Chapter 9. Reservior Planning and Storage Analysis

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RESERVOIR PLANNING AND STORAGE ANALYSIS

9.1

Reservoir Planning

9.1.1

Site Assessment and Dam Selection

The functional suitability of a site is governed by the balance between its natural physical characteristics and the purpose of the reservoir. Technical suitability on the other hand is dictated by the hydrological and geological or geotechnical characteristics of the catchment and the site. Furthermore, the assessment of anticipated environmental consequences of construction and operation of the dam is equally important. The principal stages involved in site appraisal and optimum selection of site and type of dam follow the steps mentioned in Section 1.5 and a re explained below. The statement of the objectives, the collection of data, the future projections, the project formulation and project evaluation as shown in Fig 9.1 all combined form the feasibility study. This may take 1 to 5 years and would include resource mobilization. The collection of data would include field reconnaissance surveys to collect extensive topographical, geological and hydrological survey data and also mapping of data. Feasibility studies would include options with regard to the location, height and type of dam and comparisons drawn in terms of estimated costs and construction programs. Initial dam site evaluations are then carried out through extensive investigations to confirm that the site can be developed to the desired scale at an acceptable cost. These require careful planning and considerable time and resource investment. After which the type of dam to be constructed is then confirmed. This may take 2 – 4 years. Detailed dam site investigations are then carried out. This is done to determine the engineering parameters to be safely used for foundation stability; to determine seepage patterns and parameters; to confirm the containment integrity of the reservoir basin and stability of its margins; and confirmation of the nature, suitability and availability of natural construction materials. Key features of this phase include: meticulous logging of all natural and excavated exposures and borehole records. This information is used to prepare the detailed design and may take 1-2 years. The next stage is implementation which first involves the procurement of a contractor, after which the contractor can commence; this stage lasts 2-6 years (Novak et al, 1996).

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Statement of Objectives

Collection of Data

Future Projections

Project Formulation

Project Evaluation

Project Implementation Fig 9.1 Steps in Project Development 9.1.2

Site Selection

A good site for a reservoir should have the following characteristics: i) ii)

iii)

iv)

 Large storage st orage capacity : The topography of the site should be such that the reservoir has a large capacity to store water. Suitable site for the dam : There should be good foundation for the dam and the reservoir basin should have a narrow opening in the valley so that the length of the dam is so small. The cost of the dam is often a controlling factor in the selection of a site for the reservoir. Water tightness of the reservoir : The geological conditions of the reservoir site should be such that the reservoir basin is water tight. The reservoir sites having pervious rocks are not suitable. The reservoir basin having shales, slates, gneiss, granite etc are generally suitable. Good hydrological conditions : The hydrological conditions of the river at the reservoir site should be such the runoff is available for storage. The catchment area of the river should give

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v)

vi)

vii)

viii)

ix)

high yield. There should not be heavy losses in the catchment due to evaporation, transpiration and percolation.  Deep reservoir : The site should be such that a deep reservoir is formed in the construction of the dam. A deep reservoir is preferred to a shallow one because in the former the evaporation losses are small, the cost of land acquisition low and the weed growth less. Small-submerged area : The site should be such that the submerged area is minimum. It should not affect the ecology of the region. Monuments of historical and architectural importance should not be submerged.  Low silt inflow: The life of the reservoir is short if the river water at the site has a large quantity of sediments. The reservoir site should be selected such that is excludes the water from those tributaries which carry a high percentage of silt.  No objection minerals: The soil and rock mass at the reservoir site should not contain any objectionable soluble minerals which may contaminate the water. The stored water should be suitable for its intended use.  Low cost of real estate: The cost of real estate for the reservoir site, dam, dwellings, and roads should be low.

9.2

Basic Terms and Definitions

i)

Full Reservoir Level (FRL): The full reservoir level is the highest water level to which the water surface will rise during normal operation conditions. The effective storage of the reservoir is computed up to the full reservoir level. The FRL is equal to the crest of the spillway. In case of dams without spillway gate (Fig. 9.2a, the FRL is equal to the level of the top of the crest. However if the spillway is gated (Fig. 9.2b, the FRL is equal to the level of the top of the gates. The full reservoir level is also called the  full tank level (FTL) or the normal pool level (NPL).  Normal Conservation Level (NCL): It is the highest level at which water is intended to be stored for various uses other than flood. The normal conservation level is different from the FRL as the latter may include a part of the flood. However if there is no storage for the flood up to FRL, the normal conservation level and the FRL FR L become identical.  Maximum Water Level ( MWL)  MWL): The maximum water level is the maximum level to which the water the water surface will rise to when the design flood passes over the spillway. The maximum water level is higher than the full reservoir level so that some surcharge storage is available between the two levels to absorb absorb flood. The maximum water level is also called the the maximum pool level (MPL) or maximum flood level (MFL).  Minimum pool level: The minimum pool level is the lowest level up to which the water is withdrawn from the reservoir under ordinary conditions. The minimum pool level generally corresponds to the elevation of the lowest outlet (or sluice gate) of the dam. However, in the case of a reservoir for hydroelectric power, the minimum pool level is fixed after considering the minimum working head required for the efficient running of the turbines. The storage below the minimum pool level is not useful and is called the dead storage. Useful storage : The volume of water stored between the full reservoir and the minimum pool level is called the useful storage. The useful storage is available for various purposes of the reservoir. In most of the reservoirs, the useful storage is the conservation storage of the reservoir. However in the case of multi purpose reservoirs, in which the flood control is also a designed function, the useful storage is divided into: the conservation storage  for other purposes and the  flood control storage for the flood control in accordance with the adopted plan of the operation of the reservoir. The useful storage is also known as the live storage. Surcharge storage : The surcharge storage is the volume of water stored above the full reservoir level up to the maximum water level. The surcharge storage is an uncontrolled storage, which exists only when the river is in the flood and the floodwater is passing over the spillway. The storage is available only for the absorption of flood it cannot be used for other purposes.

ii)

iii)

iv)

v)

vi)

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vii)

viii)

ix)

x)

xi)

xii)

xiii)  xiv)

 Dead storage: The volume of water held below the minimum pool level is called dead storage. The dead storage is not useful, as it cannot be used for any purpose under ordinary operating conditions.  Bank storage: If the banks of the reservoir are porous, they temporarily store some water when the reservoir is full. The stored water in the banks drains into the reservoir when the water l evel in the reservoir falls. Thus the banks of the reservoirs act like mini reservoirs. The bank storage increases the effective capacity of the reservoir above the capacity indicated by the elevation – storage curve. However, in most of the reservoirs, the bank storage is small because the banks are usually impervious. Valley storage : The volume of water held by the natural river channel in its valley up to the top of its banks before the construction of a reservoir is called valley storage. The valley storage depends upon the cross-section of the river, the length of the river and its water level. The net increase in the storage capacity after the construction of a reservoir is equal to the total capacity of the reservoir up to FRL minus the valley storage. However, this distinction between the net storage capacity and the total storage is not of much significance in a conservation or storage reservoir where the main concern is the total water available for different purposes. But in the case of a flood control reservoir, the difference between the net storage capacity and the total storage capacity is important because the effective storage for flood control is reduced due to the valley storage. The effective storage is equal to the sum of the useful storage and the surcharge storage minus the valley storage in the case of a flood control reservoir. Yield from a reservoir : Yield is the volume of water, which can be withdrawn from a reservoir in a specified period of time. The time period for the estimation of yield yield is selected according to the size of the reservoir. It may be a day for a small reservoir and a month or a year for a large reservoir. The yield is expressed as Mm3 / year for the large reservoirs. Safe Yield (Firm Yield) : Safe yield is the maximum quantity of water, which can be supplied from a reservoir in a specified period of time during a critical dry year. The lowest recorded natural flow of the river for a number of years is taken as the critical dry period for determining the safe yield. Secondary Yield : Secondary Yield is the quantity of water, which is available during the period of high flow in the rivers when the yield is more than the safe yield. There is guarantee to supply the secondary yield. It is supplied on as and when basis at the lower rates.  Average Yield : The average yield is the arithmetic average of the firm yield and the secondary yield over a long period of time.  Design Yield : The design yield is the yield adopted in the design of a reservoir. The design yield is usually fixed after considering the urgency of the water needs and the amount of water involved. The design yield should be such that the demands of the consumers are met me t with, and at the same time, the storage required is not unduly large (Arora, 2007).

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Below is an Illustration of the basic definitions:

M.W.L Surcharge Storage

Crest

F.R.L Useful Storage

Outlet

Minimum Pool Level

Reservoir

Dead Storage

Dam

Fig 9.2a Dam without a spillway gate

M.W.L Surcharge Storage F.R.L Useful Storage

Outlet

Spillway Gates (Removed during flood)

Minimum Pool Level

Reservoir

Dead Storage

Dam

Fig 9.2b Dam with spillway gates

9.3

Investigations for Reservoir Planning

The following investigations are usually conducted for reservoir planning; engineering surveys, geological investigations and hydrological investigations.

9.3.1

Engineering Engineer ing Surveys

Engineering surveys are conducted for the dam, the reservoir and other associated works. Generally, the topographic survey of the area is carried out and the contour plan is prepared. The horizontal control is usually provided by triangulation survey and the vertical control by precise levelling.

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a)  Dam Site.  For the area in the vicinity of the dam site, a very accurate triangulation survey is conducted. A contour plan to a scale of 1/250 or 1/500 is usually prepared. The contour interval is usually 1m or 2m. The contour plan should cover an area at least up to 200m upstream and 400m downstream and for adequate width beyond the two abutments. b)  Reservoir . For the reservoir, the scale of the contour plan is usually1/15,000 with a contour interval of 2m to 3m, depending upon the size of the reservoir. The area-elevation and storage –elevation curves are prepared for different elevations up to an elevation of 3 to 5m higher than the anticipated maximum water level (MWL). 9.3.2

Geological Investigations.

Geological investigations for the dam and reservoir are done for the following purposes i) Suitability of foundation for the dam; The type and height of the dam mainly depend upon the type of foundation. Subsurface explorations are carried out to determine the depth of the overburden to be removed for laying the foundation of the dam, the type of rock, the nature and extent of the fault zones, if any, present in the rock.

Depending upon the location of the bedrock, the following methods of sub-surface explorations are used. a) b) c) d)

Excavation of open pits or trenches of suitable size Drifting (or tunnelling) into the sides of the valley Driving vertical inspection shafts into the rock Core drilling to obtain the samples and to determine the configuration of the strata

If the hard rock lies far below the surface, the dam site may be suitable for a low gravity dam or an earth dam. The dam in that case may have to be founded on soil foundations. Sub-surface investigations are done to determine the type of soil, the properties of soil, the soil profile, and the location of the water table The information obtained from the geological investigations is used for determining a suitable programme of foundation treatment and grouting, if necessary ii)

Water tightness of the reservoir basin The reservoir basin should be watertight so that the stored water is not wasted due to seepage through the bed and banks; otherwise the very purpose of constructing the reservoir would be defeated. Geological investigations are conducted to detect the presence of cavernous rock formations, which have cavities and are porous. The stored water may escape through such cavities into adjacent valleys. If such formations exist in small areas, they may be treated and made watertight. However, if they are widespread, the site may have to be abandoned.

iii)  Location of quarry sites for the construction materials Large quantities of construction materials such as stones, aggregates, sand, soil, rockfill, etc. are required for the construction of a dam. Geological investigations are conducted for location of suitable quarries for stones and burrow areas for soils. The quantity and quality of the available construction materials are ascertained.

9.3.3

Hydrological Investigations.

Hydrological investigations are conducted for the following purposes: i). To study the runoff pattern and yield

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The most important aspect of reservoir planning is to estimate the quantity of water likely to be available in the river from year to year and season to season. For determination of the required storage capacity of a reservoir, the runoff pattern of the river at the dam site is required. If the stream gauging has been done for a number of years before the construction of the dam, the runoff pattern will be available from the record. It is generally assumed that the runoff pattern will be substantially the same in the future as well. The available record is used for estimating the storage capacity. The inflow hydrographs of two or three consecutive dry years when the discharge is low are frequently used for estimating the required capacity. However, if the stream gauging records are not available, the runoff and yield have to be estimated indirectly by other methods. ii). To determine the flood discharge at the site. The spillway capacity of the dam is determined from the inflow hydrograph for the worst flood when the discharge in the river is at maximum. Flood routing is done to estimate the maximum outflow and the maximum water level reached during the worst flood. The methods for estimating the maximum flood discharge are discussed under flood routing in C hapter 11.

