water resources engineering

Water Resources Engineering FCE 525

Prof. Patts M.A. Odira

Key Textbooks:

1. Water Resources Engineering by David A. Chin 2nd Edition, Pearson Prentice Hall; 2006.

2. Water Resources Engineering by Ray K. Linsley; Joseph B. Franzini; David L. Freyberg; George Tchobanoglous, 4th Edition, McGraw-Hill, Inc, 1992.

Other Reading Books:

1. Hydrology and Water Resources Engineering by S.K. Carg; 11th Edition, Khana Publishers; 2000.

2. Elements of Water Resources Engineering by K.N. Duggal and J.P. Soni; New Age International Publishers; 2007.

3. Hydrology for Engineers by Ray K. Linsley JR.; Max A. Kohler; Joseph L.H. Paulhus; 3rd Edition; McGraw-Hill International Book Co; 1984.

Course Content

Water Law:- Introduction to common law; Water Codes; Ground Water and National Law and Water policy. Planning for Water Resources development:- National, regional, catchment planning; objective of planning; planning process; introduction to multiple purpose projects; introduction to systems analysis. Economics in Water Resources planning:- Importance; costing and cost allocation; interest rates; taxed; economy studies e.g. cost-benefit, annual cost comparison. Water resources for arid and semi arid areas:- Surface and subsurface Dams; Rainwater Harvesting.

Course Content

Dams; spillways; gates and outlet works:- Focus on types of spillways, gates and outlets; scour protection and energy dissipation. Reservoir: physical characteristics; capacity and yield of reservoirs determination methods. Sedimentation in reservoirs. Hydroelectric power:- types of plants and layouts; systems and loads in power plants, arrangement and

operation of elements. Power duration curves.

CHAPTER ONE- INTRODUCTION TO WATER RESOURCES ENGINEERING

1.1. What is WRE? WRE may be defined as the conception, planning, design,

construction and operation of facilities to control and utilize water.

Water resources problems are not only the concern of engineers but may also involve other professionals e.g. Economists, political scientists, geologists, chemists, biologists etc.

Each water development project is unique in its physical conditions hence standard designs which lead to simple handbook solutions do not always apply.

The special conditions of each water resources project should be met through an integrated application of the fundamental knowledge of many disciplines.

1.2. Fields of WRE

Water is controlled and regulated to serve a wide variety of purposes.

Flood mitigation

Land drainage

Sewerage

Highway culvert design.

These are applications of water resources engineering to the control of water so that it will not cause excessive damage to property, inconvenience to the public, or loss of life.

Water supply

Irrigation

Hydroelectric power development

Navigation improvements

1.2. Fields of WRE (cont..)

These constitute utilization alternatives for beneficial purposes.

Water Quality Management is another field of WRE concerned with pollution control to preserve the water sources for beneficial use

1.3. Quantity of Water

The job of a water resources engineer may be reduced to a number of basic questions.

Since a water resources project is mainly concerned with the control or use of water the pertinent questions the engineer is faced with are:

1. How much water is needed (use socio-economic aspects)?

2. How much water is available or expected (application of hydrology)?

3. Who may use the available water (application of water law)?

1.3. Quantity of Water (Cont...)

In answer to question No. 1:

Social and economic aspects of use must be considered besides the engineering.

On the basis of an economic analysis a decision must be made on the life span for which the proposed engineering works must be adequate.

In considering water use it is important to distinguish between several aspects.

Diversion or withdrawal of water from a system.

Introduction of water into a system.

Consumption water that is evaporated or combined in a product and is no longer available for use.

1.3. Quantity of Water (cont..)

In answer to Question No. 3:

It is important to note that water flowing in a stream is not necessarily available for use by any person or group desiring it at anytime.

Where water is scarce the right to use water has considerable value.

Like all other things of value, water rights are protected by law.

The water law also controls;

oDiversion of stream flow which may cause damage to property.

oAlterations in natural flow conditions

Hence the legal implications should be taken into account in the planning of the project.

1.4. Water Quality

Available water must meet the standards of quality for the intended utilization.

Regulatory government agencies must always maintain surveillance on the water sources to

Guard against pollution by careless discharge of wastes.

Where quality does not meet the utility requirements the engineer has to provide the necessary facilities to remove impurities from the water by physical, chemical, or biological methods.

1.5. Hydraulic Structures

Control, withdrawal and utilization of water resources involve

the design and construction of major and minor hydraulic structures which must be carefully planned.

Why?

They cost a lot of money.

They may have adverse environmental impacts.

They may be a source of major disaster.

PLANNING FOR WATER RESOURCES DEVELOPMENT

2.1. What is planning? The orderly consideration of project from the original

statement of purpose (or objective) through the evaluation of the alternatives to the final decision on a course of action.

Includes all the work associated with the design of a project except the detailed engineering of individual structures.

Forms the basis for the decision to proceed with (or abandon) a proposed project.

Each individual step toward the final decision should be supported by quantitative analysis rather than estimates.

Includes the evaluation of alternatives by the principles of engineering economy.

2.2. Levels of Planning

National - Gives broad guidelines including all aspects of economic development but may specify targets for water management.

Agencies - e.g. River Valley Dev. Authorities, Ministry of Water Development deal with specific projects

Provincial level

District level

Private Developers Missions, other institutions

Community level

2.3. Phases of Planning

Reconnaissance study usually a coarse screen to identify those projects that need further study and eliminate those that are obviously infeasible (usually based on evaluation of existing information).

Feasibility Study - may be one or more. There may be a pre-feasibility study to study various aspects of proposal.

Here a thorough evaluation of the proposed activity is undertaken in order to formulate a description of the most desirable actions to be taken. E.g. for the Kerio ValleyLake Victoria inter-basin water transfer: the desirable features were navigation, hydropower, irrigation, etc.

Design of specific project production of specifications and construction drawings

2.4. Why so many phases of planning?

Sequential studies reduce planning costs by testing the weakest aspects of the project first; e.g. if the project is eliminated because of some aspect, the expense of studying all the other aspects will have been avoided.

BUT: The series of studies should not be allowed to become a series of increasingly more thorough reviews of all aspects of the project otherwise the cost will be increased because many aspects will have been repeated several times.

2.5. Objectives of Planning

Feasibility of a project e.g. water supply project, implies that it will effectively serve its intended purpose without any serious negative impacts.

To measure feasibility the project objectives or purposes need to be specified prior to planning. The rules for measuring achievement or success must also be specified.

At National level objectives are normally broad e.g.

1. Enhancement of national economic development

2. Enhancement of quality of environment

3. Increase national food production

4. Encourage regional development

5. Improve transportation etc.

2.5. Objectives of Planning (cont..)

At project level (e.g. river basin authority level) objectives may be translated into more specific goals, for example:-

Increase of food production by adopting irrigation farming; land drainage, flood protection and other non-water related actions e.g. fertilizer application, education of farmers, improved seeds, etc.

Planning process will test all these alternatives or their appropriate combinations.

2.6. Data Requirements

The problem at hand should be assessed as carefully as possible and the factors most likely to be critical in shaping the plan should be identified.

Example:

In an arid area the availability of water will be a critical factor in the feasibility of an irrigation project.

Most data required will depend on the problem at hand but generally in water resources planning these data are current data describing existing conditions of:

1. Land use must be available when needed or collected at any time prior to need

2. Population

3. Topography

4. Available water resources, hydropower potential, etc.

2.6. Data Requirements (cont..)

Also historical data including;

5. Hydrological and climatological data collected over a period of time. If this historical data is inadequate, installation of new stations must be undertaken at inception of plan so that some data will be available. This data includes:

- stream flows, water levels in lakes, etc.

- rainfall, humidity, rainfall days

- evaporation, sunshine days, solar radiation

- wind speeds and directions

6. Geological data to determine foundation conditions e.g. for dams, groundwater potentials, artificial recharge of groundwater, etc.

2.7 Project Formulation

Begins when the necessary basic data and projections for future conditions are assembled.

