: 1 : civil engg. ese mains...correction for gradient: effective gradient = 100 0.5% 1260 6.3...



01. (a)

Sol:

(1) The velocity components in x and y

directions are

y2x

))1x2(y(

xu

1x2y

))1x2(y(

yv

The resultant velocity is equal to

22 vuV

22 )1x2()y2(

At the point P(3, 4), when x = 3 and y = 4

22 )42()132(

= 9.434 m/s

(2) The value of stream function at the point P

(3,4) is

dyy

dxx

d

d = –vdx + udy

d = (2x – 1) dx + –2ydy

Integrating both sides we get

= x2 – x – y2 + C

At point P (3, 4)

= 9 – 3 – 16 = – 10 units

(b) (i)

Ans:

Methods which can be used to control

swelling of soil may include:

1) Lime stabilization: Lime stabilization

reduces liquid limit and plasticity index

and thereby expansiveness of clay.

2) Moisture control: Moisture is

controlled in and around the foundation

and wetting and drying is prevented.

3) Prewetting: It reduces future expansion.

4) Compaction control: Compaction of

expansive clays at a moisture content

slightly more than the natural moisture

content and at low density reduces

swelling of soil.

5) Replacement: Replace the foundation

soil with non swelling soil like granular

soil like coarse sand. The minimum

thickness of fill below the footings is

1m. The replacement should bevertically

below the footing and also laterally.

: 3 : Test – 16




(ii)

Ans: Advantages of Reinforced soil walls:

1. They are generally more economical if

the wall heights are large or when the

subsoil conditions are poor.

2. They can be quickly constructed

3. Require relatively simple construction

equipment

4. They can be considered as flexible

structures with greater ability to

withstand differential settlement than the

rigid retaining walls.

5. Due to the large base to height ratio, the

foundation stress distribution is nearly

uniform with little stress concentration at

the toe. This enables the construction of

high retaining structures.

(c) (i)

Ans:

The following are some of the important

methods of population forecasts or

population projections:

1. Arithmetical increase method

2. geometrical increase method

3. Incremental increase method

4. Graphical method

5. Comparative method

6. Zoning method

7. Ratio and correlation method

8. Growth composition analysis method.

1) Arithmetic increase method:

Rate of change of population with time is

assumed to be constant.

Applicable to old and large cities with no

industrial growth and reached a saturation

or maximum development.

2) Geometrical increase method:

Percentage increase in population from

decade to decade is assumed to be

constant. It gives good results for young

and rapidly expanding cities.

3) Incremental increase method:

Combination of the above two methods.

In this method , the average of increase in

population is found by arithmetic

increase method and to this is added the

average of net incremental increase.

4) Decreasing rate method: Quite rational

method for the cities whose rate of

increase goes on reduce, as they reach

saturation.

: 5 : Test – 16


5) Simple graphical method:

A graph plotted for a city between

time and population is smoothly

extended to the desired time. The

graph is called ‘ logistic curve’

This gives very approximate results,

It is very unsafe to use this method

alone.

6) Comparative graphical method:

Based on the assumption that the city

under consideration may develop

same as the selected similar cities

developed in the past.

It is based on a logical background,

precise and reliable results can be

obtained by this method

(ii)

Ans:

Factors affecting population growth:

1. Economic Factor: Such as development

of new industries, discovery of oil or

other minerals in the vicinity of the city.

2. Development Programmes:

Development of projects of national

importance, such as river valley projects

etc.

3. Social Facilities: Educational, medical,

recreational and other social facilities.

4. Communication Link: Connection of

the town with other big cities, and also to

the mandies of a agricultural products.

5. Community Life: Living habits, social

customs, and general education in the

community.

(d)

Sol:

Given:

Free mean speed Vsf = 80 kmph,

Sjam = 6.9 m

Jam density, Ki =9.6

1000

= 145 Vehicles/km(per lane)

Maximum flow qmax or capacity flow,

4

KVq jsfc

=4

14580

2900 Vehicles/ hour/ per lane

: 7 : Test – 16


(e) (i)

Ans:

The rate of evaporation is dependent the

following factors.

1. Vapour Pressure: The rate of evaporation

is proportional to the difference between the

saturation vapour pressure at the water

temperature, ew and the actual vapour

pressure in the air ea.

2. Temperature: Other factors remaining the

same, the rate of evaporation increases with

an increase in the water temperature.

3. Wind: Wind aids in removing the

evaporated water vapour from the zone of

evaporation and consequently creates greater

scope for evaporation. However, if the wind

velocity is large enough to remove all the

evaporated water vapour, any further

increase in wind velocity does not influence

the evaporation. The rate of evaporation

increases with the wind speed up to a critical

speed beyond which any further increase in

the wind speed has no influence on the

evaporation rate

4. Atmospheric Pressure: Other factors

remaining same, a decrease in the barometric

pressure, as in high altitudes, increases

evaporation.

