be capable of being
operated at variable speeds to meet A supply main starting from reservoir
the maximum and minimum is laid along the main submains are
demand and maintain sufficient connected to the main directions
residual pressures at various points along the other road divide into
of consumption. This system is not several branch times. Service
commonly preferred. connections are taken from these
branches to the individual houses.

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Suitability: For old towns and cities  Laying of mains peripherally

with irregular and unplanned increase the pressure at farthest
development. points.
Advantages: Quite simple design:  Suitable for towns and cities
Lesser number of cutoff valves; Easier having well planned roads.
expansions; Short pipe lengths are  Advantages and disadvantages are
required Cheap and economical same as Grid Iron System.
Limitations: Considerable area gets
affected during repairs,
 Water stagnation at dead ends
leads to pollution
 Water rate of supply cannot be
increased in case of fire breakouts
 Less successful in maintaining
satisfactory pressures in the
remote parts.
3.4 Radial System: The area is divided
3.2. Grid Iron System: Mains, submains into small distribution zones and in
and branches are interconnected with the center of each zone a distribution
each other. reservoir is provided. Water from
Main line runs through the centre of the these reservoirs is supplied through
area. radially laid distribution pipes
Suitability: Suitable for well planned running towards the periphery of the
cities. zone.
 Ensures high pressure and
efficient water distribution.
 Suitable for cities with radial
roads.

Advantages:
Since water reaches from different
directions, sizes of pipes get reduced. 4. Design of Distribution System:
Very small area gets affected in case of For head loss calculation, Hazen-
repairs. Due to free circulation, water Williams formula is more commonly
cannot be polluted. More water can be used.
diverted in case of fire breakout. Head loss by Hazen-Williams Formula:
Disadvantages: 10.68L Q1.852
More lengths of pipes and larger hj 
CH1.852D 4.87
number of valves
Costlier Where CH = roughness coefficient
Design is difficult hf is proportional to Q1.852

3.3 Ring system: Main pipe is laid 4.1 Conditions to be satisfied:

peripherally. 1. The inflow into each junction must
be equal to the flow out of the

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FUTURE GATE ACADEMY EE

junction, as per principle of 5.3 Standpipes: They boost the pressure.

continuity. A standpipe is a vertical cylindrical
tank resting just above the ground.
2. In each loop, the loss of head due
to flow in clockwise direction must 5.4 STORAGE CAPACITY OF
be equal to the loss of head due to DISTRIBUTION RESERVOIR:
flow in anticlockwise direction (i.e. The storage capacity of a distribution
the algebraic sum of head loss in reservoir to be provided is based on
each closed loop must be zero) the following requirements.
Important Methods for Analysis of (i) Balancing storage (or equalizing
Pipe Networks: 1. Equivalent pipe storage or operating storage)
method. 2. Hardy – Cross method. (ii) Breakdown storage or emergency
storage
4.2 Equivalent Pipe Method:
(iii) Fire storage
For the purpose of analysis the entire
(i) Balancing storage (or equalizing
network of pipes is considered to be
split up into two portions viz., (i) pipes storage or operating storage):
in series, and (ii) pipes in parallel. The quantity of water required to
be stored in the reservoir for
4.3 Hardy – Cross Method: It is a method balancing or equalizing the
of successive approximations which variable demand of water against
involves a trial and error process. the constant rate of pumping is
Hardy Cross method may be carried known as balancing storage.
out in the following two ways. The balancing storage of a
(a) Balancing heads by correcting distribution reservoir can be
assumed flows; and determined by the following tow
(b) Balancing flows by correcting methods
assumed heads. (a) Hydrograph method
Note: (b) Mass curve method
1. Clock wise flows and their (ii) Breakdown Storage: It is the
corresponding head losses are storage required to be provided in
taken as +ve a distribution reservoir to take
2. Anti clock wise flows and their care of emergencies which may
corresponding head losses are arise due to failure of pumps,
taken as –ve failure of electric supply, etc. For
TYPES: According to the situation this storage a lump sum provision
of about 25% of the total storage
with respect to ground, the
distribution reservoirs are classified in capacity of the distribution
the following three types. reservoir is provided.
(iii) Fire storage: A provision of fire
(1) Surface reservoirs (2) Elevated
storage in a distribution reservoir
reservoirs (3) Standpipes
is required to be made to provide
water for fire fighting purposes.
5.1 Surface reservoirs: Constructed at
1 to 4 lit/per/day for normal
ground level or below ground level
Indian conditions.
Surface reservoirs should be located
at high points in the distribution Total capacity of D.R = Balancing
system. storage + Break down Storage +
Fire storage
5.2 Elevated reservoirs: Constructed at
an elevation from G.I. Also called as
overhead tanks.

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FUTURE GATE ACADEMY EE

6. APPURTENANCES IN DISTRIBUTION dead ends and depressions or low

SYSTEM points in the pipeline.
1. Valves 2. Manholes 3. g) Pressure-relief Valves: Also called
Fire hydrants 4. Water meters overflow towers are provided to keep
6.1. Valves: Valves are provided in the the pressure in a pipeline below a
pipelines to control the flow of water, predetermined value, and thus
to isolate and drain sections for test, protect it against the possible
inspection, cleaning and repairs, to danger of bursting due to excessive
regulate pressures and to release or pressure. Thus these valves are
admit air. often placed at low points where the
a) Sluice Valves or Gate Valves: used pressures are high. Further a
to regulate the flow of water through pressure relief valve is usually
the pipelines. provided on the upstream side of a
b) Butterfly valves: used to regulate sluice valve so that the pipe lying on
and stop the flow especially in large the upstream side of the valve is
size conduits. relieved of water hammer pressure
c) Globe Valve: To regulate flow. resulting from the sudden closure of
These valves are normally used in the sluice valve.
pipes of small diameter (less than
100 mm) and as water taps. 6.2 Manholes: Provided at suitable
d) Check Valves: Also known as reflux intervals along the pipeline for
inspection and repairs. Usually
valves or non-return valves. A
spaced 300 to 600 m apart on large
check valve allows water to flow in pipelines.
one direction only and the flow in Their most useful positions are at
the reverse direction is summits and downstreams of main
automatically stopped by it. The valves.
reflux valve is invariably placed in a
6.3 Fire hydrants: Hydrant is an outlet
pumping main so that if the pump
provided in a pipeline for tapping
fails or stops, water is prevented water mainly for the purpose of fire
from flowing back to the pump and fighting. Also be used for withdrawing
thus pumping equipment is saved water for certain other purposes such
from possible damage. a sprinkling on roads, flushing
streets, etc. Generally fire hydrants
e) Air Valve or Air-relief Valves: The
are placed at all important road
air valve helps to admit air into the junctions and at intervals not
pipe when the pipe is being emptied exceeding about 300 m.
or when negative or vacuum
pressure is created in the pipe. Air 6.4 Water Meters: Installed in pipelines
valve operates automatically while to measure the quanity of water
flowing through them.
allowing air to escape from or to
Types:
enter a pipe. The air valves are
(i) Inferential type meters or velocity
usually located at summits and also meters: Used for large pipes.
at changes in grade to steeper (ii) Displacement type meters: Used
slopes. for small pipes and domestic
f) Scour Valves or Blow-off Valves or connections.
Drain Valves: Provided for
Location of Leaks: For locating leaks in
completely emptying or draining of
water supply pipes following methods may
the pipe for removing sand or silt be used.
deposited in the pipe and for a) By direct observations
inspection, repair, etc Located at b) By using sounding rods
c) By plotting hydraulic grade line

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FUTURE GATE ACADEMY EE

d) By using waste Hours Demand in % of

- Detecting meters (Deacon’s meter) 24 hours
demand
Sounding rods: A metal rod is inserted
into the ground at the suspected portion. 0–8 5
If there is a leak in the pipe the sound of 8 – 10 40
the water escaping through the leak can
be heard by placing the ear against the 10 – 16 20
rod, or by means of an amplifying device 16 – 20 25
such as aqua phone or sonoscope.
20 – 24 10
CLASS ROOM OBJECTIVES

1. When the source of water is a

reservoir on a mountain slope and the PREVIOUS QUESTIONS
cityis in plains then the system of
distribution is 1. Match the following:
(a) Gravity system (GATE-CE-2005)
(b) pumping system Group I
(c) Combined gravity and pumping P. Release valve
system Q. Check valve
(d) Gravity system for a part and R. Gate valve
pumping for remaining S. Pilot valve
Group II
2. Pumping system is best suited when 1. Reduce high inlet pressure to lower
(a) fire accidents occur frequently outlet pressure
(b) source of water is at low level 2. Limit the flow of water to single
(c) density of population is high and direction
space available is less 3. Remove air from the pipeline
(d) power failures are more common 4. Stopping the flow of water in the
pipeline
3. Combined pumping and gravity flow Codes:
system best suited, where P Q R S
(a) the city is in plains and source is (a) 3 2 4 1
fairly elevated (b) 4 2 1 3
(b) the city had a gentle slope and (c) 3 4 2 1
source is elevated (d) 1 2 4 3
(c) the city is on steep slopes and Ans: D
source is below
(d) any type of topography
4. At a town of present population
50,000 have average per capita water
demand of
160 l.p.c.d. water is pumped into an
overhead tank at a uniform rate
continuously for 24 hrs of the day.
The demand pattern is shown below.
Determine the required balancing
storage capacity of the overhead tank.

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FUTURE GATE ACADEMY EE

WASTE WATER ENGINEERING (SANITARY ENGG.)

1) Introduction
Terms: (a) Sanitary sewer is a sewer to carry the
1. Refuse : All waste material (a) solid sanitary or domestic sewage only.
(b) liquid (c) semi solid (b) Separate sewer is one, which carries
2. Garbage : Organic, Dry refuse → only sanitary sewage and the dry-
Vegetables, Grain, Sweepings, weather flow.
Leaves (c) Combined sewer is that which carries
3. Rubbish : Inorganic, Dry Waste in addition to the sanitary sewage,
from (a) Offices (b) Houses surface and storm water flows also.
It is combustible (d) House sewer is a sewer, which carries
4. Sewage: A liquid Waste (a) sewage from buildings or houses into
Domestic Origin (b) Industrial the public or street sewer.
origin Foul in nature [Dirty, (e) Lateral sewer or lateral is a sewer,
Smells (or) tastes unpleasant] collecting sewage from houses or flow
contains 99.9% water [Dangerous] from streets but does not receive
Domestic Sewage + Industrial sewage from any other sewer.
Sewage → Sanitary Sewage (f) Sub main sewer is a branch sewer,
5. Sullage: Waste Water (a) Bathroom which receives discharge from two or
(b) Kitchen (c) Washbasin. Less more lateral sewers.(denoted by b in
foul in nature fig )
6. Storm Water: Run off from roads, (g) Main sewer or truck sewer is one
buildings & other catchment areas which collects the flow from two or
storm drainage (or) Drainage. more sub main sewers as an outlet for
7. DWF (Dry Weather Flow): Normal a large area .(denoted by b in fig c ).
flow available in any seasons due (h) Intercepting sewer. When it is not
to sanitary sewage. It is generally desirable to have the sewers empty
1/20th to 1/25th of maximum flow directly into a body of water, the entire
during Manson. flow from a number of separate sewers
8. Sewer : Pipe carrying sewage (or) or the dry-weather flow from a
conduit number of combined sewers may be
9. Sewage: Process of Newtork of discharged into an intercepting sewer
sewers in collection + conveyance (denoted by d).
of sewage. (i) Outfall sewer is that sewer which
SEWERS & ITS TYPES carries sewage from the lower end of a
Sewer is an underground conduit used for collecting system to a suitable point of
the removal of sewage and sewerage is final discharge or to a treatment plant
general process of removing sewage. The (denoted by e).
entire system of conduits and Classification of Sewarage Systems
appantanances solved is called sewage The basic concentions for the design of a
system or sewer system sewarage system are:
Types of sewers The combined System, where the storm
Depending upon their respective function water together with the sewage is
and layout following types of sewers may conveyed by one and the same system of
be employeed. pipes and conduits to the discharge point

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FUTURE GATE ACADEMY EE

or treatment plant. During heavy storms, The Pressure Type System Small pumps
when the combined flow of sewage and discharge the house sewage from a
rainwater exceeds a certain value, diluted collection tank into the sewerage system
sewage is discharged via one or more which is entirely under pressure. Smaller
overflows directly to the surface water. diameter pipes can be used because the
The Separate System, consisting of two velocities can be higher. The depth of the
distinctly separated systems of pipes and pipes can be less than those which are
conduits, where one conveys the sewage used in a gravity scheme. This type of
to the point of destination and the other scheme can therefore be particularly
channels the storm water of the nearest advantageous in flat areas.
surface water.
The Pseudo-separate or partially – TYPES OF SEWAGE
separate System is a combination of the 1. Domestic Sewage
two. One system receives the sewage and 2. Industrial Waste Water
a part of the storm water, i.e., the run-off 3. Infiltration Water
from the roofs of the buildings which have In order to calculate the required capacity
a sanitary connection to the system. The of a sewer, it is necessary to know the
other system takes care of the remaining maximum flow that can be expected.
storm water. However, exceptionally high flow
The Vacuum Type System A sewer pipe intensities for periods of not more than a
with a relatively small diameter connects few seconds are to be disregarded, they
the appliances ( Toilets,washbasins etc) cause a wave phenomenon in the sewer
directly with a sewage collection task. In that will be rapidly reduced in height by
the tank an under pressure (vaccum) of 5 damping. An average maximum flow of
to 7m is maintained by a pump then at 10 to 15 minutes duration is taken for the
certain times value on the appliance is base of our calculations.
opened and the waste water is drawn into Domestic Sewage
the collection tank. The tank should be The domestic sewage flow depends on:
emptied from time to time by a tanker and 1. The population served by the
or special can be incorporated in the tank sewer
to pump the average to a conventional 2. The water consumption
sewerage system or to treatment works. 3. The percentage of water returned
to the sewer
Advantages of the system are: 4. The peak load coefficient
1. Relatively small sewer pipes
ESTIMATING OF DOMESTIC SEWAGE:
2. Great flexibility in alignment and
 Seasonal Maximum Rate of Flow =
gradient
3. Possibility to lift sewage up to 5 m
1.30  Average Daily Rate of Flow
4. Conservation of water when using  Monthly Maximum Rate of Flow =

special vacuum toilets 1.40  Average Monthly Rate of

This system is only used in particular Flow

cases; e.g., on ships, or to serve several  Daily Maximum Rate of Flow = 2 

isolated buildings by connecting them to a Average Daily Rate of Flow
central collection tank from where the  Hourly Maximum Rate of Flow =
sewage can be pumped away. 1.5  Average Daily Flow = 3 
Average Daily Flow

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FUTURE GATE ACADEMY EE

 Sewers are designed for carrying ESTIMATING STORM SEWAGE:

Maximum hourly flow running Rational Method:
3/4th Full. Quantity of storm water

 Peak flows for lateral sewers are CiA

Q= [Wet weather Flow]
much greater than that of large 360
trunk sewers because More flow RO
C → Coefficient of Run off →
time for trunk sewers than RF
laterals. For different types of surface areas →
Impervious (or) RO coefficient Factor C
Determination of Peak Flow by A1C1  A2C2  A3C3 .... AnCn
(or) K =
Empirical Formulae: A1  A2  A3  ..... An
5 i → Intensity of RF in mm/hr coefficients
1. Bobbit’s Formula  M=
P1 / 5 of time of concentration
M = Ratio of peak rate to average rate A → Drainage Area (or) Catchment Area →
Hectares - 104 m2
14 Time of Concentration:
2. Harmon’s Formula  M = 1
4  P1/ 5 The period after which the entire area
P = Population in thousands starts contributing run off (or) Maximum
time taken by the rain drop to travel from
18  R farthest point (or) remoted point to outlet
Some times it may be Q/f =
4 P of catchment.
Q = Maximum Flow f = Average Flow  Time of concentration tc
= Time of entry te + Time of Flow tf

