The words you are searching are inside this book. To get more targeted content, please make full-text search by clicking here.

## Water Resource Engineering_Test - 03 Solutions

Discover the best professional documents and content resources in AnyFlip Document Base.

# Water Resource Engineering_Test - 03 Solutions

### Water Resource Engineering_Test - 03 Solutions

GATE
CIVIL

WaterResource
Engineering

Test-03Solutions

WATER RESOURCE ENGINEERING
1. Match List-I (Technique/Principle) with List-II (Purpose) and

select the correct answer using the code given below the lists:
List-l
A. ∅-Index
B. Slope - area method
C. Flow duration curve
D. Dilution technique
List-II
1. Dependable flow
2. Reservoir regulation
4. Run-off volume
Codes:

ABCD
a) 3 5 1 4
b) 4 1 2 3
c) 3 1 2 4
d) 4 5 1 3

1

2. A unit hydrograph for a watershed is triangular in shape with base
period of 20 hours. The area of the watershed is 500 ha. What is
the peak discharge in m3/hour?
a) 7000
b) 6000
c) 5000
d) 4000
Solution:
The hydrograph is shown below:

From this Unit Hydrograph

QPTB = A × �1100�
2

QP×20 = 500 × 104 × 1
2 100

QP = 5000 m3/s

3. An urban area is located in plains having "average climatic

conditions". The impervious area thereof for which drainage must

be provided is 3.6 ha and the design rainfall intensity is 2.0 cm/hr.

The drains will be designed for a runoff of
a) 0.05 m3/s
b) 0.10 m3/s

2

c) 0.20 m3/s

d) 0.40 m3/s

Solution:

The runoff, Q is given by

Q = 316k.p.A

K = 1 (impervious)

∴ Q = 1 × 1 × 2 × 3.6 = 0.2 m3/sec
36

4. During a 3-hour storm event, it was observed that all abstractions

other than infiltration are negligible. The rainfall was idealized as

3 one-hour storms of intensity 10 mm/hr, 20 mm/hr and 10 mm/hr

respectively and the infiltration was idealized as a Horton curve,

f = 6.8 + 8.7 e (-1) (f in mm/hr and t in hr). What is the effective

rainfall?

a) 10.00 mm

b) 11.33 mm

c) 12.43 mm

d) 13.63 mm

Solution:

3

Horton’s equation, f = 6.8 + 8.7 e–t
At t = 0 hr
f = 6.8 + 8.7 e–0 = 15.5 mm/hr
at t = 1 hr,
f = 6.8 + 8.7 e–1 = 10 mm/hr
t = 2 hr f = 6.8 + 8.7 e–2 = 7.98 mm/hr
t = 3 hr, f = 6.8 + 8.7 e–3 = 7.23 mm/hr
The Horton’s curve is super imposed on the hyetograph.
The hatched portion indicates runoff.
Area below the Horton’s curve from t
= 1 hr to 3 hr is
= ∫13(6.8 + 8.7e−t)dt
= �6.8t + 8.7−e1−t�13
= [6.8 × 3 − 8.7 −3] – [6.8 × 1 − 8.7 −1]
= 19.97 – 3.60
= 16.37 mm (equal to infiltration loss from t = 1 hr to 3 hr)
Total rainfall depth = 20 × 1 + 10 × 1 (in 2nd hr and 3rd hr)
= 30 mm
Rainfall excess = 30 – 16.37 = 13.63 mm
5. With 'n' variables and 'm' fundamental dimensions in a system,
which one of the following statements relating to the application
of the Buckingham's pitheorem is incorrect?
a) With experience, pi terms can be written simply by inspection
of variables in a flow system.

4

b) Buckingham's pi theorem is not directly applicable in
compressible flow problem.

c) Buckingham's pi theorem yields dimensionless pi terms given
by the difference between the number of variables and the
number of fundamental dimensions.

d) Buckingham's pitheorem reduces the number of variables by
the number of fundamental dimensions involved.

