MAHARASHTRA STATE BOARD OF TECHNICAL EDUCATION

(Autonomous)
(ISO/IEC - 27001 - 2013 Certified)

WINTER–19 EXAMINATION
MODEL ANSWER 22501
Subject Name: WATER RESOURCES ENGINEERING Subject Code:

Important Instructions to examiners:

1) The answers should be examined by key words and not as word-to-word as given in the model answer
scheme.
2) The model answer and the answer written by candidate may vary but the examiner may try to assess the
understanding level of the candidate.
3) The language errors such as grammatical, spelling errors etc should not be given more Importance (Not
applicable for subject English and Communication Skills.
4) While assessing figures, examiner may give credit for principal components indicated in the figure. The
figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent
figure drawn.
5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may
vary and there may be some difference in the candidate’s answers and model answer.
6) In case of some questions credit may be given by judgment on part of examiner of relevant answer based
on candidate’s understanding.
7) For programming language papers, credit may be given to any other program based on equivalent
concept.

Q. Sub Answers Marking Total

No. Q. Scheme Marks
N.
Q.1 Attempt any Five of the following: 10
a) Define
i) Irrigation
ii) Runoff
Ans
i) Irrigation: Irrigation may be defined as the process of artificially supplying water to soil
1 M for
for full-fledged nourishment of the crops. OR each
definition
It is the science of artificial application of water to the land in accordance with crop
requirements throughout the crop period.

ii) Runoff: The amount of water which flows over the surface of the earth after
considering all losses is called as runoff. Runoff is expressed as in ‘m’ OR ‘cm’ or mm.

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Q.1 b) Enlist the Classification of Irrigation on the basis of purposes.

Ans A. Single Purpose Irrigation Project: A project planned for one or single
purpose is called as single purpose project e.g. Koyana project for
hydropower, Gangapur project for irrigation etc.
2M
B. Multipurpose Irrigation Project: A project planned for more than one for
purpose is called as multipurpose project. e.g. Bhakra Nangal project for any
irrigation & hydropower generation, Jayakwadi project for irrigation &
classifi
cation
hydropower generation etc.

OR
A. Surface Irrigation: a) Flow Irrigation : i) Perennial Irrigation ii) Flood
irrigation
b) Lift irrigation:
B. Sub-surface Irrigation: a) Natural Irrigation
b) Artificial Irrigation

( Note: Marks should be given to any type of classification)

Q.1 c) Define:
i) C.C.A 1 M for
ii) G.C.A each
Ans i) Culturable Command Area ( C.C.A.): It is the net area which can be irrigated by a
definition
canal system . It includes all land on which cultivation is possible , though all area
may not be under cultivation.
ii) Gross command area ( G.C.A.): It is the total area lying between drainage
boundaries which can be irrigated by a canal system. It includes barren land, village ,
roads and area of habitation.

G.C.A = C.C.A + Unculturable area

Q.1 d) Define Spillway and Enlist types of spillway
Ans Spillway: It is the overflow portion of every dam , over which the surplus water flows 1 M for
from the upstream side of reservoir to the downstream side of reservoir. It is the safety definition
valve of a dam. OR
It is an arrangement provided at the crest of dam to expel the excess water rises above
the full reservoir level.
Types of spillway : &
1. Main / Service spillway
i) Free fall or straight drop spillway.
ii) Ogee or overflow spillway 1 Marks
for
iii) Siphon spillway
types
iv) Side channel spillway
v) Trough / Chute spillway
vi) Shaft spillway
vii) Conduit or Tunnel spillway
2. Subsidiary or Emergency Spillway
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Q.1 e) Enlist the types of gates provided to reservoir.

Ans A. Non-automatic Gates: i) Radial ot Taintor gate
ii) Vertical gate: 1) Sliding gate 1M
2) Fixed gate each
3) Stony gate
for
iii) Rolling gate
any 4
B. Automatic Gate: i) Reynolds Gate
ii) Vishveshrayya gate
types

Q.1 f) Enlist the factors affecting on silting of reservoir.

