Friday, January 13, 2017

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.





thanks!

Saturday, October 8, 2016

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.



Monday, July 28, 2014

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)
                                                                   OR
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)

                                                                 OR
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)

                                                                   OR
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)
                                                                     OR
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)

----------------------------------------------------------------------------------------------------------------------

Thanks for your kind visit!

Sunday, June 15, 2014

Ten major Dam reservoirs of India- safety storage of India

Hi, 

 This article mentions the names of the dams in the water resources map of India.

These are the major dams which form the reservoirs that are the major water resources of India.


  1. Bhakra Dam - in Himachal Pradesh

It forms the third largest reservoir named as Gobind Sagar, with a storage capacity of 9.34 km^3. With an height of 226m, Bhakra dam is the highest vertical concrete gravity dam in Asia. 

clicked on the way to Naina Devi of Bilaspur. (source: wikipedia)






It is constructed just on the border of Himachal and Punjab, on the Sutlej river in the Bilaspur district of Himachal Pradesh.

 2. Tehri Dam  in Uttarakhand

Tehri Reservoir on 1st March 2008.



civil engineering students from NIT Hamirpur, on an industrial visit to Tehri dam on 1st march 2008 .. with CP Sir.

 With a height of 261m i
ts the highest dam in India. It's an rock and earth fill dam, constructed on the Bhagirathi river near Tehri in Uttarakhand. Reservoir has a capacity of 4.0 Km^3.

3. Sardar Sarovar (Gujrat)

With an installed reservoir capacity of 9.5 Km^3, it is one of the largest reservoir.

Sardar Sarovar dam, undergoing height extension in 2006 (source wikipedia)

Length of the dam is 1.21 Km, while the height is 128 m and growing. 

4. Rihand Dam (Uttar Pradesh)

Witha a length of 934 m and a height of 91 m, this dam create one of the biggest reservoirs with total impounding capacity of 10.6 km^3.
Jawahar Lal Nehru at Rihand Dam (source; wiki)


But due to the siltation problem, mostly the alkaline ash from the nearby many thermal power plants, it has only 8.9 km^3 as its active capacity. The dam is constructed on the Rihand river in the Sonbhadra district of Uttar Pradesh. 

5. Indira Sagar Dam (madhya Pradesh)

Constructed on river Narmada, and with a capacity of 12.22 Km^3, it creates the largest reservoir in India.

6. Hirakud Dam (Orissa)

It is the longest dam in the world with a composite length of 25.8 km. Main dam is a concrete dam of 4.8 km while the dykes of 21 km makes the total length. It has the largest artificial lake in the India.
It was constructed to control the wild floods on the river Mahanadi.

7. Nagarjuna Sagar (Aandhra Pradesh)

8. Mullaperiyar Dam( Kerala)

9. Tunga-Bhadra Dam (Karnataka)

10. Sholayar Dam (Tamilnadu)




I have the dream of visiting the other dams too. How many of them have you visited? 

Please leave a comment.

thank you :)


Sunday, January 26, 2014

GATE 2014, PSUs- irrigation Engineering - one liners - part 11

Hello there,

Here is the 11th part of our notes for the preparation of the GATE and PSUs exams.