9.3.4

Hydropower, Environmental and Social Safeguards

Investment in hydropower has high upfront capital costs and has lengthy lead times for planning, permitting and construction. However, on life cycle costing hydropower has the best performance compared to other sources of energy with running cost as low as 1% of the initial capital investment. On the medium and long term, hydropower is a cheaper way to produce electricity. With regard to climate change, hydropower tends to have a very low greenhouse gas (GHG) footprint (Altinbilek et al, 2007). Dams are designed to manipulate the flows of rivers and as a result they impact indirectly on the downstream river ecosystem by potentially affecting every part of the flow, sediment, thermal and water quality regimes. They can also impact the ecosystem directly by blocking fish passage. These are explained as follows: i) Low flows: dams may store low flows during the wet season for release downstream in the dry season. As a result the seasonal pattern of low flows may be partially or wholly reversed, eradicating conditions needed for life cycles of aquatic plants and insects to reach completion. In some rivers dry-season low flows are periodically completely eradicated by damming or direct abstraction; ii) Intra-annual floods; small and medium floods may be completely stored in reservoirs and they are known to sort riverbed sediments, help maintain and control the spread of marginal vegetation, provide depth of water for fish migrationsand enhance water quality during dry months; iii) Inter annual floods are known to maintain riparian belts , scour channels and river beds and eradicate patches of inchannel and bank vegetation enhancing diversityas new growth appears. The magnitude and frequency of these floods will be reduced, the larger floods spill over, iv) Sediment regime and the ecosystem, dams trap sediments passing down the river as well as altering flows, which may lead to degradation ( loss of sediments) or aggradation ( accumulation of sediments) depending on whether the remaining floods can move the remaining incoming sediments (Basson, 2004). Any investment in a water resources project and in particular hydropower, must comply with the environmental and social safeguard policies of the specific Government and the Development Partner concerned. Usually they are quite similar and are in the form of an environmental assessment. This can take the form of an environmental audit, an environmental scan, environmental review or a comprehensive environmental impact assessment (EIA). Some can be very complex and require in depth study of certain aspects of the environment. These are usually required for larger investment projects. Others may not be so in depth and usually apply to the smaller projects, where the impacts are known and can be readily addressed (MEMD, 2007). As mentioned in Section 1.3, in Uganda, the National Environmental Act 1995, states that EIAs are required and the EIA Regulations specify the types of projects subject to an EIA. The World Bank has a set of ten environmental and social safeguards policies for which all World Bank financed projects must be in compliance. These are; i) Environmental

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Assessment ii) Forestry iii) Natural Habitats, iv) Pest management, v) Involuntary Resettlement, vi) Indigenous Persons vii) Cultural Property, viii) Dam Safety ix) International Waterways and x) Disputed Areas. All are applicable to hydropower except for Forestry and Disputed Areas. Table 9.1 provides the potential impacts, consequences and mitigation of a hydropower project.

Table 9.1 Potential Impacts, Consequences and Mitigation of Hydropower Projects

Potential Impact Habitat Loss

Consequences Important downstream habitats may rely on regular water flows and minimum flows for their maintenance.

Loss of water to downstream users

Loss of Domestic supply and Irrigation water. Downstream users may have to relocate.

Water and Soil contamination

Lubricating oils and Greases contaminate downstream domestic water supplies and affect downstream aquatic communities Loss of life limb and injuries requiring medical attention. Loss of productivity

Personal injury as a result of accessible equipment

Mitigation Downstream surveys are required to determine the water needs of habitats and important ecosystems. Avoid schemes on water courses with significant downstream flora and fauna Schemes should be established only where downstream effects are negligible or where water is quickly returned to the water course Ensure proper handling and storage of lubricants and storage and disposal of used lubricants.

Remarks Natural habitats

Involuntary resettlement.

Ensure site and all equipment is well shielded from human interference. Safety warnings community workshops and signing

Source: (MEMD, 2007)

9.3.5

The Development of Large Dams

There has considerable discussion recently about the impact of large dams on society. To this effect, the World Commission on Dams (WCD) was established as in 1998 on the recommendation of a workshop involving participants from governments, private sector, international financial institutions, civil society and dam affected people held in Switzerland. It was an independent institution and comprised of 68 members derived from a cross section of views and interests. The workshop was convened by International Union for the Conservation of Nature (IUCN) with support from the World Bank to address growing conflicts throughout the World regarding the development of large dams. The mandate was to: i) review the development effectiveness of large dams and assess alternatives for water resources and

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energy development and ii) develop internationally acceptable criteria guidelines and standards for the planning, design, appraisal, construction, operation, monitoring and decommissioning of dams. After more than two years of consultations with stakeholders in favour and against large dams, the Commission concluded the following: i.

Dams have made an important and significant contribution to human development and the benefits derived from them have been considerable.

ii.

In many cases, an unacceptable and often unnecessary price has been secured to secure dam benefits, especially in social and environmental terms, by people displaced, by communities, by taxpayers and by the natural environment.

iii.

The lack of equity in the distribution of benefits has brought into question the value of many dams in meeting water and energy development needs, when compared to other alternatives.

iv.

By bringing together all those with rights involved and those who bear the risks associated with different options for water and energy resources development, the conditions for a positive resolution for competing interests and conflicts are c reated.

v.

Negotiating outcomes will greatly improve the development effectiveness of water and energy projects by eliminating unfavourable projects at an early stage and by offering as a choice only those options that the key stakeholders agree represent their needs.

vi.

In future there will be growing competition for water to meet the demands for agriculture, industry, and drinking water and that as demand increases, loss of water by evaporation in the dry climates will exacerbate the problem. Therefore decisions on large dams call for rethinking the management of freshwater resources, so as to minimize dramatic impacts of withdrawing water from lakes, rivers and aquifers that adversely affect the continued supply of water.

WCD recommended the following: i.

Five core values of equity, sustainability, efficiency, participatory decision making and accountability are applied to all future dam developments. In order to achieve this, development choices and decisions on dams and development should be based on public choice guided by policy and address a wide range of needs.

ii.

During planning and developing large dams, a rights and risks approach should be applied that comprises gaining public acceptance, comprehensive options assessment, addressing existing dams, sustaining rivers and livelihoods, recognizing entitlements and sharing benefits, ensuring compliance and sharing rivers for peace and development.

iii.

Establish criteria and guidelines for good practices related to strategic options, life cycle, environmental flows, risk analysis and integrity packs (WCD, 2000).

In Uganda, significant progress has been made through the Uganda Dams Dialogue (UDD), which started in 2004, to contextualize the WCD recommendations by making them serve as a reference rather than being applied in totality (UDD, 2006). This approach is partially being operationalised through the development of a Multi Objective Decision Support System (MODSS), which models the evolving relationships between decision makers, negotiators, stakeholders and their values, operational goals, control decisions, criteria and group pReferences. Here stakeholders structure their management objectives and performance indicators of some preferred management alternative using value trees (Zaake, 2007).

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9.4

Reservoir Storage Capacity

For simple determination of the storage capacity of a reservoir, several different methods can be used; the i) Mass Curve Method ii) Analytical Method iii) Sequent Peak Algorithm and iv) Flow duration curve method and v) Pondage. A sixth method, Behaviour Analysis takes into account the behaviour of the reservoir during operation. The above methods are commonly used in design practice. In a study on the applicability of twelve storage estimation methods, to determine the least bias and standard error, based on 3000 years of synthetic streamflow data, it was noted that the modified Gould and Behavior Analysis had the least standard error and are recommended for use together with simulation methods for investigating storage requirements for Irish and similar climatic conditions (Carty and C unnane, 1997).

9.4.1

Mass Curve Method

 Mass Inflow Curve : The mass inflow curve is used in the determination of the reservoir capacity. A mass inflow curve, also called a mass curve, is a plot between accumulated inflow volumes against time. A mass inflow curve is prepared from the inflow hydrograph of a river for a large number of consecutive years.

Fig 9.3 shows the hydrograph of a river at a particular site from the year 1980 to 1992. The discharge ordinates represent the annual discharge (average discharge of the full year obtained from the total, annual volume of stream flow by dividing it by (365 x 24 x 3600). The area under the hydrograph from the starting year 1980 to 1981 represents the volume of water in cumecs that has flowed through the river in that year. Similarly the areas under the hydrograph in other years represent the volume of water in corresponding years. Thus the total area of the hydrograph from 1982 is equal to the total volume from 1980 to 1982 and so on. 14000

250

12000 200

10000 150

8000 6000

100

A1

4000

50

2000 A1 0 1980

0 1981

1982

1983

1984

1985

1986

Year

Fig 9.3 A Flow Hydrograph

1980

1981

1982

1983

1984

1985

1986

Year

Fig 9.4 A Mass Curve

Fig 9.4 shows the mass inflow curve. The ordinate of the curve at the year 1980 is zero and that at the year 1981 is equal to the volume of water flowed from the year 1980 to 1982. The ordinate of the curve at the year 1982 represents the total volume of water year from the year 1980 to 1982. Likewise the other ordinates of the mass are found and plotted. The mass curve is a continuously rising curve as it shows the accumulated volume against time. If there is no flow during a certain period, the mass curve can be horizontal but it can never fall. On the other hand the mass curve can rise steeply when there is a large

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inflow. The slope of the curve at any point indicates the rate of inflow at that time. Relatively dry periods are indicated as depressions with concavity upwards, whereas the periods of high discharge are indicated as crests with convexity upwards.  Mass Demand Curve A mass demand curve is a plot of demand rate against time. The mass demand curve is determined from the demand curve in the same manner as the mass inflow curve is determined from the hydrograph.

14000

250

12000

   ) 200    s    /    3    m    ( 150    e    g    r    a 100    h    c    s    i    D 50

10000 8000 6000

A1

4000 2000

A1

0 1980 1981 1982 1983 1984 1985 1986

0 1980

1981

1982

1983

1984

1985

1986

Year Year

Fig 9.5 Uniform Demand

Fig 9.6 A Mass Demand Curve

When the demand is uniform, the demand curve is a horizontal line (Fig 9.5) and the corresponding mass demand curve is a straight line, having a slope equal to the demand rate (Fig 9.6) On the other hand, if the demand is variable, the mass demand curve is obtained from the demand curve after finding out the area of the demand curve for consecutive years, as in the c ase of a mass inflow curve. The assumptions for the mass curve method and others based on it, like the analytical method, the sequent peak algorithm are as follows: i) If N years of data are available, the inflows and demands are assumed to repeat themselves in a cyclic progression of N year cycles, which also implies that the future flows will not contain a more severe drought than historical data. ii) The reservoir is assumed to be full at the beginning of the dry period. Thus while using the mass curve method, the beginning of the dry period should be noted and the minimum storage required to pass each drought period should be calculated. Procedure Obtain the adjusted inflow Q a.  Obtain the rate of Demand, D.  

Calculate the Cumulative Qa (ΣQa)



Calculate the Cumulative D (ΣD) Plot a mass (inflow) curve using the calculated values Plot a mass demand curve (line) corresponding to the given rate of demand superimposing it with the mass inflow curve. Draw the lines AB, FG, etc such that they are parallel to the mass demand curve and tangential to the crests, A, F, etc of the mass curve. The points A, F, etc indicate the beginning of dry periods marked by depressions as shown in Fig 9.7.

  

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 

Determine the vertical intercepts CD, HJ, etc, between the tangential lines and the mass inflow curve. These intercepts indicate the volumes by which the inflow volumes fall short of demand. Determine the largest of the vertical intercepts obtained above. The largest vertical intercept represents the storage capacity required. The vertical distance such as FL between successive tangents represents the volume of water spilled over the spillway of the dam.

∑Q

Time (Years) Fig 9.7 Mass Curve Method

Example 9.1 Twenty-five years of mean monthly flows of River Moroto are recorded in the table below. A monthly seepage loss of 0.2m3 /s is estimated. Determine the required storage capacity if the demand is 80% of the mean monthly flow and a compensation flow of 0.3m3 /s must be maintained in the river. Month Q(m3 /s)

J 7.5

F 8.7

M 6.8

A 3.5

M 2.7

J 2.3

J 1.7

A 2.0

S 4.7

O 8.7

N 9.6

D 7.8

Solution: 3

Seepage loss,

S = 0.2 m  /s

Compensation,

C = 0.3 m  /s

Adjusted flow, Demand D=80% of

3

Qa = Q - (S+C) = Q – 0.5 m3 /s Mean Monthly Flow = 66

 x 80% = 4.4 m 3 / s  12

The flow Q, adjusted flow Q a, the demand D, the cumulative adjusted flow Σ Q a and cumulative demand Σ D are all tabulated in Table 9.2, after which the mass curve is drawn as shown in Fig 9.8.and the value of the intercept is read off.

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Table 9.2 Mass Curve Calculations 3

Month J F M A M J J A S O N D

Q (m  /s) 7.5 8.7 6.8 3.5 2.7 2.3 1.7 2 4.7 8.7 9.6 7.8

3

Qa (m  /s) 7 8.2 6.3 3 2.2 1.8 1.2 1.5 4.2 8.2 9.1 7.3

D 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4 4.4

∑ Qa

∑D

7 15.2 21.5 24.5 26.7 28.5 29.7 31.2 35.4 43.6 52.7 60

4.4 8.8 13.2 17.6 22 26.4 30.8 35.2 39.6 44 48.4 52.8

Mass Curve 70 60 50 storage 40

  s    /    3   m

30 20 10 0 0

2

4

8

6

10

12

14

Months mass flow curve mass demand curve

From the mass curve plotted, 3 3 6 3 Required storage capacity = 45 – 33 = 12 m  /s = 12 x 30 x 24 x 3600 m  = 31.1 x 10  m .