MINIMUM FLOW:
Length of Drain
o tf =
 At minimum flow, velocity reduces, Vel. of Drain
 Silting occurs. Emperical Formula for Intensity:
 Slope at which sewer to be laid 25.4a
i= [Time for inlet → outlet]
down → is based on permissible tc  b
minimum velocity at minimum EMPERICAL FORMULA: For areas larger
flow. than 400 hectares
 Minimum Daily Flow = 2/3 a) Dicken’s Formula: For North Indian
Catchments
Average Daily Flow
Q = Cd . A3/4
 Minimum Hourly Flow = 1/2
Cd → Constant depending on factors
Average Daily Flow = 1/3 Average A → Catchment area in Km2
Daily Flow b) Ryve’s Formula: For south Indian
catchments Q = Cr A2/3
Example: A city with a population of
1,00,000 has the rate of water supply
200lped. Assuming 75% of water
supplied reaches the sewer, the dry
weather flow in m3/sec will be
Solution: Qty of sewage = P x PCD x %
= 1,00,000 x 200 x 75/100
= 15,000,000 L/day
= 0.174 m3/sec

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Problem 1: The drainage discharge of a

town of 16 hectares area, consisting of
40% hard paved (K = 0.80), 30% unpaved Problem 4:
(K = 0.20) & remaining wooded (K = 0.10), In a certain down town residential
with a maximum rain intensity of 5 distinct population per hecta is 296.
cm/hr, would be computed by rational The estimate daily contribution of
formula, as equal to domestic sewage for each person is
(a) 0.91 Cu mec (b) 0.091 Cu 455 litres of which 341 litres are
mec (c) 9.1 Cu mec (d) discharged in a period of 16 hours.
none The quantity of storm water to be
A1C1  A2C2  A3C3 removed in an hour is the volume that
Solution: K = = would cover the district to a depth of
A1  A2  A3
25mm. What relations capacity for
(0.40  0.80)  (0.30  0.20)  (0.10  0.30) domestic sewers, storm sewers and
0.80  0.20  0.10 combined sewers.

0.32  0.06  0.03

= = 0.41
1

CiA 0.41 50 16

Q= 
360 360

= 0.91 cu mec (or) 0.91 m3/sec

Problem 2: In the above example, if the

density of population is 300
person/hectare & the rate of water supply
is 250lpcd. Calculate amount of sanitary
sewage for
(a) Separate system
(b) Partially separate system

Problem 3: Calculate the maximum

hourly flow the minimum hourly flow
and the average daily flow for w/w
from the given data below:
Population 9000 Area hectare
Average water consution 150 gpcd
Maximum population density 21
persons/acre
Rainwater reaching the sewer = 30
lpcd.
Infiltration rate = 2000 gal/day/acres
Ass 20% of water consumed return to
sewer.

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FUTURE GATE ACADEMY EE

2) Design of Sewers
The design of sewer is difference from that  Minimum Velocity to be generated
of conduit used for conveyance of water in in sewers to avoid silting = Vs Self
water supply → Normally sewers flow → Cleaning Vel.
partially full → empty space → to absorb 0.6 – 0.9 m/sec → optimum
unseen (or) unexpected sewage flow  0.75 m/sec → Generally Best
 Main Difference Water supply→  Manning’s Formula : Generally
water flows under pressure in used for design of sewers
pipes → pipes always remain full. Mean Velocity of Flow for a given
 Sewage sewers → sewage flow slope & dimension
remain gravitational force  1  2/3 1/2 1
 For must of period sewers partly V=   m .S (or) m2/3 S of
filled with sewage
n n
conduit
 Flow in a sewer is → open channel
Where n → Manning’s coefficient
flow
(or) rugosity coefficient
 Sewage contains solid particles of
S → Bed Slope
organic & inorganic nature.
m → Hydraulic Mean
They may settle down or clog.
They may also cause wearing of Wetted Area A
Depth → =
pipe material. Wetted Perimeter P
 Sewer pipes → size & gradient  Max. Velocity occurs when sewer
so as to generate “Non Silting & flow depth is 0.81 x Full Depth
Non Scouring Velocities” at diff.  Diameter of sewer should not be
possible discharges. less than 150mm
Sewer pipes must be laid at a
 1
continuous downward gradient  Q = AV = d r  .m2/3 .S 1/2
upto out fall from where it will be 4 n
lifted up, treated, disposed off. Problem: Determine the velocity of flow in
 Design period of sewer lines → 30 a circular sewer diameter 150cm, laid on
yrs. a slope of 1 in 750 while flowing full?
 Self Cleaning Velocity → Minimum Mode of cast iron pipe
velocity of flow at which no silting Solution: Hydraulic Mean Depth when it
occurs. 1 .5
is flowing fell m = d/4 = = 0.375
 The velocity which causes both 4
floating & heavy solids to get 1
transposed easily. Slope B = n = 0.020 for
750
 Larger size of the sewer → Higher
cast iron pipe
is the velocity of flow Vs α d
1 2/3
Vs = C K .d ( S  1) → Shield’s V= .m S
n
8K 1
Equation (or) Vs = ( S  1) gd = .(0.375) 2 / 3 1 / 750
f 0.020
0.52 0.52
=  =
f – Frictional Factor 0.020  27.39 0.55
C → Chazy’s Constant 0.94 m/sec
K → Sediment Characteristic
Constant determined by Express
S → Sp. Gravity of Sediment d →
Diameter of Grain

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FUTURE GATE ACADEMY EE

Permissible Maximum Velocities in  Mfs from clay & shales → Good

Sewers: resistance to sulphide corrosion
1) Cast Iron Pipes : 3.5 – 4.5 m/sec 4) Brick Sewers → Storm Wake Drains
2) Easthern Channel: 0.6 – 1.2 m/sec 5) Cast Iron Sewers : Strong → outfall
3) Brick Sewer Pipe : 1.5 – 2.5 m/sec sewers, inverted siphons rising mains
4) Stone ware sewer pipe: 3 – 4.5 m/sec at pumping station
5) Concrete sewer pipe :2.4 – 3.0 m/sec Testing of Sewer Pipes: Physical Tests
(1) Water Test → Leakage
Shape of Sewer: (2) Mirror Test → for straight men
1) Circular 2) EGG Shaped of sewer line
1) Preferred for 1) Preferred for (3) Ball Test → to detect any
separate sewerage combined system obstruction
system because because DWF (Dry
Discharge does not Weather Flow) is METHOD OF DESIGN OF SEWER:
vary too much & about 1/20 to 1/25 Sewers in a sewerage system design can
chances of sewer times the max. be treated as Open Channel. The
running at low discharge. designing involves following steps:
depth are less. 2) At low discharges 1) ZONE FORMATION: Town (or)
2) Uses min. maintains city → Divided into different zones
quantity of material hydraulic mean Layout of sewers → plotted along the
 Economical depth nearly roads
3) To carry large uniform. Each Zone → Marked Separately
discharge most 3) For Full Sewer which carry sewage & storm →
durability → Horse condition, ‘m’ is also moved separately
shoe shaped → same for circular & 2) SEWER NETWORK: 1) proposed
Structure (or) egg shaped. arrangement (or) network for different
Srisialam But is higher for zones is then worked out & marked on
underground smaller depths. the plant.
tunnel High velocity (2) Low lying areas → Marked
provides best self separately → isolated form main
cleaning velocity at sewage system → pumping is done →
low discharges. pumps are arranged in suitable
locations.
Sewer Materials: (3) Sewage → Flow under gravitation
1) Asbestos Cement Pipes → Used for at force  Sewage flow → High level
bringing down the rain water from zone → low level zone
roots or sullage from kitchen as these 3) SEWAGE QUANTITY: → Maximum
are unable to bear huge comp. & minimum quantity of sewage
stresses. calculated for each sewer.
Simplex joint → Flexible & permits 120 → Variation factors determined
deflection → Actual quantity of sewage for which
2) RCC Pipes: (1) Strong → used for sewer is designed is
branch pipes & main sewers (2) Bell & determined.
Spigot joint (3) Crown Corrosion → 4) FLOW VELOCITY: Limiting velocity,
Easily undergo corrosion by eminating self cleaning velocity & maximum
gases like H2S  Vitrified by clay lining velocity calculated.
to overcome corrosion. 5) SEWER SECTION: Area of sewer
 Corrosion is due to anaerobic Q Discharge
decomposition of sewage solids
section ‘A’ = 
V Velocity
3) Vitrified Clay (or) Stone wave
6) GRADE OF SEWERS: Slope of sewer
Sewers: House Laterals & lateral
line is determined & longitudinal
sewers

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FUTURE GATE ACADEMY EE

section of each sewer line is drawn in 05. Determine the correctness or

suitable case. otherwise of the following Assertion (A)
These sections show invert levels & all and Reason (R) (GATE-2008-2M)
sewer appearances. Assertion (A) : The crown of the
Keeping suitability of depth of sewer outgoing larger diameter sewer is
from ground level at highest point (or) always matched with the crown of
point of disposal, the laying of sewers incoming smaller diameter sewer.
can be started.
Reason (R) : It eliminates backing up
of sewage in the incoming smaller
diameter sewer.
PREVIOUS QUESTIONS (a) both A and R are true and R is
01. At the same mean velocity, the ratio of the correct explanation of A
head loss per unit length for a sewer (b) both A and R are true but R is not
pipe flowing full to that for the same a correct explanation of A
pipe flowing half full would be
(c) A is true but R is false
(a) 2.0 (b) 1.63
(d) A is false but R is true
(c) 1.00 (d) 0.61
02. An inverted siphon is a
(a) device for distributing septic tank PRACTICE QUESTIONS
effluent to a soil absorption
system 01. The flow in a sewer is
(b) device for preventing overflow from (a) pressure flow
elevated water storage tank (b) open channel flow
(c) device for preventing crown (c) laminar flow
corrosion of sewer
(d) super-critical flow
(d) section of sewer which is dropped
02. Find the minimum velocity and
below the hydraulic grade linein
gradient required to transport
order to avoid an obstable.
sewage with sand particles of 1 mm
03. A circular sewer 2m diameter has to diameter and specific gravity of 2.65
carry a discharge of 2m3/s when through a sewer of 1 m diameter.
flowing nearly full. What is the Assume K = 0.1, f = 0.03. The sewer
minimum required slope to initiate may be assumed to run half full.
the flow ? Assume Manning’s Take Manning’s n = 0.013
n = 0.015. GATE-2000-2M)
03. Calculate the diameter of a circular
(a) 0.00023 (b) 0.000036 vitrified clay sewer (n = 0.013),
(c) 0.000091 (d) 0.000014 which will just carry 0.5 cumecs,
04. An existing 300 mm diameter circular when flowing full at a slope of 1 in
sewer is laid at a slope of 1: 28 and 1000. If the flow were at 0.6 depth
carries a peaok discharge of 1728 what would be the discharge and
m3/d. Use the partial flow diagram velocity in the partially full sewer,
shown in the figure below and assume given the following data :
Manning’s n = 0.015. At the peak d/D q/Q v/V
discharge, the depth of flow and the 0.1 0.02 0.3
velocity are, respectively
0.5 0.39 0.8
(a) 45 mm and 0.28 m/s
0.6 0.54 0.88
(b) 120 mm and 0.50 m/s
0.8 0.85 0.01
(c) 150 mm and 0.57 m/s
(d) 300 mm and 0.71 m/s

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FUTURE GATE ACADEMY EE

2) Quality Characteristics of Sewage

Decomposition of Sewage: dependent upon the substances
(a) Aerobic (Oxidation) (i) occurs in contained in sewage.
presence of O2 (b) End products → Composition: Sewage contains 0.08 to 0.1
CO2, Sulphates, Nitrates. percent solid matter (by weight) Solids
(b) Anaerobic (Reduction) (i) Occurs in present may be in any of the four forms
presence of O2 (b) End products → suspended, dissolved, colloidal and
CO2, CH4, H2S, NH3. settleable, suspended solids are
Sewage: Organic → attracted by remaining dissolved in sewage just as salt
streamlines in water. Colloidal solids are finely
 Before design of any sewage treatment divided solids either in solution or in
plant, the knowledge of characteristics suspension. Settleable solids are portions
of sewage is essential. of solid matter, which settle out if sewage
is allowed to remain undisturbed for a
PHYSICAL CHARACTERISTICS: period of 2 hours. It is estimated that
Fresh Sewage → Alkaline about 1,000 kg. of sewage contains 0.45
→ 7.3 – 7.5 pH kg. of solids of which 0.225 kg. is in
Septic Sewage → Acidic solution, 0.112 kg. in suspension and
1) Colour: Yellowish, Grey or Light 0.112 kg. settleable.
Brown → Fresh Sewage
The solids in sewage comprise of both the
Indicates freshness of sewage Black
organic and inorganic matter. The
(or) Dark Brown → Stale & Septic
organic matter accounts for about 45
Sewage
percent of the total solids and consists of
→ When it is black → Decomposition
animal and vegetable wastes and
starts
proteins, carbohydrates from vegetable
Industrial sewage colour depends on
matter, sugars, startches, cellulose, fats
the chemical process of industry
and leaves from kitchens, laundries,
2) Temperature : Slightly higher than
garages shops etc; and urea. Compound
water supply.
of nitrogen formed on the decay of animal
 If sewage flows in closed conduits or plant, which later breaks up into
temp rises, resulting in the ammonia. The inorganic matter accounts
increase of viscosity & bacterial for the other 55 percent of solids and
activity. consists of minerals and salts viz., sand,
 Temp. has an effect an (a) gravel, debris, dissolved salts, chlorides,
Biological activity → increases sulphates etc.
with increase in temp.
Besides solids and liquids, sewage also
(b) solubility of gases in sewage →
contains gases obtained from atmosphere
decreases with increase in temp.
and due to the action of bacteria on
3) Turbidity: Depends on quantity of
compounds in solution and suspension.
solid matters present in it in
These gases are mostly hydrogen sulphide
suspension ___
(H2S), carbon dioxide (CO2) and methane
4) Odour : Foul small
(CH4).
CHEMICAL CHARACTERISTICS OF
SEWAGE DETERMINATION OF SOLIDS:
Chemical Characteristics: These indicate (1) Total Solids: Total Solids =
the state of sewage decomposition, its Mass of Residue left
strength are type of treatment required. mg/l
Vol. of sample evaporated
Fresh sewage is alkaline and good for
bacterial action. Stale or septic sewage is Determined by evaporating [1030C]
acidic and difficult to be efficiently known vol. of sewage sample &
treated. The chemical characteristics are

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weighting the dry residue left in the (7) Colloidal Solids → Solids between
containes. suspended & Dissolved solids
(2) Suspended Solids: Solids retained by → They do not settle by gravity due to
filter of 1μ pores & visible to naked Brownian movement.
eye. → Removed by chemical or biological
 Filtering carried out through treatments.
Gooch crucible containing a mat of DISSOLVED OXYGEN DETERMINATION
asbestos fibre. The solids
 To know the extent of pollution of
remaining are dried & weighed.
sewage
 Size → larger than 1μ. → Glass
 At least 4 ppm of DO should be
fibre filter apparatus is used
ensured while discharging sewage
(3) Dissolved & Colloidal Solids = Total into river stream.
Solids – Suspended Solids
 If temp → more DO content →
(4) Volatile Solids: Suspended solids Less
obtained after drying are burnt &
 Saturation DO content at 200C →
ignited about 5500C in an electrical
9.2 ppm
maffle furnace for 15 – 20 min
 A single rapid test to determine
 Loss of weight due to ignition →
pollution status of river water →
volatile solids weight
DO
 Fixed Solids = Suspended solids –
Living organisms require O2 to
volatile solids
maintain their metabolic process.
(5) Settleable Solids : If the sewage is
‘DO’ in important for precipitating &
allowed to settle in imhoff cone for →
dissoluting the inorganic substances
the quantity of solids settled.
in water.
 The solids which will settle in
The solubility of O2 in water depends
1hour to the bottom of the cylinder
on its temperature
of specific height.
 Analysis of Do is main key test for
 Quantity is determined
sanitary engineering.
Gravimetrically / Volumetrically
Dissolved Oxygen
 Practically, it is determined by
This is the amount of oxygen dissolved in
Imhoff cone.
sewage. The presence of dissolved oxygen
 Imhoff Cone: Conical glass of 1 lt in sewage indicates that it is fresh or
capacity, graduated at its bottom weak. It presence in the effluent of a
in ml. treatment works indicates good
 Sewage is filled in the cone, the treatment.
vol. of solids settled in the bottom The concentration of oxygen dissolved in a
after one hr is directly noted which running stream into which sewage
gives the quantity of settleable effluents discharge must be such as not
solids. to deplete the level of dissolved oxygen
 Settleable solids → indicates vol. (D.O) concentration to a level as to
of sludge. endanger the life of aquatic animals.
(6) The hormogenous & molecularly Besides, for maintaining aerobic
dispensed solids in the water → conditions in waters receiving pollutional
dissolved solids → 0.2μ to 1μ size. → matter so as to avoid an aerobic
Not visible to naked eye & not filtered. condition, resulting into the liberation of
nuisance gases and public nuisance, it is
 Dissolved Solids → Removed by
important that the D.O. concentration
precipitation using chemicals.