Ans a) Catchment area
b) Shape of catchment 𝟏⁄ M
𝟐
c) Slope of country each for
d) Nature of surface soil any 4
e) Climatic conditions
factors
f) Rainfall characteristics
g) Vegetation cover
h) Construction of check dam
Q. 1 g) State ill effects of excess Irrigation

Ans a) Water logging:

𝟏⁄ M each
Water logging renders the soil infertile and useless for cultivation. 𝟐
b) Salt efflorescence due to water logging For
Salt efflorescence due to water logging damages the soil. any 4
c) Breeding of mosquitoes effects
Breeding of mosquitoes takes place due to over irrigation.
d) Damp climate
The climate becomes cold and damp due to over irrigation of an area. This dampness
in climate causes diseases such as malaria , dengue etc
e) Excessive humidity
Excess and intense irrigation may result in increasing the humidity of the area.
f) Ecological imbalance:
g) Water flooding in the crop root zone: Water flooding in the crop root zone causes
wilting of plants.

Q.2 Attempt any THREE of the following 12

a) Describe in brief working of Symons rain gauge with neat labelled sketch.
Ans The most common type of non- recording rain gauges used in our country is the ‘Symon’s
raingauge’.
Working: A Symons Rain gauge consists of a cylindrical vessel of 127 mm internal diameter
2 M for
with an enlarged base of 210 mm diameter. Inside the cylinder, a rain collecting bottle, made
working
up of glass is placed. Over the top of this bottle, a glass funnel is inserted. The top section of
the funnel is provided with a circular brass ring of exactly 127 mm diameter. With each
And
raingauge, a cylindrical graduated measuring glass is also provided. Each graduation of this
glass reads 0.2 mm. The reading should be read correctly upto 0.1 mm. The raingauge is set
upon a concrete block. The rim of the funnel should remain at least 305 mm above the

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ground. The bottle collects the rain falling over the rim which is then measured by the
measuring glass in ‘mm’ and ‘tenth of mm’

2 M for
sketch
with
labelling

b) Compute the average rainfall over the catchment by Thiessen’s polygon method
and arithmetic mean method
Raingauge station A B C D
Area of thiessens 45 38 30 40
polygon (sq.km)
Precipitation in mm 30.8 34.6 32.6 24.6

A) Calculate Average annual rainfall

Ans

Rain gauge Area of Precipitation in A× P

Station Thiessen’s mm (P)
polygon (A)
A 45 30.8 1386
2 M each
B 38 34.6 1314.8 for
C 30 32.6 978 both
D 40 24.6 984 methods
No of Stations ∑A = 153 ∑ P = 122.6 ∑AP = 4662.8
(N) = 4

i) Theissons polygon method-

𝚺𝐀.𝐏 𝟒𝟔𝟔𝟐.𝟖
𝑷𝒂𝒗𝒈 = = = 30.476 mm
𝜮𝑨 𝟏𝟓𝟑

ii) Arithmetic mean method-

𝚺𝐏 𝟏𝟐𝟐.𝟔
𝑷𝒂𝒗𝒈 = = = 30.65 mm
𝑵 𝟒

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c) Explain the factors affecting Duty

Ans 1) Type of Crop: Different crops require different amount of water, and hence
the duties for them are different.

Duty will be less for a crop requiring more water and vice-versa 1 M each
for
2) Climate and Season: In hot season, the atmospheric temperature of the area Any 4
becomes high, the evaporation loss is more hence duty gets reduced as crop factors

requires more water.

3) Rainfall: More the rainfall less will be the requirement of irrigation water,
and hence more will be the duty.
4) Type of Soil: If soil is permeable, water lost due to percolation will be more,
hence duty will be less and vice-versa.
5) Method of Irrigation: If method of irrigation is not proper i.e. in case of free
flooding, wastage of water will be more and hence less duty.
6) Canal condition: Canal should be kept in proper condition by proper
maintenance and lining, which will reduce the seepage, leakage and thus
increase the duty.
7) Mode of Assessment: The tendency of using more water can be checked by
adopting volumetric assessment which gives more duty than area wise
assessment.
8) Method of tilling (cultivation): If proper method of tilling is adopted there
will be less wastage of water, thus duty is more.

d) Explain in brief Engineering surveys carried out for reservoir planning.

