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:

Thanks!

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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If there is any doubt, please comment below.

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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If you have any doubt, please ask in the comment box.

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.

thanks!

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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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.

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 its 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 :)

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.

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Six Points of Comparison Betweenn Kennedy's Theory and Lacey's Theory

Hello,

Here is a point-wise comparison of Kennedy's theory and Lacey's theory for the design of channels for canals 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!

Types of Open channel flows

Hello there,

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

Steady flow
Un-steady Flow

Further each is classified as:

Uniform flow
Non-uniform flow

Non-uniform flow or varied flow is further classified as:

Gradually varied flow
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.

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Gradually Varied Flow(GVF)

The Assumptions used while analyzing Gradually Varied flow(GVF)

A steady, non-uniform flow in which the depth of flow varies gradually along the length of the channel is called a gradually varied flow.
To analyze Gradually varied flow, the following assumptions are made:

Channel is a prismatic channel section and alignment remains the same.
Bed slope is small, i.e. So=Sf.
Energy loss is the same for a uniform flow at a section having the same velocity and hydraulic radius.
Energy correction factor a=1.
The roughness coefficient is independent of the depth of flow and constant throughout the channel.

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GATE questions from Hydrology - GATE PSUs preparation -part 10

Hi,
Here are few statements from Hydrology which has made their place in GATE Civil Engineering(CE). The bold letter represents the answer which are absent in actual question.

A linear reservoir is one in which storage varies linearly with outflow rate.

When there is an increase in the atmospheric pressure, the water level in a well penetrating in a confined aquifer does not undergo any change.

During a 6-hour storm, rainfall intensity was 0.8 cm/hour on a catchment of area 8.6 km^2. The measured runoff volume during this period was 2,56,000 m^3. The total rainfall that was lost due to infiltration, evaporation, and transpiration (in cm/hour) is 0.304.

Vertical hydraulic conductivity of the top soil at certain is 0.2 cm/hr. A storm of intensity 0.5 cm/hr occurs over the soil for an indefinite period. Assuming surface drainage to be adequate, infiltration rate after the storm has lasted for a very long time, shall be 0.2 cm/hr.

While applying Rational formula for computing design discharge, the rainfall duration is stipulated as the time of concentration because this leads to the largest possible rainfall intensity.

The plan area of a reservoir is 1 km^2. Water level in the reservoir is observed to decline by 20 cm in a certain period. During this period, the reservoir receives a surface inflow of 10 hectares-meters, and 20 hectares-meters are abstracted from the reservoir for irrigation and power. The pan evaporation and rainfall recorded during the same period at a near by meteorological station are 12 cm and 3 cm respectively. The calibrated pan factor is 0.7. The seepage loss from the reservoir during this period in hectare-meters is 4.6.

to be contd...

GATE, PSUs - Hydraulics, Irrigation, water resource Engg. notes - part 9

Hello there,
How have you been? Here is the 9th part of our notes for preparation of GATE and other examinations related to Civil Engineering.

The displacement thickness of a boundary layer is the distance by which the main flow is to be shifted from the boundary to maintain the continuity equation.
The range of specific speeds in MKS units for Francis turbine is 60-300, for Kaplan turbine is 300 - 1000; for Pelton with one jet is 10-35 and for Pelton with two jets is 35-60.
If the height of submerged portion of a symmetrical right circular cone is 'h', then the centre of buoyancy will be at a height of 3/4(h) from the apex on the bottom side.
Loss of head for various pipe fitting is given by (K.v^2)/(2g). The value of K in will be in increasing order for the following sequence of fitting: 45 degree elbow, 90 deg. elbow, foot valve of pump and close return bend.
In a pipe network at a point, if a pipe diverges into two other pipes of smaller diameters and if those two again converge into one at another point, then the potential drop between these two points will be same for both of the pipes.

One Liners | GATE, PSUs 2022 | Hydraulics, Irrigation, Water Resource Engineering | part 8

Hello there,
How have you been!

Here is the 8th part of our notes for preparation of GATE and other examinations related to Civil Engineering.

Vorticity and stream function exist both in rotational and irrotational flow.
The flow of water in a wash hand basin when it is being emptied through a central opening is an example of a free vortex.
Mach number has its application in the launching of rockets.
Thoma number has its application in the cavitation phenomenon.
Reynolds number has its application in motion of submarine.
Weber number has its application in capillary flow in soil.
An error of 0.5% in the measurement of the head in a V-notch causes an error of 1.25% in the discharge.
Smaller eddy sizes and large intensities of turbulence would entail a greater energy dissipation in a turbulent flow.
Shear velocity is a fictitious quantity.
A model of a weir made to a horizontal scale of 1/40 and vertical scale of 1/9 discharges 1 liter/sec. Then the discharge in the prototype is estimated as 1080 liter/s.
Laminar flow occurs between extensive stationary plates. The energy correction factor is nearly 2.0
In the steady laminar flow of a liquid through a circular pipe of internal diameter D, carrying a constant discharge, the hydraulic gradient is inversely proportional to D^4.
For laminar flow between parallel plates separated by a distance of 2h, head loss varies inversely as h^3.
A turbine works at 20 m head and 500 rpm speed. Its 1:2 scale model to be tested at a head of 20 m should have a rotational speed of nearly 1000 rpm.
In a reciprocating pump, the air vessel reduces the acceleration head and consequently reduces the effect of friction head also.
The correct sequence, in the direction of flow of water for installations, in a hydro-power plant is the reservoir, pen-stock, surge tank, and turbine.