Solution:
According to Buckingam Pi theorem
- Number of independent variables = n
- Number of fundamental dimensions = m
- Repeating variables to be selected = m which contain all

fundamental dimensions
- Dimensionless number obtained = n – m
- Reduction of variables = n – (n – m) = m
Thus method can be applied to any type of flow compressible,
incompressible, inviscid or viscous flow
6. The best estimate of runoff represented by 57 mm of runoff depth
from a basin area 3300 km2 is nearly
a) 2300 cumec-days
b) 2225 cumec-days
c) 2175 cumec-days
d) 2020 cumec-days

5

Solution:

Best estimated runoff

= Total volume generated
duration

= �57 × 10−3� × 3300 × 106 = 2177.083 ≃ 2175 cumec – days
24 × 60 × 60

7. S-Curve Hydrograph is the hydrograph

a) producing 1 cm of runoff over the basin

b) of flow from a 1 cm intensity rain of infinite duration

c) having a volume of 1 cm3

d) of the total storm duration in any single storm rainfall

Solution:

S-curve hydrograph is the hydrograph of flow from a 1 cm

intensity rain of infinite duration.

8. The probability that a 100 years flood is equaled or exceeded, at

least once in 100 years is

a) 99%

b) 64%

c) 36%

d) 1%

Solution:

The probability corresponding to 100 years return period

P = 1 = 0.01
100

6

The probability of the flood exceeding at least once in 100 years

is

R = 1 – (1 – P)100 = 1 – 0.99100

= 0.634 = 64%

9. The procedure to follow in solving for discharge when hf (head

loss), L (pipe length), D (inside diameter), u (kinematic viscosity)

and k (wall roughness) are given, is to

a) Assume an f (friction factor) compute V, Re (Reynolds

number,) DK, look up for f and repeat, if necessary

b) Assume an Re, compute f, check K etc.
D

c) Assume a V, compute Re, look up f compute V again etc.

d) Assume a, Q compute V, Re,look up f

Solution:

For turbulent flow

hf = f∙ L ∙ V2 (Weisbach equation)
D 2g

Discharge Q = A.V
= πD4 2.V
As inner dia D is given, velocity V is required to calculate

discharge Q.

Given:

L = length of pipe

7

D = inner dia
= kinematic viscosity

k = wall roughness

In the Darcy Weisbach equation, written above friction factor ‘f’

and velocity V are unknowns.

To work out velocity V, assume value of ‘f’ initially.

As a trial 1 calculate V.

Calculate Reynold’s number

Re = ρVD = VD
μ ϑ

Calculate relative roughness = k
D

Moody diagram for commercial pipes is shown below.

Now, based on the value of Reynold’s number Re and relative

roughness k calculated, choose the corresponding friction factor
D

f. if this value of ‘f’, is a not tallying with the initial value of ‘f’

assumed, go for second trail and so on, till the values of ‘f’ will

merge.

Calculate

Q = A.V

8

Where V = velocity of flow worked out in the final trial where the

value of ‘f’ assumed is almost correct.

10. A sprinkler irrigation system is suitable when

a) the land gradient is steep and the soil is easily erodable

b) the soil is having low permeability

c) the water table is low

d) the crops to be grown have deep roots

11. From the analysis of rainfall data at a particular station, it was

found that a rainfall of 400 mm had a return period of 20 years.

What is the probability of rainfall equal to or greater than 400 mm

occurring at least once in 10 successive years?

a) (0.95)10

b) 1-(0.95)10

c) 1-(0.5)10

d) (0.05)10

Solution:

The probability of occurrence of an event at least once over a

period of ‘n’ successive years is called the risk, R. it is given as

R = 1 – (1 – P)n

Where P = 1 = 1 = 0.05
T 20

∴ R = 1 – (1 – 0.05)10

= 1 – (0.95)10

9

12. Rainfall mass curve shows the variations of

a) rainfall intensity with time

b) rainfall intensity with cumulative rainfall

c) rainfall excess with time

d) cumulative rainfall with time

13. The surface tension of water at 20oC is 75 × 103 N/m. The

difference in water surfaces within and outside an open-ended

capillary tube of 1 mm internal bore, inserted at the water surface,

would nearly be

a) 7 mm

b) 11 mm

c) 15 mm

d) 19 mm

Solution:

In this problem, the bore is not defined properly. Here bore is used

h = 2σ cos θ
ρ∙g∙R

= 103 2�75 × 10−3�
× 9.81 × (1 × 10−3)

