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Roll No.:
End Semester Examination
UCE 502: Water Supply Engineering
B.E.Civil Engineering (3rd
year)
Instructor: Dr. Akepati S. Reddy
Date: 14-12-2013 Time: 3 hours (02-00 PM to 05-00 PM) Max. Marks: 100
Note:
1. Please assume if any requisite data is not given.2. Answer all parts of a question at one place
Q.1 Answer the following?
1.1 A rapid gravity filter has 0.6 m thick layer of sand. During backwashing the filter is expanding
to 0.9 m thickness. Find the filter backwash velocity and the minimum head required for the
fluidization of the filter bed during the backwashing? Assume D10, D60and D90as 0.4 mm, 0.75mm and 0.96 mm respectively. Take specific gravity of the sand medium as 2.65 and porosity
of the sand bed as 0.4. If the filter is run at 7 m/hour rate and backwashed after running for
10 hours, and if the backwashing for 20 minutes duration, find the percentage of filtered
water used in the backwashing?
Sand bed thickness (L): 0.6 m
Sand bed thickness when expanded(Lfb): 0.9 m
Porosity of the bed material (): 0.4
Specific gravity of the bed material (SG): 2.65
Porosity of the expanded bed:fb
fb LL
1
1
1-(0.6x(1-0.4))/0.9 = 0.6
D90 of the bed material: 0.96 mm or 0.00096 m
Terminal settling velocity of the D90 particle:
p
fluid
fluidparticle
d
p dC
gv
3
4
34.0324
RR
dNN
C
pp
R
dvN
Take density of water as 1000 kg/m3
Take acceleration due to gravity as 9.81 m/sec2
Take kinematic viscosity as 1.003E-6
Assume terminal velocity of particle as 0.1 m/sec.
Find NR value
Using the NR value find Cd value
Using the Cd value find terminal settling velocity of the particleUsing the calculated terminal settling velocity as assumed terminal settling velocity iterate and
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obtain the terminal settling velocity (continue iteration till assumed velocity becomes same as or
closer to the assumed velocity)
Iteration No. Assumed velocity NR Cd Calculated
velocity
1. 0.1 m/sec. 95.7 0.897 0.152 m/sec.
2. 0.152 145 0.754 0.166
3. 0.166 159 0.729 0.169
4. 0.169 162 0.724 0.169
Terminal settling velocity of the D90 particle (Vt): 0.169 m/sec.
Backwash velocity of the filter bed (Vb):5.4
fBtB VV
0.0166 m/sec. or 59.7 m/sec.
Filtration rate: 7 m/hr.
Filter run time: 10 hr.
Filtered water produced in 10 hours: 70 m3
Duration of backwashing: 20 min.
Backwash water generation: 59.7x20/60 = 19.9 m3
Percent filtrate used in the backwashing: 19.9/70 = 28.43%
Minimum head required for the backwashing: w
wmfb Lh
1 = 0.594 m
1.2 Using Hardy-Cross method,
balance water flows through the
following pipe network loop?
Flow velocity in the water supply
lines should be between 0.9 to
1.5 m/sec. Pipe sizes available are
in the increments of 25 mm.
Assume friction factor as 0.02.
Qi1 flow through flow balancing is (100+150+125+175+75)(300) = 325 m3/hr.