5. Soluble Salts: When a solute is dissolved

in water, the vapour pressure of the solution

is less than that of pure water and hence

causes reduction in the rate of evaporation.

Under identical conditions, evaporation from

sea water is about 2-3% less than that from

fresh water.

6. Heat Storage in Water Bodies: Deep

water bodies have more heat storage than

shallow ones.

(ii)

Ans:

On the basis the saturated formations are

1. Aquifer: An aquifer is a saturated

formation of each material which not only

stores water but yields it in sufficient

quantity. Thus, an aquifer transmits water

relatively easily due to its high permeability.

Unconsolidated deposits of sand and gravel

form good aquifers.



2. Aquitard: It is a formation through which

only seepage is possible and thus the yield is

insignificant compared to an aquifer. It is

partly permeable. A sandy clay unit is an

example of aquitard. Through an aquitard

appreciable quantities of water may leak to

an aquifer below it.

3. Aquiclude: It is a geological formation

which is essentially impermeable to the flow

of water. It may be considered as closed to

water movement even though it may contain

large amounts of water due to its high

porosity. Clay is an example of an aquiclude.

4. Aquifuge: It is a geological formation

which is neither porous nor permeable. There

are no interconnected openings and hence it

cannot transmit water. Massive compact rock

without any fractures is an aquifuge.

02. (a)

Sol:

1)g

pzHGL

HGL1 = m65.39)807.9)(900(

3500000

m75.34)807.9)(900(

250000)40)(sin10(HGL o2

Since HGL1 > HGL2 , the flow is upward.

2) hf = HGL1 – HGL2 = 39.65 – 37.45

= 4.90 m

3) µ = ν

= (900) (0.0002) = 0.180 kg/m.s

L128

hgdQ f

4

(Poisellei’s formula)

)10)(180.0)(128(

)90.4(100

6)807.9)(900)((

4

= 0.00764 m3/s

4100

6

00764.0

A

Qv

2

= 2.70 m/s

8100002.0

)70.2(100

6

v

VdRe

This value of Re is well within the laminar

range(less than 2000);

Hence the flow is most likely laminar.



(b)

Sol:

Given:

d (in situ) = 14 kN/m3

d (min) = 10.8 kN/m3

Relative density , ID = 85%

Consider a prism of soil of 1 m width and 1 m

length and 5 m thickness (in-situ condition).

Initial volume of soil, V = 1 1 5 = 5 m3

Let Ws be weight of solids

d(in situ) =V

WS

5

W14 s

Ws = 70 kN (in the prism considered)

(1) After compaction the volume,

V = 1 1 4.5

= 4.5 m3

After compaction,

55.155.4

70

V

Wsd kN/m

3

(2) Using the relation,

ID =mindmaxd

mindd

d

maxd

8.10

8.1055.15

55.1585.0

maxd

maxd

d max = 16.85 kN/m3

Dry unit weight in densest state,3

maxd m/kN85.16

(3) Ideal compaction produce,

d max = 16.85 kN/m3

V

Wsd

H11

7085.16

H = 4.15 m

Max possible settlement under ideal

compaction = 5 – 4.15 = 0.85 m

(c)

Sol: Correction for Elevation :

The basic length is to be increased at the rate

of 7% per 300 mm elevation above mean

sea level

Correction for elevation = 1260300

400

100

7

= 117.6 ≃ 118 m

elevationfor

correctionafterrunwayofLength

= (1260 +118) = 1378 m

Correction for temperature:

Airport reference temperature =3

TTT 121

: 11 : Test – 16


Where T1 = Mean of the average daily

temperature = 26.2oC

T2 = Mean of the maximum daily

temperature = 44.8oC

Airport reference temperature

=3

2.268.442.26

= 26.2 + 6.2 = 32.4oC

Standard atmospheric temperature at mean

sea-level = 15oC

Rise in temperature = (32.4 – 15) = 17.4oC

Applying correction at the rate of 1% for

every 1oC

Correction for temperature

= 4.171378100

1

= 239.77 ≃ 240 mCorrected length = (1378 + 240)

= 1618 m

Check:

Total correction for elevation and

temperature = (118 + 240) = 358 m

Percentage increase = %41.281001260

358

According to ICAO, this should not be more

than 35%

Correction for gradient:

Effective gradient = %5.01001260

3.6

Applying correction for the effective gradient

at the rate of 20% for each 1% effective

gradient,

Correction for gradient =1

5.01618

100

20

= 161.8 ≃ 162 mActual length of runway = (1618 + 162)