Starts with a listing of all the alternatives;

First step is definition of boundary conditions which restrict the project e.g.

o Physical limitations e.g. no possibility of navigation on torrential mountain streams

o Locational limitations flood control for an existing city whose boundary and location cannot be changed.

o Maximum useable areas e.g. land area available

o Available water may be limited to only minor changes.

o Policy restricting certain land for specific purposes e.g. game reserve or park, or recreational areas.

2.7. Project Formulation (cont..)

o Legal constraints

o Certain points of water use that exist, and which must be continued.

Constraints may eliminate some alternatives from consideration

The alternatives that might exist also include:

Engineering alternatives locations, heights, capacities

Management alternatives flood plain regulation

Alternative objectives what needs to be maximized?

Institutional alternatives who should manage?

Timing alternatives delay or speed up?

2.7. Project Formulation (cont..)

Systems analysis: Involves the optimization methods to enable the planner to select the best of all alternatives.

There are 3 levels of optimization:

1. Optimization of individual features of a project e.g. by using the cost-diameter or headloss diameter functions for a pipeline to determine the least cost ( optimum) solution;

2. Optimization of a single project; - e.g. by sub optimization of the various project units (or components) an optimum of the total project is obtained.

3. Optimization of a system of projects e.g. multiple reservoirs, levees, canals, etc. This is achieved successfully by simulation (or operational study) through a computer programme into which many alternative combinations can be quickly worked out.

2.8. Multiple Purpose Projects Why consider multiple uses of water resources or project

facilities?

Because multiple uses of project facilities may increase benefits accruing to the project without a proportional increase in costs hence economic justification for the project.

The multiple uses of a hydraulic multi-purpose project may include:-

oWater supply

o Irrigation

oHydropower

oNavigation

oRecreation

o Flood mitigation

o Sanitation and Conservation of wildlife

2.8. Multiple Purpose Projects (cont..)

What is a multi-purpose project?

Only those projects which are planned, designed and operated to serve two or more purposes should be described as multiple-purpose.

The basic factor in multipurpose design is compromise, i.e. there should be a reasonable efficiency of operation of each purpose.

Maximum efficiency is not necessarily attained for any single purpose.

The physical elements of a multiple-purpose dam project, for example, may not differ from that of a single purpose project e.g. (dam, spillway, spillways, gates, power plant).

BUT: the unique feature is the selection of the physical works and an operational plan which forms an effective compromise among the various uses.

2.8. Multiple Purpose Projects (cont..)

For example:

In allocating reservoir storage in a multipurpose dam, it can be assumed that:

1. No storage is jointly used

2. All storage is jointly used

o In case (1) storage requirements for all functions must be compounded to create a large storage requirement. This can be economically attained only when unit cost of storage is constant or decreases as total storage increases.

o In case (2) maximum economy is achieved since the required storage is not greater than that necessary for any one of the several purposes.

HOWEVER:

Situation in case (2) is rare and the multipurpose dam project is designed to fall somewhere between these extremes

2.9. Functional Requirements of a Multi- purpose Project

Success obtained in achieving joint use of say storage space in a multipurpose dam project depends on the extent to which the various purposes are compatible.

A review of the requirements of the various uses is therefore needed and also consideration be taken of ways in which uses can be coordinated.

Examples:

Irrigation:

Water requirements are seasonal with a maximum during the dry season and a minimum during the wet season.

Annual variations are minimal except during low-rainfall years, and also when project are is increased.

2.9. Functional Requirements of a Multi- purpose Project (cont..)

Irrigation storage is an insurance against drought; therefore it is desirable to maintain as much storage as possible according to current demand.

Water Supply:

Domestic requirements are constant throughout the year.

Domestic demand increases slowly over the years, hence need to plan for this increase.

Adequate reserve is needed to avoid water shortage especially during drought.

Drinking water quality preservation may preclude the reservoir from some recreational uses


Hydropower:

Demand has seasonal fluctuation depending on area served and also connection to the grid system. Hence there is flexibility in coordinating power needs with other uses.

Hydropower production is a non-consumptive use of water hence it is quite compatible with other uses.

Navigation:

Reservoirs required to sustain downstream flows in navigation channels have marked seasonal water requirements.

Peak water requirements occur towards the end of the dry season when natural river flow tends to be minimum.


Reservoirs required to replenish slack-water flow in streams for navigation must have limited height because of provision of locks to govern water depth in navigation channel.

Therefore: such reservoirs have limited storage for other water uses.

Flood Mitigation:

Basic requirement is a sufficient empty space for storage to permit the withholding of flood water during the rainy season; the reservoir must therefore be empty to receive the flood waters when they occur.


Recreation:

Recreational reservoirs should remain nearly full all the time during the recreational season to permit boating, water games, etc.

A reservoir subjected to large drawdown will be unsightly and present problems of dock or landing maintenance, beach maintenance, etc.

Fish and wildlife conservation:

Main problems involve protection, change in habitat created by reservoir construction which may eliminate some species of wildlife and increase presence of others.

Provision of fish ladders may be necessary for migratory fish to go and breed upstream.


Rapid and large fluctuations in reservoir level are harmful to fish, especially during the spawning time because eggs are laid around the edges of the reservoir.

Complete stoppage of flow downstream of the dam is also destructive to fish and wildlife.

Dams should not flood spawning areas otherwise hatcheries should be provided, to maintain economical fish runs.

Pollution Control:

Reservoirs built mainly for low-flow augmentation i.e. the release of water during the low flow season to provide dilution water to a stream receiving waste water discharges to help it to better assimilate the wastewater.


Release from reservoir may increase pollution if there are water quality changes in the reservoir especially due to salt accumulation as a result of evaporation.

Algae growth or decay of vegetative matter in the reservoir may depress dissolved oxygen in the deeper levels of the reservoir.

2.10. Compatibility of Multi-purpose Use

Examples:

Irrigation, water supply and navigation all require a volume of water which cannot be jointly used; therefore a project combining these functions must provide a separate allocation of storage space.

Since hydropower production is a non-consumptive use, any water release for other uses can be used for power generation.

BUT, a certain amount of storage should be provided in case the fluctuations in power release do not appropriately coincide with the requirements of other uses. Also a regulating dam should be provided downstream to smooth out the fluctuations.

2.10 Compatibility of Multi-purpose Use (cont..) A reservoir creating a pool for slack water navigations may

be used for power generation. Also if the dams are made a little higher this might provide space for additional storage for flood mitigation.

Flood mitigation is least compatible of all uses because it requires empty space for storage. But some space is always made above spillway level for flood mitigation; when there is no flood this space may be used for something else so long as there is enough inflow to fill it.

Recreation benefits are enjoyed as opportunity permits i.e. during the times when it is possible to maintain full reservoir levels especially during the high tourist season. Alternatively, sub-impoundments may be provided for swimming, etc.

The same conditions apply to fish and wildlife production.

2.11 Environmental Considerations

Necessity to recognize the interrelationships among water pollution, air pollution, and solid waste disposal:

The role of water supply in population dispersion;

The consequences of water project construction on local ecologic relationships;

The effect of water projects on water pollution;

Necessity to ask or critically evaluate the NEED (real) for a project.

Examples of environmental consequences of water resources projects may include the following.

Degradation of downstream channel or coastal beaches by loss of sediment trapped in a reservoir.

Loss of unique geological, historical, archaeological or scenic sites flooded by a reservoir.

2.11 Environmental Considerations (cont..)

Flooding of spawning beds for migratory fish preventing their reproduction, or destruction of spawning gravel by channel dredging or lining.

Change in stream water temperature as a result of a reservoir leading to changes in aquatic species (or biodiversity) in the streams.

Release of reservoir bottom water which may be high in dissolved salts or low in oxygen resulting in change of biodiversity.

Drainage of swamps decreasing the opportunities for survival of aquatic species.