= 15.29 mm

10

14. Match List-I (Flow problem under study) with List-II (Model
law) and select the correct answer using the codes given below
the lists:
List – I
1. Rise of gas bubbles in liquid
2. Flow of gas in a pipe
3. Flow over a spillway dam
4. Flight of supersonic jet
List – II
A. Euler number
B. Froude number
C. Mach number
D. Reynold's number
E. Weber's number
Codes:
1234
a) C E D A
b) D C B A
c) E B D C
d) E D B C

11

15. The discharge per meter width at the foot of a spillway is 10

m3/s at a velocity of 20 m/s. A perfect free hydraulic jump will

occur at the foot of the spillway when the tail water depth is

approximately equal to:

a) 4.50 m

b) 5.00 m

c) 5.50 m

d) 6.50 m

Solution:

Upstream conditions

q = 10 m3/s/m

V1 = 20 m/s

y1 = q = 0.5 m/s
V1

F1 = V1 = 20 0.5 = 8.9 > 8
�gy1 √10 ×

∴ y2 ≃ 1.41 F1y1 = 6.3 m

16. Assertion (A): In the border strip method of irrigation, the size

of the strip depends on soil characteristics, slope of the land and

discharge.

Reason (R): Border strip method is a controlled type of

subsurface irrigation method.

a) Both A and B are true and B is the correct explanation of A

b) Both A and B are true but B is not a correct explanation of A

c) A is true but R is false

12

d) A is false but R is true

17. A Field measures 40 hectares. When 8 cumecs of water was

supplied for 6 hours, 30 cm of water was stored in the root zone.

The field application efficiency is nearly.

a) 70%

b) 80%

c) 85%

d) 90%

Solution:

Total water applied

= 8 × 3600 × 6

= 172800 m3

Height of water applied

= Volume of water = 172800 = 0.432 m
Area 40 × 104

= 43.2 cm

∴ Field application efficiency

= 30 × 100 = 70%
43.2

18. Consider the following statements:

The general depth of scour calculated by Lacey's formula in a

river represents the depth below the

1. Maximum flood level in the river

2. Minimum flow water level in the river

13

3. Normal flow water level in the river

4. Existing river bed level

Which of these statements is/are correct?

a) 1, 2, 3 and 4

b) 1, 2 and 3 only

c) 2, 3 and 4 only

d) 1 only

19. A jet of water issue from a 5 cm diameter nozzle, held vertically

upwards, at a velocity of 20 m/sec. If air resistance consumes 10%

of the initial energy of the jet, then it would reach a height, above

the nozzle, of:

a) 18.35 m

b) 19.14 m

c) 19.92 m

d) 20.00 m

Solution:

Using Bernoulli’s equation

V2 = H + 0.1V2g2 [0.1V2g2 is the head loss]
2g

H = 0.9V2g2 = 0.9 × 20 × 20
2 × 9.81

= 18.32 m

14

20. A hollow cylinder made of wood (sp. gr. = 0.8) has an external
diameter of 1.0 m and an internal diameter of 0.6 m. It floats in
water with its axis vertical and is in stable equilibrium This is
possible only when the length of the cylinder is equal to or less
than:
a) 0.72 m
b) 0.95 m
c) 1.03 m
d) 1.20 m
Solution:
For floating of cylinder, the immersed length.
Weight of cylinder = weight of the liquid of the same volume as
that of cylinder.

Let cross-sectional are of cylinder be A.
lAρc = hAρωg

h = 0.8 l

where h is the submerged length.

OB = h = 0.4 l
2

OG = = 0.5 l
2

15

BG = 0.1 l

BM =
V

l → Moment of inertia of cylinder

V → Volume of submerged portion.

I = π�D4e− Di4� and
64

V = h∙ π�D2e− Di2�
4

BM = �D2e− Di2� = �1126+×00.6.82� 1
16h

= 0.10625

GM = BM – BG

= 0.10625

= 0.1l

For stable equilibrium GM > 0

l < 1.03 m

Directions: The following items consists of two statements; one

labelled as 'Assertion (A)' and the other as 'Reason (R)'. You are

to examine these two statements carefully and select the answers

to these items using the codes given below:

Codes:

a) Both A and R are true and R is the correct explanation of A

b) Both A and R are true but R is not a correct explanation of A

c) A is true but R is false

d) A is false but R is true

16

21. Assertion (A): An alluvial channel is defined as a channel in
which the flow transports sediment of the same physical
characteristics as the material in the wetted surface of the channel.
Reason (R): This ensures that the channel cross section and the
channel slope do not change.
Solution:
The Cross-section and Slope of an alluvial channel changes due
to silting and erosion at the wetted surface of the channel.