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should be maintained to a level of 4-8 Rate of change of organic matter with

mg/L at all times. time (or) organic matter present in the
DO content of water is measured by sewage at time, t.
Winkler’s Method which is color reaction dLt
followed by a Titrimetric Analysis. Lt
t
 Minimum DO required for river
Lt = L.e-kt
stream, to avoid fish kills is → 4
= L.10-KDt
ppm ultimate BOD → BOD4
BODt = L(1-e-K.t)
BOD
= L(1-10-KDt)
BIOCHEMICAL OXYGEN DEMAND L → Amount of org. matter present at
The amount of O2 required for Biological time ‘t’
Decomposition of Biodegradable org. K → Rate constant
matter under aerobic conditions at
KD → Deoxygenation constant
specified temperature & specified
duration. The graph is exponential BODt = BODint
Standard BOD → reported at 200C for 5 (1  10 KDt )
days period → 68% of total demand → BODu = ultimate BOD
180 mg/lt
When t = infinite
COD represents both biologically active &
BODu = L
inactive organic matter but
KD changes with Temperature (T)
BOD represents biologically active org.
matter only. KD(t) = KD(20) [1.047]T-20

 COD > BOD Kt(20) = 0.23/day

COD – Chem. Oxidation with potassium KD(20) = 0.1/day

permanganate (or) pot. Dichromate in Aerobic Decomposition of org. matter
acid solution. done in 2stages
 Time & temperature plays an
important Role in BAD test
 Ultimate BOD BOD4 → Final BOD
Test Method: → Initial DO measured in
beginning of sample diluted with H2O
 Sample inculcated for 5 days at
200C.
 Final DO after 5 days
1) Carbonaceous Demand: The first
BOD5 = 0.68 BOD4 (or) BOD5 = 2/3 of demand that occurs during first 20
BOD4 days due to oxidation of org matter →
 BOD = (Initial DO – Final DO) I stage BOD
Dilution factor  ‘OAB’ in the above figure → O2
Dilution Factor: Number of times sewage consumed here is not recoverable.
is diluted with distilled H2O. (or) for 2%  BOD of sewage is 90% of total
solution, DF = 100/2 = 50 BOD for first stage → 5 days at
Vol.of sample + vol. of water 300C.
DF =
Vol. of sample 2) Nitrogeous Demand : Occurs due to
biological oxidation of ammonia
 ‘AC’ in above figure.

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Dilution % of Sample: b) BOD5/COD → 0.63 and 0.68 →

1) 0.1 – 1% → strong trade waste waste water → Fully biodegradable
water Theoretical O2 demand: O2 demand
2) 1 – 1.5% → Raw (or) settled worked out theoretically
sewage TOD > COD > BOD
3) 5 – 25% → Treated effluent COD = 80% TOD
4) 25 – 100% → River Water Relative Stability
 In 5 days → about 60 – 70% of O2 Available in the effluent (DO)
org. matter decomposed Total O2 req. to satisfy 1st stage BOD
In 20 days → about 95 – 99% of
org. matter decomposed  SR for sewage whose DO equals
the total O2 required to satisfy
 During BOD testing sunlight is not
BOD is 100%
allowed to fall → to prevent growth
Determination of Chlorides: Titration with
of algae plants.
standard AgNO3 in which AgCl
If algae grow, value of BOD →
precipitates out. End of titration →
Wrong
formation of red silver chromate from
COD excess AgNO3 & pot. Chromate used as
indicator.
Chemical Oxygen Demand
The test for measure of amount of carbon (A-B)N × 35450
Chlorides (Mg/l) =
in org. matter of sewage. It is chemical Vol. of sample (ml)
oxidation with pot. Permanganate (or) pot.
A → ml of AgNO3 for sample
Dichromate in acid solution.
B → ml of AgNO3 for Blank
 It is useful in determining strength
N → Normality of AgNO3
of industrial coates in sewage,
which can’t be determined by BOD Population Equivalent: Ind. Waste water
test. are generally compared with per capita
normal domestic waste water so as to
 It determines O2 required for
charge industries properly.
chem. Oxidation of org. matter
with strong chemical oxidation. Population Equivalent =
Total std BOD5 of ind. sewage per day
 BOD require 5 days but COD
Std BOD5 of domestic sewage per person per day
determination require only 5 hrs.
Avg. Std. BOD5 → for domestic sewage →
Principle: Org. matter gets oxidized by
0.08 kg/day/person
pot dichromate [K2Cr2O7] remaining after
the reaction is filtrated with Fe(NH4)2  Specific gravity of sewage is → slightly
(SO4)2. The consumption of dichromate greater than 1
gives the O2 req. for oxidation of org.  Well oxidized sewage contains N2 in
matter. the form of → Nitrates
(a-b)N × 8000  Partially oxidized sewage contains N2
 COD = in the form of → Nitrates & Ammonia
Vol. of sample in ml
 Well oxidized sewage contains S in the
a = Fe(NH4)2 (SO4)2 → for Blank in ml form of → Sulphates
b = Fe(NH4)2 (SO4)2 → for sample in ml
 Partially oxidized sewage contains S in
N = Normalityof Fe(NH4)2 (SO4)2 the form of → H2S, Sulphites
BOD/COD Ratio:  pH of fresh sewage → more than 7
a) BOD4/COD → between 0.92 and 1  The solubility of O2 in sewage is 90%.
→ waste water → Fully When compared to solubility in D.H2O
biodegradable

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PREVIOUS QUESTIONS Find the amount of oxygen required

for this reaction. The chemical mass
balance equation for the above
01. The B.O.D5 of a surface water sample
reaction is given is
is 200 mg/litre at 200C. The value of
C6H12O6 + 6O2  6CO2+6H2O
the reaction constant is K = 0.2 day–1
with base ‘e’. The ultimate B.O.D. of 07. The dissolved oxygen (DO) in an
the sample is (GATE-1997-1M) unseeded sample of diluted waste
(a) 126 mg/litre having an initial DO of 0.5 mg/l is
measured to be 3.5 mg/l after 5
(b) 544 mg/litre
days. The dilution factor is 0.03 and
(c) 146 mg/litre the reaction state constant, k =
(d) 316 mg/litre 0.22/day (to the base ‘e’ in the decay
02. In a BOD test using 5% dilution of the curve at 300C). Estimate
sample (15 ML of sample and 285mL i) The 5 day BOD of the waste at 200C
of dilution water), dissolved oxygen ii) Ultimate carbonaceous BOD
values for the sample and dilution
water blank bottles after five days
incubation at 200C were 3.80 and 08. If the BOD5.20 of waste is 150 mg/L
8.80 mg/L, respectively. Dissolved and the reaction rate constant (to
oxygen originally present in the the base ‘e’) at 200C is 0.35/day, the
undiluted sample was 0.80 mg/L. The ultimate BOD in mg/L is
5-day 200C BOD of the sample is (a) 97.5 (b) 181.5
(a) 116 mg/L (b) 108 mg/L (c) 212.9 (d) 230.5
(c) 100 mg/L (d) 92 mg/L 09. In a domestic wastewater sample,
03. Standard 5-dayu BOD of a wastewater COD and BOD were measured.
sample is nearly x% of the ultimate Generally which of the following
BOD, where x is (GATE-1999-1M) statement is true for their relative
magnitude? 2002-1M
(a) 48 (b) 58
(a) COD = BOD
(c) 68 (d) 78
(b) COD > BOD
04. In a BOD test, 5 ml of waste is added
to 295 ml of aerated pure water. Initial (c) COD < BOD
dissolved oxygen (D.O) content of the (d) Nothing can be said
diluted sample is 7.8 mg/l, after 5 10. The theoretical oxygen demand of a
days of incubation at 200C, the D.O. 0.001 mol/L glucose solution is
content of the sample is reduced to (a) 180 mg/L (b) 192 mg/L
4.4 mg/L. The BOD of the waste water
(c) 90 mg/L (d) 96 mg/L
is (GATE-1999-2M)
(a) 196 mg/l (b) 200 mg/l
11. A waste water sample has an initial
(c) 204 mg/l (d) 208 mg/l
BOD of 222 mg/L. The first order
05. The BOD removal efficiency, in BOD decay coefficient is 0.4/days.
percentage, during primary treatment, The BOD consumed (in mg/l) in 5
under normal conditions is about. days is
(GATE-2000-1M)
(a) 150 (b) 192 (c) 30 (d) 50
(a) 65% (b) 85%
12. Water stamples (X and Y) from two
(c) 30% (d) zero different sources were brought to the
06. Consider a glucose solution (C6H12O6) laboratory for the measurement of
of molarity 1.75  103 that is dissolved oxygen (DO) using modified
completely oxidized to CO2 and H2O. Winkler method. Samples were

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transferred to 300 mL BOD bottles. 2 15. A 5-day BOD of a wastewater sample

mL of MnSO4 solution and 2mL of is obtained as 190mg/I (with k =
alkaliodideazide reagent were added to 0.01h-1). The ultimate oxygen demand
the bottles and mixed. Sample X (mg/l) of the sample will be
developed a brown precipitate, whereas (a) 3800 (b) 475
sample Y developed a white precipitate.
(c) 271 (d) 190
In reference to these observations the
correct statement is (GATE-2004)
(a) both the samples were devoid of DO 16. If the BOD3 of a wastewater sample is
75 mg/L and reaction rate constant k
(b) sample X was devoid of DO while
(base e) is 0.345 per day, the amount
sample Y contained DO
of BOD remaining in the given sample
(c) sample X contained DO while sample after 10 days in (GATE-2010-2M)
Y was devoid of DO
(a) 3.21 mg/L (b) 3.45 mg/L
(d) both the samples contained DO
(c) 3.69 mg/L (d) 3.92 mg/L
13. A portion of waste water sample was
subjected to standard BOD test (5 days, 17. A sample of domestic sewage is
200C), yielding a value of 180 mg/L. The digested with silver sulphate,
reaction rate constant (to the base ‘e’) at sulphuric acid, potassium dichromate
200C was taken as 0.18 per day. The and mercuric sulphase in chemical
reaction rate constant at other oxygen demand (COD) test. The
temperature may be estimated by kT = digested sample is then titrated with
k20)1.047)T-20. The temperature at which standard ferrous ammonium sulphate
the other portion of the sample should (FAS) to determine the unreacted
be tested, to exert the same BOD in 2.5 amount of (GATE-2012-1M)
days is (GATE-2004-2M) (a) mercuric sulphate
(a) 4.90C (b) 24.90C (b) potassium dichromate
(c) 31.70C (d) 35.00C (c) silver sulphate
(d) sulphuric acid
14. To deterime the BODs of a waste
water sample, 5, 10 and 50 mL alquots
of the wastewater were diluted to 300 18. A student began experiment for
mL and incubated at 200C in BOD determination of 5-day, 200C BOD on
bottles for Monday. Since the 5th day fell on
5 days. (GATE-2006-1M) Saturday, the final DO readings were
taken on next Monday. On
The results were as follows
calculation, BOD (i.e. 7 day, 200C) was
Sl.No. Wastewater Initial DOf found to be 150 mg/L. What would be
Volume, DO the 5-day, 200C BOD (in mg/L)?
mL Mg/ Assume values of BOD rate constant
L (K) at standard temperature of 200C a
0.23 day (base e) (GATE-2013-2M)
1 5 9.2 6.9

2 10 9.1 4.4
19. The amount of CO2 generated (in kg)
3 50 8.4 0.0 while completely oxidizing one kg of
CH4 to the end products is ________
Based on the data, the average BODs of
(GATE-2014-1M-SET-1)
the wasterwater is equal to
(a) 139.5 mg/L (b)126.5 mg/L
(c) 109.8 mg/L (d) 72.2 mg/L

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FUTURE GATE ACADEMY EE

PRACTICE QUESTIONS Waster Initial Do after

01. The correct statement of comparison water DO, mg/l 5 days,
of ultimate BOD, COD, Theoretical volume ml mg/l
Oxygen Demand (ThOD) and 5-day 5 9.2 6.9
BOD (BOD5) is
10 9.1 4.4
(a) BODu > COD > ThOD > BOD5
(b) COD > ThOD > BODu > BOD5 50 83.4 0.0

(c) ThOD > COD > BODu > BOD5

(d) COD > BODu > BOD5 > ThOD 08. The 5-day BOD of a waste water
sample is obtained as 190 mg/lL
02. The 5 day 300C BOD of a sewage with k = 0.01 h-1 (base e), the
sample is 110 mg/l. Calculate its 5 ultimate oxygen demand (mg/l) of
day 200C BOD. Assume KD(20) = the sample will be
0.1/day (base 10) (a) 3800 (b) 475
03. The BOD of sewage incubated for (c) 275 (d) 190
one day at 300C has been found to
be 110 mg/l. What will be 5 day
200C BOD. Assume KD(20) = 0.1/d
KEY – PREVIOUS QUESTIONS
(base 10)
04. The average sewage flow from a city 1. D 2. D 3. C
is 80  106 1/d. If the average 5 day
BOD is 285 mg/l. Compute the total 4. C 5. C
daily 5 day oxygen demand in Kg,
and population equivalent of sewage 6. 1.807X103 Molts/Litre
assuming per capita BOD of sewage
7. (i) 200 mg/l, (ii) 400 mg/l
per day = 75 g.
05. In a test conducted for determining 8. B 9. B 10. B
the relative conductivity at 200C, the
period of incubation is found to be 11. B 12. C 13. D
12 days, Calculate the percent of
14. A 15. C 16. C
relative stability ?
06. The BOD5 of a waste has been 17. B 18. 128.112 MG/L
measured as 600 mg/l. If K =
0.23/d (base e), what is its ultimate 19. 2.75Kg
BODu of the waste. What proportion
of the BODu would remain
unoxidised after 20 days ?
07. To determine the BOD5 of a waste
water sample, 5, 10 and 50 ml
aliquots of the waste water were
diluted to 300ml and incubated at
200C in BOD bottles for 5 days. The
result were as follows. Based on the
data, the average BOD5 of the waste
water is equal to
(a) 139.5mg/l (b) 126.5mg/l
(c) 109.8mg/l (d) 72.2mg/l