2.11. Environmental Considerations Cont........

Change in water quality as a result of drainage from an irrigation project which may encourage growth of algae in the receiving stream or lead to change in biodiversity due to increase in salinity of the receiving stream.

Creation of a barrier to normal migration routes of land animals by a reservoir.

CHAPTER THREE - RESERVOIRS

Reservoirs are of 2 types:

1. Storage or conservation reservoirs

2. Distribution reservoirs e.g. Elevated tanks in distribution networks

The storage or conservation reservoirs e.g. dams built across rivers can also be used for flood control, irrigation purposes, power generation, fish farming, etc.

Reservoir can be multipurpose i.e. used for the above purposes simultaneously or single purpose designed to serve only one purpose

3.1. Physical Characteristics of a Reservoir

(w.r.t. storage reservoirs only)

3.1.1. Storage Capacity

o For a regular shaped reservoir any formula for volume of solids will give the storage capacity of the reservoir.

o For natural reservoirs several methods are used for estimating the storage capacity (or volume).

Contour maps of the reservoir site

Surveyed cross-sections of the reservoirs

o From the surveyed cross-sections of the reservoir one can plot the area elevation curve which when integrated produces the elevation storage (or capacity) curve.

area elevation

curve

area

Elevation

900

Elevation

w.r.t

(m.s.e)

area- elevation

curve(capacity curve)

storage volume

Elevation

3.1.1. Storage Capacity Continue.......

3.1.1. Storage Capacity Continue.......

o The capacity curve of a reservoir defines the volume available for storage below a certain elevation; the curve is the integral of the area-elevation curve.

o The increment of storage between 2 elevations is usually computed by multiplying the mean of the areas at the 2 elevations by the elevation difference. The summation of these increments below any level is the storage volume below that level.

POOL LEVEL AT DESIGN FLOOD

Sluice

way

STREAM BED MINIMUM POOL

LEVEL

DAM

NATURAL STREAM SURFACE BEFORE DAM CONSTRUCTION

USEFUL STORAGE

NORMAL POOL LEVEL

SURCHARGE STORAGE

3.1.2. Normal Pool Level

It is the maximum elevation to which the reservoir surface will rise during ordinary operating conditions. For most reservoirs the normal pool level (or elevation) is determined by the elevation of the spillway crest or top of spillway gates

3.1.3. Minimum Pool Level

o It is the minimum level to which the pool is to be drawn under normal operating conditions. This level may be fixed by the elevation of the lowest outlet in the body of the dam. Alternatively, in the case of hydropower reservoirs, by the conditions of operating efficiency of the turbines e.g. the minimum head required.

3.1.4. Useful Storage o It is the storage volume between the minimum and the

normal pool levels. In multipurpose reservoirs the useful storage is divided into:

1. Conservation Storage

2. Flood control storage

o According to the plan of operation, during the floods, discharge over the spillway may cause the water level to rise above the normal pool level. This is known as the surcharge storage and it is normally uncontrolled. The water level will rise to the maximum pool level determined by the design flood.

3.1.5. Dead Storage

It is the volume below the minimum pool level. It is storage not available for any useful purpose

3.1.6. Bank Storage

o Reservoir banks are usually permeable and water enters the soil when the reservoir fills and drains out as the water level is lowered.

o This bank storage effectively increases the capacity of the reservoir above that indicated by the storage-elevation curve.

o The amount of bank storage depends upon the local geological conditions and may amount to several percentages of reservoir volume.

3.1.7. Valley Storage

o When the flow in the stream increases, the depth of flow must also

increase. As the depth increases a quantity of water is stored temporarily until the run off weight decreases. This is known as valley storage.

o Thus considering a reach of channel in a time t:

o Inflow volume into the reach = outflow volume from the + valley storage.

o Therefore over a period of time, the water in a natural stream channel occupies a variable volume of valley storage.

o The net increase in the storage capacity resulting from the construction of the reservoir = total capacity natural valley storage.

o In flood control reservoirs, the effective storage = useful storage + surcharge storage valley storage corresponding to the rate of inflow into the reservoir.

3.2. Reservoir Yield

The yield is the amount of water which can be supplied from the reservoir in a specified interval of time e.g. 1 day for small distribution reservoirs, or a year or more for large storage reservoirs.

The yield of the reservoir depends on the inflow and hence varies from year to year.

3.2.1. Safe or Firm yield (Draft)

Is the maximum quantity of water which can be guaranteed during a critical dry period

3.2.2. Secondary Yield

Is the water available in excess of safe yield during periods of high flow

Power commitments to domestic or other essential uses must be made on the basis of firm yield only unless a standby or supplementary unit e.g. a diesel or steam power plant is available.

Power generated from the secondary yield may be used for irrigation and any other purposes.

The average yield = Firm + Secondary over a long period of time

Yield Yield

3.3. Selection of Capacity of a Reservoir

The process of selecting the capacity for a reservoir is also known as

Operational Study. This can be demonstrated by the following example.

Given below are the monthly inflows during the critical low water period in a dry year at the site of a proposed dam; the corresponding monthly pan evaporation and precipitation at a nearby station; and the estimated demand for water. On consideration of water priorities it is essential to release the full natural flow or 100,000m3/month, whichever is less. Assuming that 80% of the rainfall on the land area is to be flooded by the reservoir is immediately available and using an average net increased pool area of 4km2 in your computations:

What would be the required useful storage (capacity) of the reservoir for the period of 6 months from October to March both inclusive? The pan coefficient is 0.7.

3.3. Selection of Capacity of a Reservoir Cont.............

Month Oct. Nov. Dec. Jan. Feb. March

Inflow vol. m3

2*106

30,000

10,000

5,000

1,000

3,000

Pane evaporation (cm)

9 15 17 14 8 6.5

Precipitation (cm) 12 3 5 2 0 14

Water demand m3 40,000 80,000 130,000 140,000 130,000 30,000


1 2 3 4 5 6 7 8 9 10 Month Flow vol

(m3) Pan evap

cm Precip.

cm Demand (m3)

D/s priority

Req. (m3)

Evap. Vol. (m3)

Precip. Vol. (m3)

Adjusted flow vol. (m3)

Req. Capacity (m3)

Oct Nov Dec Jan Feb March

20x105 30x103 10x103 5x103 1x103 3x103

9 15 17 14 8 6.5

12 3 5 2 0 14

40x103 80x103 130x103 140x103 130x103 30x103

100x103 30x103 10x103 5x103 1x103 3x103

252x103 420x103 466x103 392x103 224x103 182x103

384x103 96x103 160x103 64x103 0 448x103

2032x103 -324x103 -306x103 -328x103 -224x103 266x103

0 404x103 436x103 468x103 354x103 0

TOTAL 1662X103


Solution tips:

Col (6): fill in the lesser of natural flow and 100 x 103m3 the downstream priority requirement.

Col (7): given that the increased pool area is 4 km2 (meaning that when the reservoir is full the flooded area is 4 km2), therefore evaporation volume is calculated as:

Col(3) m x 0.7 x (4x 106) = evaporation vol. in m3

100

Given that the proportion of precipitation available = 80%,

Col. (8) = Col(4) m x 0.8 x (4x 106) = precipitation vol. in m3

100


Adjusted flow volume is given by:

Col. (9) = Col. (2) + Col. (8) - Col. (7) - Col. (6)

The required storage capacity (useful storage) is calculated from:

Col. (10) = Col. (5) - Col. (9) when result is +ve; or = 0 if result is ve.

The determination of the required capacity of a reservoir is called an operation study and is a simulation of the reservoir operation for a period of time following a set of rules.

The operation study may analyze only a selected critical period of low flow; in this case the study will define the capacity required during the selected drought.

Alternatively the study may use a long period of synthetic data record to estimate the reliability of reservoir of various capacities.


Operation study can be performed with annual, monthly, or daily time intervals. For small reservoirs, daily or weekly data are used because the sequence of flow within a month may be critical.

Monthly data are commonly used but for large reservoirs which carry-over storage for many years, annual intervals are used.