22. A check dam is a
a) flood control structure
b) soil conservation structure
c) river training structure
d) water storage structure
Solution:
Check dam is used for flood control.

23. In a ski-jump bucket provided in an overflow spillway, the lip
angle is 30o, and the actual velocity of flow entering the bucket is
30 m/s. The maximum vertical height attained by the trajectory
of the jet, measured above the lip of the bucket, is nearly
a) 45 m
b) 35 m
c) 22 m
d) 11 m

17

24. A hydro-electric reservoir can supply water continuously at a

rate of 100 m3/s. The head is 75 m. The theoretical power that can
be developed is:
a) 1000 mhp
b) 1,00,000 mhp
c) 7500 mhp
d) 7500 kW
Solution:

Theoretical power developed = ρgHQ
= 10 × 75 × 100
= 75,000 kW
1 m ph = 746w = 0.746 kW ≃ 0.75 kW
∴ 75000 kW = 1,00,000 mhp
Here m is used for metric so m hp means metric horse power
25. Consider the following statements:
1. Flow is established in a pipe when the boundary layer
thickness is equal to the radius of the pipe.
2. For laminar flow, the friction factor in Darcy Weisbach
equation varies inversely as the Reynold number.
3. For turbulent flow, the friction factor in Darcy Weisbach
equation varies inversely as the square of Reynold number.
4. When the boundary layer is rough, friction factor varies with
the relative roughness of the pipe.

18

Of the statements:

a) 1, 2 and 3 are correct

b) 1, 2 and 4 are correct

c) 2, 3 and 4 are correct

d) l, 3 and 4 are correct

Solution:

In turbulent flow, friction factor f

f= 0.316 104 < Re < 105
For smooth turbulent flow� 1 Re1/4 − 0.80 for all Re
=
√f 2 log�Re√f�

For rough turbulent flow

1 = log �∈ � + 1.74
√f

∈ → roughness of pipe

26. Consider the following statements:

An aqueduct is a cross drainage work in which

1. a canal is carried over the drainage channel.

2. a drainage channel is carried over the canal.

3. both drainage channel and canal are at the same level

Which of these statements is/are correct?

a) 1 only

b) 1 and 2 only

c) 2 and 3 only

d) 1, 2 and 3

19

Solution:
An aqueduct and syphon aqueduct are cross drainage works in
which the canal is taken over the natural drain, such that the
drainage water runs below the canal either freely or under
syphoning pressure. Super passage and syphon are cross drainage
works in which the drain is taken over the canal such that the canal
water runs below the drain either freely or under syphoning
pressure. Level crossing is a cross drainage work in which the canal
water and drain water are allowed to intermingle with each other.
A level crossing is generally provided when a large canal and a
huge drainage (such as a stream or a river) approach each other
practically at the same level.
27. The water utilizable by plants is available in the form of

a) gravity water
b) hygroscopic water
c) capillary water
d) chemical water
Solution:

Gravity water ----- Superfluous wat
Hygroscopic water ---- Available water
Capillary water ------ Water stored due to capillary action

20

28. The yield of a well depends upon
a) permeability of soil
b) area of aquifer opening into the wells
c) actual flow velocity
d) all of the above
Solution:
Yield of a well means the discharge from the all excavated
through given aquifer
Q = nvaA = KiA
Where
n is porosity
va is actual flow velocity of ground water
A is area of the aquifer opening into the well
K is permeability of soil

29. A tropical cyclone in a northern hemisphere is a wind stream
with
a) High pressure zone of anticlockwise rotation.
b) High pressure zone of clockwise rotation
c) Low pressure zone of anticlockwise rotation
d) Low pressure zone of clockwise rotation

21

30. A rectangular channel of 2.5 m width and 2 m depth of water

carries a flow of 10 m3/s. The specific energy for the flow is given

by:

a) 1.18 m

b) 3.4 m

c) 2.0 m

d) 2.2 m

Solution:

E = y + v2
2g

V = Q = 10 2 = 2 m/s
by 2.5 ×

E = 2 + 2×2 = 2.2 m
2 × 10

22

FRCA Toluna News Août
Ajánló 2019. aug-okt