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FUTURE GATE ACADEMY EE

Treatment of Sewage
Grit Chamber removes the inorganic 2. Communicators & Morenators.
suspended solids & PST removes organic  These are the units which are being
suspended matter in sewage. employed for breaking down the
smaller impurities by cutting and
Disposal of inorganic matter is much grinding actions respectively into
convenient in comparison to that of lighter ones.
organic matter; hence these are being  These units can be before and after
removed separately in the treatment plant screens. (Coarser- before, fines-offer)
(grit chamber & PST respectively.
3. Flow Equalization chamber:
Biological treatment removes the organic i) This method is employed to overcome
matter present in sewage, by inducing the operational troubles observed due to
biological activity in it which is being variation in the rate of flow of waste
carried out by bringing the contact water which affects the efficiency of
between micro-organism and the organic downstream biological process.
matter. ii) These are the units which are being
employed to damper the fluctuations
This contact can be brought by any of the in the discharge of wastewater in
following mechanism order to reduce cost and size &
increase efficiency of biological
1. By suspending biomass in process is shock loadings is diastrous
wastewater (suspended growth for micro- organism carrying out the
system )- SP, OP, ST biological activities.
2. By passing wastewater over the bio
moss layer which is attached to 4. Grit chamber/Detritus tank:
medium. (Attached growth system ) i) Grit chambers are used to remove the
– TF, RBC. inorganic suspended particles like
clay, still, sand, glass, egg shelves and
10 treatments normally remove 60% of to pass forward organic suspended
suspended impurities and also satisfy 30- particles present in oil for the removal
40% BOD associated with it. in primary settling tank.
ii) These tanks are in the form of long
Secondary treatment satisfies 85-95% narrow channels which may be
BOD associated with the wastewater. rectangular or parabolic in shape.
iii) Velocity control devices such as
Anaerobic sludge/ 10 sludge/raw proportional (sutro) weir and partial
sludge/primary sludge is treated flume are employed of the end of these
anaerobically in Anaerobic digestor and chambers.
secondary or biological sludge is treated iv) Proportional weir is used if
aerobically in digestor. rectangular section is adopted &
partial flume is used if parabolic
The conc. of organic matter in PST is section of got proportional well as
more than SST. Thus the growth of micro- head loss is smaller in this case.
organisms may be uncontrollable in PST. v) These units are designed generally in
To avoid this (favourable conditions in the form of two chambers one to carry
presence of O2) the 10 sludge is treated overage discharge and second to carry
anaerobically. variation of the discharge in average
discharge.
1. Screening: Screens are used to vi) These tanks are designed to settle
remove heavy suspended impurities inorganic particles of size greater than
present in wastewater.
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FUTURE GATE ACADEMY EE

0.2 mm and pass away organic other and concentration of particles,

particles. (Stokes low is not following types of settling may be
applicable). For removal of these observed in sedimentation tank.
particles the overflow rate is
1. Discrete settling (Type -1 settling)
approximately 2160m3/m2/day which
is calculated using transition law. - Shape, size and mass of particle
vii) Detention time is normally in the remaining same
range of 30-60sec. Generally 60sec. - Settling is independent of another
adopted. particle.
viii) Horizontal velocity of flow is in the - Stokes law is applicable.
range of 0.15-0.3 m/Sec. - Settling velocity can be also calculated
ix) Depth of tank is approximately 1- using transition law.
1.8m. Free board at 0.3m is provided Eg. Settling in grit chamber.
over this depth (to avoid overflow)
x) The length of tank is increased 2. Flocculants settling (Type II
approximately by 25-30% in order to settling)
consider the turbulence at inlet and - Size, shape and mass changes during
outlet of chamber. settling
- Theoretically settling velocity cannot
5. Skimming tank: be calculated.
i) It is used to remove oil, grease and - Flocks are formed
soap present in water. - This occurs when conc. of particles in
ii) It is generally used before PST. If not waters is very less (1000 mg/l)
used oil and grease present in wastewater Eg: Primary settling tank where organic
inhibits growth of micro-organisms which particles settle.
affects efficiency of biological treatment.
iii) In this method compressed air is 3. Zone of Hindered Settling (Type
passed in the wastewater from the bottom III settling)
of tank which coagulates oil & grease - This type of settling is observed when
particle & carries them along with it to conduct of the particles in water is in
surface of tank from where it can be medium range such that velocity fields of
easily skimmed off. the particles during their settlement
overlaps each other. (Conc. 1500mg)
iv) In order to increase efficiency of
skimming tank chlorine is also used also - Particle settles in form of zone in order
used along with compressed oil to destroy to maintain their relative position with
colloidal effects of proteins which keep oil respect of each other.
and grease particles in emulsion form. Eg: Settling in secondary settling tank
v) These tanks are not used in India as followed by activated sludge process is
co-agulation of oil and grease particles this type of settling.
does not take place at higher
temperature. 4. Compression Settling (Type IV
Settling)
vi) Instead of using skimming tanks - The concentration of particles is very
evacuators can also be used for removable high such that they are in physical
of oils and grease. (by creating negative contract with each other (>1800mg/L).
pressure on top and pulling oil and grease - Bottom layers of these particles
particles). support the weight of top layers of
particles due to which any further settling
6. Sedimentation :
results by compression of entire particles
 Depending upon tendency of in the medium accompanied by squeezing
suspended particles to interact with each out of water from voids of particles.

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FUTURE GATE ACADEMY EE

PREVIOUS QUESTIONS 4. g = 9.80 m/s2,

01. A typical biological process in treating  = 1.002  10-3 N – s/m2 at 200C.

waste water using aerated lagoon Asuming that the Stokes law is
can be described by one of the valid, the largest diameter particle
following schematic diagrams that would be removed with 100
(GATE-1997-1M) percent efficiency
(a) 0.04 mm (b) 0.21 mm
(c) 1.92 mm (d) 6.64 mm

05. In aerobic environment, nitrosomonas

convert (GATE-2005-1M)
(a) NH3 to NO2 (b) NO2 to NO3
(c) NH3 to N2O (d) NO2 to HNO3

06. A circular primary clarifier processes

an average flow 5005m3/d of
municipal wastewater. The overflow
rate is 35m3/d. The diameter of
clarifier shall be
(a) 10.5 m (b) 11.5 m
02. A municipal waste treatment plants to
work with average and peak loading (c) 12.5 m (d) 13.5 m.
rates of 4,000 and 8,000 m 3/day
respectively. Design a primary KEY
clarifier to remove 65% suspended
matter at average flow. An average
overflow rate of 35m3/m2 day is 1. C 2.70.735 m3/m2/day
expected to correspond to 65%
3. B 4. B
suspended matter removal
efficiency. Obtain the diameter, side 5. A 6. D
wall depth, detention time and
calculate the overflow rate at peak
condition. (GATE-1997-2M)

03. The unit in which both sedimentation

and digestion processes of sludge
take place simultaneously is
(a) Skimming Tank
(b) Imhoff Tank
(c) Detritus Tank
(d) Digestion Tank

04. The following data are given for a

channel-type grit chamber of length
7.5 m.

1. flow-through velocity = 0.3 m/s

2. the depth of wastewatger at peak
flow in the channel = 0.9 m
3. specific gravity = 2.5
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4) Activated Sludge Process

(Aerobic Suspended Growth System)
The entire process of biological Micro –organisms in the system (Mixed
decomposition is believed to take place in Liquor volatile suspended solids)
four phases.
I. Lag Phase In activated sludge process secondary of
II. Log Phase biological sludge (sludge coming from
III. Stationary Phase SST) which mostly consist comprises of
IV. Endogenous Phase living micro-organism also termed as
I. Lag Phase Active micro-organisms is recirculated
 In this phase micro-organisms get back to aeration process to carry out
accumulated to food and environment oxidation of or decomposition organic
given to them. Rate of growth lags matter. Hence process is termed as
(Com law) activated sludge process.
 If the micro-organisms are already
familiar with the environment and Activated sludge process is suspended
food given to them then duration of growth culture in which sludge return is
Lag Phase is less. done by following any of the flow
 The growth of biomass in this phase is (mechanism)
very less hence it is termed as Lag
Phase. A) Complete Mix Process:

II. Log Phase (i) It is adopted for plants having

 In Log Phase, micro –organisms capacity less than 25 MLD
reproduces rapidly by cell division (ii) In this process incoming wastewater is
leading to the rapid increase in the completely, mixed with activated
growth of biomass at corresponding sludge by inducing mixing in the
decrease in the organic matter in aeration tank.
wastewater. (iii) In complete mix process square or
 In ASP, we try to maintain system in circular tanks are used proved with
log phase. under normal treatment mechanical aerators.
condition. (iv) Operational Stability of this process
III. Stationary Phase: with regard to shock loading is very
 In this phase growth of biomass is high as complete mixing is induced in
slightly abstracted due to endogenous the aeration tank.
reparation resulting from scarcity of (v) F/M ratio and oxygen demand is
food in the bottom layer of the system. uniform throughout the tank in this
IV. Endogenous phase: system (due to mixing)
 In this phase endogenous process of (vi) This system is capable of holding high
respiration starts due to depletion of MLSS concentration.
organic matter from system. B. Plug flow Process:
MLSS: (i) It is conventional method used for
Mixed Liquor Suspended Solids is plants of capacity more than 300MLD.
generally taken as index for active micro – (ii) Plug how process represents the grad
organisms present in wastewater but it u at how at wastewater along the
also comprises of dead cell mass and length of tank in which activated
other inorganic impurities present in sludge is mixed at the inlet.
wastewater. Hence to remove the effect of (iii) Long narrow channels are adopted in
inorganic impurities considered in MLSS. plug how process. In this system F/M
The MLYSS is used to represent active. ratio and oxygen demand gradually
decreases along the length of tank.

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(iv) This process locks operational Re-circulation Ratio (R):

stability with respect to shock
loadings is comparatively.
X
(v) In plug how process decomposition of R
organic matter upto the nitrate level XU  X
takes place as sufficient time is
provided for the nitrifying bacteria Sludge Volume index (SVI)
which satisfies nitrogenous BOD. MLSS concentration in aeration tank is
(vi) More efficiency of complete mix dependent on circulation ratio which is
process than this method. determined on the basis of sludge volume
index.
Design Parameters Used in ASP: Sludge volume represents physical state
of sludge and signifies concentration of
Hydraulic retention time (HRT): sludge in aeration tank which helps in
It is defined as volume of aeration tank to deciding sludge recirculation rate to
the rate of flow of wastewater excluding obtain the desired MLSS and F/M ratio
recirculation. for the given degree of treatment.
Sludge volume index is defined as volume
V
HRT = (hours) occupied in ml by 1 gm of solids when
Q0 allowed to settle for 30 min. (m-1/gm)
Organic loading rate (OLR):
It is defined as ratio of mass of BOD the
system to the volume of aeration tank.
Q0 S0
OLR = kg/m3/day
V
Specific food utilization (U):
It is defined as ratio of BOD removed in
the system to the mass of biomass in the
aeration tank.
Qo ( S0  S )
U=
VX
Sludge Age (  c ):
It is the overage time for which sludge
remains in the system. It is defined as
ratio of mass of MLSS in the aeration tank
to the mass of MLSS leaving the system
per day.
VX
c= days
Qw X u
F/M Ratio : It is defined as ratio of BOD
applied to system to the mass of biomass
in aeration tank. Efficiency pf tank
depends on F/M ratio (less F/M ratio
more efficiency)
Q o .S0
F/M = day -1
V .X
1
Y U = KER +
Qc

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PREVIOUS QUESTIONS 1. A crucible was dried to a constant

mass of 62.485 g
2. 72mL of a well-mixed sample was
01. An activated sludge aeration tank
taken in the crucible
(length 30.0m; width 14.0m;
3. The crucible with the sample was dried
effective liquid depth 4.3 m) has the
to a constant mass of 65.020 g in a
following parameter : (1998-2M)
drying oven at 1040C
Flow 0.0796 m3, soluble BODs after
4. The crucible with the dried sample was
primary settling 130 mg/L; mixed
placed in a muffle furnace at 600C, for an
liquor suspended solids (MLSS)
hour. After cooling, the mass of the
2100 mg/L; mixed liquor volatile
crucible with residues was 63.145 g.
suspended solids (MLVSS) 1500
The concentration of organic fraction of
mg/L; 30 minute settled sludge
solids present in the return sludge sample
volume 230 mL/L; and return
is
sludge concentration 9100 mg/L.
(a) 8800 mg/L (b) 25000 mg/L
Determine the aeratin period, food
(c) 33800 mg/L (d) 42600 mg/L
to microorganisms (F.M) ratio,
sludge volume index (SVI), and
06. Bulking sludge refers to having
return sludge rate.
(a) F/M < 0.3d
(b) 0.3/d F/M < 0.6/d
02. Chlorine is sometimes used in
(c) F/M = zero
sewage treatment (GATE-1999-1M)
(d) F/M > 0.6/d
(a) to avoid flocculation
(b) to increase biological activity of
Common Data for Questions 07 & 08
bacteria
A completely mixed activated sludge
(c) to avoid bulking of activate
process is used to treat a wastewater flow
sludge
of 1 million litres per day (1 MLD) having
(d) to help in grease separation
a BOD5 of 200 mg/L. The biomas
concentration in the aeration tank is 2000
03. Critical factors for the activated
mg/L and the concentration of the net
sludge treatment process are
biomass leaving the system of 50 mg/L
(GATE-2000-1M)
the aeration tank has a volume of 200 m3.
(a) maximum hourly flow rate
(GATE-2007-2M)
(b)maximum and minimum flow rate
07. What is the hydraulic retention time
(c) maximum hourly flow rate
of the wastewater in aeration tank ?
(d) minimum hourly flow rate
(a) 0.2 h (b) 4.8 h
(c) 10 h (d) 24 h
04. Settling test on a sample drawn
08. What is the average time for which
from Aeration Tank liquor of ASP
the biomass stays in the system ?
(MLSS = 2800 mg/I) was carried out
(a) 5 h (b) 8 h
with 1 litre sample. The test yielded
(c) 2 days (d) 8 days
a settled volume of
200 ml. The value of sludge volume Common Data for Questions 09 & 10
Index shall be (GATE-2003-2M) An activated sludge system (sketched
(a) 14.0 (b) 34.2 below) is operating at equilibrium with the
(c) 71.4 (d) 271 following information Wastewater related
data :
05. An analysis for determination of
solids in the return sludge of
Activated Sludge Process was done
as follows : (GATE-2004-2M)