All information is learned through books and practical exercises.

References (Also the best books for GATE and PSU preparations):

One Liners | GATE, PSUs 2022 | Hydraulics, Irrigation, Water Resource Engineering | part 7

Hello there,
How have you been!?

Here is the 7th part of our notes for preparation of GATE and other examinations related to Civil Engineering.

As compared to gravity dams, earthen dams require less skilled labor.
The most suitable material for the central impervious core of a zoned embankment type dam is clay mixed with fine sand.
Seepage through embankments in the earthen dam is controlled by drain trenches.
Seepage through the foundation in an earthen dam is controlled by providing impervious cut-off.
The flow of water after spilling over the weir crest in the chute spillway and side-channel spillway respectively are at a right angle and parallel to the weir crest.
The discharge passing over an Ogee spillway is given by CLH^(3/2) where L is the effective length of spillway crest and H is the total head over the spillway crest including velocity head.
The coefficient of discharge of an Ogee spillway depends on the depth of approach and upstream slop and also on downstream apron interference and downstream submergence.
Ogee spillway is least suitable for earthen dams as compared to chute spillway, side-channel spillway, and shaft spillway.
In the case of the non-availability of space due to topography, the most suitable spillway is the shaft spillway.
In the case of chute spillway, the flow is usually super-critical.
For the upstream face of an earthen dam, the most adverse condition for stability of slope is sudden draw-down.
If there are two canals taking off from each flank of a river, then there will be two divided walls and two under-sluices.
Generally, the weir is aligned at right angles to the direction of the main river current because, it ensures less length of the weir, gives better discharging capacity, and is economical.

The main function of the divider wall is to separate the under-sluices from the weir proper.
A divider wall is provided at the right angle to the axis of the weir.
As compared to the crest of the normal portion of the weir, the crest of the under-sluice portion of the weir is kept at a lower level.
Silt excluders are constructed on the river bed upstream of the head regulator.
If 'h' is the ordinate of hydraulic gradient line above the top of the floor and G is the specific gravity of floor material, then the thickness of the floor is given by the formula h/(G-1).
According to Khosla's theory, the exit gradient in the absence of a downstream cutoff is infinity.
The minimum size of stone that will remain at rest in a channel of longitudinal slope S and hydraulic mean radius R is given by 11 RS.
The ratio of average values of shear stress produced on the bed and the banks of a channel due to flowing water is greater than 1.
If the critical shear stress of a channel is Tc, the average value of shear stress required to move the grain on the bank is 0.75Tc.
A watershed canal avoids the cross drainage works.
A canal that is aligned at right angles to the contours is called a side slope canal.
Garret's diagrams are based on Kennedy's theory.

All information is learned through books and practical exercises.

References (Also the best books for GATE and PSU preparations):

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GATE, PSUs - Hydraulics, Irrigation, water resource Engg. notes - part 6

Hello there,
How have you been? Here is the 6th part of our notes for preparation of GATE and other examinations related to Civil Engineering.

The Hardy Cross method of hydraulic analysis of pipe networks, besides satisfying the continuity and energy principles, must also satisfy the condition that the algebraic sum of the head losses around any closed loop is zero.
Given that, S = Slope of channel bottom, Se = Slope of the energy line, F= Froude number, the equation of gradually varied flow is expressed as dy/dx = (S-Se)/(1-F^2).
A fluid motion in which stream lines are concentric circles is known as a vortex flow.
A fluid motion is free vortex flow when the fluid particles moving in concentric circles may not rotate about their mass centre.
In a Sutro weir, the rate of flow for all flows above the rectangular base of width W and depth 'a' is proportional to the head above a datum a/3 above the crest.
In a steady laminar flow through a circular pipe, shear stress is zero at the centre, velocity is maximum at the centre and hydraulic gradient varies directly with the velocity.
Boundary layer thickness is the distance from the boundary where velocity is 99% of uniform velocity.
Displacement thickness is the distance from the boundary by which the main flow can be assumed to be shifted.
Turbulent boundary layer is the distance from the boundary where from the flow ceases to be laminar.
Laminar boundary layer is the region near the boundary where viscous stress is also present.
An irrigation canal has a steady discharge Q at a section where a cross - regulation(gated) is provided for control purposes. If the gate of the regulator, which is normally fully open, is suddenly lowered down to a half open position, then a rapidly varied unsteady flow results. In such a case, it would take the form of a +ve surge moving u/s and a -ve surge moving d/s. [ES 93].
The specific energy 'E' in a critical flow at depth Yc occurring in a triangular channel is given by 1.25*Yc. [ES 94].
When no external energy is imposed, Energy line always falls in the direction of flow and Hydraulic gradient line never rises in the direction of flow.
Stream lines - Tracing of motion of different fluid particles
Streak lines - Identification of location number of fluid particle.
Path lines - Tracing of motion of any one fluid particle.
Equipotential Lines - Location of equal piezometric heads.
As the depth of the immersion of a vertical plane surface increases, the location of pressure comes closer to the centre of gravity of the area.