When lengthy synthetic data are analyzed, computers are employed using the sequent peak algorithm. Here the cumulative sum of inflows less withdrawals (including evaporation and seepage) are calculated.

The first peak (local max. of cumulative net inflow) and the sequent peak (next following peak > the first) are identified.

Net storage for the interval is the difference between the initial peak and the lowest trough in the interval.

Sequent Peak

Sequent Peak

Storage

(

Flo

w-D

em

and

)

Sequent Peak Algorithm

Maximum Storage

Peak 1

(+ve)

(-ve)

Time


3.4. Mass Curve Technique (or RIPL Diagram)

A cumulative plotting of net reservoir inflow (i.e. inflow adjusted for evaporation and required releases for downstream users).

The slope of the mass curve at any time is the measure of the inflow at that time.

Demand curves representing a uniform rate of demands are straight lines.

If demand curves are drawn tangent to the high points of the mass curve they represent rates of withdrawals from the reservoir.

Capacity to satisfy demand

Spill over

Capacity to satisfy demand

Demand /Year

Capacity

1 Year

1 2 3 4

500

400

300

200

100

Cumulative rates flow

Year

A

B

C

3.4.Mass Curve Technique (or RIPL DIAGRAM) Cont..................

3.4. Mass Curve Technique (or RIPL DIAGRAM) Cont..................

Assuming that the reservoir is full whenever a demand line intersects the mass curve, the maximum departure between the demand line and the mass curve represents the reservoir capacity required to satisfy that demand.

The vertical distance between successive tangents represents water spilled over the spillway.

If demand is not uniform the demand line becomes a curve known as a mass curve of demand but the analysis in the same.

BUT

The demand line for non-uniform flow must coincide chronologically with the mass curve i.e. June demand must coincide with June inflow.

3.5. Reservoir Site Selection It is virtually impossible to locate a reservoir site having

completely ideal characteristics

General rules for choice of reservoir site are as follows:-

1. A suitable dam site must exist; the cost of dam construction is a controlling factor in site selection;

2. The cost of relocation of infrastructure (e.g. roads, railway, cemetery, housing, schools, market centres, cultural sites, etc) for the reservoir must also not be excessive;

3. Site must have adequate capacity for intended operation of the reservoir;

4. A deep reservoir is preferable to a shallow one because of lower land cost/unit of capacity; it has less evaporation losses, and less weed growth.

3.6.Reservoir Sedimentation

The ultimate destiny of all reservoirs is to be filled with sediment.

If sediment inflow is large compared to the total capacity of a reservoir, the useful life of the reservoir will be very short.

Reservoir planning must include consideration of the probable rate of sedimentation in order to determine the useful life of a reservoir before it is constructed.

Sedimentation occurs as a result of 2 processes;

1. Erosion of soil

2. Transport of the eroded soil (sediment) by water.

1. Erosion of soil may be due to several causes but the major ones are

o Rain

o Flowing water

3.6.Reservoir Sedimentation Cont...................

For water flowing through a channel at depth D, the shear stress exerted on the channel bed due to the motion of water is given by:

= Ds

Where is the unit weight of water

S is the longitudinal slope of the channel bed

D is the depth of channel


V

D

J

S


The shear stress is a function of the soil properties, cover on the channel bed and the slope of the channel.

2. Transport of sediment by water;


Concentration-

Distribution Curve

Finite depth

a

Ca y

D

V

E

V

C


The concentration of soil sediment in moving water in a natural channel may be expressed at any depth y as:

Where C = concentration of sediment of sediment in the water at a depth y

Ca = concentration at any finite depth a

D = depth of channel

=

=


Where w = fall velocity of the sediment particles in the sediment water mixture

k = von Karmans constant U* = shear velocity = And = shear stress at the channel bed = mass density of the sediment water mixture Depending on the particle size the total sediment load is

classified as:

1. Bed material load the load scoured from the channel bed

2. Wash load the sediment collected as the water flows along


The transport of the bed material load is due to the fluid transmitted stresses, while the transport of wash load is due to minor turbulence only.

Depending on the mode of movement of sediment particles, the total sediment load can be divided into:

1. Bed load which moves by:

o Sliding or rolling along the bed

o Saltation hopping and rolling

2. Suspended load always moving in suspension as the water flows along

A state of equilibrium is always aimed at by the flow in a river.

At a particular discharge Q, a river tends to maintain a channel slope s and will have a capacity to transport a certain load W of sediment of a particular mean size d50.


If any of these factors changes, the other factors are also affected.

Therefore the equilibrium may be achieved after some interval of time in which case at any particular time the river stays at a quasi equilibrium state.

The total amount of sediment that passes any section of a stream is known as the sediment yield and is given by:

The total load = Bed load + suspended load

(100%) (5-20%) (80-95%)

3.6.1. Estimation of Amount of Sediment in a Channel This is done by establishing a sediment rating curve for the

particular channel

The relation between suspended sediment transport Qs and stream flow Q is represented by a logarithmic plot:

Qs = kQn n = 2 to 3 and k is very small

Continuous records of steam discharges are maintained by gauging stations and also sediment load gauging stations can be installed.

Flow synthesizing techniques are used to estimate future flows assuming this relationship will be holding in future years.

3.6.1. Estimation of Amount of Sediment in a Channel Cont.............

1

2

3

4

5

Use simposons rule to estimate

discharge Q

Log Qs Amount of sed

Log Q flow rate

3.6.1. Estimation of Amount of Sediment in a Channel Cont...... The relation between suspended sediment transport Qs and

stream flow Q is represented by a logarithmic plot:

Qs = kQn n = 2 to 3 and k is very small

Continuous records of steam discharges are maintained by gauging stations and also sediment load gauging stations can be installed.

Flow synthesizing techniques are used to estimate future flows assuming this relationship will be holding in future years.

If Q increases Qs will increase but this relationship will not be maintained throughout i.e. Qs vs Q is an average relationship.

3.6.1. Estimation of Amount of Sediment in a Channel Cont...... WHY? With the first rise of a flood in a river,

most of the erosion occurs hence there will be a large amount of sediment to be transported; as the flood continues the sediment content of the water will decrease.

For a channel of discharge Q, the power available in the water which is imparted to the soil to cause erosion is given by: P = Q s

3.6.2. Measurement of Qs (sediment flow)

Region of bed load flow

10-20 cm

3.6.2. Measurement of Qs (sediment flow) Cont..

Sub-divided into:

a) Bed load measurements

b) Suspended load measurements

c) Total load measurements

For measurement of bed-load samples are taken from bedload flow region which is located 1020cm above the channel bed. For small rivers, samples are obtained by pumping from this region.

3.6.2. Measurement of Qs (sediment flow) Cont.......

For larger rivers a scoop type of sample is used. The scoop is placed at the channel bed and left there for some time interval for the sediment to collect inside; the scoop is then removed and the accumulated sediment measured.

The sediment load measured is expressed in parts per million (ppm) or mg/l.

Due to discontinuous flow of sediment along the channel bed and the interaction between water and channel bed, the sediment forms ripples on the channel bed. As flow changes further, the ripples change to dunes i.e. with the increase in Froude No. of flow, i.e.


a) ripples are formed

b) dunes are formed

c) flat bed results (occasionally) then

d) anti dunes begin to form

Because of the non-uniform flow or movement of dunes along the channel bed, there is usually a difficulty in obtaining samples from the channel bed.


a scoop placed in the trough of the

dunne collects little sediment scoop placed here is completely

submerged by the sediment

3.6.3 Suspended Load Measurements

a) Point integrating samplers

kept at a fixed depth

a) Depth integrating samplers

lowered and removed at a known constant velocity

; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; , ; ,

sediment flow sample kept at

a fixed depth

3.6.3. Suspended Load Measurements Cont.......... By keeping the sampler at different points of the

channel depth the sediment concentration at each depth can be obtained.