Ans In Engineering survey, various type of surveys such as plane table survey, traverse
survey, aerial and photographic surveys etc. are carried out .
2 M for
Purpose or Aim of engineering survey: To prepare Contour map and Topographical purpose
map of the site. or aim
The contour interval may be taken as 1 to 3 m for fairly plain area, and 5 to 10 m
for hilly area.
For Dam site careful triangulation survey is required where contour interval is less
than 1 m.
Horizontal scale may be 1 in 500 to 1 in 1000.
Necessary information from contour plan:
From the contour map of reservoir area, the water spread of the reservoir at any

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elevation may be directly determined by measuring the area with help of

planimeter.
2 M for
Following information can be obtained from contour map. informatio
1. Water spread area from Area-elevation curve n
2. Arrangement of lines of communication. ( Any 2
3. Capacity of reservoir from storage elevation curve. points)
4. Site for waste weir and outlets.
5. Suitable dam site.
From the contour map, the storage capacity of the reservoir should be fixed by
assuming the height of the dam. Then the valuable land and properties which may
get submerged in the reservoir can be found out.

Q. 3 Attempt any THREE of the following 12

a) State the reasons for which elementary profile of dam is not possible to construct in
practice.
Ans
Reasons:
a) For constructing a road at the top of dam, provision of top width is
must but the shape of elementary profile is right angle triangle. 1M
b) Free board is essential to be kept to avoid spillage of water over the each
body of dam but in elementary profile, there is no provision for free for 4
reasons
board.
c) In elementary profile, only pressure due to water and self-weight is
considered but the other forces such as seismic pressure , silt
pressure ice pressure , uplift pressure wind pressure , tail water
pressure are neglected.
d) In elementary profile, the provision of drainage galleries is not
considered.
b) Draw a labelled sketch of Earthen dam

2 M for
sketch

&

2M for
labelling

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c) Differentiate between earthen dam and gravity dam with respect to seepage ,
foundation , construction and maintenance

Ans Parameters Earthen dam Gravity dam

Seepage More seepage Less seepage 1 M for
Foundation Suitable almost on any Suitable only when foundation is each
parameter
foundation of solid rock having no fissures ,
cracks , cavities etc.
Construction Construction materials are Construction materials are stone ,
stone, earth containing silt, brick .Concrete etc
clay and sand.
Maintenance More costly Less costly

d) Explain the requirement of site for percolation tank

Ans 1) The bed of tank should be pervious so that the water will percolate
and then join the ground water table.
1 M each
2) There should be sufficient number of wells and bore wells existing in
for
the command area.
any 4
3) The nalla or stream should have sufficient discharge in monsoon. points
4) The flanks on the both sides of the nalla should rise with steep slope.
5) Construction material , labour, machinery should be available near the
site.
6) A good agricultural land should be available near each well for
irrigating the crops.

Q. 4 Attempt any THREE of the following 12

a) State the necessity and advantages of Drip irrigation scheme.

Necessity: 1 M each
Ans 1) Drip irrigation is application of water, at a very slow rate, to the root zone of for
a plant. any 2
2) This method is very useful in arid regions where water is scarce. necessitie
3) It enables application of fertilizers along with irrigation water. s
4) It ensures optimal growth, better fruiting and growth of crops with optimum
quantity of water
5) It is applicable to a variety of row crops – from widely spaced fruit crops to
closely spaced vegetable crops

Advantages:
(i) It eliminates evaporation losses. ½M
(ii) It requires less water to mature a crop. each for
(iii) Frequent ploughing is not required. any 4
(iv) It is best method of irrigation with ground water in arid zones. advantages
(v) It is very useful in reclaiming and developing desert area.
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b) Discuss the different activities undertaken under Jalyukt Shivar Abhiyan

Ans Jalyukt Shivar Abhiyan was launched by Maharashtra Government in 2015 in a bid to 2M
make Maharashtra draught free by 2019. The main aim of the project is to provide for
relief to farmers of draught affected areas of Maharashtra. It includes checking the descriptio
problems and implementing new water conservation and irrigation techniques. ns

Activities undertaken under Jalyukt Shivar Abhiyan include: 1 M each

1. Deepening and widening of streams. for
2. Construction of earthen and cement stop dams any 2
3. Work on nullahs activities
4. Digging of Farm Ponds

c) Enlist types of weir and explain any one in detail.