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9.4.2

Analytical Method

This is one of the techniques based on the mass curve method (Rugumayo and Kizza, 2001) and is similar to the sequent peak algorithm method except that in this case the number runs is equal to the number of years of data available, whereas in the latter case, the number of runs is twice the number of years of data available. The capacity of the reservoir is determined from the net inflow and demand and storage is required when the demand for that period exceeds the net inflow. Procedure The stream flow data at the reservoir site is obtained, Q (monthly or annual low rate)  Determine the direct precipitation volume falling on the reservoir during the month (P)  Estimate the evaporation losses, which would occur from the reservoir (E)  Determine the compensation required during period(C)  Ascertain the demand during various months (D)  Calculate the adjusted inflow during each month (Qa)  Qa = Q + P- E - C Compute storage capacity for each month (S)  S = Qa - D Obtain Cumulative (Q a – D)  Obtain the maxima and minima for each cycle  Note  The maxima represents a full reservoir  The minima represents the minimum reservoir level (maximum depletion)  A cycle is the period from when the reservoir is at maximum level to the next time it returns to maximum or from when it is at the minimum level to the next time it is at minimum again To obtain the amount of water that should be stored in any cycle, obtain the difference between the  maxima and the subsequent minima .i.e, (9.1) Storage = Max (Σ(Q-D)) - Min (Σ(Q – D)), in a cycle.  

The maximum value of the storage values gives the necessary capacity of the reservoir To obtain the amount of spill i, add the storage of the cycle to the minimum (to fill the reservoir) and subtract from the subsequent maxima, or obtain the maxima of a cycle and subtract the sum of the storage and the minima. (9.2) Spill = Max ( Σ(Q-D)) -[ Storage + Min ( Σ(Q-D))]

Example 9.2 Same Question as in Example 9.1, but use the Analytical Method. Solution: In this case the first four columns are the same as for the mass curve; iethe month, the flow Q, the adjusted flow Q a and the demand D. This is followed by the fifth column, where there is the difference between the adjusted flow and the demand ( Qa-D) and then the cumulative values ∑(Qa-D) in the sixth column. In the seventh column inflexion points are tabulated from, which the storage can be estimated.

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Table 9.3 Analytical Method Calculations Month

Q (m3 /s)

Qa (m3 /s)

D(m3 /s)

Qa-D(m3 /s)

∑(Qa-D) (m  /s)

Inflexion pts 3 (m  /s) 2.6

3

J

7.5

7.0

4.4

2.6

2.6

F

8.7

8.2

4.4

3.8

6.4

M

6.8

6.3

4.4

1.9

8.3

A

3.5

3.0

4.4

-1.4

6.9

M

2.7

2.2

4.4

-2.2

4.7

J

2.3

1.8

4.4

-2.6

2.1

J

1.7

1.2

4.4

-3.2

-1.1

A

2.0

1.5

4.4

-2.9

-4.0

S

4.7

4.2

4.4

-0.2

-4.2

O

8.7

8.2

4.4

3.8

-0.4

N

9.6

9.1

4.4

4.7

4.3

D

7.8

7.3

4.4

2.9

7.2

Storage (m3 /s)

8.3

5.7

-4.2

12.5

7.2

11.4

∑Q = 66 The maxima is 8.3, while the minima is -4.2 . Taking the difference between these values gives: 3

6

3

Storage = (8.3 – (- 4.2)) = 12.5m /s = 12.5 x 30 x 24 x 3600 = 32.4 x 10  m .

9.4.3 Sequent Peak Algorithm Method As mentioned in 9.4.2, this method is similar to the analytical method except that in this c ase, the analysis is carried out for a period of 2N, where N could be months or years or any other period (Thomas 1963). Let the data be available for  N   consecutive periods not necessarily of uniform length. These periods can be year, month, day or hours depending upon, the problem. In the ith period let xi = inflow volume and  Di = demand volume. The surplus or deficit of storage in that period is the net—flow volume given by. Net-flow volume = Inflow volume - Outflow volume (9.3) = xi - Di In the sequent peak algorithm, a mass curve of cumulative net-flow volume against chronological time is used. This curve in Fig. 9.9.), will have inflexion points; peaks (local maximums) and troughs (local minimums). For any peak P, the next following peak of magnitude greater than P is called a sequent peak. Using two cycles of N periods; where  N is the number of periods of the data series, the required storage volume is calculated by the following procedure: . – 1. Calculate the cumulative net-flow volumes viz. t 

∑ ( X  − D )  for t = 1, 2, 3 ……., 2N i

i

(9.4)

2. Locate the first peak P1 and the sequent peak P2 which is the next peak of greater magnitude than P 1 (Fig. 9.9)

15

HWRE AR 2010

3. Find the lowest trough T 1 between P1 and P2 and calculate (P1 —T 1). 4. Starting with P2 find the next sequent peak P3 and the lowest through T 2 and calculate (P2 —T 2). 5. Repeat the procedure for all the sequent peaks available in the 2N periods. j.e. determine the sequent peak P j, the corresponding T j) and the jth storage. (P j — T  j) for all j values. 6. The required reservoir storage capacity is S = maximum of (P j —T j) values. (9.5) (Subramanya, 2001)

Base length = 2N years Where N = No. of years of record

Sequent peak, P3 Sequent peak, P2

   )   e   v    i    t    i   s   o    P    (

   )    i    D      i    X    (        Σ

 ,    e    m   u    l    o    V   w    o    l    F      t    e    N    e   v    i    t    a    l   u    m   u    C

First peak, P1

   s    t    i    n   u    e    m   u    l    o    V

Lowest trough, T2

Time (months) Lowest trough, T1    )   e   v    i    t   a   g   e    N    (

Fig 9.7 The Sequent Peak Algorithm Method Example 9.3. Same Question as in Example 9.1, but use the Sequent Peak Algorithm Method All the seven columns are the same as in the analytical method, however the data set is repeated as shown in Table 9.4.

16

HWRE AR 2010

Table 9.4 Sequent Peak Algorithm method

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As can be seen, P1 – T1 = P2 – T2 = 12.5 m  /s, which is the same value as obtained in the 6 3 Analytical Method, Example 9.3. The storage therefore is 12.5 x 30 x 24 x 3600 = 32.4 x 10  m 9.4.4

Flow Duration Curve Method

A Flow Duration Curve is a plot between the discharge as the ordinate and the percentage of time that the discharge is exceeded as abscissa. The data used could be daily weekly, ten daily or monthly values. Before the Flow Duration Curve is drawn, the streamflow data has to be collected. We can assume that the average daily flow data are available for one year (365 days) for, which the flow duration curve is to be drawn The following procedure can then be used for plotting the flow duration curve from the daily flow: The streamflow data is then arranged in a descending order of discharges, using class intervals especially since the number of individual values is very large. If n  is the number of data points are used in the listing, the plotting position of any discharge (or class value) Q is

17

HWRE AR 2010

P = (m/n) x 100%

(9.6)

where m is the order number of the discharge or class value and P is the percentage probability of the flow magnitude being equaled or exceeded. The plot of discharge against Q and P is the flow duration curve. Arithmetic scale paper, semi-log or log-log paper is used depending upon the range of data and use of the plot. The storage can be estimated and will be equal to the area between the demand line and the flow duration curve, and below the demand line as shown in Fig 9.10. Flow duration curves are used widely in water resources planning as follows: i. Estimating dependable flows in water resources engineering projects. ii. Evaluating characteristics of the hydropower potential of a ri ver. iii. In the design of drainage systems iv. In flood control v. In estimating sediment load and dissolved solids of a stream vi. In the extension of streamflow data to adjacent catchments, Example 9.4 Same as Question in Example 9.1, but use the Flow Duration Curve Solution: The adjusted flow values are ranked in descending order in column 1 of Table 9.5. The number of times this amount is equaled or exceeded is given in column 2. Column 3 gives the percentage time exceeded. Table 9.5 The Flow Duration Curve Adjusted mean monthly flows (in descending order) ↓Qa (m3 /s)

Number of times equaled or exceeded ( m)

Percentage of time P = ( m n  X  1 0 0 )

9.1

1

8.3

8.2

3

25.0

8.2

3

25.0

7.3

4

33.3

7.0

5

41.7

6.3

6

50.0

4.2

7

58.3

3.0

8

66.7

2.2

9

75.0

1.8

10

83.3

1.5

11

91.7

1.2

12

100.0

From the Flow Duration Curve plotted as shown in Fig 9.10 The storage capacity required = Hatched area A

18

HWRE AR 2010

Area A is approximately

(1 00 − 5 7 ) x (4 .4 − 0 .2 ) = 1 2  x 10 0

= 0.93 m 3 / s / month   = 0.93 x12 = 11.2 m 3 / s  = 11.2 x 30 x 24 x 3600 = 29.0 3 x10 6 m 3 From the example above, it can be clearly seen that though the value of the answers obtained are close to each other, the flow duration curve method gives a lower value (compared to the values obtained using the mass curve, analytical and sequent peak algorithm methods.

Flow Duration Curve 10 9 8 7    )   s    /    3   m    (   a    Q  ,   e    t   a   r   w   o    l    F

6 5 4 3 2 1 0 0

20

40

60

80

10 0

Percentange o f t im e e qu al ed or e xc ee de d

Fig 9.8 The Flow Duration Curve

19

HWRE AR 2010

Example 9.5 Determine the Flow Duration Curve for the data provided in the following table. Month

Year 1

Year 2

Year 3

J

110

180

193

F

102

118

109

M

97

88

97

A

84

79

91

M

70

56

82

J

62

52

74

J

45

47

68

A

67

35

43

S

82

60

30

O

134

75

48

N

205

D

142

98

49

127

66

Solution Flow Duration Curve The data is arranged in descending order, divided into class intervals and the midpoint of each class interval determined. The frequency of each class interval is determined from, which m  is obtained as shown in the Table below. The flow duration curve is then plotted as shown in Fig 9.11.

Class interval

Mid point

Frequency

Cumulative frequency (m)

Percent of time P = (m/n)x100%

201 – 210

205

1

1

2.8

171 – 200

185

2

3

8.3

141 – 170

155

1

4

11

101 – 140

120

6

10

27.8

71 – 100

85

11

21

58.3

41 – 70

55

13

34

94.4

0 – 40

20

2

36

100.0

n = 36

20

HWRE AR 2010

250

200

   )   s    /    3   m 150    (   e    t   a    R 100   w   o    l    F 50

0 0

20

40

60

80

100

Perc entage of time exc eeded or equalled

Fig 9.9 Flow Duration Curve Example 9.6 A dam is to be constructed in the River Sezibwa Catchment to store water for a proposed irrigation scheme. The mean daily flows for the river have been measured for the critical year and are summarised in the Table 9.6. A compensation flow must be maintained in the river and hence no water can be abstracted until the flow exceeds 15% of the average daily flow. In addition it is considered uneconomical to abstract water at a rate greater than three times the average daily flow. Determine the volume of water that could be abstracted from the river per year. Table 9.6 Daily Flows Mean Daily flow range

Number of days

0.10 – 0.99

144

1.00 – 1.99

63

2.00 – 2.99

60

3.00 –3.99

29

4.00 – 7.99

32

8.00 –11.99

18

12.00 –1 5.99

8

16.00 – 23.99

7

24.00 -= 35.99

2

36.00 – 48.00

2

Solution: The Flow Duration Curve Method can be used to solve this problem, because we are given frequencies for particular flow ranges.

21

HWRE AR 2010

First, we determine the average daily flow and then construct a flow duration curve from which the required volume of water is estimated. Column 2 in Table 9.7 gives the mid value, which corresponds to the range of the flow values for the frequency interval. Column 3 shows the number of days and the fourth column is the mid value multiplied by the frequency in days and the number of seconds in a day to give the volume for the number of days given. Table 9.8 is then developed based on the flows and their frequencies. Table 9.7 Flow Frequency and Volumes Mid value-q (m 3 /s)

Frequency (f)(days)

Volume = qf x 86400 3 6 (m x 10 )

0.10 – 0.99

0.545

144

6.781

1.00 – 1.99

1.495

63

8.138

2.00 – 2.99

2.495

60

12.934

3.00 –3.99

3.495

29

8.757

4.00 – 7.99

5.995

32

16.575

8.00 –11.99

9.995

18

15.544

12.00 –1 5.99

13.995

8

9.673

16.00 – 23.99

19.995

7

12.093

24.00 – 35.99

29.995

2

5.183

36.00 – 48.00

42.00

2

7.258

365

102.936

Mean daily flow range 3 (m  /s)

6

3

The volume discharged from the river in one year is 102.936 x 10  m .