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Flow rate = 500 m3/hour, 03. The MLSS concentration in an

influent BOD = 150 mg/L, aeration is 2000 mg/l and the
sludge vol, after 30 min of settling in
effluent BOD = 10 mg/L.
a 1,000 ml graduated cylinder is
Aeration tank related data: 176 ml, then SVI is
hydraulic retention time = 8 hours, (a) 88 ml/gm (b) 176 ml/gm
mean-cell-residence time = 240 hours, (c) 200 ml/gm (d) zero
volume = 4000 m3,
Common Data for Questions 04 - 09
MLSS = 2000 mg/L
An average operation data for
conventional activate sludge treatment
09. The food-to-biomass (F/M) ratio (in plant is as follows:
kg BOD per kg biomass per day) for (i) Waster water flow = 35,000 m 3/d
the aeration tank is (ii) vol. of aeration tank = 10,9000 m3
(a) 0.015 (b) 0.210 (iii) influent BOD = 250 mg/l
(iv) effluent BOD = 20 mg/l
(c) 0.225 (d) 0.240
(v) MLSS = 2,500 mg/l
(vi) effluent suspended solids = 30 mg/l
10. The mass (in kg/day) of solids (vii) waste sludge suspended solids =
wasted from the system is 9,700 mg/l
(a) 24000 (b) 1000 (viii) quantity of waste sludge = 220 m3
(c) 800 (d) 33 Based on this information
04. Aeration period is (hrs)
11. The dominating microorganisms in
(a) 4.5 (b) 6.4 (c) 7.47 (d) 8.43
an activated sludge process reactor
are
05. Food to microorganism ratio F / M
(a) aerobic heterotrophs (kg/d BOD/kg MLSS) is
(b) anaerobic heterotrophs (a) 0.18 (b) 0.32
(c) autotrophs (d) phototrophs (c) 0.48 (d) 0.52

06. Percentage efficiency of BOD

PRACTICE QUESTIONS removal is
(a) 80% (b) 88%
01. A town produces sewage of 50  106 (c) 92% (d) 96%
l/d with BOD of 180 mg/l. Taking
permissible volumetric organic 07. Sludge age in days is
loading as550 gm of BOD per 1 (a) 8.58 (b) 10.42
cu.m of volume, the volume of (c) 12.86 (d) None
aeration tank required is
(a) 16363 m3 (b) 14360 m3 08. Assuming that the supernatant is
(c) 2800 m 3 (d) 8423 m3 clear of suspended solids one liter of
activated sludge containing 1000
02. A town produces sewage of 50  106 mg/l MLSS occupies a volume of
l/d with BOD of 180 mg/l. Taking 200 ml after settling for 30 minutes
in a measuring cylinder, the value of
F
 0.5d 1 and MLSS as 1800 mg/l SVI will be
 (a) 50 (b) 100 (c) 200 (d) 400
the volume of aeration tank required
is 09. What is the hydraulic retention time
(a) 16363 m3 (b) 14360 m3 of the wastewater in aeration tank ?
(c) 2800 m 3 (d) 10000 m3 (a) 0.2h (b) 4.8h (c) 10h (d) 24h

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KEY FOR PREVIOUS QUESTIONS

1. 0.0238 m3/ sec 2. C 3. C 4. C 5. B

6. A 7. B 8. D 9. C 10. C 11. A

Design data for ASP:

Process Flow MLSS F/ HRT c R  kgO2 reqd MLVGS

Regime (mg/L M (hrs)
day kgO2 removed MLSS
)
s
Average Plug flow 1500- 0.3- 4-6 5-8 0.25- 85- 0.8-1 0.8
Flow 3000 0.4 0.5 95%
Process
Complete Complete 3000- 0.3- 4-5 5-8 +0.25 85- 0.8-1.0 0.8
mix mix 4000 0.5 -0.8 92%
process
Extended Complete 3000- 0.1- 12- 8-12 0.8-1 96 1-1.2 0.8
aeration mix 5000 0.1 24 to
8 98%

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5) Trickling Filters
(Aerobic Attached growth system)
 As the wastewater trickles through organisms. Thus biomass formed will
filler medium, biomass layer grows clog their filter thus reducing rate of
and attaches to the medium surfaces removal of organic matter.
making filter ready for operation  Trickling filter used to carry out
within 2-3 weeks removal of organic matter is generally
 When the wastewater flows through of two types.
this biomass layer, organic matter  Standard rate trickling filter
present in it comes in contact with the  High rate trickling filter
micro-organism present in the
biomass layer which carries out A) Standard rate trickling filter :
oxidation of organic matter resulting (i) In these types of tickling filter
in formation of biomass which gets hydraulic loading rate is less as there
attached over filter medium. is no provision of re-circulation in this
case. (we can manipulate hydraulic
 The layer in which this process takes
loading rate not organic loading rate)
place is termed as slime layer.
(ii) Distribution of wastewater on the filter
 The thickness of slime layer varies in wastewater is done by either
from 0.1 -2 mm. rotatary distribution method or by
 In the top surface of this layer aerobic spray nozzle method.
process takes place (0.1 -0.2 mm) and (iii) Rotatory distribution system is more
in remaining anaerobic process takes effective. No odorous gas forms in this
place. Over a period of time scarcity of system.
food & oxygen takes place in bottom
layer due to increase in thickness of Operational troubles in Standard rate
slime. Leading to endogenous trickling filter:
respiration in bottom layer which
leads to increase in conc. of dead cell 1. Fly Nuisance:
mass in layer. (i) As TF is open –to atmosphere, insects
 The presence of dead cell mass are generated over its surface which
weakers bond between medium attracts files (larvae of insects serves
particles and biomass layer. Resulting as food for files)
in its sloughing removal due to (ii) This problem can be avoided by
continuous flow of wastewater spraying insecticides like DDT over
through filter. surface of TF (Dichloro Diphenyl
Trichloro Ethane –DDT)
 This sloughed biomass is finally token
to SST for removal
2. Odour Problem:
 The Fate of removal of organic matter (i) As the hydraulic loading rate is less in
in system depends upon following case of SRTF, the decomposition of
factors. organic matter takes place for longer
1) Hydraulic loading rate duration leading to evolution of
2) Organic loading rate odorous gases.
(ii) To avoid this, hydraulic loading rate is
3) Temperature (high temperature
increased.
increases rate of removal of
3. Pending Problem :
organic matters)
(i) Due to the growth of fungi and algoe
 Large organic loading rate will
in the trickling filter chocking of voids
produce more amount of organic
takes place leading to standing on
matter removed by more micro

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pending of waste water above tricking Design data for trickling filter:
filter. Q0
(ii) It can be overcome by addition of any  Surface area =
HLR
oxidising agent like lime copper
(excluding recirculation)
sulphate etc.,
Q0  QR
 Surface area =
In standard rate trickling filter, oxidation HLR
of organic matter upto nitrate level takes (including recirculation)
place (nitrification) as hydraulic loading Q0 S 0
rate is less in this case which provides  Volume =
OLR
sufficient time for nitrifying bacteria
(excluding recirculation)
(outolrops for satisfying nitrogenous BOD.
Q0 S 0 Q R .S R
 Volume =
High rate trickling filter: OLR
In high rate trickling filters hydraulic (including recirculation)
loading rate is increased by recirculating
the portion of the treated wastewater Efficiency of Trickling filter:
discharge. As per GOI manual.
100
(iv) The efficiency of high rate trickling (a) for SRTF:  (%) =
1  0.0044 u
filter is more than standard rate
Where u = unit organic loading rate
trickling filter as the contract between
(kg/Ha.m/day)
micro-organism and organic matter is
brought more than once.
(b) HRTF : for stage I,
(v) In HRTF fly nuissance odour & 100
I =
ponding problem is not formed due to W1
high loading hydraulic rate & reduces 1  0.0044
V1 F1
cabbaging of filter. It increases
where W1 – amount of BOD entering
sloughing of bio moss layer and keeps
into stage 1 (kg/day) (Q0 –S0)
filter ventilated.

V1 – volume of filter in stage I (Ham)

(vi) BOD fluctuations and shock loadings
F1 – Recirculation factor for Stage I
are also dampened in HRTF (more
operational stability) due to mixing of
recirculated waste water (indirectly (1  R) QR
F1 = =R=
flow equivilisation is developed) (1  0.1R) 2 Q0

(vii) In HRTF nitrogenous BOD is not HRTF : for stage II ,

satisfied as sloughing takes place
before the action of outotrops.
100
 II =
(viii) The recirculation of Sewage in 0.0044 W2
1
HRTF can be performed in various 1   I V2 F2
stages
where W2 – BOD entering into stage II
(ix) Classification of high rate trickling
Overall efficiency of biological process
filter on the basis of method of
 = 1 + (1- I)  II
recirculation of wastewater.

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Design data for trickling filter:

Design Parameters SRTF HRTF Super HRTF

Hydraulic loading rate 1-4 10-40 40-200
(m3/m /day)
2 (Including
recirculation)
Organic loading rate 0.08 - 0.32 0.32-1 0.6-0.8
(kg/m3/day) (Excluding
recirculation)
Depth (m) 0.8-2.5 0.9-3 4.5-12
QR 0 0.5 -3 1-4
Recirculation ratio (R = )
Qo

Bio-Towers
 These are super high rate trickling filters in which plastic medium is used which
provides more surface area for the growth of biomass layer resulting in increased
efficiency.
 The porosity of this plastic medium is very high which ensures higher hydraulic
loading rate. Leading to the increased sloughing in filter medium which results in
higher rate of removal of organic matter.
 Odour problem, pending and fly nuisance is not observed in these types of filters.
 A well separated bio-towers result in the decomposition of organic matter upto the
nitrate level.
 BOD of the effluent coming out of these filters can be computed using Eckenfedler’s
equation.
e kD / Q
n
Se = S o
Where,
Se - BOD of effluent in mg/L
So - BOD of influent in mg/L
D – depth of tank in m
K – Treatibility constan (min-1)
Q – hydraulic loading rate (m3/m2/min)
n = constant depending on type of medium
(n = 0.5, for plastic medium)

Rotating Biological contractors (Aerobic attached growth system)

 In this case rotating discs are used which are closely placed with each other as a
medium for growth of biomass layer
 This biomass layer comes in contact with organic matter when disc is emerged in
wastewater & utilises o2 when exposed to atmosphere to oxidise organic matter
resulting in formation of biomass which gets itself attached to rotating disc.
 Over a period of time thickness of biomass layer over the rotating disc increases and
is sloughed off sheared off due to turbulence created by rotation of disc in
wastewater and is finally taken to SST for its settlement.
 The depth of emersion of disc is kept to be 40% size of disc in wastewater.
 This system takes the advantage of both attached and the suspended growth system.
 It is highly effective method of treatment of organic matter which oxidises the
biological solids upto nitrate level.

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PREVIOUS QUESTIONS PRACTICE QUESTIONS

01. T trickling filter is designed to 01. The BOD of a sewage entering PS is
remove (GATE-2002-1M) 200 mg/l. If the effluent of the
(a) Settleable Solids sewage of TF is 40 mg/l, then the
efficiency of the TF is
(b) Colloidal Solids
(a) 40% (b) 85%
(c) Dissolved Organic Matter
(c) 80% (d) zero
(d) None of the above

02. A TF is designed with an unit

02. Which of the following sewage organic loading of 0.175 kg/m 3/day.
treatment methods has inherent If the influent BOD of the sewage is
problems of odour, ponding, and fly 150 mg/l, then the effluent BOD is
nuisance ? (GATE-2003-1M)
(a) 43.35 mg/l
(a) UASB system (b) 23.3 mg/l
(b) Activated sludge process (c) 126.7 mg/l
(c) Trickling filters (d) None of the above
(d) Stabilization ponds
03. From a primary settling tank sewage
03. Match the following : flows to a standard trickling filter at
(GATE-2009-2M) a rate of 6 MLD having a 5-day BOD
Column – I: of 150 mg/l, Find

P. Grit chamber (a) the depth and volume of the

filter adopting a surface loading of
Q. Secondary settling tank
2500 l/m2/day and an organic
R. Activated sludge process loading of 175 gm/m3/day.
S. Trickling filter (b) the efficiency of the filter unit
Column – II: using NRC equation
1. Zone settling (c) the B.O.D of effluent
2. Stoke’s law
3. Aerobic 04. Calculate the volume of the single
4. Contact stabilization stage Trickling filter required to yield
an effluent of BOD5 of 20mg/l when
The correct match of Column – I with
treating settled domestic sewage
Column – II is
with BOD5 of 120 mg/l. The sewage
(a) P – 1, Q – 2, R – 3, S – 4 flow is 2200 m3/day and the
(b) P – 2, Q – 1, R – 3, S – 4 recirculation is constant at 4000
(c) P – 1, Q – 2, R – 4, S – 3 m3/day.
(d) P – 2, Q – 1, R – 4, S - 3

KEY:

1. C 2. C 3. D

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6) Oxidation Ponds
(Aerobic suspended growth system)
Oxidation pond is in the form of long Latitude (N) Organic
earthen channels which provides Loading Rate
comparatively large detention time during kg/Ha/day
which waste water gets oxidised by action 36 150
of micro – organisms. 32 175
28 200
 In this pond special relationship 24 225
exists between aerobic micro- 20 250
organisms and algae in aerobic zone. 16 275
14 300
 Oz released by algae during 12 325
photosynthesis is utilised by micro 8 350
organisms to carry out 4 375
decomposition of organic matter
resulting in biomass which again BOD entering
serves as nutrient for algae. Plan Area =
Organic laoding rate
 Such mutual beneficial relationship Q0 S 0
=
is termed as the symbiotic OLR
relationship. This type of relationship
(iv) Pathogenic bacteria removal efficiency
also exist between aerobic micro-
is a approximately 99% while BOD
organisms in top zone and anaerobic
removal efficiency is 95%.
micro organisms in bottom zone.
(v) The effluents of oxidation pond are nol
discharged and are used for sewage
 Gases released by anaerobic micro –
forming as it is sufficiently clarified.
organism during decomposition rise
to surface and act as food for aerobic (vi) Rate of accumulation of sludge varies
micro-organisms and biomass between 2.5 cm/year.
formed by aerobic micro-organism (vii) Due to overloading odour problem
settles down to bottom layer to act as may persist in pond. To avoid ti
nutrient for anaerobic micro- sodium nitrate is added which is
organism. oxidising agent because that act as
- removes odorous gases
 In real terms oxidation pond is
- serves as nutrient for
facultative process (practically)
growth of algae.

Design Parameters for oxidation pond: (viii) Detention time of oxidation pond
can be computed empirically by
(i) Depth of tank is in the range of 1-1.8m following relation
(1-1.8 m) 1  L 
Ld = log   days
KD Lr
(ii) Detention time is approximately 2-3
weeks (14-21 days) Where L – influent BOD in mg/L
Y = BOD removed in mg/L
(iii) Organic loading rate depends upon
temperature of locality where pond is
Oxidation pond is generally provide for
to be constructed and temperature of
small communities having no source of
locality and in turn depends on
power.
latitude.