Assume flow rates and flow directions in different pipes through flow balancing at each of the
nodes. Show flow rates and flow directions in the figure as
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A B
C
D
E
100 m3/hr
450 m
500 m
400 m
700 m
250 m
350 m
150 m3/hr
Qi1
Qi2
50m3/ho
ur
50 m3/hour150 m3/hour
75m3/hour
Find pipe diameters by taking flow velocity as 1.5 m/sec. and adjust the diameter to the available
higher pipe diameter for ensuring the velocity to be
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B C -52.087 0.109 0.125 350 18937 -3.964 274.00
C D 98.719 0.154 0.175 500 5030 3.782 137.94
D A -76.281 0.133 0.15 400 8698 -3.905 184.30
-0.124 692.16 0.323
B E 50.806 0.109 0.125 250 13527 2.694 190.90
E C -74.194 0.133 0.15 700 15221 -6.465 313.70
C B 52.087 0.109 0.125 350 18937 3.964 274.00
0.193 778.59 -0.447
iteration-3
Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)
A B 149.042 0.188 0.2 450 2322 3.980 96.13
B C -51.317 0.109 0.125 350 18937 -3.848 269.95
C D 99.042 0.154 0.175 500 5030 3.807 138.39
D A -75.958 0.133 0.15 400 8698 -3.872 183.52
0.067 687.98 -0.176
B E 50.359 0.109 0.125 250 13527 2.647 189.22
E C -74.641 0.133 0.15 700 15221 -6.543 315.59
C B 51.317 0.109 0.125 350 18937 3.848 269.95
-0.048 774.75 0.112
Iteration-4
Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)A B 148.866 0.188 0.2 450 2322 3.971 96.02
B C -51.605 0.109 0.125 350 18937 -3.891 271.46
C D 98.866 0.154 0.175 500 5030 3.794 138.14
D A -76.134 0.133 0.15 400 8698 -3.890 183.94
-0.017 689.56 0.044
B E 50.471 0.109 0.125 250 13527 2.659 189.64
E C -74.529 0.133 0.15 700 15221 -6.524 315.11
C B 51.605 0.109 0.125 350 18937 3.891 271.46
0.026 776.21 -0.061
Iteration-5Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)
A B 148.911 0.188 0.2 450 2322 3.973 96.05
B C -51.499 0.109 0.125 350 18937 -3.875 270.91
C D 98.911 0.154 0.175 500 5030 3.797 138.20
D A -76.089 0.133 0.15 400 8698 -3.886 183.83
0.009 688.99 -0.024
B E 50.410 0.109 0.125 250 13527 2.652 189.41
E C -74.590 0.133 0.15 700 15221 -6.534 315.37C B 51.499 0.109 0.125 350 18937 3.875 270.91
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-0.007 775.69 0.015
Iteraction-6
Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)
A B 148.887 0.188 0.2 450 2322 3.972 96.03
B C -51.539 0.109 0.125 350 18937 -3.881 271.11
C D 98.887 0.154 0.175 500 5030 3.795 138.17
D A -76.113 0.133 0.15 400 8698 -3.888 183.89
-0.002 689.21 0.006
B E 50.425 0.109 0.125 250 13527 2.654 189.47
E C -74.575 0.133 0.15 700 15221 -6.532 315.31
C B 51.539 0.109 0.125 350 18937 3.881 271.11
0.004 775.89 -0.008
Iteraction-7
Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)
A B 148.893 0.188 0.2 450 2322 3.972 96.04
B C -51.524 0.109 0.125 350 18937 -3.879 271.04
C D 98.893 0.154 0.175 500 5030 3.796 138.18
D A -76.107 0.133 0.15 400 8698 -3.887 183.88
0.001 689.13 -0.003
B E 50.417 0.109 0.125 250 13527 2.653 189.44
E C -74.583 0.133 0.15 700 15221 -6.533 315.34
C B 51.524 0.109 0.125 350 18937 3.879 271.04
-0.001 775.82 0.002
Iteration-8
Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)
A B 148.889 0.188 0.2 450 2322 3.972 96.03
B C -51.530 0.109 0.125 350 18937 -3.880 271.07
C D 98.889 0.154 0.175 500 5030 3.796 138.17
D A -76.111 0.133 0.15 400 8698 -3.888 183.89
0.000 689.16 0.001
B E 50.419 0.109 0.125 250 13527 2.653 189.45
E C -74.581 0.133 0.15 700 15221 -6.533 315.33
C B 51.530 0.109 0.125 350 18937 3.880 271.07
0.000 775.84 -0.001
Iteration-9
Pipe Flow
rate
(m3/hr)
D
calculate
(m)
D
chosen
(m)
Length
(m)
K
value
hL hL/Q correctio
n to flow
(m3/hr)
A B 148.890 0.188 0.2 450 2322 3.972 96.03B C -51.528 0.109 0.125 350 18937 -3.880 271.06
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C D 98.890 0.154 0.175 500 5030 3.796 138.18
D A -76.110 0.133 0.15 400 8698 -3.888 183.88
0.000 689.15 0.000
B E 50.418 0.109 0.125 250 13527 2.653 189.44
E C -74.582 0.133 0.15 700 15221 -6.533 315.34
C B 51.528 0.109 0.125 350 18937 3.880 271.06
0.000 775.83 0.000
Marks: 2 x 10 = 20
Q.2 Answer the following?