Average daily flow

=

102.936 × 10 6 m 3 60 × 60 × 24 × 365s

= 3.26m 3 / s ∴Compensation flow

= 115

Maximum rate of extraction

100

 x 3 .2 6

= 3 .7 5 m 3 /s 3

= 3 x 3.26 = 9.78 m  /s

22

HWRE AR 2010

Table 9.8 Flows and Frequencies 3

Flow (m  /s) (Mid value)

Days

Cumulative days (m)

Percent of time p=m/nx100%

42

2

2

0.55

30

2

4

1.10

20

7

11

3.01

14

8

19

5.21

10

18

37

10.14

6

32

69

18.90

3.5

29

98

26.85

2.5

60

158

43.29

1.5

63

221

60.55

0.55

144

n =365

100.00

45

3500 3000 2500    )    W2000    K    (    r    e   w    o 1500    P

   )    s    /

40

3139.2

35

2746.8

30

2354.4

   3

25    m    (    Q  ,   w 20    o    l    F

1962 1569.6

15

1177.2

10

784.8

1000 500

5

0

0

A

0

B

10

392.4

20

30

40

50

60

70

80

90

100

Percentage of time equalled or exceeded

Fig 9.10 A Flow Duration Curve and Power Duration Curve The volume of water that could be abstracted from the river can be obtained from the area of the hatched area.

23

HWRE AR 2010

Hatched area

= Area A + Area B

=

(9.78 − 3.75) x (10 − 0)

100 = (0.603 + 0.4 82)

(9.78 − 3.75) x (26 − 10) + 1 2 x 100

= 1.0854 m 3 / s 

∴Volume of water that could be abstracted in a year = 1.085 m 3 /s x 60 x 60 x 24 x 365 s 6

3

= 34.22 x 10  m , which is about a third of the available flow. A Power Duration Curve  can be plotted by making the power as the ordinate and the percentage of time a particular power amount is equaled or exceeded as the abscissa. Since the power generated is a product of the discharge and head, a flow duration curve can be converted into a power duration curve by changing the ordinate from discharge into power if the head remains constant according to the equation: P= ρQgHη (9.7) P= Q(ρgHη) 3 2 Where Q is the flow, ρ is the density of water (1000kg/m ), g is the acceleration due to gravity (9.81m/s ), H is the constant head available (assumed 10m) and η  is the efficiency of the turbines and power generating equipment (assumed 80%). The bracketed portion would be constant and the corresponding values of power are shown on the additional scale of ordinates on the left of Fig 9.12. Like in the case of the flow duration curve, the area under the entire curve represents the average yield of the river, the area under the corresponding power duration curve gives the average yield of power from the power plant (Arora, 2007).

9.4.4.1 Some Basic Power Terms

In a hydropower plants, the electrical energy cannot be economically stored on a large scale. The generation of power depends upon the electrical use or the demand for power which keeps on varying with time during different hours of the day. The  Base Load is the energy required to be supplied continuously for most of the time for which the electrical generator has to be run. The pattern of energy or load variation with time is called  Load Curve. From the load curve it may be noted that at a certain hour of the day, the demand or the requirement for power reaches the maximum or peak  value. The Peak Load is generally defined as that part of the load, which is carried out at rate greater than 1.33 times the average load so as to supply the fluctuation portion of the load or energy requirement. The Average Load  is taken as the mean of the load for the period considered. The Load Factor  is the ratio of the average load to the peak load over a given period. Depending upon the mode of variation of load, the load factor maybe calculated daily, weekly, monthly or yearly. Since the area under the load curve represents the energy consumed in kilowatt hours, the load factor may also be defined as the ratio of the energy consumed to the peak demand, if assumed to last for 24 hours of the day. Load Factor =

Energy consumed in 24 hours

24

HWRE AR 2010

Peak demand or load x 24 hours

(9.8)

It should be noted that, while the peak load determines the capacity of the generating units, the load factor gives an idea of the degree of utilization of the capacity. Thus, a load factor of 70 per cent would mean that the electrical machines are producing only 70 per cent of their maximum production capacity. The Capacity Factor   is a measure of the plant use. It is defined as the ratio of the average load to the installed capacity of the plant. For instance, a plant with an installed capacity of 60,000 KW and producing on output of 4.5 x l0 KWH while working for a period of 150 hours would have capacity factor = 4.5 x 106 = 0.5 60,000 x150. If the plant were so operating that 60000x150 the peak load becomes equal to the plant installed capacity, load factor is identical to the capacity factor. The Utilization Factor   is a measure of plant use as affected by the water supply. More specifically, it is the ratio of the quantity of water actually utilized for power generation to that available from river source. When there is sufficient water to run the plant at capacity, the utilization factor is the same as the capacity factor. However, depending upon the shortage of water supply, output may be curtailed, which may either decrease or increase the utilization factor, according to the plant load factor. It may, in practice, vary from 0.40 to 0.90. The Gross Head in a hydroelectric power plant is the total difference in elevation between maximum surface in the reservoir at the dam and the water level in t he river where the tail race is to be located. The  Net Head (also called effective head) is the head available for energy production after deducting losses in friction, entrance and unrecovered velocity head in the draft tube. The  Firm Power (also called  primary power) is the maximum annual rate at which energy in a hydroelectric plant can be generated without interruption. This is a completely dependable power and corresponds to the minimum streamflow available for all times. Firm power is thus available to consumers for all the twenty four hours of the day. The Secondary Power is the surplus or non-firm power which is available intermittently at off-peak hours. Its use is primarily to relieve the load of overworked power plants in an interconnected grid system and thereby ensure some economy or efficiency to the system.(Nagpal, 2000, Duggal, Soni, 2007) 9.4.5

Pondage

While storage refers to large reservoirs to take care of monthly or seasonal fluctuations in the river flow,  pondage usually refers to the small storage at the back of a weir, in run-of river plants, for temporarily storing water during non-working hours, idle days and low load periods for use during hours of peak load demand. These fluctuations may arise because of sudden change in the load demand on the turbines or on the natural inflows of the year. The sudden increase in load demand would necessitate an instantaneous increase in the inflows to the turbines, which can be met by stored water available in the pond created for that purpose, or the excess natural inflows can be temporarily stored as pondage to cater for increased water requirements of peak load demand. Run-of-river plants are feasible for streams which have a minimum dry weather flow or receive flow as regulated by any storage reservoir upstream (Raghunath, 2005, Duggal and Soni, 2007).

25

HWRE AR 2010

The pondage factor is the ratio of the total inflow hours in a given period (say a week) to the total number of hours of working of the power plant in that period (a week). For example, assuming constant stream flow, if a power plant operates for 6 days in a week at 10 hours per day, then the pondage factor would be

7 × 24 6 × 10

 = 2.8, and if the plant works only for 5 days in a week, the pondage factor would be

7 × 24 5 × 10

=

3.36 and the pondage required in the latter case would be (2x24 for idle days, 14x5 for idle hours)

48 + 70 24

  x daily flow volume =

59 12

  of daily flow-volume. Thus the pondage factor serves as a rough

guide of the amount of pondage required when the stream flow is constant and the plant works only for a part of the period. Pondage is needed to cover the following four aspects: I. To store the idle day flow. II. For use during hours of peak load. III. To balance the fluctuations in the stream flow. IV. To compensate for wastage (due to leakage) and spillage. Example 9.7 The available flow for 97% of the time (i.e., in a year) in a river is 36cumec. A run-of-river plant is proposed on this river to operate for 6 days in a week round the clock. The plant supplies power to a variable load whose variation is given below:

Period (hr)  Load during period 24-hr average load

ratio

0-6

6 - 12

12 - 18

18 – 24

0.6

1.4

1.5

0.5

The other relevant data are given below; Head at full pond level

= 18 m

Maximum allowable fluctuation of pond level

= 1.5 m

Plant efficiency

= 85%

Pondage to cover inflow fluctuations

= 20% of average daily flow

Pondage to cover wastage and spillage

= 10%

Determine: I. The average load that can be developed II. Daily load factor III. Plant capacity IV. Weekly energy output V. Pondage required and VI. The surface area of the pond for satisfactory operation Solution: I. 7 days flow has to be used in 6 da ys

∴ Average flow available for power development 26

HWRE AR 2010

Q = 36

×

7 6

= 42 cumecs

Since maximum allowable fluctuation of pond level is 1.5 m, average head H=

18 + 16.5 2

= 17.25 m

The average load that can be developed P= =

gQH 

× ηo

1000 9.81 × 42 × 17.25 1000

× 0.85  = 6.04 MW

II. Daily load factor :

Average load = daily load factor = 1 = 0.67 Peak load 1.5 III. Plant capacity = 6.04 x (peak factor)1.5 = 9.06 MW IV. Weekly energy output = Average load in kW x No. of working hours 6 = (60.4 x 1000) (6 x 24) = 8.698 x 10  kWh It should be noted that the installed capacity has to be equal to the peak load and the number of units (kWh) generated will be governed by the average load. V. Pondage required 6 3 a. To store the idle day’s flow = 36 x 60 x 60 x 24hrs = 3.1104 x 10 m b. To store the excess flow during low loads to meet the peak load demand. Since power developed is proportional to discharge (assuming constant average head of 17.25m), flow required during peak load periods of 6.00 to 12.00 hr is (1.4 – 1) 42 cumec and from 12.00 to 18.00 hr is (1.5 – 1) 42 cumec.

∴Pondage to meet peak load demand = (0.4 + 0.5) 42 cumecs for 6 hr = (0.9 x 42) (6 x 60 x 60) 5 3 6 3 = 8.1648x 10 m , or 0.81648 x 10 m c. Pondage to cover inflow fluctuations = (0.20 x 36) 86400 5 3, 6 3 = 6.2208 x 10 m  or 0.62208 x 10 m 6 3 Total of a, b, and c = 4.549 x 10 m Add 10% for wastage and spillage = 0.4549 x 10 6m3 Total pondage required = 5.0039 Mm 3 or 5.0039 x 10 6 m3 Since the maximum fluctuation of pond level is 1.3 m 3 6 2 2 VI. The surface area of pond = 5.0039 Mm = 3.336 x 10  m  = 3.336 km , or 334 ha 1.5 m3

27

HWRE AR 2010

9.4.6

Behavior Analysis

This method as mentioned earlier simulates the behavior of a reservoir during operation. In the Behavior Analysis of a finite reservoir, storage content changes are calculated using the mass storage equation below (McMahon, Mein, 1978): Sn+1 = Sn + In – Dn - ∆En - Ln

(9.9)

Subject to; 0 ≤ Sn+1 ≤ C th

th

Sn+1= storage at the n  time period and the beginning of the n+1  time period. Sn = storage at the beginning of the nth time period. th In = inflow during the n  time period. th Dn = release during the n  time period. ∆En = net evaporation loss from the reservoir during the nth time period. th Ln = other losses during the n  time period. C= active storage capacity (where sedimentation reduces storage capacity during reservoir life, this should be considered). The time period is often taken to be one month, but this can be varied accordingly. Also, net evaporation is taken as the difference between evaporation from the proposed reservoir and the evapotranspiration from the proposed reservoir site and depends on the surface area of the water in the reservoir. Other losses are considered negligible. To construct a behavior diagram, the following steps must be followed: • Firstly, choose a reservoir of active capacity C, and assume it to be full initially i.e. SO = C.

•

Apply the mass storage equation (1) given above for every month or so depending on the time period chosen. The release D n  can either be constant, assumed to have a seasonal variation, or vary as a function of S n. Inclusion of net evaporation may be done only when necessary.

•

Plot a graph of S n+1 against time.

•

Consequently, compute the probability of failure by dividing the number of time periods for which the reservoir is empty by the total number of time periods (total number of months in this case).

•

If the probability derived above is unacceptable, a new C value should be chosen and the steps repeated (the process is thus iterative to determine the storage size for a particular design probability of failure).

Conversely, the storage size for which a reservoir empties only once for the period of historical data can be set. Calculation of this storage could be done using the mass curve and residual mass curve methods. During the calculations the following assumptions are made; i. The reservoir is initially full. ii.

The historical data sequence is representative of future river flows.

However, the following limitations are prevalent;

28

HWRE AR 2010

i.

Handling of non continuous records is difficult because of complexities in applying the initial reservoir condition after a break in the stream flow data.

ii.

Due to the fact that analysis is based on historical records, the sequencing of flows may not be representative of the population of flows.

iii.

The significance of the assumption that the reservoir is initially full on storage size can be checked through examining a behavior diagram for various starting conditions. Analysis based on generated data suggests that at least 100 years of stream flow are required for some rivers before the effect of this assumption can be ignored.

iv.

Drafts (demands) related to growth rates in time for example through increased urban water demand because of population increase are not easily projected because it is hard to relate the demand in the future year to a specific year in the historical flow record.

Despite all these limitations, the following attributes are noticeable; i. Application of the procedure to data based on any time interval is possible. ii.

The procedure takes into account serial correlation, seasonality and other flow parameters insofar as they are considered in historical flows used in analysis.

iii.