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PREVIOUS QUESTIONS PRACTICE QUESTIONS

01. Design an oxidation pond to treat
01. Sewage treatment in an oxidation waste water from a community of
pond is accomplished primarily by 10,000 people with a percapita BOD
(a) alga-bacterial symbiosis contribution of 40 g/day water is
supplied at a rate of 100 lpcd and
(b) algal photosynthesis only
80% of this is converted to sewage.
(c) bacterial oxidation only Determine the detention time and
(d) chemical oxidation only also whether the effluent is suitable
for irrigation of 80% of BOD removal
02. From amongst the following sewage is achieved in the pond
treatment options, largest land Assume :
requirements for a given discharge (a) ponds L : B :: 4 : 1
will be needed for (GATE-2003-1M) (b) Depth of the pond : 2m
(a) trickling filter
(c) permissible BOD  100 mg/l for
(b) anaerobic pond irrigation water
(c) oxidation ditch (d) The permissible BOD loading
(d) oxidation pond rate to oxidation pond is 200 kg
BOD/hect/d

02. Design an oxidation pond for the

KEY: following data :
Population served: 10,000
1. A 2. D
Sewage flow : 150 lpcd
Influent BOD : 300 mg/l
Effluent BOD : 30 mg/l
Organic loading rate: 300 kg/hec/d
Pond removal constant KD = 0.1/d
L : B :: 4 : 1

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7) Sludge Digestion & Drying

Sludge produced in sedimentation tank is  These micro organisms are very
a considerable hazard to the environment delicate and work in the narrow pH
hence must be treated before the range of 6.5 to 7.5 (reason for avoiding
disposal. The treatment of sludge can be fluctuations in sewers).
carried out by any of the following
 Acid formers respond very quickly to
method.
food and convert it into acids and
alcohols but methane forms are not
1. Aerobic digestion
quick enough to convert this acids
and alcohols into methane resulting
 This method is generally adopted
in the decrease in pH of system and if
for secondary or biological sludge
this pH exceeds tolerance limit for
 In this method, sufficient
methane forms the formation of
availability of O2 is maintained in
methane ceases, leaving
the system to carry out removal of
decomposition of organic matter in
organic matter in sludge.
complete.
 Due to scarcity of organic matter in
secondary sludge, the endogenous  The prime function of anaerobic
respiration starts, leading to the digestion is to reduce the organic
increased conc. of dead cell and matter into as much of liquids (acids
other cell fragments in digested and gases) and gases (methane &
sludge which is difficult to dewater. carbon dioxide) and produce very less
quantity of biomass.
2. Anaerobic Digestion :
Factors affecting anaerobic digestion:
 Anaerobic digestion is generally
Temperature :
adopted for primary sludge which
With increase in temperature
consists of large quantity of organic
deoxygenating constant (Kd) increases
matter that may induce rapid
thus increasing biological activity.
growth of micro-organisms it
With increase in temperature of digestion
treated aerobically.
process, period of digestion decreases.
 Wastewater consists of wide variety
The digestion is mostly carried out in
of organic matter wide variety of
Mesophilic range.
micro organics required for its
decomposition.
pH: The optimum range of the sludge
 Micro-organisms which generally
digestion is from pH 6.5 -7.5
leads to anaerobic decomposition
are as follows:
Nuissance Bacteria:
In anaerobic digestion sulphate reducing
(a) Acid Formers :
bacteria are the nussiance bacteria as
 These are the micro-organisms
they reduce sulphate present in system
which may either be facultative or
into sulphide which are disastrous to
anaerobic that solubilise organic
methane farmers
matters through hydrolysis which
is further fermented to acids and
The remedy to the above problem is to
alcohols of lower molecular weight.
add join to system leading to removal of
sulphate from system.
(b) Methane farmers :
 These are the micro-organisers
which are strictly anaerobic that
convert acids and alcohols along
with Co2 and water into methane.

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FUTURE GATE ACADEMY EE

Digestion of sludge is carried out in a  It acts as thermal insulator and does

tank termed as sludge digestor, which are not allow heat produced during
generally of two types. digestion to escape from digestor.
 Standard rate sludge digester ( SRSD)
 High rate sludge digester (HRSD)  It reduces the extent of evolution of
 During digestion process sludge is odorous gases, out into the
digested into following components . atmosphere.
o Gas
o Supernatant liquids B. High rate sludge digestor (HRSD)
o Digested sludge  In high rate sludge digester entire
 Volume of digested sludge is process is carried out in two stages.
approximately 1/3 volume of original
rd

sludge (raw sludge)  Complete mixing is induced in 1st

stage which creates the homogenous
A) Standard rate sludge digestor environment bringing about contact
(SRSD) between micro-organisms and organic
 Raw sludge is sent/fed in matter thereby increasing rate of
digester intermittently in digestor sludge digestion.
where it is being acted upon by
anaerobic micro-organism that  Operational stability of system is more
breaks sludge into gases, in comparison to that of standard rate
supernatant liquids & digested digester in case of shock loading due
solids. to the mixing induced in first stage.

 No mixing is induced in the  No change in the volume of sludge

digestor due to which the takes place in 1st stage, as no
stratification takes place init. (the dewatering is permitted in this stage.
only mixing present in digestor is
due to the rising of the air  The entire stage I in high rate digester
bubbles of gases formed during can be considered as active zone.
digestion )
Design data for sludge diestor (SRSD)
 Digested sludge formed settles of  Diameter of tank - 6-18m
the bottom and is sent for diying  Depth of tank - 6-12 m
on sludge drying beds before its  Diameter to depth ration -1.5-4
further use.  Digestion period (Mean cell
residence time ) – 30 days
 Gases formed during the process  Volume of the tank is computed
which is collected from the top to by following relation as per GOI
be utilised as fuel. manual.
2
 Supernatant liquid is recycled V= [V1 - (V1 – V2)] t
3
back to PST as large conc. of
suspended solids is observed in it 2
(V1 –V2) = parabolic
which are induced due to lasing 3
air bubbles of gases passing V1  V2
through this layer. -linear
2
Where , t = Digestion period
 Scum layer formed at the top due
V1 = Volume of raw sludge
to anaerobic digestion serves the
V2 = volume of digested
following purposes.

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Design data for high rate digestor PREVIOUS QUESTIONS

(HRSD):
Common Data for Questions 01 & 02
 For stage I, volume of tank = V1. t1
V1 = Volume of raw sludge A conventional Activated Sludge Plant
t1 = mean cell residence time for 1st stage treating 1000m3/d of municipal waste
and is approximately 10-12 days. water disposes of its anaerobically
 Volume of tank digested sludge on relatively impervious
farmland. Use the following data
2
VII = [V1 - (V1 – V2)] t2 + V2T (GATE-2003-2M)
3 1. Raw sewage
t2 - mean cell residence time/digestion
VSS = 225mg/l (70% volatile)
period approximately 10-15 days.
BOD = 190 mg/l
 all other parameters are same as
(Excess activate sludge returned to
standard are sludge digestor.
primary)
Gas production in digestor (anaerobic ): 2. Primary setting
VSS – 50% removal
 Of all the solids that come into
BOD – 30% removal
digestor 7 0% are volatile and 30%
3. Excess Activated Sludge
are fixed.
VSS = 0.4g VSS produced per g BOD
 Of all the volatile solids, 65% are
applied (80% Volatile of total)
reduced to gases and 35% forms the
4. Anaerobic Digester:
digested sludge.
VSS reduced 50%
 Volume of the gas produced is 0.6 Digested Sludge Concentration: 60%
m3/kg of total volatile solids entering
5. Application farmland : 2m / ha.d
into the tank or 0.9m3 /kg of total
volatile solids reduced to gases.
01. Total volatile suspended solid to be
 Of the total gas produced 65% to anaerobically digested (kg/d VSS)
70% is methane (CH4) 30% is carbon shall be
dioxide (Co2) and left is other gases.
(a) 133 (b) 168
 Calorific value of methane is 8600 (c) 233 (d) 245
Kcal/m3 (By burning 1 m3 of
methane – 8600 Kcal energy) 02. Area requirements (ha) for disposal
 Calorific value of CO2 is zero. of the sludge on farmland shall be
(a) 2.95 (b) 1.95
(c) 0.95 (d) 0.55

03. Match the following (2005-2M)

Group – I
P. Thickening sludge
Q. Stabilization of sludge
R. Conditioning of sludge
S. Reduction of sludge
Group – II
1. Decrease in volume of sludge
2. Separation of water by heat or
chemical treatment
3. Digestion of sludge
4. Separation of water by flotation
or gravity

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Codes : PRACTICE QUESTIONS

P Q R S
01. For the same solid content if the
(a) 4 3 1 2 quantity of sludge with moisture
(b) 3 2 4 1 content of 98% is X, then the
(c) 4 3 2 1 quantity, of sludge with moisture
content of 96% will be
(d) 2 1 3 4
(a) X/4 (b) X/2 (c) X (d) 2X

04. A aerobic reactor receives

wastewater at a flow rate of 02. A sedimentation tank is treating 4.5
500m3/d having a COD of MLD of sewage containing 275 ppm
2000mg/L. The effluent COD is 400 of suspended solids. The tank
mg/L. Assuming that wastewater removes 55% of the suspended
contains 80% biodegradable waste, solids. Calc ulate
the daily volume of methane i) weight of sludge produced per day
produced by the reactor is assuming moisture content of the
sludge as 96%

(a) 0.224m2 (b) 0.280m3 ii) if sp, gravity of the sludge is

1.02, calculate sludge quantity in
(c) 224m3 (d) 280m3
bulk (volume)

Statement for Linked Answer

03. The moisture content of a sludge in
Questions 05 & 06
reduced from 98% to 92% in a
The sludge from the aeration tank of the sludge digestion tank. Find the %
activated sludge process (ASP) has solids decrease in volume of the sludge
content (by weight) of 2%. This sludge is
put in a sludge thickener, where sludge
volume is reduced to half. Assume that 04. Design the volume required for a
the amount of solids in the supernatant sludge digestion tank for the
from the thickener is negligible, the primary sludge with the following
specific gravity of sludge solids is 2.2 and data :
the density of water is 1000 kg/m3 i) Average sewage Flow = 20 MLD
(GATE-2011-2M) ii) Total suspended solids in raw
05. What is the density of the sludge sewage = 300 mg/l
removed from the aeration tank ? iii) Assume that 65% of the
(a) 990 kg/m3 (b) 1000 kg/m3 suspended solids settle in the
(c) 1011 kg/m3 (d) 1022 kg.m3 sidementation tank
iv) moisture content of undigested
sludge = 95%
06. What is the solids content (by
weight) of the thickened sludge ? v) moisture content of digested
sludge = 85%
(a) 3.96% (b) 4.00%
vi) Specific gravity of raw sludge =
(c) 4.04% (d) 4.10%
1.02
vii) Detention time 30 days
KEY

1. A 2. B 3. C

4. D 5. C 6. B

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FUTURE GATE ACADEMY EE

8) Disposal of Sewage Effluents

Disposal of effluent after treatment can be ZONES OF POLLUTION IN RIVER STREAM
done either on land or disposing it by
dilution (into river) Zone of degradation:

1. Disposal by dilution in river:  This zone is found after a certain

When the effluents are disposed after distance downstream of point of
treatment in the river self purification of application of sewage effluent of river.
river water takes place by following  The end of this zone is characterised by
mechanism. 60% depletion of saturated DO.
(a) Dilution and dispersion:
 Algae is not found inn this zone that
Dilution and dispersion is not in real
fishes may survive
terms a method of self purification.
 (at 200C, Sal D0 = 10 mg/L
It only reduces the potential hazards
 Remaining D0 = 4mg/L after 60%
which may be caused due to disposal
depletion
of effluents in the river.
 Water in this zone is dark and turbid
(b) Sunlight:
In the presence of sunlight oxygen is Zone of active Decomposition:
released in water due to  This is zone of heavy pollution in which
photosynthesis, which is further water becomes darker than in zone of
utilised by micro organism to carry degradation.
out decomposition of organic matter  DO may even fall upto zero levels.
present in it.  Aerobic micro organisms are replaced
by anaerobic microbes in this stage.
(c) Sedimentation : Leading to release of acids, gases &
Over o period of time organic solids alcohols which results in the formation
present in water get settled on the of dirty scum layers on the surface of
surface of river-bed (due to river.
sedimentation) hence are removed  Both algae and fishes don’t survive in
from river water. this zone. In some cases they may
These solids get in contact with o2 in survive in upper layer.
water to get stabilize and hence are  Do in this zone may again reach up
finally removed from water system. 40% of saturated value at the end of
this zone.
(d) Oxidation and reduction :
Due to oxidation & reduction organic Zone of recovery:
matter present in waste water get
decomposed either by aerobic, In this zone as the demand for oxygen
anaerobic. is satisfied, BOD level falls down. Do
approaches its saturated value, fishes
Dilution Degree of treatment
& algae reappear in this zone.
Factor
>500 No treatment
 Oxygen deficit at any given time for
300-500 Plain sedimentation
given polluted stream can be computed
(SS<150ppm)
using Streeler –Pheleps relation.
150-300 Sedimentation (SS<
60PPM)
K D L   kD .t
 10 RL  + D0
K
<150 Complete treatment (SS< Dt = 10
30ppm & BOD < 20ppm) K R .K D  
x 10-KD.t
(equation of oxygen deficit curve)

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FUTURE GATE ACADEMY EE

Where , b) Productivity of lakes :

Do =initial oxygen deficit
KD = deoxygenation constant (day-1) a) The ability of lakes to support food
KR = reaoxygenation constant (day-1) chain or ability of lakes to promot
L = Ultimate BOD of mixture/total organic growth of algee is termed as
matter productivity of lakes.
b) On the basis of productivity of lakes ,
 With the increase in temperature they can be classified as..
solubility of oxygen in water - Oligotrophic lakes
decreases (amount of oxygen) but the - Mesotrophic lakes
rate of reoxygenation (KR) with - Eutrophic lakes
increase in temperature (but their - Senescent lakes
value may be unknown).
Oligotrophic lakes :
 The increase in de-oxygenation (KD)
will be more than increase in (1) They are least productive lakes in
reoxygenation (KR) with increase in which no algae is formed.( fresh water
temperature (but their value may be lakes )
unknown) (2) Sunlight in these lakes can penetrate
KR up to Hypoliminion layer.
= Self purification constant =f (3) These lakes are generally found in
KD Antartico.
(4) Oxygen is sufficiently available in
Disposal in lakes: these types of lakes , hence aerobic
a) stratification of lakes decomposition may take place.
 Study of lakes is important to
understand the phenomenon of self Mesotrophic lakes :
purification , known as ( Timinology ) (1) These are the lakes of medium
 In temperate zone water in lakes and productivity level. Hence support
reservoir transfers heat by the medium growth of algae.
procees known as lake stratification. (2) Though O2 depletion takes place in
 In summers water present in lakes or these lakes, conditions are considered
reservoirs divides into upper layer of to be aerobic.
warm circulating water, in which Eutrophic lakes :
oxygen is readily available and is (1) These are lakes of fairly high
known as Epiliminion . productivity levels in which sunlight
 Bottom layer remain essentially can penetrate up to Epilimilion layer
unmixed & hence deficiency of only.
oxygen. It is termed as Hypoliminion. (2) Anaerobic conditions exist in this loke
 Both layers are seoerated by steep due to deficiency of oxygen.
temperature gradient known as Senescent Lakes:
Thermoccline or Metaliminion. It represent old lake which is
 This stratification will be interrupted practically being converted into
in winters & in springs (due to some marshy area.
density in springs & ice layer formed Eutroficaiton of lakes:-
in winters )  It is a natural process under which
 Maximum killing of oquatic life takes lakes get infested with alage over a
place during this mixing as they are period of time, and gradually silts
not accumsted with the sudden upto become shallower, due to the
change in temperature. entry and cycling of the nutrient like
phosphorous, nitrogen, sulphur,
carbon.