All information is learned through books and practical exercises.

Reference:

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GATE, PSUs - Hydraulics, Irrigation, water resource Engg. notes - part 5

Hello there,
How have you been? Here is the 5th part of our notes for preparation of GATE and other examinations related to Civil Engineering.

A normal shock wave occurs when the approaching flow is supersonic.
While sub-sonic flow through a converging duct, velocity increases and density decreases.
While sub-sonic flow through a diverging duct, velocity decreases and pressure density, temperature increases.
While a supersonic flow through a converging duct, density increases and velocity increases.
While a supersonic flow through a diverging duct, velocity increases and pressure decreases.
A fluid is a substance which can not remain at rest when subjected to shear.
The viscosity of a fluid varies with temperature.
The locus of elevations that water will rise in a series of pitot tubes is called the energy grade line.
For a developed turbulent flow in a horizontal pipe, the pressure gradient varies linearly with distance.
A streamlined body is a body about which the flow separation is suppressed. [GATE 87].
If the velocity distribution is rectangular then the kinetic energy correction factor will be equal to unity. [GATE 90].
At room temperature, the dynamic and kinematic viscosity of water are respectively greater than and less than of air. [ES 93].
While choosing repeating variables in dimensional analysis, repeating variables should contain all primary units used in describing the variables in the problems and should not contain the dependent variables.

All information is learned through books and practical exercises.

Reference:

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GATE, PSUs - Hydraulics, Irrigation, water resource Engg. notes - part 4

Hello there,
How have you been? Here is the 4th part of our notes for preparation of GATE and other examinations related to Civil Engineering.

Apart from inertia force, Viscous force is most important force for the motion of submarines under water.
In case of capillary waves in channels viscous force is unimportant.
In case of a flow through a long capillary tube, inertia force would be unimportant.
For the resistance to motion of a ship's model through water, the basic similitude criteria is Reynold's Law and Froude's Law
The causes of cavitation are high suction lift and high pump speed.
An impulse turbine operates by initial complete conversion to kinetic energy.
Water turbines may be put in the decreasing order of specific speeds as Propeller turbine, Reaction turbine, Impulse turbine.
Two geometrically similar units are homologous if they have similar streamlines.
Impulse turbine is ideal for high head development.
A reciprocating pump does not need priming.
A centrifugal pump can run at high speed.
Pumps in increasing order of specif speed : Centrifugal pumps, mixed flow pump, axial flow pumps.
The specific speed of a turbine is defined as the speed of a unit of such a size that it produces unit power for unit head.An isentropic process is always frictionless and adiabatic.
Momentum and continuity equations are used to produce Rayleigh lines.
For isentropic flow of air: Critical pressure ratio = 0.528; Critical Temperature Ratio = 0.833; Critical Density Ratio = 0.634

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GATE, PSUs - Hydraulics, Irrigation, water resource Engg. notes - part 3

Hello there,
How have you been!

Are you preparing yourself for some Engineering exams such as GATE or for PSUs?

A multipurpose reservoir is one that is planned and constructed to serve various multi-purposes.
The useful storage is the volume of water stored in the reservoir between minimum pool level and normal pool level.
The water stored in the reservoir below the the minimum pool level is called dead storage.
For a flood control reservoir, the effective storage is equal to Useful Storage+ Surcharge Storage - Valley Storage.
Trap efficiency of a reservoir is a function of capacity/inflow ratio.
The force considered for the analysis of the elementary profile of gravity dam under empty reservoir conditions is the dam's self-weight.
The uplift pressure on a dam can be controlled by :

constructing cutoff under upstream face
constructing drainage channels between the dam and its foundations
by pressure grouting in the foundation

The uplift pressure on the face of a drainage gallery in a dam is taken as two-third of hydro-static pressure at toe plus one-third of hydro-static pressure at the heel.
Horizontal acceleration due to earth-quake results in hydro-dynamic pressure and inertia force into the body of the dam.
Hydro-dynamic pressure due to earthquake acts at a height of 4H/3.pi. above the base.
The major resisting force in a gravity dam is self-weight.
Total force due to wave action on a gravity dam acts at a height of 3/8*hw above the reservoir surface, where hw = water depth.
When the reservoir is full, maximum compressive force in a gravity dam is produced at the toe.
The maximum permissible eccentricity for no tension at the base of a gravity dam is B/6.
The presence of tailwater in a gravity dam decreases the principal stress and shear stress.
The elementary profile of a dam is a right-angled triangle.
In the empty condition of the reservoir and with the elementary profile of a dam, the vertical stress at heel and toe respectively are given by 2W/B and 0.

References (Also the best books for GATE and PSU preparations):

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