solutions of end semester paper-2013

8/13/2019 solutions of end semester paper-2013

1/12

Page 1of 12

Page 1of 12

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


2/12

Page 2of 12

Page 2of 12

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


3/12

Page 3of 12

Page 3of 12

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


4/12

Page 4of 12

Page 4of 12

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


5/12

Page 5of 12

Page 5of 12

-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


6/12

Page 6of 12

Page 6of 12

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


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


7/12

Page 7of 12

Page 7of 12

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%


8/12

Page 8of 12

Page 8of 12

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


9/12

Page 9of 12

Page 9of 12

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


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.


10/12

Page 10of 12

Page 10of 12

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


11/12

Page 11of 12

Page 11of 12

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


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?


12/12

Page 12of 12

Page 12of 12

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


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

Documents

solutions of end semester paper-2013