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FUTURE GATE ACADEMY EE

 To avoid this entry of the nutrients in PREVIOUS QUESTIONS

the lakes must be restricted, and it 01. The ‘sag’ in the dissolved oxygen curve
can be removed by help of copper results because (GATE-1997-1M)
sulphate and lime. (a) It is a function of the rate of
addition of oxygen to the stream
3. Disposal in Sea Water : (b) it is a function of the rate of
 Sea Water consists 20% less depletion of oxygen from stream
saturated DO in comparison to river (c) it is a function of both addition
at some temperature. and depletion of oxygen from the
 But the dilution available in sea, is stream
much large in comparison to that in (d) the rate of addition is linear but
river water, hence sea is more the rate of depletion is non-linear
appropriate source of disposal than
rivers. 02. Secondary effluent from a municipal
waste plant is discharged into a
 Disposal of sewage in sea may leads stream at the rate of 12,000 m3/day
to the formation of sludge and gases at 200C with a BOD5 of 50 mg/litre.
due to its reaction with solids The stream flow is estimated to be
present in sea water. 40,000 m3/day, and the water
 Sewage is always disposed in the sea quality, parameters in the stream
away from shore at the time of low upstream of the effluent outfall are :
tides.
4. Disposal on land : Bod5 of 3mg/litre,
 Effluent irrigation and sewage dissolved oxygen 7mg/litre 200C
forming are two techniques employed Assume a decay constant for the
for disposing effluent from mixture to be K = 0.23 (to the base
wastewater treatment on land. ‘e’ in the decay curve).
 In effluent irrigation disposal of Estimate (GATE-1997-2M)
effluent is prime concern, whereas in (a) BOD of the mixture
sewage farming growing of crops is (b) Ultimate BOD
important. Hence proper treatment is (c) DO of the mixture
ensured before utilising the effluent
for sewage forming. 03. Match the following : (2003-2M)
 Standard values of parameters on Group – I:
the basis of source of disposal to be (Characteristics of sewage discharged
present in effluent are given below. into inland waters)
P. BOD5 Q. COD
Parame Disposal Disposal Disposal R. Oil and Grease
ter in river in Sea on land S. Total Suspended Solids
BOD 20 100 500 Group – II: (Allowable limit, mg/I)
(mg/L) 1. 250 2. 30 3. 20
Ph 5.5-9 5.5-9 5.5-9
SS(mg 30 100 2100 4. 10 5. 5 6. 3
/L) Codes :
A B C D
(a) 2 5 4 2
(b) 4 1 6 4
(c) 3 1 4 2
(d) 2 1 6 3

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FUTURE GATE ACADEMY EE

04. In a certain situation, wastewater PRACTICE QUESTIONS

discharged into a river, mixes with the
01. Secondary effluent from a municipal
river water instantaneously and
waste water plant is discharged into
completely. Following is the data
a stream at a rate of 12,000 m3/d
available:
with a BOD5 of 50mg/l. The stream
Wastewater DO = 2.00 mg/L flow is estimated to be 40,000 m3/d
Discharge rate = 1.10 m3/s and the B.O.D2 of water in the
River water DO = 8.3 mg/L stream is 3mg/l. The B.O.D of the
Flow rate = 8.70 m3/s mixture is
Temperature = 200C (a) 6.8 mg/l (b) 13.85 mg/l
Initial amount of DO in the mixture of (c) 17.65 mg/l (d) 10.55 mg/l
waste and river shall be
(a) 5.3 mg/L (b) 6.5 mg/L 02. A reactive chemical plant disposed
(c) 7.6 mg/L (d) 8.4 mg/L by dilution in a river at the uniform
rate of dissipation of 0.12 mg/l per
05. A waste water stream (flow = 2m3/s, hour. The water discharged per day
ultimate BOD = 90 mg/l) is joining a is 30  106 liters having chemical
small river (flow = 12 m3/s, ultimate concentration of 25 mg/l. The flow
BOD = 5 mg/l). Both water streams of river above the sewer out fall is 58
get mixed up instantaneously, cross- m3/sec. If the river has zero
sectional area of the river is 50 m2. chemical concentration, calculate
Assuming the de-oxygenation rate the distance on down stream side
constant, k = 0.25/day, the BOD (in upto which the chemical residue
mg/l) of the river water, 10 km persists. Assume mean velocity of
downstream of the mixing point is river flow as 20 cm/sec.
(a) 1.68 (b) 12.63
(c) 15.46 (d) 1.37 03. A town disposing 8 Mld of the
effluent on land capable of
consuming 80000 lit/ha/day of
06. Which one of the following solid waste sewage. Find area of land required ?
disposal methods is ecologically most
acceptable ? (IES-CE-1995) 04. A town disposes sewage by land
(a) Sanitary landfill treatment. It has a sewage farm of
(b) Incineration area 140 ha. The area includes an
(c) Composting extra provision of 40% for rest and
(d) Pyrolysis rotation. It produces 4.5 Mld of
waste water. Find consuming
07. A polluted stream undergoes self- capacity of soil ?
purification in four distinct zones :
1. Zone of clear water
2. Zone of active decomposition
3. Zone of degradation KEY
4. Zone of recovery 1. C
The correct sequence of these zones is 2. (a) 13.85mg/lit (b) 20.26 mg/lit
(a) 3, 4,2, 1 (b) 2, 3, 4, 1 (c) 5.85 mg/lit
(c) 2, 4, 3, 1 (d) 3, 2, 4, 1 3. C
4. C
5. C
6. C
7. D E

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FUTURE GATE ACADEMY EE

9) Solid Waste Management

Solid waste is mixture of solid and - Anaerobic – Acids & Alcohols
semi-solid material released by formed initially
various activities in communities. - Formation of methane- by
 Domestic solid waste action of CH4 formers.
(Municipal solid waste) MSW
(Refuse)  For optimum decomposition of
 Industrial solid waste refuse, moisture content in cell is
kept below 60%.
1. Domestic Solid Waste:
 Solid waste generated due to  When decomposition of solids
domestic activities is termed as takes place height of the land
refuse or municipal solid waste. filling is reduced to 20-40% of its
 Domestic solid waste can be original height.
classified as
- Garbage  In the initial stages decomposition
- Rubbish takes place in the mesophilic
- Ashes range and in final stages it is
A) Garbage: shifted to Haemophilic range.
It is the biodegradable waste having
density in the range of 450-900 kg/  The entire process of digestion if
m3 released due to domestic activities. completed in 2-32 months.

B) Rubbish:  Only disadvantages associated

It is non –biodegradable waste having with it is release of gases and
density 50- 400 kg/ m3 generated formation of leachate. (density of
from domestic activities. CO2 is more than water, thus it
settles down and contaminates
C) Ashes: groundwater making it acidic)
It is incombustible waste i.e., left after
the burning in furnaces of either  Alcohols, non-biodegradable waste
houses or industries. and water which may be ca-
Its density varies between 700 – 800 cinogenic. It is formed in rainy
kg/ m3. season when water seeps through
cells.
Disposal of solid waste
Sanitary land filling:  The cells may be covered by
In this method refuse in dumped into selands (to avoid escape of gases
low lying area into the layer of depths into atmosphere) where water
approximately 1.5 each layer of refuse is seeps through cells.
termed as cell.
 Each layer is compacted properly and  The cells may be covered by
left for 7 days (approx) for its aerobic selands (to avoid escape of gases
decomposition & it covered with layer of into atmosphere) which may be of
mud in order to carry out further process rubber, PVS Aesbestos, Bitumin
as cell. etc.,
 The entire process of decomposition in 1. Incineration & Thermat Pyrolysis
this method is biological and takes place  Incineration 15 method in which
in three stages. refused to between at higher
- Aerobic temperature.

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FUTURE GATE ACADEMY EE

 It is generally adopted for those waste during solid decomposition

wastes which posses high fuel & in anaerobic treatment solid
value waste is left without any mixing.
 Thermal pyrolysis is method in
which waste is between in limited
 Aerobic method of treatment is also
supply of O2 to avoid its complete
termed as Indore process and
burning resulting in formation of
anaerobic process is Bangalore
charcoal like product.
process.

2. Pulverisation :
It is the process in which heavier  In composting process C/N ratio
solids are broken into finer/higher should be properly maintained in
ones which are further disposed by order to carry out optimum
any of the above mentioned methods. digestion.

3. Barging, refuse into sea:

 For optimum digestion ca/n ratio
Disposal of solid waste into sea is
is 30-50 (as micro-organism
done only at the times of law tides for
utilising carbon are 30-50 time
away from coasts (15-20 km away
more competent than one using
from share) 30m below sea-level.
nitrogen)
4. Autoclave:
 It is method of disposal of
biochemical waste in which it is  If C/N ratio is more than
being brought in contact with the permissible value carbon will be
steam for the sufficient duration left in system & nitrogen will be
and for sufficient design conditions exhausted leading to the
in order to stabilise the incomplete decomposition of
biochemical solids. organic matter.

 It is a low heat process used for  If C/N ratio is less than

dis-infecting the bio-chemical permissible value carbon is
waste. exhausted prior to nitrogen & this
left out nitrogen reacts with
Disposal of composting : hydrogen & water in system
 It is also a biological method in leading to formation of ammonia
which solid waste is decomposed which is disastrous to CH4
either aerobically or anaerobically. formers that against leads
decomposition of organic matter
incomplete.
 The end products formed in this
method are termed as compost,
humus or manure

 In the initial stages decomposition

is carried out by micro-organisms
in mesophilic range which is
gradually shifted into thermophilic
range.

 Aerobic conditions are generated

by continuous mixing of solid

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FUTURE GATE ACADEMY EE

PREVIOUS QUESTIONS 04. Two biode gradable components of

municipal solid waste are
01. A synthetic sample of water is
(a) plastics and wood
preparted by adding 100 mg
Kaolinite (a clay mineral), 200 mg (b) cardboard and glass
glucose, 168 mg NaC1, 120 mg (c) leather and tin cans
MgSO4, and 111 mg CaCl2 to (d) food wastes and garden
1 litre of pure water. The trimmings
concentrations of total solids (TS)
and fixed dissolved solids (FDS)
respectively in the solution in mg/L PRACTICE QUESTIONS
are qual to (GATE-2006-1M) 01. The following composition of a solid
(a) 699 and 599 waste is given :
(b) 599 and 399 Component % by Energy
(c) 699 and 199 mass KJ/kg
(d) 699 and 399
Food waste 15 4,650

Paper 45 16,750
02. The composition of a certain MSW
sample and specific weights of its Card board 10 16.300
various components are given below
Plastics 10 32,600
: (GATE-2006-2M)
Garden 10 6,500
Component Percent Specific
trimmings
by Weight
Weight (kg/m3) Wood 5 18,600

Food waste 50 300 Tin cans 5 700

Dirt and Ash 30 500 (i) Compute the unit energy content
(as discarded)
Plastics 10 65
(ii) determine the energy content on
Wood and 10 125 a dry basis, if the m.c. of the waste is 21%
Yard waste
(iii) determine the energy content on
Specific weight (kg/m3) of the MSW an ash – free dry basis, assuming
sample is the ash content as 5%
(a) 319 (b) 217
(c) 209 (d) 199 02. 50g of CO2 and 25g of CH4 are
produced from the decomposition of
03. 50 g of CO2 and 25g of CH4 are municipal solid waste (MSW) with a
produced from the decomposition of formula weight of 120g. What is the
municipal solid waste (MSW) with a average per capita green house gas
formula weight of 120g. What is the production in a city of 1 million
average per capita green house gas people with a MSW production rate
production in a city of 1 million of 500 ton/day ?
people with a MSW production rate (a) 104g/day (b) 120g/day
of 50 ton/day ? (GATE-2007-2M) (c) 208g/day (d) 313g/day
(a) 104 g/day (b) 120 g/day
(c) 208 g/day (d) 313 g/day KEY TO PREVIOUS QUESTIONS
1. D 2. B
3. D 4. D

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FUTURE GATE ACADEMY EE

10) Air Pollution & Control

Atmosphere:
Based on temperature profile, the layers 1. Oxides of Sulphur:
are classified  So2 is an irritant gas that causes
1. Troposphere (Lower Atmosphere ) deficiency of O2 and increase
2. Stratosphere (Ozone layer) breathing rate (lungs & respiratory
3. Mesosphere (Temperature decreases track)
due to less heat from stratosphere)  So2 is generally released in
4. Ionosphere (Ionic reactions) atmosphere from refineries,
chemical plants, steel plants,
90% Atmosphere included in troposphere power plants generally where
&10% in stratosphere. burning of carbon takes place.

The air above the ground is headed up by 2. Oxides of Carbon: (Co)

the earth surface, as we more up the  Co has approximately 200 times
temperature will decrease. more affinity for Haemoglobin than
O2 hence it replaces O2 from
Volume of N2 conc. – 2,80,000 ppm Haemoglobin causing its deficiency
O2 conc. - 2,09,000 ppm in body.
CO2 - 320 ppm  If 50%of O2 is replaced from body
Excessive concentration of foreign matter Haemoglobin by Co, death is
in the air which adversely affects the well certain.
being of individual (or) cause damage to  It is primarily released from
property is called Air Pollution. automobiles & furnaces.
(Plants, Animals & Buildings)  It present in high conc. attacks
 Industrialization & urbanisation – central nervous system and causes
common reasons heart attack.
 Common gase contaminants - Co,
O3,So2 & oxides n-n2 Hydrocarbons & 3. Oxides of Nitrogen :
Aldehydes  High conc. of these oxides causes
 Co2 is not considered as contaminant eyes , nose irritation
because of its role in photosynthesis of  These are generally released in
plants. atmosphere from automobiles and
 By breathing of average adult – 0.017 burning of fuels.
m3 per hr of co2 released im3 of gases on
burning -0.017 im3 of co2 released. Secondary Air Pollutants:
 The presence of one or more When one or more primary air
contaminants in such conc. and for pollutants combine with each other
such duration that it starts affecting present in atmosphere with water vapours
biosphere life is termed as “Air leads to the formation of secondary
Pollution”. pollution in presence of sunlight.
 Air pollutants can be classified into - H2SO4
primary & secondary air pollutions. - O3
- Formaldehyde
Primary air Pollutants : - PAN (Peroxyacetyl nitrate )
 Oxides of sulphur (SO2 & SO3) peroxy Acetyl Nitrate
 Oxides of carbon (CO & CO2)
 Oxides of nitrogen (No, No2)
 Lead
 Hydrocarbon
 H2S,H2F,Ethyl & methyl Mercaptan