Ans Types of weirs are

a) Based on materials of construction
1) Masonry weir
2) Rock-fill weir
3) Concrete weir 2 M each
b) Based on use and function
1) Storage weir for any 2
2) Pick-up weir types of
3) Diversion weir weirs
4) Waste weir
c) Based on design
1) Gravity weir
2) Non gravity weir

1. Masonry Weir
i. It consists of a vertical drop wall or crest wall with or without crest gates and
an impervious floor
ii. Cut off piles are provided at the upstream and downstream ends of the
impervious floor
iii. Aprons are also provided at both upstream and downstream side to prevent
scouring
iv. The raised masonry crest does the most ponding and a small part is done by
the crest shutter
v. These weirs are suitable on any foundation 2 M for
explanatio
2. Rock fill weir
n
i. This weir consists of a body wall and dry stones are laid on the upstream and
downstream side
ii. It requires large quantity of stones
iii. These weirs are unsuitable for soft foundation where difference between weir
crest and downstream river bed does not exceed 3m
iv. An example of this type of weir is Okhala weir on Yamuna river near Delhi

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3. Concrete Weir
i. This is comparatively new type of weir
ii. When water passes over it, hydraulic jump is formed on the downstream
glacis to dissipate energy of flowing water
iii. They are exclusively used nowadays, especially on permeable foundations

4. Pick-up Weir
i. It is a solid weir with crest gates constructed in concrete or stone masonry
ii. It is constructed some distance downstream of dam to form a large reservoir
to raise the water level upto FSL of canal
iii. Pick-up weirs are constructed under following situations:
a) Command area is not near the reservoir
b) Canal has to run idle
c) The terrain is geographically difficult so that canal alignment is very
costly or impossible

d) Differentiate between weir and barrage

Ans
S. No. Weir Barrage
1 Initial cost of weir is low Initial cost of barrage is high.
2 Area of submergence is more due to Area of submergence less due to
large afflux. less afflux
1M each
3 Due to crest there is problem of silting There is good control over silt
for
entry into canal.
4 The raising and lowering of shutter is The raising and lowering of gate is
any 4
not convenient it requires more time convenient as it can be operated
points
and labour as it is done manually. mechanically
5 The control over flood is not possible. There is good control over flood
situation.
6 It is difficult to inspect and repair. These provide better facilities for
inspection and repair of various
structures
7 Roadway is not possible across river Road way can be provided across
the river
8 Storage of water is done by crest and In barrage most of water storage is
very little by or nil portion of water is by done by shutter and very less or nil
gate portion of water is by crest

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e) Draw a layout of diversion headwork

Ans

2 M for
correct
sketch

and

2 M for
correct
labelling

Q.5 Attempt any TWO of the following 12

a) (i) Calculate the MFD from catchment area 950 km2. Use Dicken’s formula
(Assume C=24)

Ans Given:
A = 950 km2
C = 24 1M each
Solution: for
Q = C x A(3/4) correct
= 24 x 950(3/4) formula,
= 4106.80 m3/sec substituti
on and
Answer: MFD = 4106.80 m3/sec answer

a) (ii) Explain any three factors affecting runoff.

1. Rainfall characteristics:
a. More the rainfall, runoff will be more.
b. More the intensity of rainfall more will be the runoff.
1M each
2. Topography:
for any 3
a. It depends upon smoothness and roughness of the surface
factors
b. Steep slopes – Heavy runoff will reach the valley quickly, reducing
losses giving more runoff.
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c. If catchment is mountainous, runoff will be more

d. If catchment is in windward direction, runoff will be more
3. Shape and size of catchment:
a. Catchment area – Larger the area, more runoff.
b. Fan shaped catchment gives greater runoff.
4. Characteristics of catchment :
a. Rocky strata – heavy runoff
b. Compacted strata - heavy runoff
c. Sandy strata – reduced runoff
d. Cultivated area - less runoff
e. Presence of vegetation covers reduces the runoff during smaller storm.
5. Meteorological characteristics :
a. Low temperature – greater runoff
b. High temperature– less runoff
6. Geological characteristics :
a. Pervious soil – less runoff
b. Porous and fissure rock – very low surface runoff

b) Derive relation between duty and delta

Ans Let, D – Duty in hectares / cumec

Δ - Delta for crop in meter
B – Base period of crops in days
If 1 cumec flowing for base period (B) and irrigates field of (D) hectares then total
volume is given by,
Volume = 1 x ( 24 x 60 x 60) B m3
2M
= 8.64 x 104 x B cubic m