= 1780 m



03. (a)

Sol: Given:

12 hr UHG ordinates

To find:

4 hr UHG ordinates

D = 12 hr

T = 4 hr

T < D

T = nD

4 = n(12)

n =3

1

Time (h)Ordinatesof 12 hrUHG (m3/s)

S-curveadditives(m3/s)

S-curveordinates(m3/s) (SA)

S-curve laggedby 4 hr (m3/s)(SB)

Differencegraph(SA – SB)

4hr UHGordinates

(m3/s)n

SS BA

0 0 - 0 - 0 0

4 6.7 - 6.7 0 6.7 20.1

8 33.3 - 33.3 6.7 26.6 79.8

12 76.7 0 76.7 33.3 43.4 130.2

16 120 6.7 126.7 76.7 50 150

20 136.7 33.3 170 126.7 43.3 129.9

24 123.3 76.7 200 170 30 90

28 90.7 126.7 217.4 200 17.4 52.2

32 56.3 170 226.3 217.4 8.9 26.7

36 31.3 200 231.3 226.3 5 15

40 15.7 217.4 233.1 231.3 1.8 5.4

44 6.7 226.3 233 233.1 0.1 0.3

48 1.7 231.3 233 233 0 0

52 0 233.1 233.1

56 233 23360 233 233

: 13 : Test – 16


(b)

Sol:

(1) BOD loading based on total pond area:

Total pond area of both cells joined in series

= 60,000 m2 + 30,000 m2

= 90,000 m2 = 9 hectares

Total BOD per day = 200 kg/day

BOD loading in kg/ha/day

9

200 kg/day/ha = 22.2 kg/ha/day

BOD loading based on area of larger cell

only:

Area of larger cell = 60,000 m2 = 6 ha

BOD = 200 kg/day

BOD loading in kg/ha/day

6

200 kg/day/ha = 33.3 kg/ha/day

(2) To calculate the number days of storage

between Water level 0.6 m and 1.5 m we

have

Depth available for storage = 1.5 – 0.6

= 0.9 m

Total area = 90,000 m2

Volume of storage available

= 90,000 0.9

= 81,000 m3

Daily inflow of sewage = 900 cu.m/day

The sewage volume, which percolates and

evaporates daily = 2.5 mm depth

m100

1

10

5.2surface area of tanks

2m000,90m1000

5.2 = 225 m3

Net effective daily inflow of sewage

= (900 – 225) m3 = 675 m3/day

Winter storage available as days

day/minlowinfsewagenetDaily

minstorageof.Vol3

3

days120days675

000,81

(c)

Sol: Given data:

C = 20 kPa, = 25o

3 = 200 kPa

1 – 3 = 110 kPa

1 = 100 + 3

= 110 + 200 = 310 kPa

Let ‘u’ be the pore pressure

Using the plastic equilibrium equation,

245tanC2

245tan 231

245tanC2

245tanuu 231



2

2545tan202

2

2545tanu200u310 2

310 – u = (200 – u) 2.464 + 62.79

1.464u = 245.59

u = 167.75 kPa

(d)

Sol: Given:

Number of impulse turbine, ni = 5

Number of runners, nr = 2

Number of Nozzles, nn = 4

Total discharge, QT = 40 m3/s

Head, H = 250 m

Coefficient of velocity, CV = 0.985

To Find:

Diameter of Jet:

Discharge through one nozzle

NozzleofnumberTotal

eargdischTotal

t

T

n

QQ

nt = ni × nr × nn = 5 × 2 × 4

nt = 40

40

40Q ; Q = 1 m3/s

Velocity of jet gH2CV V

25081.92985.0

V = 68.985 m/s

Discharge = Area of jet × Velocity of jet

Vd4

Q 2

985.68d4

1 2

Diameter of jet,

d = 13.58 cm

04. (a) (i)

Ans:

The important Primary air pollutants are

1. Oxides of sulphur, particularly the

sulphur dioxide (SO2)

2. Oxides of carbon like carbon monoxide

(CO) and carbon dioxide (CO2),

particularly the carbon monoxide (CO);

3. Oxides of nitrogen like NO, NO2, NO3

(expressed as NOx);

4. Volatile organic compounds, mostly

hydrocarbons and

5. Suspended particulate matter (SPM)

The important secondary pollutants are:

1. Sulphuric acid (H2SO4);

2. Ozone (O3);

3. Formaldehydes and

4. Peroxy-acyl-nitrate (PAN); etc.

: 15 : Test – 16


(ii)

Ans:

Sound is usually divided into three types;

(1) Continuous (2) Intermittent (3) Impulsive

as explained below

(1) Continuous noise: Is an uninterrupted

sound level that varies less than 5 dB during

the entire period of observation. A running

fan is an example of such a sound.