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Dispersion of air pollutants in Negative Lapse Rate:

atmosphere: The temperature of air increases with
Dispersion of pollutants in the increase in its height then the system is
atmosphere depends on the prevailing said to be in “inversion” and the rate at
temperature, pressure & wind condition. which temperature increases with
increase in height is known as “Negative
Temperature : Lapse Rate”.
 The ambient temperature of  Inversion is generally found in winters
atmosphere in troposphere decreases  This inversion is favourable only when
with increase in height form earth the pollutants are released at
surface. particular height from earths surface
 The role at which temperature when they don’t hamper above layers
decreases with increase in height is of atmosphere as well as the life on
called lapse rate/Environmental Lapse earths surface.
rate (ELR)  When earth cools much faster then
 Normally it is equal to 6.50C/km surrounding air, in water the system
 The lapse rate at particular height can is called Radiation inversion.
be computed by sending rising air  Negative lapse rate is also formed in
balloon in atmosphere with system when high pressure region is
thermometer. surrounded by low pressure region.
 The rate of decrease of temperature This types of inversion is known as
with increase in height or mass of air subsidence inversion.
pollution is known as adiabatic rate.
As there is no heat exchange that takes Plume:
place between the pollutant and The rising air pollutant is known as
atmosphere during change of plume and the source from which this
temperature adiabatic. originates is called stack.
Dry adiabatic lapse rate = - 9.80c/km
1. Looping plume :
Types of Environment  This type of plume is observed in
unstable-environment where
1. Unstable Environment : ELRDALR ELR>ALR
 The system is which ELR> ALR is  This how is characterised by rapid
termed as unstable system as in this mixing requiring more height of
case rising parcel of air pollutant keep stacks.
rising into the atmosphere or falling
parcel keeps descending towards the 2. Neutral Plume :
surface of earth.  This type of plume is observed in
neutral environment where ELR =
 Environmental lapse rate existing in
ALR, which rise vertically upward
system is termed as super adiabatic
into the system.
lapse rate.
2. Stable Environment ALR>ELR 3. Coning plume :
 In this case, environment is said to be  This type of plume is observed in
stable as prevailing ELR is less than stable environment when
ALR due to which pollutants always ALR>ELR. Where wind velocity is
remain heavier than surrounding it. less than 32km/hr.
 No mixing (rapid) is observed
 Environmental lapse rate in this case
is termed Sub adiabatic Lapse rate
(ALR>ELR)

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4. Fanning plume : PREVIOUS QUESTIONS

 This type of plume is observed in
01. The mean indoor airborne
extreme id version conditions in
Chloroform (CHC13) concentration
which emissions will be spread will
in a room was determined to be
be spread only horizontally.
0.4  g/m3. Use the following data: T
 Higher stacks are need in this
= 293K, P = 1 atmosphere, R =
case.
82.05  10-6 atm.m3/mol-K. Atomic
weights: C = 12, H = 1, C1 = 35.5.
5. Lofting Plume :
This concentration expressed in
 It is most ideal plume behaviour
parts per billion (Volume basis,
which occurs when there exist
ppbv) is equal to
strong super adiabatic lapse rate
above surface inversion.
 Such plume has minimum (a) 1.00 ppbv (b) 0.20 ppbv
downward mixing as it is prevented (c) 0.10 ppbv `(d) 0.08 ppbv
by inversion layer below stack.
02. The dispersion of pollutants in
6. Fumigating Plumes : atmosphere is maximum when
 This type of system occurs when (a) environmental lapse rate is
inversion is present above super greater than adiabatic lapse rate
adiabatic lapse environment. (b) environmental lapse rate is less
 It is not desired as the pollutant than adiabatic lapse rate
cannot escape into the atmosphere (c) environmental lapse rate is equal
and remains in troposphere. to adiabatic lapse rate
7. Trapping Plume : (d) maximum mixing depth is equal
 This plume is formed when super to zero
adiabatic environment is formed in
between two inversion layers. 03. Two electrostatic precipitators
 Pollutants will neither be able to go (ESPs) are in series. The fractional
upward nor more downwards thus efficiencies of the upstream and
termed as trapping plume. downstream ESPs for size dp are
80% and 65%, respectively. What is
the overall efficiency of the system
for the same dp? (GATE-2007-2M)
(a) 100% (b) 93%
(c) 80% (d) 65%

04. Two primary air polluatants are

(a) sulphur oxide and ozone
(b) nitrogen oxide and
peroxyacetylnitrate
(c) sulphur oxide and hydrocarbon
(d) ozone and peroxyacetynitrate
05. Particulate matter (fly ash) carried in
effluent gases from the furnaces
burning fossil fuels are better
removed by
(a) Cotton bag house filter
(b) Electrostatic precipitator (ESP)
(c) Cyclone (d) Wet scrubber

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06. Match List-I with List-II and select m3/min. Samping continued for
the correct answer by using the 24 hours. The airflow after 24 hours
codes given below the lists : was measured to be 1.4 m3/min.
List-I The dry weight of the filter paper
A. Coriolis effect after 24 hour sampling was 10.283
B. Fumigation g. Assuming a linear decline in the
C. Ozone layer air flow rate during sampling, what
is the 24 hour average TSP
D. Maximum mixing depth
concentration in the ambient air ?
List-II
1. Rotation of earth (a) 59.2  g/m3
2. Lapse rate and vertical (b) 118.6  g/m3
temperature profile
(c) 237.5  g/m3
3. Inversion
4. Dobson (d) 574.4  g/m3
Codes :
A B C D 10. Elevation and temperature data for
(a) 2 1 4 3 a place are tabulated below :

(b) 2 1 3 4 Elevation, m Temperature 0C

(c) 1 3 2 4 4 21.25
(d) 1 3 4 2
444 15.70

Based on the above data, lapse rate

07. An air parcel having 400C
can be referred as :
temperature moves from ground
level to 500 m elevation in dry air (a) Super-adiabatic
following the ‘adiabatic lapse rate”. (b) Neutral
The resulting temperature of air (c) Sub-adiabatic
parcel at 500m elevation will be (d) Inversion
(GATE-2010-1M)
(a) 350C (b) 380C 11. The two air pollution control devices
(c) 410C (d) 440C that are usually used to remove very
fine particules from the flue gas are
(GATE-2014-1M-SET-2)
08. Consider four common air
pollutants found in urban (a) Cyclone and Venturi Scrubber
environments, NO, SO2, Soot and (b) Cyclone and Packed Scrubber
O3. Among these which one is the (c) Electrostatic Precipitator and
secondary air pollutant? Fabric Filter
(a) O3 (b) NO (d) Settling Chamber and Tray
(c) SO2 (d) Soot Scrubber

09. Total suspended KEY TO PREVIOUS QUESTIONS

particulate matter (TSP)
concentration in ambient air is to be 1. D 2. A 3. B
measured using a high volume
sampler. The filter used for this 4. C 5. B 6. D
purpose had an initial dry weight of
7. A 8. A 9. C
9.787 g. The filter was mounted in
the sampler and the initial air flow 10. A 11. C
rate through the filter was set at 1.5
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11) Noise Pollution

SOUND & it’s characteristics: normal healthy ear and the value
is 0.00002 N/m2 (or) 20 Pa.
 Sound is a form of energy which is
emitted by a vibrating body and on  When sound is louder, pressure
reaching the ear causes the increases.
sensation of hearing through
 Acoustic Reflux: The mechanism
nerves.
by which ear habitually responds
 Sound is produced by alternating to a sound when it is heard.
pressure changes in air, when part
NOISE & it’s characteristics:
of the atmosphere compressed
suddenly, when the air is elastic,  When sound waves are non
the particles originally disturbed periodic, irregular & short
causing disturbance of direction, produce a displeasing
neighboring particles. effect causing unwanted sound
termed as Noise.
 Ultimately, compression is
propagated or spread away from  A noise problem generally consists
source. The sound thus travels in of three inter-related elements- the
the form of waves & when these source, the receiver and the
waves come near to ear drum, a transmission path.
sensation of hearing is created.
 This transmission path is usually
 When sound waves are periodic, the atmosphere through which the
regular & long direction – produce sound is propagated, but can
pleasing effect and that sound is include the structural materials of
considered as Musical Sound. any building containing the
 Sounds produced by all vibrating receiver.
bodies are not audible.

 The frequency limits of audibility

are from 20 HZ to 20,000 HZ.
 Sounds of frequencies less than 20
HZ are called infrasonics.
 Sounds of frequencies greater than
20,000 HZ are called ultrasonics.
 Amplitude of sound waves
expressed using the units: N/m2.  Noise that pose greatest threat to
 Audible sound: The pressure human body is with high pitch,
waves which the human ears high amplitude, poorest tone,
detect are considered as Audible longest duration.
sound and it is in the range of
 Noise is measured by the units:
0.00002 N/m2 - 200 N/m2
decibels (dB). 1dB unit is the
 Threshold of hearing (or) approximate smallest change of
audibility: The minimum sound sound intensity able to hear by
pressure which is audible for human ear.
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 Range of audible sound to painful  Nervous system: It causes pain,

noise : 1 to 130dB ringing in the ears, feeling of
tiredness, thereby effecting the
Types of Noise: functioning of human system.
 Sleeplessness: It affects the
Noise is classified into 3 types sleeping there by inducing the
based on duration of occurrence. people to become restless and lose
concentration and presence of
Continuous Noise: Uninterrupted
mind during their activities
noise less than 5dB during the  Damage to material: The
entire period of observation. Ex: buildings and materials may get
Running Fan damaged by exposure to infrasonic
/ ultrasonic waves and even get
Intermittent Noise: Noise that collapsed.
continues more than 1 sec and
interrupted for more than 1 sec.
Noise Effects observed
Ex: Drilling machine produced by
level
dentist

Impulse Noise: Change of sound 0 – 1dB Threshold of audibility

pressure > 40dB in time < 0.50
second Ex: 120 dB Pain threshold
Firing of a weapon
190 dB Major permanent
Effects of Noise: damage, if prolonged

 Annoyance: It creates annoyance

to the receptors due to sound level
fluctuations. The aperiodic sound Typical sound levels:
due to its irregular occurrences
causes displeasure to hearing and Location Noise
causes annoyance. level
City Traffic in busy
 Physiological effects: The 70 dB
street
physiological features like
Loud Noise
breathing amplitude, blood 120 dB
accompanying
pressure, heart-beat rate, pulse lighting
rate, blood cholesterol are affected.
Aero plane Noise at 130 dB
Leads to fatigue & also decreases
distance of 3m
efficiency of persons.
 Loss of hearing: Long exposure to Boiler Factory 110 dB
high sound levels cause loss of
hearing. This is mostly unnoticed, Large Factory 90 dB
but has an adverse impact on
hearing function.
Noise Standards: According to
 Human performance: The
ministry of environment and forestry,
working performance of
noise standards during day and night
workers/human will be affected as
for different zones of the city are given
they'll be losing their
as follows:
concentration.

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FUTURE GATE ACADEMY EE

Noise released from different

Type of the Day Night
sources:
zone Noise Noise
Combined effect of different sound
Sensitive
50 dB 40 dB pressures from different sources is
worked out using various systems of
Residential
55 dB 45 dB statistical concepts.
1) Addition of sound levels:
Commercial 65 dB 55 dB
The effective sound levels form two
Industrial 75 dB 70 dB or more sources cannot be simply
added algebraically. For example,
Measurement of NOISE: the effective sound level from two air
conditioners 60 dB each, say is not
Sound Pressure Level:
60 + 60 = 120 dB, but 60 + 3 = 63
Noise released from single source is dB. Similarly, the effective sound
measured as Sound Pressure Level. level of 57 dB, 63 dB, 63 dB, 66 dB
and 69 dB is 72 dB. The
Noise (or) Sound is measured as
computation is illustrated below.
sound pressure level (SPL)

SPL = 20 log 10 (P/Pref)

Where, P : Pressure of
sound wave (N/m (or) Pa)
2

Pref : Reference pressure = 0.00002

N/m2 (or) 20 Pa
Noise released from two sources: 2) LTotal Concept:

 When two sources emit different When several sources emitting

sounds at a time at a given place, different sounds at a time at a given
the resultant noise is measured place, the resultant noise is
according to Thumb Rule. measured by LTotal Concept.

 Consider the noise from sources as LTotal = 20 log 10 (P/Prms)

L1 and L2. L1 > L2 and based on
Where, Prms = Root mean square
difference in L values, determine
pressure, Equivalent pressure of
resultant noise as per below
above sine wave is called as root mean
values.
square Pressure Prms

L1 – L2 Add to L1 Resultant L 3) Average Noise Level (or) Average

Sound Level:
0-1 3 dB L1 + 3 dB
When several sound pressure levels
2-3 2 dB L1 + 2 dB
recorded at a particular place over a
given period, the noise level is
4-8 1 dB L1 + 1 dB measured using Average Noise Level
concept.
9/more 0dB L1

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1 CONTROL OF NOISE:
LP = 20 log10
N
 (10) LN/20 1) Design of Doors and Windows:
Sound Insulation is achieved by
1 construction glazed windows with
LP = 20 log10 [10L1/20 + 10L2/20 +
N double or triple panes of glass.
10L3/20 + ------ + 10Ln/20 ]
2) Vibration Damping
4) LN Concept: Noise level (or) sound
pressure level that will be equaled or 3) Planting of Trees
exceeded for N% of measuring time. 4) Enclosures like shields, barriers to
For example, sound pressure level of cent off same sound waves while
50db equaled or exceeded for 70% propagating.
measuring time, then LN is L70. L70 =
50dB
5) Leq concept: Sound pressure (or)
Noise level equivalent to a number of
different sounds produced at a place
for different time intervals.
in
Leq = 10 log10 [ 
i 1
10Li/10 x ti]

Leq :Equivalent sound pressure level

in dB
n : Total No. of sound pressure levels
recorded
Li : Values of sound pressure levels
recorded in db with i = 1,2,3.
ti : Time duration

Source located at different places:

With the increase of distance, noise

levels decreases. When a source is
placed at two different locations, noise
is measured using the formula:
L2 = L1 - [20 log10 (r2/r1)]
Noise measuring instruments:

The various equipment used in noise

measurement include Sound level
meter, Impulse meter, Frequency
analyzers, graphic recorders, Noise
dosimeters etc.

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FUTURE GATE ACADEMY EE

PREVIOUS QUESTIONS PRACTICE QUESTIONS

01. The cumulative noise power 01. If P represents the pressure of
distribution curve at a certain sound wave and Pref represents the
location is given below. The value of reference pressure, then sound
L40 is equal to (GATE-2006-2M) pressure level (SPL) is equal to
(a) 20 log10 (P/Pref)
(b) (1/20) log10 (P/Pref)
(c) 20 log10 (Pref/P)
(d) (1/20) log10 (Pref/P)

02. A 60 dB re : 20  Pa noise is
accompanied with another 60 dB re
: 20  Pa noise. Then the total noise
(a) 90 dBA (b) 80 dBA level in dB re : 20  Pa is
(c) 70 dBA (d) 60 dBA (a) 120 (b) 63
(c) 84.85 (d) 60
02. The reference pressure used I the
determination of sound pressure 03. A source emitting 80 dB and other
level is emitting 60 dB if put in the same
(a) 20  Pa (b) 20 db location will produce a noise of
(c) 10  Pa (d) 10 db (a) 140 dB (b) 80 dB
(c) 70 dB (d) 60 dB
03. According to the Noise Pollution
(Regulation and control) Rules,
2000, of the Ministry of 04. While recording weighted sound
Environment and Forests, India, the levels, 4 readings have been taken at
day time and night time noise level a site a different times of a day.
limits in ambient air for residential These reading are 20, 56, 66, and 42
areas expressed in dB (A) Leg are dB re : 20  Pa . Then the average
(a) 50 and 40 (b) 55 and 45 sound level is
(c) 65 and 55 (d) 75 and 70 (a) 56.8 dB (b) 46 dB
(c) 66 dB (d) none

KEY: 05. A noise level of 80 dB lasting for 10

minutes is followed by 60 dB for 80
1. B 2. A 3. B
minutes and 100 dB for 5 minutes
one after the other. What is the
equivalent continuous equal energy
level Leq for the 95 minutes period?

06. The sound pressure level for a jet

plane on the ground with sound
pressure of 2000  bar should be
(a) 60 db (b) 100db
(c) 140db (d) 180db

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