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JNTU B.Tech III Year WATER RESOURCES ENGINEERING - I Previous Question Paper (August - 2022 - R13)

B.Tech | III Year | I Semester | R13 | August - 2022

Answer PaperPaper Code: 115ACMaximum Marks: 75Time: 3 hours

Time: 3 hours

Max. Marks: 75

Answer any five questions

All questions carry equal marks

Main

115 marks
a

What do you understand by precipitation? Explain various types of precipitation.

7 marks

Answer

Precipitation is any form of water that falls from the atmosphere to the Earth's surface. It is a crucial component of the hydrological cycle, providing the primary source of fresh water for various ecosystems and human activities.

The main types of precipitation are:

  • Rain: Liquid water droplets, larger than 0.5 mm in diameter.
  • Drizzle: Very light rain with small droplets (less than 0.5 mm in diameter) that appear to float.
  • Snow: Ice crystals that form in clouds and fall to the ground without melting.
  • Sleet: Raindrops that freeze into ice pellets before reaching the ground.
  • Hail: Lumps of ice that form in thunderstorms and fall to the ground.
  • Freezing Rain: Rain that falls as a liquid but freezes upon contact with surfaces that are at or below freezing point.
b

A catchment has five rain gauge stations. In a year, the annual rainfall recorded by the gauges is 78.8 cm, 90.2 cm, 98.6 cm, 102.4 cm and 70.4 cm. For a 6 % error in the estimation of the mean rainfall, determine the additional number of gauges needed.

8 marks

Answer

To determine the additional number of gauges needed, we will use the formula for the optimum number of rain gauges and the concept of allowable percentage error.

The formula for the optimum number of rain gauges (N) is given by:

N = (C_v / E)^2

Where:

  • N = Optimum number of rain gauges required
  • C_v = Coefficient of variation of rainfall (standard deviation / mean rainfall)
  • E = Allowable percentage error in the estimation of mean rainfall

First, we need to calculate the mean rainfall and the standard deviation from the given data.

Given rainfall data (in cm): 78.8, 90.2, 98.6, 102.4, 70.4

Number of existing gauges (n) = 5

Mean rainfall (R_bar) = (78.8 + 90.2 + 98.6 + 102.4 + 70.4) / 5 = 440.4 / 5 = 88.08 cm

Now, we calculate the standard deviation (s):

Variance (s^2) = [Σ(x_i - R_bar)^2] / (n-1)

x_i - R_bar: -9.28, 2.12, 10.52, 14.32, -17.68

(x_i - R_bar)^2: 86.1184, 4.4944, 110.6704, 205.0624, 312.5824

Σ(x_i - R_bar)^2 = 86.1184 + 4.4944 + 110.6704 + 205.0624 + 312.5824 = 718.928

Variance (s^2) = 718.928 / (5-1) = 718.928 / 4 = 179.732

Standard deviation (s) = sqrt(179.732) ≈ 13.406 cm

Coefficient of variation (C_v) = s / R_bar = 13.406 / 88.08 ≈ 0.1522

Allowable percentage error (E) = 6 % = 0.06

Now, calculate the optimum number of gauges (N):

N = (C_v / E)^2 = (0.1522 / 0.06)^2 = (2.5367)^2 ≈ 6.435

Since the number of gauges must be an integer, we round up to the nearest whole number. So, the optimum number of gauges required is 7.

The number of additional gauges needed = Optimum number of gauges (N) - Existing number of gauges (n)

Additional gauges needed = 7 - 5 = 2

Therefore, 2 additional rain gauges are needed.

215 marks
a

What are the methods of computing run-off from a catchment area? Give various formulae stating clearly the area for which each is applicable.

7 marks

Answer

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b

What are the methods for computation of average rainfall over a basin? Explain.

8 marks

Answer

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315 marks
a

Briefly explain the components of a single peaked hydrograph.

7 marks

Answer

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b

A peak of a flood hydrograph due to a six-hour storm is 470 m3/sec. The average depth of rainfall is 8.0 cm. Assume an infiltration loss of 0.25 cm/ hour and a constant base flow of 15 m3/sec. Estimate the peak discharge of 6 hour unit hydrograph for this catchment.

8 marks

Answer

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415 marks
a

Explain the method of determining direct run-off from a given storm hydrograph.

7 marks

Answer

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b

What do you mean by synthetic unit hydrograph, write a note on it?

8 marks

Answer

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515 marks
a

Write a note on ‘storage coefficient’ and ‘coefficient of transmissibility’.

7 marks

Answer

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b

Distinguish clearly between shallow well and a deep well. How does a deep well in confined aquifer?

8 marks

Answer

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615 marks
a

Explain the percussion method of drilling a tube well.

7 marks

Answer

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b

The discharge available from a tube well is 150 m3/hour. Assuming 4000 hours of working for a tube well in a year, estimate the culturable area that this tube well can command. The intensity of irrigation is 45% and average depth of Rabi and Kharif crop is 38 centimeters.

8 marks

Answer

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715 marks
a

Write a detail note on method of furrow irrigation.

6 marks

Answer

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b

A water course has culturable commanded area of 2600 hectares, out of which the intensities of irrigation for perennial sugar-cane and rice crops are 20% and 40% respectively. The duty for these crops at the head of water course is 750 hectares/cumec and 1800 hectares/cumec respectively. Find the discharge required at the head of water course if the peak demand is 120% of the average requirement.

9 marks

Answer

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815 marks
a

Explain Lacey’s silt theory.

6 marks

Answer

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b

Explain the procedure of designing a channel with Kennedy’s theory.

9 marks

Answer

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This page contains the question paper of WATER RESOURCES ENGINEERING - I from JNTU for B.Tech CE III Year I Semester, August - 2022, R13.

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