(2) Intermittent noise: Is a noise which

continues for more than 1 second and is then

interrupted for more than 1 second. A drilling

machine used by a dentist produces such type

of sound.

(3) Impulse Noise: Is a characterized by a

change of sound pressure of atleast 40 dB

within 0.5 second with a duration of less then

one second. The noise produced from firing

of a weapon would fall in this category.

(iii)

Sol:

Assume 100 kg of sludge

m.c of sludge = 95%

Weight of total solids = 5 kg

Weight of water in sludge = 95 kg

5 kg of total solids contain 65% of volatile

solids

Weight of volatile solids = 5 kg 65%

= 3.25 kg

Weight of fixed solids = 5 – 3.25 = 1.75 kg

Specific gravity of volatile solids = 1.00

Specific gravity of fixed solids = 2.2

Specific gravity of total solids

5

2.275.100.125.3 = 1.35

Specific gravity of wet sludge

95 kg specific gravity of water

kg100

solidstotalofgravityspecifickg5

100

35.1500.195 = 1.0175

(b) (i)

Sol: Design of a lined canal:

Most generally lined canal will be

trapezoidal cross section with rounded

bottom.

D

B = given = 6

A

QV

A = BD + D2 ( + cot )

= D2

cotD

B

90o 90o

D

B



2

3cot

= 0.588 rad

588.0cot588.06D 2 = 088.8D088.26D 22

P = B + 2D ( + cot )

=

cot2D

BD = D [6 + 2(2.088)]

= D [10.176]

D795.0176.10D

088.8D

P

AR

2

From Manning’s formula & continuity

equation

A

QSR

N

1V 2/13/2

088.8D100

5000

1D795.0

016.0

12

2/13/2

D8/3= 16.30

D = 2.85 m

B = 6D = 17.1 m

: 17 : Test – 16


(ii)

Sol:

b = 8 – 1.4

= 6.6 m

a = 8 – 1.7

= 6.3 m

K = 1 10–4 m/s

Annual rainfall = P = 85 cm

606024

drainperareaP%1QDrain

606024

LP01.0

[Area per drain

per unit length = L 1]

Also, Qdrain = dL

Where d = dia of drain

d

22

Q

abk4L

606024

L100

5.801.0

3.66.6104 224

8501.0

1006060243.66.6104L

2242

L = 125.43 m

Provide spacing of 125 m between two

consecutive drains

(c)

Sol:

From figure, in a triangle ABC

451102

AB45110Cos

222

(1)

From triangle BCD

0011102

105100110cos

222

(2)

(1) = (2)

451102

AB45110

1001102

105100110 222222

0.5034 =9900

AB45110 222

AB2 =1102 + 452 – 0.5034 9900

1.4 m

6.3 m

1.7 m

6.6 m8 m

LC

110 m

105 m

B

D

A

45 m

55 m



AB = 95.610 m

The angle ‘’ which the line CD makes

with the line AB.

From triangle ABC

45610.952

11045610.95cos

222

= 0.1085

= 96.23o

(d)

Sol:

Consider a tank of uniform cross-section A,

through an orifice of area ao, located in the

base of the tank. An unsteady state mass

balance on the tank for no inflow of liquid

relates the rate of flow from the tank through

the orifice to the change of capacity of the

tank.

dt

dHAgH2aC od

Rearranging, the time taken for tank

drainage from a head H1 to H2 can be found

by integration.

2

1

H

H

2

1

od

t

o

dHHg2aC

Adt

Completing the integration

2

1

12

1

2od

HHg

2

aC

At

For total drainage, H2 = 0. Therefore

g

H2

aC

At 1

od

The initial liquid level above the orifice, H1.

is obtained from the volume of the tank Vt,

where dt is the diameter of the tank

21

t1 d

V4H

21

14

= 1.27 m

Therefore

g

27.12

4

02.06.0

4

1

t2

2

= 2120 s = 35 minute 20seconds

Total drainage is found to take 35 minutes

and 20 seconds.