2.1
For the water with the above ionic composition (expressed in mg/L as CaCO3) suggest the
treatment scheme for water softening? If the water has 25 mg/L of dissolved carbon dioxide,
find the chemical (lime, soda ash and carbon dioxide) requirement for the softening? Indicate
where or at which step how much of each of the chemicals should be dosed?
Since Mg2+ is 41.2 mg/L as CaCO3 (or 0.824 meq/L), excess lime process should be used. The
process involves the four steps:
Step-1: Precipitation by dosing 1.25 meq./L of excess lime, raising pH to 11.0 and settling to remove
precipitates
Step-2: Recarbonation to neutralize hydroxides and reduce pH to 10.3.
Step-3: Precipitation by dosing soda ash and settling to remove precipitates
Step-4: Recarbonation to neutralize carbonates to bicarbonates and reducing pH to 8.5 to 9.0Lime requirement in the step-1:
Ionic species Concentration Chemical dose as meq/L (mg/L
as Ca(OH)2Mg/L as CaCO3 meq./L
Lime
CO2 25 1.136 1.136
Ca(HCO3)2 90.2 1.804 1.804
CaSO4 9.6 0.192 ---
MgSO4 41.2 0.824 0.824
Na2SO4 19.1 0.382 ---
NaCl 6.6 0.132 ---
KCl 8.9 0.178 ---
Dose 1.136+1.804+0.824+1.25 = 5.014
= 5.014x28 =140.39 mg/L as CaO
Carbo Dioxide
Mg(OH)2 0.2 0.2
Ca(OH)2 1.25 1.25
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Dose 1.47 = 1.47x22 = 32.34 mg/L as
CO2
Soda ash
CaSO4 9.6 0.192 0.192
MgSO4 41.2 0.824 0.824
Dose 1.016 = 1.016x53 = 53.848 mg/L
as Na2CO3
Recarbonation
MgCO3 0.2 0.2
CaCO3 0.6 0.6
Dose 0.8 = 0.8x22 = 17.6 mg/L as CO2
2.2 Using the settling profile graph given, design
a clari-flocculator for the clarification of 250
m3/hour flow of water with 65% TSS removal
efficiency?
If TSS of the water is 300 mg/L and if the
consistency of the sludge is 4% find the daily
quantity and volume of the sludge
generated?
Estimation of overall TSS removal efficiency by different times of settling (HRT) and for different
overflow rates:
S.No. Tine (HRT)
(min.)
Overflow
rate (m/hr)
Removal components Overall
removal %
1. 20 7.50 = 40% +
10%x(2.5+0.75)/2/2.5 = (6.5%) +
10%x(0.75+0.4)/2/2.5 = (2.3%) +
10%x(0.4+0.24)/2/2.5 = (1.28%) +
10%x(0.24+0.12)/2/2.5 = (0.72%) +
20%x(0.12+0)/2/2.5 = (0.48%
51.25%
2. 40 3.75 = 50% +
10%x(2.5+1.15)/2/2.5 =(7.3% ) +
10%x(1.15+0.65)/2/2.5 = (3.6%) +
10%x(0.65+0.35)/2/2.5 = (2%) +
20%x(0.35+0.0)/2/2.5 = (1.4%)
64.3%
3. 60 2.50 58%+
2%x(2.5+2.25)/2/2.5 = (1.9%) +
+10%x(2.25+1.25)/2/2.5 = (7%) +
73.9%
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10%x(1.25+0.75)/2/2.5 = 4% +
20%x(0.75+0.0)/2/2.5 = 3%
4. 80 1.875 66% +
4%x(2.5+2.1)/2/2.5 = (3.68%) +
10%x(2.1+1.2)/2/2.5 = (6.6%)+20%x(1.2+0.0)/2/2.5 = (4.8%)
81.08
Theoretical HRT and overflow rate for 65% overall removal efficiency from the graphs is 42 min and
3.57 m/hr respectively.