The procedure is fairly simple and displays clearly the behavior of stored water. The behavior diagram can be readily understood by non technical people.

iv.

Not only can seasonal drafts be easily taken into account, but also complicated operating policies can be modelled.

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Example 9.8 Using behavior analysis, compute the storage required for a river whose data is given below to supply a 6 3 uniform draft of 77.1% of the stream flow with a probability of failure of 6.4%. (All values are x 10  m ) year

Jan

Feb

Mar

Apr

May

Jun

Jul

Aug

Sep

Oct

Nov

Dec

total

1951 1952 1953

5

3

3

18

21

11

19

52

27

12

80

48

299

35 14 9 42

27 16 44 19

22 25 199 14

12 2 130 3

27 22 84 49

29 47 183 46

32 41 159 42

45 58 195 68

131 91 218 83

125 85 263 58

56 19 316 31

13 29 99 28

554 449 1899 483

98 34 57

22 52 30

64 42 25

87 76 76

90 112 51

62 149 52

101 387 110

63 235 139

110 306 201

126 239 181

72 159 113

53 54 57

948 1845 1092

26 14 4 40 57 39

12 16 24 22 33 21

14 7 59 35 22 33

17 14 47 30 23 28

64 16 44 46 67 32

39 56 91 80 69 78

64 42 489 253 59 95

63 154 312 337 87 117

39 146 174 276 156 174

76 101 122 330 168 236

46 89 109 165 232 238

17 30 78 94 75 86

477 685 1553 1708 1048 1177

35 41 28

49 25 16

67 19 22

139 36 49

44 100 120

49 159 496

79 297 372

118 324 207

164 250 472

226 279 265

167 126 349

73 63 241

1210 1719 2637

84 59 53 139 54 43

48 65 56 64 46 22

28 28 53 88 28 17

26 30 31 481 12 21

44 46 48 414 14 89

58 64 121 521 65 105

178 68 180 505 109 151

253 149 576 456 89 271

297 129 495 402 99 105

319 117 449 352 132 226

195 249 241 278 60 104

84 153 123 116 53 72

1614 1157 2426 3816 761 1226

42 23 77 32 32 17 32 15 98 7 87

26 5 57 19 24 14 15 14 45 2 54

16 11 32 12 6 10 14 15 36 1 28

34 20 13 16 7 22 16 12 19 6 42

23 118 5 141 9 29 19 35 8 76 45

33 132 8 109 15 79 22 44 4 128 84

72 207 5 86 36 367 54 68 12 58 178

86 269 87 132 85 287 67 112 35 222 132

195 223 128 119 111 289 152 207 52 155 197

211 208 95 104 137 409 78 232 91 319 103

96 146 65 90 105 154 62 201 45 152 75

36 95 57 43 54 97 48 138 10 73 62

870 1457 629 903 621 1774 579 1093 455 1199 1087

1954 1955 1956 1957 1958 1959 1960 1961 1962 1963 1964 1965 1966 1967 1968 1969 1970 1971 1972 1973 1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984

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Solution: − mean flow = 101.59 x 106 m3 /month

− −

−

6

6

3

draft = 0.771 x 101.59 x 10  = 78.33 x 10 m  /month 6 3 assuming the reservoir is initially full with a storage of 1000 x 10  m , and applying the equation St+1= St + It - Dt - ∆Et - Lt, month by month for the whole period. Evaporation ( ∆Et) and other losses (Lt) are considered negligible in this case because of their extremely small values. S o = 1000 x 10 6 m3. From calculations shown in the tables given, a Summary of the results is given below; Storage estimate (106  m 3 )

Probability of failure

1 2

1000 900

3.43 5.64

3 4 5

850 870 865

6.86 5.88 6.37

Trial

−

Therefore from the iterations performed as shown for the steady draft (release) given with the reservoir initially full, the required storage capacity is 865 x 10 6m3 as shown in Fig 9.13, the behavior diagram

Fig 9.11 A Behavior Diagram for Example 9.8

A study was carried out on two catchments in Uganda (Rugumayo and Kizza, 2001); one is of bimodal rainfall (Chambura) and the other is of unimodal rainfall (Nyagak) on their response to within year storage and relate it to reliability of the yield, the Behavior analysis method was applied for three levels of reliability (90%, 95%and 99%) while the draft selected ranged from 031

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100% of the mean annual flow (MAF). The results showed that for zero storage a yield of nearly 30% was possible for all reliability levels and was higher for Chambura , that receives bimodal rainfall, as shown in Fig 9.14. Therefore for run- of- river schemes the demand would fall in this region. Furthermore, the smaller the capacity of the reservoir, the greater the reliability and as the yield increases so does the reservoir capacity increase until some assymptotic value of between 50%-70% of the MAF. The shape of the storage yield curve, gives an indication of the relative costs of reservoir development on the respective rivers. For instance, a steep yield storage curve means it will require less storage to meet a particular increase in demand, when it is assumed that material and labour costs are comparable at both sites. The smaller the reservoir capacity, the less the exposed surface area and hence the lower the evaporation from the surface. Evaporation losses can be a significant factor during site selection especially in the Equatorial region.

   F    A    M    %    t    f    a    r    D

   F    A    M    %    t    f    a    r    D

Capacity (% MAF) Capacity (% MAF)

Fig 9.12 Reservoir capacity yield curves Summary In order to plan for the sustainable water resources, aspects of site assessment and dam selection must be considered. This should include the functional and technical suitability of a site and an assessment of anticipated environmental consequences of the construction and operation of the dam. The above aspects and the principal stages involved in site appraisal and optimum selection of site and type of dam are explained in this chapter.   The challenges in the development of large dams together with the environmental and social safeguard policies of the main stakeholders are mentioned.   The commonly used methods for the estimation of the storage capacity of a reservoir are discussed and examples given.

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References 1.

Altinbilek,D., Abdel-Malek, R., Devernay,J-M., Gill, R., Leney,S., Moss,T., Schiffer,H.P., Taylor,R.M.,  Hydropower’s Contribution to Energy Security , World Energy Congress, Rome, International Hydropower Association, 2007, London, UK. 2. Arora, S,O. Water Resources Hydropower and Irrigation Engineering , Standard Publishers and Distributors, 1996,New Dehli, India. 3. Basson, G.,  Hydropower Dams and Fluvial Morphological Impacts – An African Perspective , Proceedings United Nations Symposium on Hydropower and Sustainable Development, 27 th – 29th October 2004, Beijing, China. 4. Carty, J.G., Cunnane,C.,  An Evaluation of Some Methods of Determining the Storage Yield  Relationships for Impounding Reservoirs , Water and Environment Journal, 1997,Vol 4 (1) pp35-45. 5. Duggal,K.N., Soni,J.P.,  Elements of Water Resources Engineering, New Age International (P) Limited 2007, New Dehli, India.  Environmental and Social Management Framework , A Field Manual, Rural Electrification 6. Agency, Ministry of Energy and Mineral Development 2007, Kampala, Uganda. 7. McMahon T.A., Mein R.G.,  Reservoir Capacity and Yield, Elsevier Scientific Publishing Company, 1978, Amsterdam, Netherlands. 8. Nagpal, G.R. Power Plant Engineeering , Khanna Publishers, 2000, New Dehli, India. 9. Novak P., Moffat A.I.B., Nalurri C., .Naranyanan R.,  Hydraulic Structures, Second Edition, E&FN Spon,1996, London, UK 10. Raghunath H.M.,  Hydrology, Principles, Analysis and Design , New Age International (P) Ltd, 2005, New Dehli, India. 11. Rugumayo A.I., Kizza M.,  Reservoir Yield Reliability Analysis, Proceedings of the 25th WEDC Conference, 2001, Lusaka, Zambia. 12. Subramanya, K.  Engineering Hydrology, 2nd  Edition, Tata McGraw-Hill Publishing, 2001, New Dehli, India. The National Environment Act 1995 , Uganda Printing and Publishing Company, Entebbe, 13. Uganda. The Scoping Report on the Decision Making Processes in Dams and Development in Uganda , 14. Uganda Dams Dialogue, DDP/GTZ/UNEP, 2006, Kampala, Uganda. 15. Thomas, H. A.,Burden, R.P., Operations Research in Water Quality Management , Havard Water Resources Group 1963, Havard, USA. 16. World Commission on Dams (WCD),  Dams and Development a New Framework . Earthscan Publications Ltd, 2000, London, UK/Sterling VA, USA. 17. Zaake B., Report on Stakeholder Meeting:  Enhancing applicability of innovative methodologies and tools to improve engagement and participation of stakeholders and decision makers at every stage of the decision process in the management of dams along the Victoria Nile in Uganda , 19th – th 20  July 2007, Directorate of Water Resources Management, Entebbe, Uganda. Further Reading 1. Chadwick, A., Morfett,J., Hydraulics in Civil Engineering, Allen and Unwin, 1989, London, UK. rd 2. Dake, J.M.K., Essentials of Engineering Hydraulics, 3  Edition, Macmillan,London UK. 3.  Design of Small Dams, United States Department of the Interior, A Water Resources Technical Publication, 1974, Washington, USA. 4. Garg S.K. Hydrology and Water Resources Engineering, Khanna Publishers, 1998, New Dehli, India. 5. Linsely and Franzini: Water Resources Engineering , McGraw, 1979, London, UK 6. Viessman, W., J.W. Knapp, G.L. Lewis & T.E. Harbaugh (1977). Introduction to Hydrology , Harper Row, New York, USA th 7. Wilson E.M., Engineering Hydrology , 4  Edition, Macmillan, 1996, London UK.

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Questions 1. April 6.7 39.6

The flow (in m3 /s of River Aswa over 17months (each of 30 days) was: May 16.8 68.8

June 77.1 145

July 102.4 216

Aug 238 162

Sept 247

Oct 216

Nov 63

Dec 26.3

Jan 20.6

Feb 16.8

Mar 6.5

A hydroelectric plant using a net head of 24m, and developing 15.2MW at 88% efficiency is to take water from a reservoir on this river. An exceptionally wet season proceeding the first April would have left the reservoir full. What reservoir capacity would be necessary to supply 15.2MW, continuously? What volume of water will escape over the spillway in the above period? Use the i) mass curve method, ii) analytical method and iii) flow duration curves. 2. A dam is to be constructed to embark on large-scale irrigation in the Kafu plains. Given the following monthly minimum flows, for River Kafu, over a 15month period, determine the reservoir capacity required, if the demand will be 75% of the average minimum flow. Seepage losses are estimated at 3 12,500m  /day, evaporation losses are averaged at 3.00mm/day throughout the year and downstream compensation requirements are 1.0m3 /s. The reservoir surface area is approximately 400 hectares.  Month  Inflow volume x 106  m 3

1 12

2 24

3 40

4 12

5 5

6 4

7 7

8 4

9 3

10 3

11 7

12 23

13 37

14 8

15 6

Use i) mass curve ii) analytical and iii) flow duration curve iv) sequent peak algorithm 3. Which of the above methods do you think gives the most accurate results? 4. Discuss the characteristics of a good reservoir site.

5. Define the power terms below: i. Peak load ii. Net head iii. Utilization factor 6. What is entailed in the investigations for reservoir planning. 7. Define the term  pondage and explain its use.