Volume calculated from delta Δ cm of depth required in base period

Volume = Δ x 104 and for D hectares
Volume = 104 x Δ x D cubic m
Equating above equations, 2M
1 x ( 24 x 60 x 60) B = 104 x Δ x D

Therefore,
𝟖.𝟔𝟒 𝑩
∴D =
∆ 2M

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c) Fix the full reservoir level (FRL) and Top of Dam level (TDL) from the following data:
i) Dead storage level = 112 m
ii) Crop water requirement = 8500 m3
iii) Tank losses = 1500 m3
iv) Free board = 1.5 m
v) Flood lift = 1.6 m

Control R. L. (m) 112 114 116 118 120 122

Capacity (m3) 1000 2900 5000 6000 9000 12000

Ans Effective live storage = 8500 m3

Tank losses = 1500 m3

Total live storage = Effective live storage + Tank losses

= 8500 +1500
= 10000 m3
1M
Dead storage = 1000 m3 (corresponding to RL 112)

Gross storage = 10000 m3 + 1000 m3

= 11000 m3 1M

(122 – 120) (11000 – 9000)

FRL = 120 +
(12000 – 9000)
= 120 + 1.33
1M
FRL= 121.33 m

HFL = FRL + flood lift

= 121.33 + 1.6

HFL = 122.93 m 2M

TDL = HFL + free board

= 122.93 + 1.5
1M
TDL = 124.43

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Q.6 Attempt any TWO of the following 12

a) Draw layout of lift irrigation scheme and explain in brief component parts of the
sam

2 M for
sketch
with
labeling

Components Parts

1) Intake well - A channel is constructed to divert the flow of water into inlet
chamber
2) Inlet chamber - It avoids the entry of silt and debris into Jack well
4 M for
3) Jack well - It is provided to facilitate the location of an engine house above any 4
high flood level and allows pumping during floods componen
4) Inlet pipe - It is used to convey water from inlet chamber to Jack well. It is ts
provided with a proper gradient
5) Engine house - It is a small storage room to accommodate the engines and the
pumps to be installed
6) Rising main - It is a pipe used for carrying water from well to delivery chamber
7) Delivery chamber - The water from rising main is collected here and then it is
allowed to flow in field ditches
8) Water distribution system - A system which distributes water from delivery
chamber to field channels

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Design economical trapezoidal section of canal to carry 30m3/sec of water with bed
b)
slope 1 m. per km and side slope 3H:2V. Take N=0.012.

Given:
Ans
Q = 30 m3/sec S = 1/1000 n = H/V = 3/2 = 1.5 N = 0.012

For most economical channel,

1. R = d/2
1M
2. b + 2nd = 2d√(1 + 𝑛2 )
Substituting the value of n in the second condition
b + 2 x 1.5 d = 2d √(1 + 1.52 )
b + 3d = 3.605d
b = 0.605d

Area = bd + nd2
= 0.605 d2 + 1.5d2 1M
Area = 2.105d2

Velocity (v) = 1/N x R2/3 x S1/2 1M

= (1/0.012) x (d/2)2/3 x (1/1000)1/2
Velocity (v) = 1.66d2/3

Substituting the values of Q, A and v in the equation for discharge 1M

Q=Axv
30 = 2.105d2 x 1.66d2/3
30 = 3.49 d8/3
30 / 3.49 = d8/3
8.596 = d8/3
Hence
d = 2.24 m
b = 0.605 d 1M
= 0.605 x 2.24
b = 1.35 m
1M

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c) Draw a neat labeled sketch of the following:

i) Aqueduct ii) Super passage iii)Level crossing

Ans i) Aqueduct

2M

ii) Super passage

2M

Iii ) Level Crossing

2M

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