Corrected HRT and overflow rate values are 63 minutes and 2.38 m/hr respectively.
Settling area of the primary clarifier for 250 m3/hr flow is 250/2.38 = 105 min or 1.75 hours.
Settling zone volume of the primary clarifier is 250x1.75 = 437.5 m3.
TSS of the water is 300 mg/L and removal efficiency is 65%
Sludge generation rate is 250x0.3x0.65 x24 = 1,170 kg/day or 29.25 m3/day volume of sludge at 4%
consistency .
2.3 Water demand pattern of a township is as given below:
Time4
AM
5
AM
6
AM
7
AM
8
AM
9
AM
10
AM
11
AM
12
Noon
1
PM
2
PM
3
PM
Demand
(m3/min) 0.4 0.8 1.8 3.2 2.2 1.2 07 0.6 1.0 2.8 2.1 0.9
Time4
PM
5
PM
6
PM
7
PM
8
PM
9
PM
10
PM
11
PM
Mid
night
1
AM
2
AM
3
AM
Demand
(m3/min)
1.0 1.7 2.8 3.0 1.4 0.6 0.5 0.4 0.3 0.25 0.2 0.3
Water to the township is supplied from an elevated service reservoir under gravity. A
pumping operating at constant flow rate for 14 hours/day (6 AM to 2 PM and 6 PM to
midnight) supplies water to the elevated service reservoir. Find water storage capacity of the
service reservoir required?
0
10
20
30
40
50
60
70
80
90
0 20 40 60
0
10
20
3040
50
60
70
80
90
0 2 4 6 8
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TimeInflow Outflow Net inflow
Cumulative net
inflow
m3/hr m
3/min. m
3/hr m
3/hr m
3
4:00 AM 0 0.4 24 -24 -24
5:00 AM 0 0.8 48 -48.00 -72.00
6:00 AM 129 1.8 108 21.21 -50.79
7:00 AM 129 3.2 192 -62.79 -113.57
8:00 AM 129 2.2 132 -2.79 -116.36
9:00 AM 129 1.2 72 57.21 -59.14
10:00 AM 129 0.7 42 87.21 28.07
11:00 AM 129 0.6 36 93.21 121.29
12 Noon 129 1 60 69.21 190.50
1:00 PM 129 2.8 168 -38.79 151.71
2:00 PM 0 2.1 126 -126.00 25.71
3:00 PM 0 0.9 54 -54.00 -28.294:00 PM 0 1 60 -60.00 -88.29
5:00 PM 0 1.7 102 -102.00 -190.29
6:00 PM 129 2.8 168 -38.79 -229.07
7:00 PM 129 3 180 -50.79 -279.86
8:00 PM 129 1.4 84 45.21 -234.64
9:00 PM 129 0.6 36 93.21 -141.43
10:00 PM 129 0.5 30 99.21 -42.21
11:00 PM 129 0.4 24 105.21 63.00
mid night 0 0.3 18 -18.00 45.00
1:00 AM 0 0.25 15 -15.00 30.00
2:00 AM 0 0.2 12 -12.00 18.00
3:00 AM 0 0.3 18 -18.00 0.00
Total water over the day = 1809 m3(sum of the outflow column)
Duration of pumping: 14 hours
Water pumping rate: 1809/14 = 129.214 m3/hr.
Inflow is by pumping between 6 AM and 2 Pm and between 6 Pm and midnight (shown in the inflow
column)
Water stoage capacity of the service reservoir = 190.50 + 279.86 = 470.36 m3.
Marks: 3 x 8 = 24
Q.3 Answer the following?