  • Fertility of a soil is adversely affected, when the pH value is more than 11.
  • Optimum depth of kor watering for rice crop is nearly 19 cm.
  • Average delta of rice crop is nearly 120 cm.
  • The duty of a crop is 432 hectares/cumec, when base period of the crop is 100 days. Delta for the crop will be 200.
  • Water consumed in irrigation, when compared with the total water used for all purposes in our country, is about 90%.
  • Water consumed for producing one tonne of wheat and one tonne of rice will be of order 2000 tonnes and 4000 tonnes.
  • Lime concrete lining is used when velocity of flow is below 2 m/sec, irrigation channel with capacity upto 200 cumec and where economy is required.
  • Thickness of concrete lining, for discharge upto 200 cumec varies from 10 to 15 cm.
  • Force considered for the analysis of an elementary profile of a gravity dam under empty reservoir condition is self weight.
  • Uplift pressure on a dam can be controlled by pressure grouting in foundation, constructing drainage channels between dam and its foundation and by constructing cut-off under upstream face.
  • In a gravity dam total force due to wave pressure acts at a height of 0.375.hw above the still water level.
  • Horizontal acceleration due to earthquake results in hydro-dynamic pressure and inertia force in the body of the dam.
  • Vertical acceleration due to earthquake results in increase in the effective weight of the dam and also decrease in the effective weight of the dam.
  • In the elementary profile of a dam having empty reservoir condition, vertical stress at heels and toe are respectively given by 2W/B and 0.
  • In gravity dam, main overturning force is water pressure.
  • For economical design of a gravity dam, shear friction factor should be 0.65.
  • In Ogee shaped spillway, discharge is proportional to H^(3/2).
  • Garrets diagrams are based on Kennedy's theory.
  • Discharge co-efficient of an Ogee-shaped spillway is 3.7.
Thanks for visit!

Saturday, January 25, 2014

Comparison of Kennedy's and Lacey's Theories

Hello,

Here is a brief comparison of Kennedy's theory and Lacey's theory for the design of channels for canal etc.

  • The concept of silt transportation is same in both the cases, both agree that the silt is carried by the vertical eddies generated due to friction of the flowing water against rough surface of canal. Kennedy considered a trapezoidal channel section and, therefore, he neglected eddies generated from the sides. For this reason, Kennedy's critical velocity formula was derived only in terms of depth of flow(y).  Lacey considered that an irrigation channel achieves a cup-shaped section(semi-ellipse) and that entire wetted perimeter of the channel contributes to the generation of silt supporting eddies. He, thus, used hydraulic mean radius(R) as a variable in his regime velocity formulas instead of depth(y).
  • Kennedy stated all the channels to be in state of regime provided they did not silt or scour. But Lacey differentiated between two regime conditions, i.e. initial regime and final regime.
  • According to Lacey, grain size of material forming the channel is an important factor, and should need much more attention than what was given to it by Kennedy. He connected grain size(d) with his silt factor(f) as f= 1.76(dmm)^0.5.
  • Kennedy used Kutter's formula for determining actual generated channel velocity. The value of Kutter's rugosity coefficient(n) is again a guess work. Lacey, on the other hand, has produced a general regime flow, after analyzing huge data on regime channels.
  • Kennedy has not given any importance to bed width and depth ratio. Lacey has connected wetted perimeter(P) as well as area(A) of the channel with discharge, thus, establishing a fixed relationship between bed width and depth.
  • Kennedy did not fix regime slopes for his channels, although, his diagrams indicate that steeper slopes are required for smaller channels and flatter slopes are required for larger channels. Lacey, on the other hand, has fixed the regime slope, connecting it with discharge.
Thanks!

Monday, January 20, 2014

Types of Open channel flows

Hello there,

Open channel flow can be classified in the following broad categories and then the subsequent sub-categories:

  1. Steady flow
  2. Un-steady Flow
Further each is classified as:
  1. Uniform flow
  2. Non-uniform flow
Non-uniform flow or varied flow is further classified as:
  1. Gradually varied flow
  2. Rapidly varied flow
Now let's discuss them briefly here,
  • Steady and un-steady flows  When depth of flow and velocity does not vary with time, flow is called steady flow and if depth and velocity vary with time then this is un-steady flow.
  • Uniform and Non-uniform Flow When the depth,slope,cross-section and velocity remain same/constant over given length of channel, flow is called uniform flow and if they vary then this is called non-uniform flow.
  • Rapidly varied flow(RVF) is the flow when the flow conditions changes significantly in a relatively short distance of the channel.
  • Gradually varied flow(GVF) is the flow when the flow conditions changes gradually over a long distance of the channel.  e.g. flow behind a dam at a channel transition.
Thanks for visit!