05. (a)

Sol: Given:

Side slope, m = 1

Longitudinal slope So = 0.001

Discharge Q = 0.2 m3/s

Manning’s n = 0.015

Assume y = depth of flow

For the given triangular channel

Area of flow A = 22 ymy

1m

my my

y

: 19 : Test – 16


Top width T = 2my = 2y

Wetted perimeter P = 22 y)my(2

= y22

Hydraulic radius, R =P

A

22

y

y22

yR

2

At Critical depth yc:

2

y

y2

y

T

A

g

Q 5c

c

6c

c

3c

2

m382.0

81.9

2.02

g

Q2y

5/125/12

c

At Normal depth yn:

2/1o

3/2 SARn

1Q

2/1o

3/8n

2/1o

3/2

n2n

Syn

5.0

S22

yy

n

1

18974.0

001.05.0

2.0015.0

S5.0

nQy

2/12/1o

3/8n

yn = 0.536 m

Since yn > yc , the channel has a mild slope

for this discharge. If y is the depth of flow:

For

M1 curve y > 0.536 m

M2 curve 0.536 m > y > 0.382 m

M3 curve y < 0.382 m

(b)

Ans:

The geotechnical report is a professional

document used to communicate the site

conditions and design and construction

recommendations. Site investigations for

construction projects have the purpose of

providing specific information on subsurface

soil, rock, and water conditions. This

information should be presented in a project

geotechnical report.

` The information contained in this report is

referred to often during the design period,

construction period, and even after

completion of the project. Therefore, the

report should be as clear, concise, and

accurate as possible.

While the geotechnical report content may

vary by project size and producing agency, all

geotechnical reports should contain certain

basic essential information, as follows:

M1

M2

M3

NDC (yn)

CDC (yc)



Summary of all subsurface exploration

data, including subsurface soil profile,

exploration logs, laboratory or in

situ test results, and ground water

information;

Interpretation and analysis of the

subsurface data;

Specific engineering

recommendations for design;

Discussion of conditions for solution

of anticipated problems.

(c) (i)

Ans:

Comparison between conservancy and water carriage system

Conservancy System Water Carriage systemThe system is unhygienic since every thing isvisible

The system is hygienic. Sewers are laid below theground and hence excreta etc, is not visible

Due to putrefication, there is a lot of foulsmell

No chances of putrefication, and hence no foulsmell

Compact house design is not possible Compact design is possible

Large labour force is required Labour force is negligibly small

Water consumption is small Requires high water consumption

Initial cost is small, though the running costsare high

High initial cost. Running costs small

No technical persons require Technical persons required for operation andmaintenance

Acute pollution problem Pollution problems are rare

Risk of spread of epidemic No such risk

Large land required for the disposal ofuntreated sewage

Small land required for the disposal of treatedsludge

Final disposal into streams etc not free fromrisks

Final disposal easier because of treatment works

Good quality manure available from the endproducts

The sludge has small manure value. However,treated waste water can be used for irrigated etc.

The system is more suitable for ruralconditions

The system is better suited for urban conditions.

: 21 : Test – 16


(ii)

Sol:

The sewage flow is equal to 75% of rate of

water supply

Hence sewage flow will be equal to 0.75

200 = 150 litres/capita/day

Sewage flow (D.W.F)

1000

1

606024

15050000

= 0.0868 litres/seconds

The rainfall intensity is given by

bt

a4.25R i mm/hour

Here t = 50 minutes, a = 40; b = 20

2050

404.25R i

= 14.5 mm/hour

= 1.45 cm/hour

The W.W.F is given by

Q =60.

3.05.1440

360

AIR = 0.483 m3/sec

Hence the design discharge is given by

Q = 2 (D.W.F) + W.W.F

= 2(0.0868) + 0.483 = 0.6566 lit/sec

Comment: Ratio of D.W.F and W.W.F

18.0483.0

0868.0

Since this ratio is not very large, it is

preferable to use a combined sewer system.

(d) (i)

Sol:

Calculation of area by meridian distance

method:

Meridian distance of any line = Meridian

distance of preceding line +2

1departure of

preceding line +2

1 departure of present line

Meridian distance of line AB = m1 = 50 m

Meridian distance of line BC = m2 =50 +50 + 50

= 150 m

Meridian distance of line CD = m3 =150 +50 – 50

= 150 m

Meridian distance of line DA = m4 =150 –50 – 50

= 50 m

Line M.D (m) Latitude (L) m L

AB 50 –100 –5000

BC 150 +100 +15,000

CD 150 +100 +15,000

DA 50 – 100 –5,000

A = mL 20,000 m2

Area = A = 20,000 m2



(ii)

Sol:

When the staff held vertical

D = KS cos2 + C cos

V =2

KSsin 2 + C sin

Since tacheometer is fitted with anallactic

lenses constants K=100 and C= 0

CA = 100 (2.190 – 1.250) cos2 3o 50

= 93.580 m

CB = 100 (1.945 – 1.095) cos2 0o 12

= 85 m

V1= m270.60532sin940.01002

1 o

m297.02102sin850.01002

1V o2

Assuming R.L of instrument axis as ‘h’

We have

R.L of A = (h + V1 – r1)

R.L of B = (h + V2 – r2)

