Hydraulics, Irrigation, Water Resource Engineering and Hydrology

Solved Example - Height/Elevation of Pipe for Cavitation to occur

Problem:

(a) Compute the discharge rate (m^3/s) of the water from the bowl if h = 30 cm, dia = 5 cm, H1 = 2 m, H2 = 7 m, p(atm) = 101. 3 kN/m^2, pv = 2.5 kN/m^2

(b) Compute the pressures inside the hose at points 3, 4, and 5

ANS: Q=0.023 cms; p3 = -68.64 kPa, p4 = -88.25 kPa; p5 = p3

Solution:

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Solved Example - Cavitation and Discharge in hose/pipe

Problem:
Water is pumped from the river through a 45-mm-diameter hose having a length of 3 m . The friction factor is f = 0.028 for the hose, and the gage vapor pressure for water is -98.7 kPa. Suppose that h = 2 m . (Figure 1)

Determine the maximum volumetric discharge from the hose at C so that cavitation will not occur within the hose.

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Hydrostatic forces and tension in cable holding the Gate - SOLVED Example

Example:
The retaining wall in Figure 1 holds back a combination of water and mud. Considering it to have a dimension of 1 m in the perpendicular direction (i.e. out of the page) determine what the force in cable BE and the reaction forces at D will be.

Solution:

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Venturimeter vs Orificemeter, loss of energy, cost and uncalibrated accuracy.

Hi,

So what are the differences between Venturimeter and Orificemeter, in terms of loss of energy, cost and their uncalibrated/ un-calibrated accuracies. Let's find out.

Instrument	Loss of Energy	Initial Cost	Accuracy variation with calibration (uncalibrated accuracy)
Venturi	Venturi Meter minimizes the form friction and therefore less energy loss. About 90% is recovered.	Initial cost is higher.	They are more accurate than the orifice meter, because they are generally less affected by temperature and corrosion etc. Their discharge coefficient varies with Reynold's number and variations may be 0.5- 1.5%
Orifice	Due to the form friction, a large amount of energy is lost and unrecovered. Only 60-70% is recovered.	They are in-expensive, because it is just a thin plate with orifice.	There are many factors including the sharpness of edges, corrosion, temperature pressure etc. which affect the performance of orifice. If all combined the inaccuracy may range up to 10%. So they are less accurate when uncalibrated.

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Laminar flow between two parallel plates (Navier-Stokes Equation)

Hi,

When the fully laminar fluid flow is developed between two parallel plates along the x-direction, the Navier-Stokes equation is applied and can be written and solved for the fluid velocity u as shown in the images below.

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Energy line and Hydraulic Grade Line

Hi,

Energy Line

Energy line represents the the sum of the Pressure Head, velocity and elevation Head corresponding to an assumed datum.
If there are no losses in a system, Energy head remains constant.

Hydraulic Grade Line

Hydraulic Grade line represents the Energy head subtracted by the velocity head. In other words it represents the sum of the pressure head and the elevation head along a system of fluid flow.

Example: If an Venturimeter is attached to a pipe connected to a vessel at the bottom, the Hydraulic grade changes significantly. See the image below, calculations are also given.

Note: EL = Energy Line = h ; and HGL = hydraulic Grade Line becomes negative at the throat of the venturimeter.

Water Resources and System Engineering (CE- 6003)- B.Tech. 6th sem

Hi,
Please find herewith the Question Paper of Water Resources and System Engineering(CE-6003), set by HPU/HPTU (Himachal Pradesh University/ Himachal Pradesh Technical University) for the year 2014.

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B.Tech. 6th Semester Examination

Water Resources and System Engineering

CE-6003

Time: 3 Hours Max Marks: 100

The candidate shall limit their answers precisely within the answer-book(40 pages) issued to them and no supplementary continuation sheet will be issued.

Note: Attempt five questions selecting one question from each sections A,B,C and D. Question 9 is compulsory, All questions carry equal marks. Non- programmable calculator is allowed.

Section - A

1. (a) Describe the functional requirements of various users in a multipurpose water-resources project. What is the compatibility of these users in the project?

(b) Discuss inter-basin transfer of water in the context of our country.

(15+5 = 20)

2. (a) The annual runoff data over the catchment area of a reservoir for a successive number of years are given below:
Year: 1 2 3 4 5 6 7 8
Runoff (cm) 98 143.5 168.3 94 95.3 152.4 110 131.3
Determine (i) the average yield from the catchment and (ii) storage capacity of the reservoir to use the source fully. Solve analytically. Given, Catchment area = 1675 km^2.

(b) Draw a diagram showing the various zones of storage in a reservoir.
(15+5 = 20)

Section - B

3. The data pertaining to a flood protection project to provide full safety against floods up to 50 years frequency are as follows:
Cost of project = Rs. 50 lacs.
Life of the project = 50 years.
Interest rate = 6.5%
Maintenance cost = 2% of the capital cost.
The following additional information is available
Flood frequency (years) 0 5 10 15 30 40 70
Annual damages Rs.(*10^4) 2 25 40 45 61 71 75
Find (i) Annual cost (ii) Annual benefits and (iii) Benefit-cost ratio of the project.
(20)

4. A 1000 mm diameter pipeline can be installed for Rs. 2 lacs. The annual operating and maintenance cost is estimated at Rs. 40000/-. An alternative 750 mm diameter pipeline can be installed at Rs. 1.6 lacs. Its annual operating and maintenance cost is estimated at Rs. 60000/-. Either pipeline is expected to serve for 35 years, with 7% salvage when replaced. Compare the two pipelines assuming a 15% rate of interest. (20)

Section - C

5. (a) Explain, (i) General structure of a linear programming problem
(ii) Feasible space and (iii) Initial basic feasible solution.

(b) Discuss risk and uncertainty in project evaluation. (15+5 = 20)

6. Solve the following linear programming problem graphically:
Maximize z= (3.x1 + 5.x2) subject to
(x1 + 2.x2) <= 2000
(x1 + x2) <= 1500
x2<= 600 and x1, x2 >=0 (20)

Section - D

7. Four water resources projects are to be allocated from limited funds in a small district. These projects produce net independent returns as shown below. using dynamic programming, determine the optimal allocation of 1 million rupees
Investment Rs.(*10^6) NET RETURNS FROM A PROJECT Rs. (* 10^4)
Project 1 Project 2 Project 3 Project 4
2 4 2 6 6
4 0 3 12 1
6 6 4 12 6
8 9 5 12 15
10 10 6 12 12
(20)

8. (a) Discuss the application of system engineering in water resources projects.

(b) Explain the use of mathematical models in forecasting hydrological events. (10+10 = 20)

Section - E (Compulsory)

9. (i) Discuss the role of water in the development of water resources.

(ii) What is watershed management and what are its elements?

(iii) Explain reservoir sedimentation.

(iv) Enumerate the various steps involved in the planning of a water resources project.

(v) Differentiate between micro and macro economics.

(vi) Explain the term capital recovery factor.

(vii) Discuss system engineering.

(viii) What is dynamic programming? How it differs from linear programming?

(ix) Describe principle of optimality.

(x) What is simulation and what are its limitations? (10*2 = 20)

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