The depth sampler is lowered into the water and then lifted up. The sampler is designed such that whatever depth it is at, it will allow sediment inflow at a constant rate. The depth sampler gives an average sediment concentration along the channel bed directly.

The equi-transit-rate (ETR) method is a type of depth sampling in which the sampler is lowered and removed at a uniform rate. Sediment sampling is fast and the average concentration is obtained directly.

3.6.4. Errors inherent in sediment measurements

Reasons for errors in sediment sampling in streams are as follows:

1. The presence of the sampler itself is sufficient to disturb the flow pattern and as such the intensity of sediment concentration is not correctly obtained.

2. Often it is difficult to give the sampler a correct vertical and horizontal alignment with respect to the direction of motion of the sediment flow.

3. No bed-load sampler can collect all the sizes (i.e. from the coarsest to the finest) of the bed load. The fraction of the actual bed load caught by a sampler is influenced by the type of sampler used and the relation between the geometry of the sampler and the size and geometry of the bed form i.e. ripples, dunes, etc.

3.6.4. Errors inherent in sediment measurements Cont....................

4. The accuracy of bed-load measurements by using radio-active tracers is affected by the amount of background concentration of the tracer in the stream and the degree of mixing of the tracer with the sediments between the point of introduction and point of sampling.

5. Accuracy in suspended load measurements is affected by the following factors;

a) The suspended load samplers do not traverse a region of about 10 cm above the channel bed. Since sediment concentration is maximum near the channel bed a significant proportion of the suspended sediment may remain unmeasured.

3.6.4. Errors inherent in sediment measurements Cont..............

b) Experimental evidence indicates that the standard deviation of the depth integrated suspended load concentration at a vertical may be 10% or more of the mean.

c) The depth integrated concentrations between 2 verticals in the x-section measured on the same day may differ by more than 100%.

d) Verticals having maximum and minimum concentrations in a x-section change position with time.

e) In a cross-section the verticals for maximum concentration of different size ranges may not coincide.

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heavy particles settle

down first in the form

of a delta

finer particles continue

to move on the water forming

density currents

as time passes the whole of this

area may be filled up with sed, hence

reverser capacity is reduced

some sediment may

be sucked down a sluice way

S

3.6.5. Sediment inflow into a reservoir

3.6.5. Sediment inflow into a reservoir Cont.......

As reservoir silts up, the already deposited silt gets

consolidated and the density of this silt may vary between 0.64 1.76 g/cm3

Freshly deposited sediment has a density of 0.80 0.88 g/cm3

whereas old deposits may have a density of 1.04 1.12 g/cm3.

The mean density for sediment is usually 0.96 g/cm3

The ratio of sediment trapped in reservoir to the amount of sediment brought into the reservoir is known as TRAP EFFICIENCY.

Trap eff. = amount of sediment trapped in the reservoir

amount of sediment brought into reservoir

3.6.5. Sediment inflow into a reservoir Cont.......

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Capacity/inflow ratio

3.6.6. Capacity Inflow ratio

This is the ratio of reservoir capacity to the total inflow of water into it.

It has been found that the trap efficiency is a function of capacity inflow ratio

3.6.6. Capacity Inflow ratio Cont........

0

20

60

40

80

100

Trap eff

% sediment

trapped

0.001 0.01 0.1 10 100 1.0

3.6.6. Capacity Inflow ratio Cont........

From curve we can see that if capacity reduces (with constant inflow) trap efficiency reduces, and hence lesser sediment is trapped.

Silting rate in the reservoir will be more in the beginning, and as its capacity reduces due to silting, the silting rate will also reduce hence complete reservoir silting may take a longer period.

3.6.7. Reservoir Sedimentation Control In real life sedimentation of reservoir cannot be prevented; it

may only be retarded. So the rules of planning require that you:

1. Select site where sediment inflow is low naturally; some basins are more prolific sources of sediment than others due to soil type, land slopes, soil cover, and rainfall characteristics. Avoid these prolific sites if there is an alternative.

2. Provide a large enough capacity to create a useful life sufficient to warrant construction of the reservoir.

6.6.7. Reservoir Sedimentation Control Cont.......

3. Use soil conservation methods within the drainage or catchments basin

o Terracing

o Strip cropping

o Contour ploughing, etc. to reduce erosion

o Check dams in gullies to retain some sediment and reduce sediment flow into stream

o Vegetal cover to reduce impact force of raindrops and minimize erosion.

Note: if stream is denied its normal sediment load it will tend to scour its bed or cave its banks. Therefore, stream bank protection by revetment, vegetation, etc, is necessary.

4. Prevent sediment accumulation in reservoir by providing means of discharge of some sediment.

CHAPTER FOUR FLOOD ROUTING 4.1. Reservoir Routing A process that shows how a flood ware can be reduced in

magnitude and lengthened in time (attenuated) by the use of storage in the reach between two points of a stream.

Routing techniques are used to compute the hydrograph which will result from a specified pattern of rainfall excess.

4.1. Reservoir Routing Cont..........

Discharge Q

Inflow rises very fast

out flow

(routed flow)


The flood is caught up in the reservoir and then released at a slower rate and intensity. The effect of the flood downstream is therefore reduced.

In some instances this delayed flow of the flood waters into the downstream channel from the reservoir can facilitate arrangements for evacuation before flooding occurs.

Reservoir routing is based upon the continuity equation, which states that in time t: i.e. (1)

Total Inflow into = total outflow + change in the reservoir from the reservoir reservoir storage


I1 = rate of inflow at the beginning of time t

Q1 = rate of outflow at the beginning of time t

I2 = rate of inflow at the end of time t

Q2 = rate of outflow at the end of time t

S1 = storage in the reservoir at beginning of t

S2 = storage at the end of time t

i.e (1)


1. Reservoir characteristic curves

a) Capacity curve storage-elevation curve

b) Outflow elevation curve (discharge curve)

2. Anticipated incoming flood inflow hydrograph

(2)


elev elev or 1

Storage capacity Storage capacity


elev 2 3

Inflow hydrograph

Q


From curves 1 and 2 we plot the outflow-storage curve (choice of t depends on total time of analysis).

Out flow (O)

= routing internal

Storage (S)


1. Find the out-flow rate at the beginning of 1st t; then mark the point on the ordinate of the out-flow-storage graph;

2. Plot (I1 + I2) t parallel to the abscissa from the S Q t curve;

2 2

1. Erect a vertical from this point up to the S + O t curve (call this point A); 2

2. From point A move backwards to the ordinate (// to the abscissa) and read off the outflow O2 at B;

3. Using the elevation discharge curve, find the water storage level in the reservoir at the end of t;


6. Move to the corresponding point on the S O/2 t curve i.e. point C;

7. From point C, plot (I1 + I2) t for the next time interval; -Alternative method of Plotting the D S curves

Suppose t = 6 hrs

Therefore: O t = O (6x60x60)

i.e. the distance between the S-O/2t and S + O/2t curves

= O( x 24( (60 x 60)

= O (24x60x60)

= O quarter cumec days.


storage in m3 = x 24 x 60 x 60

If t = 24 hrs the storage scale is plotted in 1 cumec-day


F

B

Outflow

2 2

1 (I +I

2

)

2 1 1 (I +I

2

)

G

C

E

A

4.2. Routing in Channels

Routing in a natural river channel is complicated by the fact

that storage is not a function of outflow alone.

There tends to be a greater storage for a given outflow during the rising stages of a flood than during the falling stages.

OThe Muskingums Method of River Routing

The method was advanced by McArthy of the U.S Corps of Engineers

The same formula as the one used in Puls method is used i.e.

4.2. Routing in Channels Con.........

In Puls method it is assumed that:

O = f(S)

But in Muskingums method of river routing:

O = f(S, I)

i.e the outflow is a function of both storage and inflow.

(3)


Considering a reach between two sections of a natural channel, the storage is found to be expressed as:

Where b, m = depend upon the stage discharge characteristics of the two control sections at the ends of the reach.

a, n = depend on the stage-volume characteristics of the two control sections

x = is a factor defining the weightage of the inflow/storage to the outflow. I.e. x is a constant that expresses the relative importance of inflow and outflow in determining storage.