Difference in R.L of A and B

= (h + V1 –r1) – (h + V2 – r2)

= (V1 –V2) – (r1 – r2)

= (6.270 – 0.297) – (1.720 – 1.520)

= 5.773 m

Horizontal distance AB can be determined

from

CBCA2

ABCBCA0540cos

222o

8558.932

AB8558.93 222

AB = 62.813 m

Gradient of AB =cetandisHorizontal

DifferenceL.R

092.0813.62

773.5

i.e 1 in 10.88

(e)

Sol:

b1 = 15 m, b2 = 25 m, d = 5 m, H = 4 m

2

11 2221

3d

b11

A

B

40o 50C

25 m15 m

4 m

E

D

5 m

C

: 23 : Test – 16


2

5131 22

5d

b22

= 4.13

97.02

11 2221

1

Uplift pressures:

P at E :

11cos

H = 2.633 m

P at C : m991.11

cosH 11

P at D: m302.2cosH 11

In units of kPa:

P = Wh

PE = 26.33 kPa

PC = 19.91 kPa

PD = 23.02 kPa

06. (a)

Sol: Power P = o QH

For the prototype:

45Q

1000

99881.982.06750

= 361.265 Q

Discharge Qp = 18.685 m3/s

Using the suffixes m and p to denote model

and prototype parameters respectively

ratioScale8

1

D

D

p

m

As, Dp = 3.0 m, Dm = m375.08

0.3

Hp = 45 m, Hm = 9.0 m,5

1

45

9

H

H

p

m

Speed:p

pp

m

mm

H

DN

H

DN

Nm = Speed of the model2/1

p

m

m

pp H

H

D

DN

= 300 (8)2/1

5

1

= 1073.3. rpm

Discharge:3pp

p

3mm

m

DN

Q

DN

Q

Qm = Model discharge

3

p

m

p

mp D

D

N

NQ

3

8

1

300

3.1073685.18

= 0.13056 m3/s

Power: 5p

3p

p5m

3m

m

DN

P

DNP

Pm = Model power



5

p

m

3

p

mp D

D

N

NP

53

8

1

300

3.10736750

= 9.433 kW

Specific speed: Since the model and the

prototype are similar we expect them to have

the same specific speed.

Hence4/5

m

mmsm H

PNN

4/59433.93.1073

= 211.5 rpm

4/5p

pp

sp H

PNN

4/5456760300

= 211.5 rpm

It is seen that Nsm = Nsp, as expected. This is

a check on the calculations.

(b) (i)

Sol:

The curve plotted on the basis of the above

data

From the curve, we find that break point

occurs at point D, at which the applied

chlorine = 1.0 mg/l

Break point dosage = 1.0 mg/l

Chlorine demand at break point

= 1.0 – 0.18 = 0.82 mg/l

It is observed that since the slope of curve C

is 45o, the chlorine demand (= 0.81 mg/l)

remains constant after the break point, since

all additional chlorine added after point D

appears as free chlorine.

Even from the data, at a dose of 1.2 mg/l, the

residual chlorine = 0.38 mg/l

Hence residual chlorine = 1.2 – 0.38

= 0.82 mg/l

(ii)

Ans:

The following valves are used in the water

distribution system:

1. Sluice valves or gate valves

2. Air valves

3. Reflux valves

4. Relief valves

5. Altitude valves0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0.2 0.4 0.6 0.7 0.8 0.9 1 1.2 1.4 1.6Res

idua

l Chl

orin

e (m

g/lit

)

Line B

C Line C

45o

0.18

Applied Chlorine (mg/lit)

: 25 : Test – 16


6. Scour valves

7. Pressure-relief Valves

1) Sluice Valves or Gate Valves: Used to

regulate the flow of water through the

pipelines.

2) Butterfly valves: Used to regulate and

stop the flow especially in large size

conduits.

3) Globe Valve: To regulate flow. These

valves are normally used in pipes of small

diameter (less than 100 mm) and as water

taps.

4) Check Valves: Also known as reflux

valves or non-return valves. A Check

valve allows water to flow in one

direction only and the flow in the reverse

direction is automatically stopped by it.

5) Air Valve or Air-relief Valves: The air

valve helps to admit air into the pipe when

the pipe is being emptied or when

negative or vacuum pressure is created in

the pipe.

6) Scour Valves or Blow-off Valves or

Drain Valves: Provided for completely

emptying or draining of the pipe for

removing sand or silt deposited in the pipe

and for inspection, repair, etc located at

dead ends and depressions or low points

in the pipeline

7) Pressure-relief Valves: Also called

overflow towers are provided to keep the

pressure in a pipeline below a

predetermined value, and thus protect it

against the possible danger of bursting

due to excessive pressure.