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10 SEDIMENTATION Sediments are produced in the catchment of a river by erosion. Rivers carry a large amount of sediment load along with water. These sediments are deposited in the reservoir on the upstream of the dam because of reduction of velocity. The deposition of sediment in the reservoir is known as reservoir silting or sedimentation. 10.1

River Classification

Sediment yield is the amount of sediment passing a specified channel location. It is influenced by a number of geomorphic processes. It may be substantially less than the amount actually eroded in the basin. Sediment transport is influenced primarily by the action of wind and water. Deposition occurs in a number of locations where energy for transport becomes weak to carry eroded sediments. Classification, in the strictest sense, means ordering or arranging objects into groups or sets on the basis of their similarities or differences. There are currently four main methods of river classification (Rowntree, 1999). These are: (a) Whole River System Classification (b) Zonal (Segment) Classification (c) Morphological Classification (d) Hierarchical Classification. Whole River System Classification ; River systems are composed of a hierarchy of catchments with small catchments nested within larger ones. The magnitude of a river system at any point within the larger river system can be described and compared to other systems. Topographic indices such as basin relief, relief ratio, basin shape and the hypsometric curve also describe and compare the relief and shape of catchments. ii)  Zonal Classification: this is classification based on the concept of zonation down the long profile and is widely used. The earliest zonal classification, done in the 19 th  Century subdivided the channel and adjacent catchment in terms of the gradient. The steep headwater zones were termed “youthful” with high potential energy and active degradation processes. This is characterized by steep gradients and giving rise to gorges or narrow steep sided valleys. The fast flowing sterams have little need to deposit their load and there is no flood plain. Most of the silt comes from the catchment due to soil erosion. This can also be termed as the rocky stage  as shown in Fig 10.2, since the beds and banks are rocky and is ideal for the construction of dams .

i)

The foothill areas were termed “mature” with more gentle slopes, less active degradation and a tendency to equilibrium between erosion and sediment deposition. The valley is both deeper and wider. This can also be termed as the boulder stage as shown in Fig 10.2. In this stage, the bed and the banks are usually composed of large bouldes, gravels and shingles. During floods, the river transports boulder and shingles downstream however, when the flood recedes,these materials are deposited in the river bed. The river section is well defined. The velocity is high but less than in the rocky stage. The lower zones characterized by low gradients and low potential energy were termed “old age” and were dominated by low velocity flows, deposition and low rates of catchment denudation. This can be considered as both the trough, alluvial and deltaic stages as shown in Fig 10.2. The river meanders and the cross section is made of alluvial sand and silt. The river carries sediments which are the same as the bed and the bank. In the deltaic stage, the last stage before it discharges into the sea, the bed slope and velocity are further reduced. The river is unable to carry sediments and consequently drops them

35

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and gets divided into channels on either side of the deposited sediments and forms a delta (Arora, 2007). A fluvial system can be further classified in terms of sediment activity into three zones: an upper  zone of sediment production (source ) (youthful), a middle zone (transfer) (mature) essentially in equilibrium, and a   lower zone (sink or deposition area) , (old) where controls are base level and diastrophism. The  fluvial cycle is the name given to the series of progressive changes in landforms, which results from the development of a river system. All the processes of erosion and deposition connected with rivers combine to bring these changes and in areas of moderate rainfall, they are among the dominat factors shaping the landscape (Read and Watson, 1972). iii)  Morphological Classification : This approach considers that the basis of any classification system should be the river reach, a homogenous reach being a stretch of a river of variable length within which controls of channel form such as hydrology, geology and adjacent catchment conditions are sufficiently uniform to result in relatively uniform channel morphology. More comprehensive classifications were developed that differentiated between straight, meandering and braided channel patterns. iv)  Hierarchical Classification: this links large regional scales (ecoregions) with small microhabitat scales. The basic assumption for the development of the hierarchical stream classification is the geomorphologic premise that the structure and dynamics of the stream are determined by the surrounding catchment.

Fig 10.1 Stages in river development Source: (Arora, 2007)

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A study (Rugumayo et al, 2004) was carried out on three rivers in Uganda, namely, Kagera, Sio and Katonga, to estimate the amount of sediment load they carried and then develop a simple classification of rivers in this region, based on the sediment load. The hierarchical system of classification was applied, because it deals with large spatial variability and can relate to fewer variables than most other systems. There are three major sources of sediment input into L. Victoria. These are direct runoff from the catchment, sediment loads from the rivers and from physiochemical and biological processes within the lake as a result of external factors. Agricultural related activities by the communities living in the river and Lake Basin, are the most influential cause of sediment inflow into the lake. The sediment load of rivers carried by moving water is subdivided into two categories namely; suspended load and bed load. The study focused on the suspended load since according to previous studies (Mwebembezi et al, 2001), approximately 94% of the sediment yield was found to comprise of suspended load, hence the significance of the suspended load analysis. A simple classification model was developed using the Rating Curve method, the Quasi Maximum Likelihood Estimator and the Smearing Estimator to assist in estimating sediment load per unit area. Typical characteristics are as shown in the Table 10.1. Table 10.1 A Simple Sediment Transport Classification Model Ratio of Sediment Input per unit area (g/m3 /day) 0.005 – 0.0019

Classification Description Low

0.200 – 0.399

Medium

0.400 - 0.600

High

>0.600

Very High (excessive)

Typical characteristics Low flow velocity Generally flat topography, well vegetated Swampy surroundings Fairly high flow velocity Hilly areas with farming activities High flow velocity Area with intense farming Usually sparsely vegetated Little or no vegetation Slope failures, landslides.

10.2 Land Degradation Land degradation is the loss of utility or potential utility, or the reduction, loss or change of features or organisms, which cannot be replaced. In Uganda there are two forms of la nd degradation; soil degradation and vegetation degradation. In this section we shall confine ourselves to soil erosion, which is a form of soil degradation. There are three major causes of soil erosion in Uganda; surface erosion, gully erosion and soil mass movement. Surface erosion involves the detachment and subsequent removal of soil particles by wind and water. In Uganda, water is the main agent of erosion with wind erosion commonly found in the drier areas of less than 800 mm of rainfall per year such as north eastern Uganda. Rill erosion occurs when running water concentrates and scours the sides of the channels. Rill erosion is the first stage of gulley formation and rills can be obliterated by ordinary tillage, but it does not replace the lost soil. Gulley erosion is normally the advanced stage of rill erosion and involves scouring and the movement of ether individual soil particles or large aggregates in a channel by a large concentration of running water. Gully formation normally begins with deeper incision in the soils of the lower reach of hill slopes that expand uphill.

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The main causes of soil erosion in Uganda are i)  Improper land use or inappropriate farming practices ; complete forest removal, overgrazing, exposure of bare agricultural land, ii) Climatic factors, especially the amount and intensity of rainfall. The characteristics of rain in many parts of Uganda render it highly erosive and put the soil at great risk of degradation. The dominant factor on amount of runoff and soil loss is usually the frequency of high intensity storms rather than the total amount of rainfall. Most often, heavy downpours are received in very short durations, iii) Other physical causes include  a) soil erodibility: In general the soils are inherently erodible and prone to erosion and erosion induced degradation, b) length and gradient of slope. In general the magnitude and severity of soil erosion is a function of slope length and steepness. These vary widely across the ecozones of Uganda ranging from 0% to 30% and so are the effects; c) mass wasting can occur in natural forests free form anthropogenic perturbations due to earthquakes., iv)  Non agricultural human activities ; for instance the construction of roads or buildings that leave the ground unprotected generates large volumes of surface runoff that cause both on site and off site erosion and especially since most roads are murrum and there is a housing boom, v) land tenure and land fragmentation;   this occurs more in the heavily populated districts of Kisoro, Mbale , Kapchorwa and Bushenyi. The areas most seriously affected by soil erosion are the steep slopes of Kabale, Kisoro, Bundibugyo, Kasese, Kabarole, Kapchorwa an Mbale. Although this problem is well recognized, the actual quantities lost per annum and the offsite damage it causes are not known. Several studies have been done to quantify soil loss from runoff plots and extrapolate them to arable lands but the discrepancies are high (Magunda and Tenywa, 2001). Recent work (Bamutaze et al, 2006) in the Mt Elgon region on major cropping systems and three landscape positions shows that runoff and soil erosion are occurring at catastrophic proportions and is more severe in annual crops, like beans and maize as compared to perennial crops like bananas. Furthermore, the greatest runoff and soil loss occurred in the lowest landscape position, whereas the lowest runoff and soil loss occurred in the middle landscape position. They recommend the promotion of perennial land use as opposed to annual crops and the sensitization of communities on land use practices.

10.3 The Process of Sedimentation There are two types of sediment load namely the suspended load and the bed load explained below: Suspended load : The suspended load is that part of the sediment load that is held in suspension against gravity by the vertical component of eddies of the turbulent flow. The suspended load usually consists of fine material dispersed throughout the river cross section.  Bed load : The bed load is that part of the sediment load, which remains in contact with the bed when moving with water. It consists largely of relatively coarser materials. The Process When the water reaches a reservoir in the vicinity of a dam, the velocity and the turbulence are considerably reduced. The larger suspended particles and bed load get deposited in the head reaches of the reservoir and a delta is formed. Some very fine particles may remain in suspension for a much longer period and may finally escape from the dam along with water discharged through the sluice gates, turbines, spillway, etc.

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Turbid inflow

Floating Debris

Water Surface Relatively clear water

Delta

Density Current Fine sediments

Sluiceways

Fig 10.2 Sediment accumulation in a typical reservoir

The sediment load of the river depends upon the following factors: i) Nature of soil in the catchment : If the soil in the catchment is loose and easily eroded, the sediment load is large. On the other hand, if the soil is hard and non er odible, the sediment load is quite small. ii) Vegetal cover : If the catchment area has no vegetal cover, the soil is easily eroded and the sediment load is large. iii) Topography of the catchment : In case of catchments having steep slopes, the sediment load is large because of high velocity of water. iv) Intensity of rainfall : If the intensity of rainfall is high, the discharge in the river is increased and the sediment load is large. v) The size of a water body:   has a strong implication for its hydrological and sedimentological characteristics. In small lakes, wind-induced re-suspension is the only important sediment transport mechanism, while in oceans current-mediate transport mechanisms assume an equally important role to wind induced distribution mechanisms. A study of the pattern and mechanisms of sediment distribution in Lake Victoria (Azza, 2006) suggests that for large lakes, with relatively shallow basins, surface wave action together with currents can predict the sediment distribution. He further suggested that morphometric, hydrological and meteorological factors come together to mix and disperse sediments via the epilimnion and produce alongshore and cross- isobath currents that transport sediment northwards and northeastwards from the western shore. A related study (Kimaro et al, 2006) carried out on the northern and southern fringes of Lake Victoria, showed that the nature of field parcels and overall farming techniques has an overall effect of protecting the soil structure against raindrop impact, encouraging water infiltration, obstructing runoff and eventually minimizing soil loss from the gardens. They further noted that the presence of settlements is associated with very high sediment yield, indicating that future unchecked expansion is likely to contribute to sedimentation of Lake Victoria. Although aquatic weeds are effective buffers, excessive sedimentation is likely to reduce filtering capacity and therefore make the lake more vulnerable.

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10.4

Effects of Sedimentation

The deposition of sediments will reduce the water storing capacity of the reservoir and if the process of deposition continues longer, a stage is likely to reach when the whole reservoir may get silted up and become useless. Other effects include increased flood levels upstream of the reservoir, retrogression of the riverbed and water levels downstream of the dam, elimination of nutrients carried by the fine sediments, and the deterioration of the water quality. Useful Life of a Reservoir  The deposition of sediments gradually reduces the available capacity of a reservoir. As more and more sediments are deposited in the reservoir a stage comes when the reservoir is not able to serve its intended purpose and its useful life is over. If the annual sediment flow is compared with the reservoir capacity, the useful life of the reservoir would be very short. While planning a reservoir, it is essential to consider the rate of sedimentation to know whether the useful life of the proposed reservoir will be sufficiently long to  justify the expenditure of its construction. The rate of sedimentation in the reservoir depends on the trap efficiency. Trap efficiency: The trap efficiency (ηt) is defined as the percent of the total inflow sediment, which is retained in the reservoir. Thus;

Trap efficiency

(10.1)

From the observations of the rate of sedimentation of existing reservoirs, it has been found that the trap efficiency of a reservoir depends upon the capacity/inflow ratio. Brune (1953) gave the curves relating the trap efficiency and the capacity-inflow ratio on the basis of a large number of existing reservoirs in the USA as shown in Fig 10.3. He also developed an empirical relationship from which the curves are drawn; E = 100 x (0.97

0.19log(C/I)

)

(10.2)

Where C is the capacity and I is the Inflow. In general the greater the capacity–inflow ratio, the greater the trap efficiency, thus sedimentation is higher in relatively larger reservoirs. The trap efficiency decreases with the age of the reservoir, because the available capacity gradually decreases. It also depends on other factors like shape of reservoir basin, type of outlets, method of operation, size and grading of the sediment and the behaviour of finer sediment fractions under various conditions. Churchill (1948) developed a relationship between the percentage of incoming sediment through a reservoir and a reservoir sedimentation index, which is defined as the ratio of the period of retention to the mean velocity of flow through the reservoir. Murthy (1980) developed guidelines for use of the above methods. He recommended the using of Brune method for large storage or normal ponded reservoirs and the Churchill method for settling basins, small reservoirs, flood retarding structures, semi-dry reservoirs or reservoirs that are continuously sluiced. A study was carried out on the River Nile Basin Sedimentation Problems (Siyam et al, 2005), which gave the Table 10.2 on the capacity/inflow ratio and estimate of the trap efficiency using Brune median curve. The results showed a range between 78% and 98%. There is ongoing research, which intends to develop a Nile Basin Reservoir Information System, (NBRIS) that will be used to plan, monitor and manage reservoirs using the various analysis of spatial and non spatial attribute overlaying, simulation and modeling and integrated analysis of spatial and non spatial attribute data. With the availability of remote sensing satellite data that provide 1-4 m resolution and the centimetre level accuracy of GPS, it would be possible to update the data regularly.

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HWRE AR 2010

The following information could also be added to the data base; • Downstream hazard potential of a dam, which can identify the hazard to the downstream area resulting from failure or failed operations of the dam. • Identification and review of the measures that can be taken to reduce sedimentation of reservoirs. • Consideration of the options for the removal of sediment, along with the associated environmental concerns and consideration of the past and future consequences of reservoir storage loss. • List of references of research in sedimentation problems in reservoirs.