For a simple reservoir x = 0 (inflow has no effect). If inflow and outflow are equally effective x = 0.5. For most streams, x = 0 to 0.3; with a mean value of 0.2

(4)


For a simple reservoir x = 0 (inflow has no effect). If inflow and outflow are equally effective x = 0.5. For most streams, x = 0 to 0.3; with a mean value of 0.2.

In a uniform rectangular channel storage would vary with the first power of stage (S =bgm, where m = 1) and the discharge would vary as the 5/3 power (Manning formula) q = agn

In the Muskingum method it is assumed that m/n = 1 and if we let b/a = K then equation (4) becomes:

S = K (xI + (1-x)O) (5)

K = storage constant time of travel of the flood ware through the reach.

x varies from 0 1.0 and for moderate size reaches x 0.2


K and x are assumed to be constant throughout the routing period.

To determine K and x for a river reach we need the following

o Inflow hydrograph

o Corresponding outflow hydrograph

From S = K (xI + (1-x)O)

K = _____S___ in hrs (6) (xI + (1-x)O)


a) During every routing interval t, compute the storage = cumulative inflow cum. Outflow

b) Compute 5 from equation (4) for an assumed or estimated value of K and several assumed values of x

c) Plot the 2 values of S obtained from (a) and (b) against storage; and draw the storage loops


Weighted

Storage

XI+(1-x)0

m 3 s

Storage

x=0.15 1/4 days


x=0.2 S

WS


x=0.3 S

WS


x=0.4 S

WS


WS

b

a

a K=b


Choose the loop which is closest to a straight line and take the corresponding value of x for the river reach. The inverse slope of the line closest to the loop gives the value of K

d). Put the obtained value of x into equation (6) and compute K. Using this value of K find another value of x by the graphical procedure; then calculate an average value of x.

With the determined values of K and x for the river reach, an inflow hydrograph is routed by rewriting equation (4) as follows:

S2 S1 = K(x (I1 I2) + (1 x) (O2 O1))

O2 = CoI2 + C1I1 + C2O1 (7)


Where;

When equations (i), (ii) and (iii) are combined we obtain:

C0 + C1 + C2 = 1

Co = (i)

Co = (ii)

Co = (iii)


t is the routing period in the same units as K.

All the 3 coefficients C0, C1 and C2 must be positive for equation (7) to give valid results.

The routing procedure is simply a solution of equation (7) with the O2 of one routing period becoming the O1 for the succeeding period


Inflow

hydro graph

Outflow

hydro graph

when x=1

Outflow

hydro graph

when x=0


Small k value

Large k value

Q


TUTORIAL EXAMPLES

1. The inflow hydrograph of a river reach is tabulated below. The storage constant K = 10hr and x

= 0 for the reach. Find graphically the outflow peak in time and magnitude. What would be the effect of making x > 0? Assume that the outflow at hour 11 is 28.3m3/s and is starting to rise.


TIME (h)

I (m3/5

Time (h)

I (m3/5)

0 5

10 15 20 25 30 35

28.3 26.9 24.1 62.3 133.1 172.7 152.9 121.8

40 45 50 55 60 65 70

90.6 70.8 53.8 42.5 34.0 28.3 24.1


a) This assumption is nearly correct because the river is still at a low stage and the flow will nearly steady.

b) The first equation to be solved is therefore

O2 = CoI2C1I1+C2OI

Now Co = = = 0.2 =

Now C1 = = = 0.2 =

Now Co = = = 0.6 =


With first value of 01 = I1 = 28.3m3/s: O2 = 0.2x26.9+0.2x28.3+0.6x28.3 = 28.02 m3/s

This value of O2 becomes O1 for the next calculation and the values are as tabulated.


If x>0; say x = 0.1

C2 = = = 0.57

C0 = = = 0.13

C1 = = = 0.30


Hrs 0.2 I2 m3/s

0.2 I1 m3/s

0.6 01 m3/s

O2 m3/s

I m3/s

0 5 10 15 20 25 30 35 40 45 50 55 60 65 70

- 5.38 4.82 12.46 26.62 34.4 30.58 24.36 18.12 14.16 10.76 8.50 6.80 5.66 4.82

- 5.66 5.38 4.82 12.46 26.62 34.4 30.58 24.36 18.12 14.16 10.76 8.50 6.80 5.66

- 16.98 16.81 16.21 20.09 35.5 57.91 73.73 77.20 71.81 62.45 52.42 43.01 34.99 28.47

28.3 28.02 27.01 33.49 59.17 96.52 122.89 128.67 119.68 104.09 87.37 71.68 58.31 47.45 38.95

28.3 26.9 24.1 62.3 133.1 172.7 152.9 121.8 90.6 70.8 53.8 42.5 34.0 28.3 24.1

The effect of increasing the value of X to x=0.1 is to reduce the peak of the outflow hydrograph and delay the outflow time slightly.


EXAMPLE 2 The following table gives

the spillway discharge reservoir storage relationship for a reservoir.

Spillway discharge m3/s

Reservoir Storage (1/4 cumec days)

0 100 200 300 400 500 600 700 800 900 1000

10,000 10,200 10,400 10,600 10,800 11,000 11,240 11,500 11,800 12,080 12,400


The water level in the reservoir is at the spillway crest level when the flood tabulated below inflows into the reservoir.

Time (Hrs)

Flood inflow m3/s

0 6 12 18 24 30 36 42 48 54 60

0 80 200 500 900 800 600 400 200 100 0


Using Puls Method of reservoir routing with 6 hr-routing interval, find the flood hydrograph over the spillway crest. Sketch the inflow & outflow hydrographs on the same axes.

What would be the maximum head over the spillway crest if the spillway is 10m long and has a co-efficient of discharge = 4.0?

Time O2 0-6 6-12 12-18 18-24 24-30 30-36 36-42 42-48 48-54 54-60

18 65 180 395 565 910 575 475 350 215


1. Construct the storage versus discharge graph i.e. the S-O curve and obtain the S-O/2 t and the S+0/2 t curves. The routing graphs.

2. From the routing graphs the following values for discharge over the spillway are obtained.

3. A plot of the following outflow hydrograph reveals that the maximum discharge = 610m m3/sThe maximum head over the spillway crest occurs when the discharge over the spillway is a maximum.

Qmax = Cd L Hmax3/2

Hmax = Qmax/(Cd. L)2/3


EXAMPLE 3

What reservoir capacity is required to assure a yield of 565x1010 m3/year, for the inflows shown in the table below?


Month 1990 1991 1992 1993 January February March April May June July August September October November December

40x1010 80x1010 120x1010 200x1010 240x1010 400x1010 720x1010 880x1010 960x1010 1040x1010 1090x1010 1120x1010

1140x1010 1150x1010 1560x1010 1760x1010 2140x1010 2200x1010 2240x1010 2260x1010 2270x1010 2275x1010 2278x1010 2280x1010

2320x1010 2400x1010 2440x1010 2445x1010 2480x1010 2520x1010 2560x1010 2570x1010 2580x1010 2590x1010 2610x1010 2610x1010

2650x1010 2760x1010 3160x1010 3280x1010 3320x1010 3760x1010 4080x1010 4240x1010 4320x1010 4360x1010 4480x1010 4560x1010

CHAPTER FIVE SPILLWAYS, GATES AND OUTLET WORKS

Spillways: discharge floods downstream and protects the dam from being damaged by overtopping.

Gates & sluiceways: Provided on the spillway crest to permit outlet works the operator to control the release of water downstream fro various purposes

5.1 Spillways

regarded as the safety valve for the dam Designed to have the capacity to discharge major foods at the

same time as keeping the reservoir level below same pre-determined maximum level.

Of 2 types:

o Controlled provided with crest gates or other facility so that the outflow rate can be adjusted.

o Uncontrolled free discharging once the water level rises above crest level.