(c)

Sol:

The relationship between Rainfall (P) and runoff can be expressed as

R = aP + b

S.No Rainfall (P) Runoff(R) P2 R2 PR

1 90.5 30.1 8190.25 906.01 2724.05

2 111.0 50.2 12321 2520.04 5572.2

3 38.7 5.3 1497.69 28.09 205.11

4 129.5 61.5 16770.25 3782.25 7964.25

5 145.5 74.8 21170.25 5595.04 10883.4

6 99.8 39.9 9960.04 1592.01 3982.02

7 147.6 64.7 21785.76 4186.09 9579.72

8 50.9 6.5 2590.81 42.25 330.85

9 120.2 46.1 14448.04 2125.21 5541.22

10 90.3 36.2 8154.09 1310.44 3268.86

11 65.2 24.6 4251.04 605.16 1603.92

12 75.9 20.0 5760.81 400 1518

P=1165.1 R=459.9 P2=126900.03 R2=23092.59 PR=53143.6

R = a P + nb ______ (i)

RP = a P2 + b P ______ (ii)

Upon solving equation (i) and (ii), we get a

and b

459.9 = a × 1165.1 + 12 × b

53143.6 = a × 126900.03 + 1165.1 × b

a = 0.616

b = 21.5

R = 0.616 P 21.5

for, P = 100 cm R = 0.616 × 100 21.5

R = 40.1 cm

Coefficient of correlation,

])R()R(N[])P()P(N[

)R()P()PR(Nr

2222

])9.459(59.2309212[])1.1165(03.12690012[

9.4591.11656.5314312r

22

r = 0.978

: 27 : Test – 16


07. (a)

Sol:

Porosity, 375.06.01

6.0

e1

en

Assume the soil to be fully saturated (S =

100%)

Let ‘hc’ be the capillary head in cm

Use the relation :

c112

21

22 hh

n.S

k2

tt

xx

In the first stage:

c22

h5.7375.01

k2

5.8

5.112

3.127 = k (7.5 + hc) (i)

In the second stage:

c22

h5.22375.01

k2

10

1321

5.1 = k(22.5 + hc) (ii)

Dividing equations (i) & (ii)

c

c

h5.22k

h5.7k

1.5

127.3

0.6131 (22.5 +hc) = 7.5 + hc

6.295 = 0.387 hc

hc = 16.27 cm

Substituting hc = 16.27 cm into equation (i)

3.127 = K(7.5 + 16.27)

K = 0.1316 cm/min

(b)

Sol:

Applying the Bernoulli equation between the

free surface of the liquid (1) and the suction

point (subscript s) in the pump.

Ls

2ss

11 Hz

g2

v

g

Pz

g

p

Rearranging, the suction pressure head

L

2s

s11s H

g2

Vzz

g

p

g

p

The minimum suction head before

cavitation occurs is

g

p

g2

vNPSH

g

p v2ss

Thus, for cavitation not to occur

L

2s

s11v

2s H

g2

vzz

g

p

g

p

g2

vNPSH

Rearranging, and noting that the pressure on

the free surface is taken as 0.94 of the

standard atmospheric pressure, the depth

that can be pumped before cavitation occurs

is

Ls1v

1 Hzg

ppNPSHz

5.03

g970

103.10194.0185

3

= 0.385 m

Corresponding to a volume of



385.05.234

V 2 = 14.95 m3

The time taken to reach this level is

therefore

Q

Vt

5

95.14 = 3 hours

The quantity of liquid delivered is therefore

15 m3, taking 3 hours before caviation occurs.

(c)

Sol:

Volume of sediment (Vsed)

=weightspecificIts

entdimseofWeight

33

Mm3.0m/kN12

kN3600000 annually

Sediment trapped = average Vsed

Life of reservoir

=trappedentdimSe

yearperfilledbetoVolume

=trappedentdimSe

year/Mm6 3

I = 60 M m3 ( = annual flood inflow)

C = Initial capacity of reservoir = 30 Mm3

%

capacity

Capacity

C(Mm3) IC from

given tableaverage

Sediment

trapped (Mm3)Life in years

100 30 0.5 0.96

0.9575 0.2872528725.0

6 = 20.8877

80 24 0.4 0.955

0.9525 0.2857528575.0

6 = 20.9973

60 18 0.3 0.95

0.94 0.28202820.0

6= 21.276

40 12 0.2 0.93

0.90 0.27002700.0

6=22.22

20 6 0.1 0.87



Total life of Reservoir,

= 20.887 + 20.9973 + 21.226 + 22.2222

= 85.38 years

= 85.4 years

(d)

Sol:

Vehicle velocity = 100 km/h = 0.278 100

= 27.8 m/s

(1) On level surface: SSD =gf2

vvt

2

=35.081.92

8.275.28.27

2

= 69.5 + 112.5 = 182 m

(2) Upward gradient = 1.98%:

SSD = fg2v

vt2

98.101.035.081.928.27

5.28.272

= 69.5 + 106.5

= 176 m

(3) Downward gradient of 2% :

SSD = 201.035.081.928.27

5.28.272

= 69.5 + 119.5

= 189 m

08. (a)

Sol:

Unconfined compression strength of clay,

qu = 120 kPa

Cohesion, C = kPa602

120

2

q u

Nc = 9

= 0.75

F = 2.5

Bo = 2S + d

= 2 1 + 0.25

= 2.25 m

(a) Ultimate group capacity based on

individual pile failure mode:

C.ACN.AnQ scbgi

=

C..L.dCN.d4

n c2

6075.01025.096025.04

9 2

= 3419 kN

(b) Ultimate group capacity based on block

failure mode

Qgb = AB C Nc + As . C

= C.LBLCNB o1c2o

= 2.252 60 9 + 4 2.25 10 60

= 8133.75 kN

Bo



Ultimate group capacity, Qg = smaller of

Qgi and Qgb

= 3419 kN

Safe group capacity, =F

Qg

5.2

3419

= 1367.6 kN

(b) (i)

Sol:

1) Calculation of flying height of aircraft

Scale of photograph = S =hH

f

450H

3.0

000,15

1

Flying height = H = 0.3 15,000 + 450

= 4950 m

2) Calculation of No. of photographs

N = N1 N2

N1 = No of photographs in each flight

=

1

L

L1

000,15

12.0

6.01S

P1L1

= 1200 m

N1 =

11200

100040 = 34.3 ≃ 35

N2 = No of flights =

1

W

W1

15000

12.0

3.01s

wP1W w1

= 21, 000 m

N2 = 1.181000,21

100036

≃ 19

Total no. of photographs = N = 35 19

= 665

3) Spacing of flights

m2000119

000,36

1N

Wd

2

1

4) Exposure interval

sec6.1936001000220

1200

V

Lt

(ii)

Sol:

A) Geostationary satellites: A satellite which

stationary with respect to a given position of

the earth is known as a geostationary

satellite. A satellite launched on the

equatorial plane and travelling along the

same angular velocity as that at which the

earth rotates and in the same angular

velocity as that at which the earth rotates and

in the same direction will always remain

above the same point on earth at all the

: 31 : Test – 16


times. This is possible only when the orbital

altitude is 36,000 km. The orbital plane of

geostationary satellites coincides with the

equatorial plane and the orbit is circular.

These satellites are deployed for

communications and metrological

observation. Examples are satellite of the

INSAT series.

B) Sun synchronous satellites: When the

orbit plane of a satellite is nearly polar and

rotates at the same rate as the mean rotation

rate of the earth around the sun, the satellite

is known as sun-synchronous satellite. A

satellite in a sun-synchronous orbit always

passes over the same point on the earth at a

given local solar time. These satellites are

placed on relatively low altitudes (300 -1000

km) and hence facilitate a good resolution.

However, as these satellites are subjected to

deceleration by terrestrial atmosphere, the

orbital parameters need to be corrected

periodically. Similarly, the altitude of the

satellite is to be permanently controlled. The

examples are IRS, etc.

(c)

Sol: Assuming grade compensation on B.G as

equal to 0.04 percent per degree of curve.

Compensation allowed = (0.04 4) = 0.16percent

1 in 20 = 0.50

Actual gradient allowed = (0.50 – 0.16)

= 0.34 % (or) 1 in

294 will be the ruling gradient for the curve.

(d)

Sol: Flow velocity,A

Qvs

Here Q = 500 litres/hour

6060

10500 3

m3/sec

A = 1 m2

36001

10500v

3

s

= 0.139 m/sec

(a) For 0.05 mm particles:

Assume S = 2.65

Settling velocity,

mm1.0d;100

)70T3(d)1S(418v 2s

vs = 418 (2.65 – ) (0.05)2 = 1.72 mm/sec = 0.172 cm/sec

% settled %100100139.0

172.0

= 124% > 100%

Hence all the particles of 0.05 mm diameter

will settle

(b) For 0.02 mm particles:

Settling velocity,

mm1.0d;100

)70T3(d)1S(418v 2s

vs = 418 (2.65 – 1) (0.02)2 = 0.276 mm/sec= 0.0276 cm/sec

% settled = 100139.0

0276.0 = 19.86%

: 1 : civil engg. ese mains...correction for gradient: effective gradient = 100 0.5% 1260 6.3...

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