Table 10.2 Trap Efficiency Coefficients for different Nile reservoirs estimated by Median Curve of Brune Method

3

Reservoir

Capacity C(Mm )

Koka Roseiris Girba Angerib Nasir Nuba

1650 3000 1300 5.28 162000

Source: (Siyam et al 2005) Total Inflow (I) C/I (Mm3) 29478 0.0560 48744 0.0615 12000 0.1083 27 0.1852 84000 .929

Brune Trap Efficiency% 78.38 79.64 86.00 90.23 98.12

Location Awash River Ethiopia Blue NileSudan Atbara river Sudan Angerib River Ethiopia Main Nile Egypt Sudan

   )    %    (    d   e   p   p   a   r    t    t   n   e   m    i    d   e    S

Capacity: inflow ratio (capacity/volume of annual inflow)

Fig 10.3 Curves relating the trap efficiency and the capacity-inflow ratio Source: (Brune, 1953)

Therefore, the complete sedimentation of the reservoir shall be more at the beginning and as its capacity reduces due to silting, the silting rate will reduce. It can also be concluded that for small reservoirs on large rivers, the trap efficiency is very low, because the ca pacity inflow ratio is very small. Such rivers silt

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very little and most of their sediment is passed downstream. On the other hand, large reservoirs on smaller rivers silt significantly and almost complete deposition of sediment may take place. 10.5

Measures to Control Reservoir Sedimentation

i) Minimising sediment input:  This is by far the most effective measure and can be achieved by optimal choice of the location of the reservoir, the prevention of erosion in the catchment by soil conservation methods (afforestation, vegetation cover, terracing etc), the trapping of sediment in traps / check dams or by vegetative screens on the tributaries upstream of the reservoir, or by bypassing heavily laden flows during floods, from an upstream diversion structure to downstream of the dam. ii) Maximising sediment throughflow: This requires flow regulation during floods and /or flushing during reservoir drawdown. Under certain conditions the sediment laden inflow does not mix with the water in the reservoir but moves along the old river bed as density current towards the dam, where it can be drawn off by suitably located and operated outlets. In principle, the development of density currents requires a significant difference between the density of the incoming flow and the water in the reservoir, a large reservoir depth, and morphological conditions (steep, straight old river bed).The formation and sediment transport of density currents can be predicted and should be applied to all reservoirs under various operational conditions. Turbidity sensors should be applied at several elevations upstream of the dam wall for management of density currents by releasing high sediment concentrations through low level outlets. iii) Recovery of Storage : This can be achieved by flushing deposited sediment, a technique, which is very effective when combined with a substantial reservoir drawdown, by siphoning or dredging; in the latter case either conventional methods, particularly a suction dredger with a bucket wheel, or special techniques can be used. Effective flushing requires excess water, suitably large low level outlets, a steep narrow reservoir basin and judicious operation. Dredging has been carried out only to a limited extent worldwide, mainly because of the high costs and environmental problems associated with the disposal of dredged sediments and as a result should be considered as a last resort (Basson, 2004). iv) Sedimentation Compensation Measures : There are a number of techniques which can be applied, which are discussed as follows; a) Dam raising; this provides an economical solution to regain storage capacity lost due to sedimentation. The rasing options considered are fixed uncontrolled spillways, crest radial gates, automatic crest gates of fuse gates, b) New dams; dam sites should be selected in regions with relatively low sediment yields. The upper reaches have relatively high runoff, with low sediment loads, however this is not always possible due to the location of power demand centresand availability of dam sites, c) Design for sedimentation: designs should acknowledge the experience in reservoir sedimentation, whereby 80% occurs in the live storage zones rather than the current practice of designing for withdrawals from the dead storage zones. d) Augmentation from adjacent catchments; regulation of runoff and sedimentation control requirements in a reservoir are often in conflict. Transfer of water from adjacent catchments can provide a solution to sediment control if it is economically feasible and it can provide sufficient excess runoff (Basson, 2004). v) Simulation Techniques: The Reservoir Conservation RESCON model is designed to assist in the selection of a reservoir management strategy that is technically feasible, which maximizes the net economic benefits of a reservoir using four explicit options of sediment management namely: Flushing; Hydrosuction; Traditional Dredging and Tracking. In addition t he do nothing alternative (i.e. no sediment removal), where eventual decommissioning is required is also analysed. The RESCON requires project specific technical and economic data in addition to environmental and social safeguards parameters. The important parameters with respect to flushing are representative flushing discharge; duration of flushing events in addition to the parameters related to reservoir geometry, outlet capacities, sediments and water flow. Before detailed calculations, the flushing scenarios are checked for technical feasibility, by comparing the amounts of sediments evacuated with respect to amount of sediments inflow in a year. The

42

HWRE AR 2010

tracking and dredging options are always considered as technically feasible. For the baseline case for no sediment removal, the model assumes two alternatives: run of river and eventual decommissioning. All options are taken through an optimization routine to find economically viable options for each. The results are compared and ranked for final decision by the user (Palmeri et al, 2003, Kawashima et al, 2003). This model was applied to the Tarbela dam in Pakistan and it was found that the reservoir can achieve a sediment balance ratio of more than 1 even with a flushing for 20 days in a year with a water 3 depth of 417 m above mean sea level and a discharge of 3,787 m  /s.(Tinsanchi and Khan, 2006). An empirical reservoir classification system was developed (Basson, 2004) based on a plot of the storage capacity- mean annual runoff (MAR) against the capacity-sediment yield ratio gives some useful results. Most reservoirs have a capacity- MAR ratio of between 0.2 to 3 and a life span of 50 to 2000 years when considering reservoir sedimentation. When the capacity- MAR ratio is less than 0.03 especially in semiarid regions, sediment sluicing or flushing, should be carried out during the floods and through large bottom outlets. When capacity – MAR ratios are however, larger than 0.2, not enough excess water is available for flushing and the typical operational model is storage operation. The operating rules for a reservoir need not be inflexible, but can change with different stsges of storage loss Storage operation may be continued in reservoirs with large capacities relative to the sediment loads, while sluicing  /flushing operation can be introduced once the loss of storage capacity reaches a certain stage. This is shown in Fig 10.4.

Fig 10.4 Empirical reservoir classification system in terms of storage runoff and sediment yield Source: Basson, 2004

10.6

Sediment Load Formulae

10.6.1 A Sediment Rating Curve Segments of a sediment rating curve are usually approximated by the power relationship of the form; Qs = mQn

(10.3)

43

HWRE AR 2010

where: Qs = suspended sediment transport (Mtons/day) 3 Q = water discharge (m  /s) m = coefficient n = exponent The exponent n for many rivers ranges about a mean of 2 (Garde and Ranga Raju, 1985). They could for instance have one for the rising flood stage and the other for the falling stage. Observations indicate that for higher discharges, the exponent will diminish and approach the value of unity. Sediment rating curves are usually based on short term data unlike water discharge measurements. However, long term sediment load estimates are required for reservoir sedimentation. As a consequence Miller (1951) developed a method for finding the average sediment yield by combining the short term sediment rating curve, with the long term flow duration curve to become the Flow Duration Sediment Rating Curve Method. Experience indicates that the flow duration sediment rating curve is most reliable when i) the recording period is long, ii) sufficient data at high flows are available and iii) the sediment rating curve shows considerable scatter (Julien, 1995). It is possible to calculate the total load or sediment load from the sum of the bed load and suspended load. Separate equations are available for bed load and for suspended load. However, experimental data are still rather sparse, and it is very difficult to separate bed and suspended load from these data. For this reason, some researchers have tackled directly the problem of total load. Three examples of total load formulae are outlined below. 10.6.2 Ackers and White Formula Ackers and White (1973) applied dimensional analysis techniques to develop a general sediment transport function in terms of three dimensionless groups, G gr, Fgr and Dgr: G gr is the sediment transport parameter, which is based on the stream power concept. For bed load, the effective stream power is related to the velocity of flow and to the net shear force acting on the grains. Suspended load is assumed to be a function of total stream power, P. The particle mobility number, F gr, is a function of shear stress/immersed weight of grains. The critical value of Fgr (i.e. the magnitude representing inception of motion) is denoted by A. Finally, a dimensionless particle size number, D gr, expresses the relationship between immersed weight of grains and viscous forces (Chadwick and Morfett, 1989). Ackers and White showed that the transport of fine materials is best related to the total shear velocity and the transport of coarse sediments is related to the mean velocity. The equations are then as follows: n

 F gr   G gr  = C  = − 1  sD  V    A   q S  Dm U ∗ 

F gr 

=

m

 

  V    gD [s − 1]  32 log (10 D m  D )  U ∗n

(10.4) 1− n

 

(10.5)

1

  g [s − 1]  3  D gr  =  D   2   ν     

(10.6)

44

HWRE AR 2010

Where

qS = volume of sediment transported per second per unit channel width = flow discharge Dm = mean flow depth D = representative grain size (Usually D35, that is the size of sediment for which 35% of the sample is finer) U ∗ = shear velocity q

V  = mean cross-section velocity s  = grain specific gravity =  ρ S   ρ   ρ S = density of sediments  ρ   =

density of water

ν   = Kinematic viscosity

The index n does have a physical significance, since its magnitude is related to D gr. For fine grains n = 1, for coarse grains n = 0, and for transitional sizes n = f (log Dgr). The values for n, m ,  A,  and C are as follows: For Dgr > 60 n = 0, m = 1.5,  A = 0.17 , C  = 0.025 For 1 < Dgr < 60

= 1 − 0.56 log  D gr  m = 1.34 + 9.66 /  D gr  n

 A = 0.14 + 0.23 /   D gr 

log C  = 2.86 log  D gr  − (log  D gr  )

2

− 3.53

10.6.3 Yang’s Formula Yang (1979) hypothesized that the unit stream power, VS, defined as the time rate of potential energy dissipated per unit weight of water, is the dominant factor in determining the sediment concentration in alluvial channels. The relationship between the unit stream power and the total sediment concentration, Ct, is expressed by log C t 

Where

=  I  +  J  log   

VS  

   w  

 J  = 1 .780

− 0.360 log

wD 50

 I  = 5 .165

− 0.153 log

wD 50

v

v

− 0.480 log

U ∗

− 0.297 log

U ∗

C t 

= total sediment concentration in ppm

U ∗

= shear velocity

w

w

(10.8, 10.9,10.10)

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HWRE AR 2010

V  S 

= mean cross-section velocity

= water slope = fall velocity w VS  w = dimensionless unit stream power = Kinematic viscosity v  I   and J  are parameters determined from multiple regression analysis of a large range of experimental data.

10.6.4 Karim and Kennedy’s Formula Karim and Kennedy (1983) used dimensional analysis to develop a power relationship between the total sediment discharge, qt, and the flow velocity, V, shear velocity, U ∗ , the fall velocity, w, and the mean particle diameter, D50

  V  qt  = k 1    g (s − 1) D50  Where

qt   is

2.97

1.47

 U ∗       w  

3 g (s − 1) D50

 

(10.11)

the total sediment discharge per metre width in m 2 /s and k 1   = 0.00139 is a coefficient of

proportionality. 10.6.5 Discussion of Sediment Transport Formulae In a study Karamisheva et al (2006) investigated nine sediment load transport formulae with experimental data obtained during inbank and overbank flows in compound channels with straight and meandering planforms. The comparison between measured and predicted sediment discharge gave the following results: • The formulae, which gave good prediction for sediment discharge for inbank flows also predicted the sediment discharges for overbank flow with good accuracy. • The Yang formula gave very good results for all the data studied. • The Karim and Kennedy formula could be a good predictor of sediment discharge for inbank and overbank flows in channels with straight and meandering planforms except that the constant of proportionality needs calibration. • The Ackers and White formula gave a very good prediction for the flood channel facility and a good overall mean discrepancy ratio, but over predicted the sediment discharge in the smaller experimental channel. 10.7

Sediment Estimation Examples

 Example 10.1 (after Garg, 1999) The following information is available regarding the relationship between trap efficiency and capacity inflow ratio.