Selection of capacity is related to the degree of protection required to be provided to the dam

This in turn depends on:

o Type of dam

o Location of dam

5.1 Spillways Continue..........

o Consequences of failure of the dam

E.g. a high dam storing a large volume of water, located above or upstream of an inhabited area should have a much higher degree of protection as compared to a low dam storing a small quantity of water and whose demonstration reach is uninhabited.

The probable maximum flood is commonly used in design for the high dam while a smaller flood based on frequency analysis is suitable for the low dam.

Determination of the area which would be flooded if the dam were to fail assists in assessing the acceptable risk.

5.1 Spillways Continue..........

TYPES OF SPILLWAYS

There are different types of spillways as follows:

Overflow spillways

Chute spillways

Side channel spillways

Shaft spillways

Siphon spillways

5.1.1. Overflow Spillways

A section of dam designed to permit water to pass over

its crest;

Widely used on gravity, arch and buttress dams;

For earth dams, the overflow section is normally of concrete gravity construction designed to serve as a spillway;

It is important that the overflowing water is guided smoothly over the crest with minimum turbulence;

If the overflowing water breaks contact with the spillway surface, a vacuum will from at the point of separation and cavitation may occur.

5.1.1. Overflow Spillways Cont...........

Cavitation plus the vibration from the alternate making

and breaking of contact between the water and the face of the dam may result in serious damage to the dam structure.

In order to reduce the occurrence of such cavitation the ideal spillway should take the form of the underside of the nape of a sharp-crested weir when the flow rate corresponds to the maximum design capacity of the spillway.

5.1.1. Overflow Spillways Cont........... The reverse curve on the downstream face of the spillway

should be smooth and gradual.

The discharge of an overflow spillway is given by the weir equation.

Q = CwLH3/2

Where: Q = discharge in m3/s

Cw = coefficient of discharge

L = Length of crest (m)

H = head on the spillway (equal to the vertical distance from the crest of the spillway to the reservoir level.

Cw varies with the design and head (1.4 2.3) but experimental models are often used to determine the coefficient in any particular situation.

5.1.1. Overflow Spillways Cont........... If the x-sectional area of the reservoir just upstream from the

spillway is less than 5 times the area of flow over the spillway, the approach velocity may increase the discharge to noticeable extent. The effect of approach velocity can be accounted for by the equation:

Q = Cw L (H + V02)3/2

2g

Where: V0 = approach velocity.

End contractions on the spillway often reduce the effective length below the actual length L.

Square cornered piers disturb the flow considerably and reduce the effective length by the width of the piers plus about 0.2h for each pier.

Streamlining the pier or flaring the spillway entrance minimizes

the flow disturbance.

5.1.2 Chute Spillways Constructed of concrete slabs in the form of a steep-sloped

open channel or trough

The slabs are 250-500 mm thick.

Adapted to earth or rock fill dams

Normally located around the end of the dam as topography permits, such locations is preferred especially for earth dams to prevent possible damage to the embankment.

The Chute may be of constant width but is usually narrowed in some sections for economy and then widened near the end to reduce discharge velocity.

As far as possible the slope should be steep enough to maintain flow below critical depth so that unstable flow conditions do not set in.

5.1.2 Chute Spillways Cont......................

The side walls of the chute must be of adequate height to accommodate bulking of water caused by entrainment of air in the high velocity flow.

Expansion joints should be provided in chute spillways at intervals of 10m, and they should be as watertight as possible to avoid percolation of water under the slab that may cause uplift forces under the slab.

As additional insurance against uplift, rock fill or perforated steel pipe drains should be provided under the spillway.

5.1.3. Side Channel Spillway

A spillway in which the flow, after passing over the crest, is carried away in a channel running parallel to the crest

The crest is usually a concrete gravity section, but it may consist of pavement laid on an earth embankment or natural ground surface.

Usually used in narrow canyons where sufficient crest length is not available for overflow or chute spillways.

After passing through the side channel, the water may be led out through a chute or tunnel.

5.1.4 Shaft Spillways (Morning Glory)

The water drops through a vertical shaft to a horizontal

conduit that conveys the water past the dam.

Often used in cases where there is inadequate space for other types of spillways, its usually undesirable to carry a spillway over or through an earth dam.

However, if topography prevents the use of chute or side channel spillway around the end of the dam, a shaft spillway through the foundation material provides the best alternative.

For low dams where the shaft height is small, no special inlet design is necessary, but for high dams, a flared inlet, referred to as a morning glory is used.

May be constructed of metal, concrete pipe or clay tile for small spillways

5.1.4 Shaft Spillways (Morning Glory) Cont......

May be constructed of metal, concrete pipe or clay tile for small spillways

For large dams, the vertical shaft is constructed of reinforced concrete, while the horizontal conduit is tunnel in rock.

Frequent a diversion tunnel is provided so that it may be used for the spillway outlet.

There are 3 possible conditions of flow in a shaft spillway.

i) At low head, the outlet conduit flows partly full, the perimeter of the inlet serves as a weir and the discharge of the spillway is given by;

Q h13/2


ii) As the water level rises, water level in the shaft rises, and the outlet may flow partly full ( weir flow) or full (orifice flow), when the shaft is completely filled and the outlet submerged under those conditions the discharge is given by;

Q h21/2

where: h2 = hL + V2

/2g = total head on the outlet.

- At this point, the increase in h2 results in only a very slight increase in discharge, and therefore this poses a limit on the capacity of the shaft spillway.

- The graph with the solid line depicts the relation between the flow rate and the water surface elevation for a properly designed shaft spillway.


iii) When design is not proper there will be a throttling of the flow as flow changes to pipe flow.

An abrupt transition between the shaft and outlet conduit may result into cavitations hence a smooth transition should be planned especially for large structures.

Complete hydraulic analysis of a shaft spillway is difficult and therefore models are employed to simulate flows and to determine the coefficients needed.

A serious problem with shaft spillways is their potential to clog with debris. Therefore, they should be protected with trash racks, floating booms etc.


Flow conditions in a shaft spillway

5.1.5 Siphon Spillways Usually provided for cases where spillway location space is

limited and where the discharge capacity desired is not large.

They have the advantage of automatically maintaining the water surface elevation within very close limits.

At higher flows, after the siphon has primed, the discharge is given by;

Q = CdA2gh; where Cd = coefficient of discharge ( 0.9)

If the outlet of the siphon is not submerged, the head h is the vertical distance from the water surface in the reservoir to the end of the siphon barrel.

When the siphon outlet is submerged then h is the difference in elevation between the headwater and the tail water.

5.1.5 Siphon Spillways Cont......

If air is prevented from entering the outlet end of the siphon, the flow through the siphon will entrain and remove air at the crown and prime the siphon.

Entrance of air can be prevented by deflecting the flow across the barrel in such away as to seal it off, or by submerging the outlet.

The siphon action will continue until the water level in the reservoir drops to the elevation at the upper lip of the siphon.

As soon as a siphon is primed, a vacuum forms at the crown. In order to prevent cavitation, the siphon should be designed so that this vacuum never exceeds three fourths of atmospheric pressure (3/4 atm).

5.1.5 Siphon Spillways Conti......

Hence at sea level, the vertical distance form the crown of the siphon down to the hydraulic grade line should not exceed 7.5m.

Trash racks can be provided to avoid clogging by debris.

Cross-section through a siphon spillway

5.2. Dynamic Forces on Spillways

Newtons 2nd Law of motion states that force equals the time rate of change of momentum.

The resultant forces on an element of water are then given by;

F = Q V (1)

Where = density of water

Q = flow rate

V= change in velocity

In vector form equation (1) can be re-written as;

Fx = Q(V2x V1x) and Fy = Q(V2y V1y) (2)

Where x and y represent any convenient system of coordinates

5.2. Dynamic Forces on Spillways Cont........

Equation 2 can be used to calculate the dynamic forces exerted by water on spillways, deflectors, turbine blades, pipes and bends, etc.