3.1 Differentiate water softening (by ion exchange process) from de-alkalization (by ion-
exchange process) and from de-cationization (by ion-exchange process)?
Water softening involves exchange of calcium, magnesium and other divalent cations with sodium
by passing the water through a strongly acidic cation exchange resin bed. The resin be is regenerated
by sodium chloride solution.
Dealkalization involves exchange of calcium and magnesium like cations with H+ ions by passing the
water through a weakly acidic cation exchange resin bed. Water coming out from the resin bed is
subjected to degasification. This results in the removal of alkalinity from water. The resin isregenerated with acids like hydrochloric acid.
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Decationation involves exchange of all the cations including sodium with H+ ions by passing the
water through a strongly acidic cation exchange resin bed. The resin is regenerated by acids like
hydrochloric acid. Output water from the resin bed is degasified to remove the alkalinity.
3.2 Write note on chlorine gas application systems and state how the chlorine dose is
controlled?
Chlorine from the liquid chlorine bullets is
vapourized into chlorine gas. Source for the
latent heat required is provided.
The gasous chlorine is dissolved in water to
prepare chlorine solution of
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n e al increase
1951 98,000
1961 125,000 27,000
1971 161,00036,000 9,000
1981 220,000 59,000 23,000
1991 270,000 50,000 -9,000
2001 342,000 72,000 22,000
2011 405,000 63,000 -9,000
Average 51,167 7,200
2
1030
YnnnXPP
P30 = 405,000 +3x51,167 + 3x(3+1)x7,200/2 = 6,01,701P40 = 405,000 +4x51,167 + 4x(4+1)x7,200/2 = 6,81,668
Marks: 4 x 6 = 24
Q.4 Answer the following?
4.1 Write note on breakpoint chlorination?
4.2 Write note on bio-sand filters?
4.3 Write note on the laboratory experimentation required for the coagulation-flocculationtreatment of water?
Laboratory experimentation involving use of jar test apparatus is required.
The experimentation involves finding rough coagulant dose, optimum pH, optimum coagulant dose
and appropriate dose of polyelectrolyte.
For rough coagulant dose, specific volume of sample is taken, pH is adjusted to 6 and then dosed
with coagulant in small increments. After each dosing the sample is flash mixed for a minute and
slow mixed 1 minute and then observed for visual flocs. The dose at which visible flocs are formed is
taken as the rough dose.
For optimum pH, the sample is taken in the jars of the jar test apparatus, pH is adjusted to different
values in different jars, rough dose of coagulant is applied to all the jars, flashedmixed for 1 to 3minutes, then slow mixed for 12-15 minutes and then left for settling for 30 minutes. the
supernatant is tested for turbidity. The pH at which the turbidity is lowest is taken as the optimum
dose.
For optimum coagulant dose the experiment is repeated this time with optimum pH in all the jars
and varying coagulant dose around the rough dose. The dose at which turbidity is the lowest is taken
as the optimum dose.
For the appropriate polyelectrolyte dose also the experiment is repeated at the optimum pH and
optimum coagulant dose with one difference. Polyelectrolyte is dosed after flash mixing and before
slow mixing. The dose at which desired level of turbidity is achieved is taken as the appropriate dose
of the polyelectrolyte.
4.4 Write note on the adsorption isotherms?
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Langmuir Isotherm
e
e
KC
KbC
M
X
1
X is the mass of the solute adsorbed on the adsorbent (mg)
M is mass of the adsorbent used
Ce is equilibrium concentration of the solute in the water/wastewater after adsorption
K and b are constants.
Freundlich Isotherm
n
ekCM
X /1
K is constant.
4.5 Write note on the classification of waters?
Marks: 5 x 4 = 20
Q.5 Answer the following?
5.1 Write note on Chicks law?
5.2 Write note on Darcys equation?
5.3 Write note on solar water stills?
5.4 What is Baringo curve?
5.5 List the techniques that can be used for the removal dissolved solids from water?
5.6 Write note on fecal coliform bacteria?
Marks: 6 x 2 = 12