The forces Fx and Fy are those acting on a significant free body of fluid and include gravity forces, hydrostatic pressures, and the reaction of any object or surface in contact with the fluid body.

5.3. Hydraulics of Outlet Works

The discharge through the dam outlet (Sluiceway) can be calculated from the equation:

Q = CdA2gH (3)

Where A = area of the outlet sluice

Cd = coefficient of discharge whose value depends on the type of gate, trash rack, conduit friction, transitions etc.

H = differential head causing flow (usually the difference in elevation between u/s and d/s.

5.3. Hydraulics of Outlet Works Cont........ The total headloss in the conduit include losses in

o trash racks

o conduit entrance

o conduit friction

o gates and valves

o transitions

o bends

Entrance loss is taken as 0.5v2 for square edged entrance. For

2g

a bell-mouthed entrance the loss is taken as 0.04 v2

2g

5.3. Hydraulics of Outlet Works Cont..... Conduit friction loss is calculated from the standard pipe formulae

e.g.

Gate loss for a fully open gate and butterfly value is taken as 0.2v2

2g

Headloss through trash racks are found to follow the table below. v2

= or = (4)

5.3. Hydraulics of Outlet Works Cont.....

Velocity thro Trash rack (m/s)

Head loss (m)

0.15 0.20 0.30 0.40 0.45 0.50 0.60 0.62

0.006 0.01 0.03 0.05 0.09 0.09 0.13 0.15

If discharge is calculated from the net effective head then;

Heff = Differential head H minus headloss

Q = CdA2g Heff (5)

5.4. Protection against Scour

Water flowing over a spillway or through a sluiceway is capable of causing severe erosion of the stream bed and banks below the dam.

The type of protection against erosion that should be provided depends on the degree of damage expected from this erosion.

The time required to develop serious erosion depends not only on the character of the stream bed material and the velocity distribution but also on the frequency with which scouring flows occur.

Often model studies are conducted during the planning period to determine the relevant factors.

5.4. Protection against Scour Cont.........

Solid rock is often resistant to erosion although if the rock has bedding planes it may not resist high velocity flow. Also if the rock has a rough, jagged surface, cavitation may assist in its erosion.

Loose earth and rock are vulnerable to the erosive action of flowing water, and may scour at velocities as low as 0.6 - 1.0m/s

Movablebed model studies with gravel, sand, or powdered coal to simulate the river bed may be used to predict velocity distributions.


For a horizontal rectangular channel the hydraulic jump is expressed as:

Where y1, and y2 are the depths of flow before and after the jump.

(6)


The approximate depth of flow at the toe of the spillway y1 may be found by applying the energy equation along a streamline between point A on the surface of the reservoir and point B at the toe of the spillway.

Neglecting friction and velocity of approach, the energy equation is:

(7)


The average flow velocity V1 at the toe of the spillway is

By substituting values of h, and Q in the energy equation, the corresponding values of y1 and V1 can be found.

The hydraulic jump equation can then be used to find the sequent depth y2

The energy dissipated in the jump is equal to the difference in specific energy before and after the jump ( )

A fully developed hydraulic jump below an ogee spillway is particularly effective as an energy dissipator in situations where h is small compared with the height of the spillway.

For a jump to occur, the flow must be below critical depth, and this condition is satisfied in almost al cases where there is potential for scour.


Typical scour protection works


Appurtenances in a stilling basin

CHAPTER SIX WATER LAW

6.1 Introduction Where there is inadequate water to meet the needs of

potential users, water is a commodity of considerable value.

Hence a system of laws is necessary to determine who has the right to water abstraction and use.

Water law can also play a major role in the economic aspects of water development since limitation on who may develop water resources will also control how it is developed and utilized.

6.2 Common Law

Each country has its own legislation that governs the

exploitation of water resources that are appropriate to the country depending on the availability of water.

However, there is a common law upon which the legislation is built. This common law hinges upon rights.

There are two distinct categories of rights viz; Riparian rights, and Appropriative rights.

Taken from French Civil law by two American jurists Story and Kent

6.2 Common Law Cont...........

English courts adopted it in early 19th century as part of its common law.

Later accepted in the Americas which adopted the English Law

Under the concept of riparian rights the owner of the land adjacent to a stream known as (riparian land) is entitled to receive the full natural flow of the stream without change in quantity and quality.

Hence the riparian owner is protected against diversion of water upstream from his properly or from diversion of excess floodwaters towards his property.

No upstream owner may materially lessen or increase the natural flow of the stream to the disadvantage of a downstream owner.


In modern society riparian doctrine has serious defect because it does not provide for actual use of water by the riparian owners fro irrigation and other intensive user.

Riparian concept has therefore been modified to permit reasonable use of water which allows riparian owners to divert and use stream flow in reasonable amounts for beneficial purposes.

No priority can exist between riparian owners i.e. all riparian owners have equal rights to their reasonable share of water. No owner can exercise his rights to the detriment of other owners.

Riparian rights inhere in the land and are not affected by use or lack of use. But it can be voided by the process of law.

Riparian rights can also be lost due to upstream adverse use.


If riparian property is sold, the right is automatically transferred to the new owner.

If the property is sub-divided, the portion not adjacent to the stream loses its riparian status unless the rights are specifically preserved in the conveyance.

Riparian rights do not attach to land outside the stream basin, even though this land is contiguous to riparian land in the basin; thus riparian owners can not transport water from the riparian land into the land outside the riparian area.


Brought to the New World by the Spaniards who adopted it from the Roman Civil Law

Profoundly influenced by the developments in the mining industry in USA (California) During the Gold rush of 1849 appropriation of water became a very active system of water rights.

Under the doctrine of appropriation, water is allocated or appropriated on the basis of declared beneficial use on a first come first in rights basis.


Under the exclusive system of appropriative rights, all water in natural water courses is subject to appropriation. And an appropriator may store water in reservoir for use during periods of drought but the amount stored is limited by the terms of storage appropriation.

Usually direct use and storage appropriations are kept separate.

6.3. Water Act 2002 Water Act 2002 was enacted to eliminate the inherent

weaknesses in the previous Water Act Cap 372. These include:

Neglect of water resources management at the expense of water services owing to lack of separation of the sub sectors.

Weak apportionment and allocation practices for water resources.

No centralized coordination of water use among different sectors: energy, industry, environment, agriculture and others

Lack of recognition of the role of communities in water management.

Lack of standards for water services

6.3. Water Act 2002 Cont........ Water Act 2002 was therefore developed upon the following

principles:

State ownership of all surface and groundwater resources. Exploitation of such resources requires authority granted through issuance of a water permit

Stakeholder involvement in management of water resources

Management of water resources on catchment basis and not administrative boundaries

Equitable allocation of water for all Kenyans

Recognition of the economic value of water

Social objectives including supplying the poor with water be achieved by other means including Government subsidy

6.3. Water Act 2002 Cont........

Accelerating supply and distribution of water in rural areas through special funding

Ring-fencing of water service operations

Development of water sector strategies for management and development of the sector

Protection of the quality of water resources

Cost recovery as a means of sustainable service provision

6.4. Issues and challenges

Kenya with a current population of 35 million and a projected population of 43 million by 2015 faces enormous challenges in management of its limited water resources.

The magnitude of the issues and challenges and severity of the water crisis, that currently face Kenya cut across most sectors of the economy making water resources management a high priority requiring urgent attention

6.4.1. Water Scarcity Kenya is classified as a water-scare country. The

natural endowment of renewable freshwater is currently about 21 BCM (billion cubic meters) of 647m3 per capita per annum.

A country is categorized water-scarce if its renewable freshwater potential is less than 1,000 m3

per capita per annum.

6.4.2. Water resources underdeveloped

About 40% the renewable freshwater has potential for development and this represents the safe yield. The remaining 60% are required to sus

water resources engineering

Documents

control of water

water resources development

water resources project

water sources

water policy

water resources problems

quantity of water

fields of wre water