46

HWRE AR 2010

Capacity inflow ratio

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Trap efficiency percent

87

93

95

95.5

96

96.5

97

97

97

97.5

Find the probable life of the reservoir with an initial reservoir capacity of 25 million cubic meters, if the average annual flood inflow is 50 million cubic meters and the avera ge annual sediment inflow is 200,000 tons. Assume a specific weight of the sediment equal to 1.2 gm per c.c. The usual life of the reservoir will terminate when 80% of its initial capacity is filled with sediment. Solution Average annual sediment inflow = 200,000 tons = 2×105 tonnes= 2×1011gm Volume of average annual sediment inflow

= =

2 × 1011 1.2

c.c =

2 × 1011 1.2 × 10

6

m3

=

0.2 1.2

× 10 6 m 3

1  M .m 3 6

= 0.1667 x 106 m3 Initial Reservoir Capacity = 25 M.m3 3 Annual flood inflow=50 M. m 3 Let us assume that 20% of the capacity,i.e 5 M. m  is filled up in the first interval . Capacity inflow ratio at the start of the interval =

25 50

= 0 .5

Trap efficiency at the start of the interval =0.96 Capacity inflow at the end of the interval

20

=

50

= 0. 4

Trap efficiency at the end of interval=0.955 Average trap efficiency during the interval

=

0.96 + 0.955 2

= 0.9575

Volume of sediment deposited annually till the 20% c apacity is filled =

1 6

× 0.9575 M .m3 6

3

= 0.1596 x 10 m 3 Therefore number of years during which 20% of the capacity, i.e 5 M. m  shall be filled up

=

5 0.1596

= 31.3 years

Similarly, in the 2nd interval Capacity inflow ratio at the start =

20 50

= 0 .4 47

HWRE AR 2010

Capacity inflow at the end

=

15 50

= 0.3

Trap efficiency at the start = 0.955 Trap efficiency at the end = 0.95 Average trap efficiency = 0.9525 6 3 6 3 Volume of sediment deposited annually during this interval = 0.1667 x 10 m x 0.9525 = 0.15875 x10 m Number of years during which the next 20% of capacity shall be filled up 5 = = 31.5 years 0.15875 rd

Similarly, in the 3  interval Capacity inflow ratio at the start = Capacity inflow at the end

=

15 50 10 50

= 0.3 = 0.2

Trap efficiency at the start = 0.95 Trap efficiency at the end = 0.93 Average trap efficiency = 0.94 Volume of sediment deposited annually during this interval = 0.1667 x 10 6m3x 0.94 = 0.1567 x 10 6 m3 Number of years during which the next 20% of capacity shall be filled up =

5 0.1567

= 31.9 years th

Similarly, in the 4  interval Capacity inflow ratio at the start = Capacity inflow at the end

=

10 50 5 50

= 0.2 = 0. 1

Trap efficiency at the start = 0.93 Trap efficiency at the end = 0.87 Average trap efficiency = 0.90 Volume of sediment deposited annually during this interval = 0.1667 x10 6 x 0.9 = 0.15 x 10 6 m3 Number of years during which the next 20% of capacity shall be filled up =

5 0 . 15

=

33 . 3  years

Total probable life till 80% capacity gets filled up = 31.3 + 31.5 + 31.9 + 33.3 = 128 years The above calculations of dividing the entire capacity into intervals (20% each in the above case can also be carried out in a tabular form, as shown below.

48

HWRE AR 2010

Capacity % Capacity Vol.  M.cum

Capacity/inflow=

100

25

0.5

80

20

0.4

60

15

0.3

40

10

0.2

20

Col.(2)

50 Mcum

0.1

Trap  Av. Trap Sediment Years reqd. to efficiency eff. trapped per  fill up 20% η   During  year ;Col 5x capacity(6 the  Av. Annual  Mcum) interval sediment 6 = inflow=Col Col .( 6) 5x (1/6)  M.cum 0.96 0.9575 0.1596 31.3 0.955 0.9525 0.1588 31.5 0.95 0.94 0.1567 31.9 0.93 0.90 0.15 33.3 0.8 =128.0 ∑  years

 Example 10.2 A proposed reservoir has a capacity of 400 ha-m. The catchment area is 100km 2 and the annual stream flow averages 12 cm of runoff. If the a nnual sediment production is 0.03 ha.m, what is the probable life of the reservoir before its capacity is reduced by 10% of its initial capacity by sedimentation? The relationship between the trap efficiency η  (%) and capacity inflow C/I is given below: C/I η  %

0.01 43

0.02 60

0.04 74

0.06 80

0.08 84

0.1 87

0.2 93

0.3 95

0.5 96

0.7 97

Average annual stream flow = 12 cm of runoff Area of catchment = 100 km 2 = 100 × 10 6 m 2 12

6 3 Thus Annual flood inflow = (100 × 10 ). 100 m

12 × 10 6 m 3

Annual sediment inflow

= 12 M .m 3 2

= 0.03 ha-m/km  of the catchment

= 0.03 × 100ha − m = 0.03 × 100 × 10 4 m 3 = 0.03 Mm 3 It means that 0.03 Mcum of sediment flows every year into the dam/reservoir site, but the qua ntity of this, which is trapped in the reservoir, depends on the average trap efficiency, in turn depends on the C/I ratio. In the question, the total capacity to be filled up by sediment is 10% of the initial reservoir capacity, i.e. 10% x 4 Mcum=0.4 Mcum

49

HWRE AR 2010

Capacity of reservoir at the end = 3.6 Mcum

=

Capacity/inflow ratio at the start

η 

12 Mcum

= 0.3333

at start =95%

Capacity/inflow at the end = η 

4 Mcum

3. 6 12

= 0.30

at the end of the interval = 95%

Average η   = 95% Sediment trapped per year = 0.0375x 95% = 0.035625 Number of years during which 0.5 Mcum 0f sediment will be trapped =

0.4

 years = 11.2 years 0.035625

Hence, after 14 years, the 10% reservoir will get filled up.

 Example 10.3  Ackers and White Formula In a flow of these characteristics Depth, Dm Mean velocity, V  Surface gradient, S

2.0 m 1.5 m/s 0.001

Over a bed characterized by Representative grain size, D 35 0.0003 m Grain specific gravity, s 2.65 What rate of sediment transport should we expect per metre width of flow?

Solution 1

  g [s − 1]  3  = 6.954 First determine  D gr  =  D   2   ν    2

o

(Use ν   = 0.00000114 m  /s, appropriate to 15 C) Next determine the four Dgr – dependent parameters n, m, A, and C n = 1 − 0.56 log  D gr  = 0.5283 m

= 1.34 + 9.66 /  D gr 

= 2.729

 A = 0.14 + 0.23 /   D gr  C  = 10

2.86 log  D gr  − (log  Dgr  )

2

−3.53

= 0.2272 = 0.01477

With these we can get the dimensionless particle mobility, F gr:

F gr 

=

  V    gD [s − 1]  32 log (10 D m  D )  U ∗n

1− n

50

HWRE AR 2010

We need the shear velocity U ∗ , which we calculate from the standard formula for parallel flow

U ∗

So

F gr 

= =

g Dm S   = 0.1401 m/s 0.3540   1.5     0.06968  27.29 

0.4717

And hence we get the mean concentration

= 1.293 qS  as

n

 F gr   G gr  = C  = − 1  sD  V    A   c Dm U ∗  n

CsD   V     F gr 

    − 1 c=  Dm  U ∗     A 

m

m

= 0.000005872 X 3.5 X 67.90 = 0.001396 The units of this are strictly dimensionless that is to say Kg/Kg, and must be multiplied by  ρ  , density of water, to convert to mass / volume, the usual expression of concentration. Therefore to convert to the customary mg/l we multiply by 1000000 (  ρ   in mg/l) and obtain  ρ  c  = 1396 mg / l

To express sediment transport we use

qs

= cVDm ρ 

Working entirely in Kg-m units we get

qS  = 0.001396 X 1.5 X 2.0 X 1000 = 4.19 Kg/m/s  Example 10.4  Ackers and White Formula A stream has a mean flow depth of 0.815m and mean cross section velocity of 1.57ms-1. The sediment in -1 this stream has a representative grain size of 1.14mm, shear velocity of 0.098ms , and a density of 3 -2 2.65g/cm . Taking acceleration due to gravity, g = 9.81ms  and kinematic viscosity of water, υ = 1.04 x -6 2 10  m  /s, determine;

i) Particle size number, Dgr

−

Representative grain size, D = 1.14 x 10 -3m

−

Density of sediment,

−

Specific gravity of sediment, s = ρs / ρ = 2.65/1 =2.65

ρs = 2.65g/cm3, density of water, ρ = 1 g/cm3

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HWRE AR 2010

ii)

iii)

−

Dgr =

−

Dgr = 28.09

= 1.14 x 10-3(9.81(2.65-1)/( 1.04 x 10 -6)2)1/3

Particle mobility number, Fgr

−

Fgr =

−

Shear velocity of sediment,

−

mean cross section velocity, V = 1.57ms

−

mean flow depth, Dm = 0.815m

−

n = 1 – 0.56logD gr = 1 – 0.56 log 28.09 = 0.189

−

Fgr =

−

Fgr = 0.562

-1

= 0.098ms -1

Sediment transport parameter, Ggr

−

Ggr =

− − − −

m = 1.34 + (9.66/D gr) = 1.34 + 9.66/28.09 = 1.684 A = 0.14 + (0.23/ √ Dgr) = 0.14 + (0.23/ √28.09) = 0.183 2 2 Log C = 2.86 log D gr – (log Dgr)  – 3.53 = 2.86 log 28.09 – (log 28.09)  – 3.53 C = 0.0327

−

Ggr = 0.0327

−

Ggr = 0.111

Summary The deposition of sediment in a reservoir is known as reservoir silting or sedimentation. During sedimentation, when the water reaches a reservoir in the vicinity of a dam, the velocity and the turbulence are considerably reduced. The deposition of sediments will reduce the water storing capacity of the reservoir and if the process of deposition continues without being addressed it will lead to a significant reduction in reservoir capacity with unfavourable consequences. In this chapter, river classification based on sediment yield is discussed, followed by land degradation and the processes of sedimentation.This is

52

HWRE AR 2010

followed by the different effects of sedimentation, measures to control reservoir sedimentation and various methods of estimating sediment load with examples in the region.

References 1. Ackers P., White W.R. Sediment Transport: A New Approach and Analysis , Journal of the Hydraulic Division, ASCE, 1973, 99, No NY 11 2041-2060, New York, USA. 2. Arora S.O. Water Resources Hydropower and Irrigation Engineering , Standard Publishers and Distributors, 1996, New Dehli, India. 3. Azza N., The Dynamics of Shoreline Wetlands and Sediments of Northern Lake Victoria PhD Thesis, UNESCO-IHE Delft 2006, The Netherlands. 4. Bamutaze, Y.,Tenywa, M.M.,Mwanjalolo,M., The Effect of Land Use on Runoff and Soil Loss  from Wanale Micro- Catchment , Mt Elgon, Uganda, , International Sediment Initiative Conference, 12-15 November 2006, UNESCO Chair in Water Resources, Khartoum, Sudan. 5. Basson, G.,  Hydropower  Dams and Fluvial Morphological Impacts – An African Perspective , Proceedings United Nations Symposium on Hydropower and Sustainable Development, 27 th – 29th October 2004, Beijing, China. 6. Brune, G.M. Trap Efficiency of Reservoirs , Transactions of American Geophysical Union; 1953, Vol 344, No 3. 7. Chadwick A., Morfett J., Hydraulics in Civil Engineering , Allen and Unwin, 1989, London, UK. 8. Churchill M,A.,  Discussion of Analysis and Use of Reservoir Sedimentation Data , by L.C. Gottschalk, Proceedings, Federal Inter Agency Sedimentation Conference 1948, Denver, Colorado, USA. 9. Das G,, Hydrology and Soil Conservation Engineering , Prentice Hall, 2002, New Dehli, India. 10. Garde, R.J., Ranga Raju, K.G.,  Mechanics of Sediment Transportation and Alluvial Stream rd Problems 3  Edition, 1985,Wiley Eastern, New Delhi, India. 11. Garg, S.K.  Hydrology and Water Resources Engineering , Khanna Publishers, 1998, New Dehli, India. 12. Julein P.Y. Erosion and Sedimentation , Cambridge University Press, 1995 Ca mbridge UK. 13. Karamisheva R.D. Lyness J.F., Myers W.R.C., Cassells J.B.C. O’ Sullivan J. , Sediment Transport Formulae for Compound Channel Flows,  Proceedings of the Institution of Civil Engineers, Water Management 159, September 2006 Issue WM3, Pages 183-193 London,UK. 14. Karim M.F., Kennedy, J.F. Computer based Predictors for Sediment Discharge and Friction Factor of Alluvial Streams , Iowa Institute of Hydraulic Research, University of Iowa,1983, Report No 242, Iowa, USA. 15. Kawashima S., Johndrow T. Annandale G.W. Shah F. Reservoir Conservation Vol II: Rescon Model and User Manual, The World Bank 2003, Washington D.C. USA. 16. Kimaro, D., Isabirye, M., Semalulu, O., Magunda, M.K., Poesen, J.,Deckers, J.,  Magnitude of  Lake Victoria Sedimentation: Evaluation of Sediments and Agricultural Land Use on a Shoreline  Microcatchment , International Sediment Initiative Conference, 12-15 November 2006, UNESCO Chair in Water Resources, Khartoum, Sudan. 17. Magunda, M.K., Tenywa, M.M., Soil and Water Conservation in Agriculture in Uganda , Vol 1, General Information, 2001, Fountain Publishers/National Agriculture Research Organisation, Technical Centr for Agricultural and R ural Cooperation, Kampala, Uganda 18. Miller C.R., Analysis of Flow Duration , Sediment Rating Curve Method of Computing Sediment Yield, United States Bureau of Reclamation,1953, Virginia USA 19. Mwebembezi L, Kyoburungi G,  A Report on Sediment Sampling 2001 , Water Resources Management Department, Entebbe, Uganda.

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20. Novak, P., Moffat, A.I.B., Nalurri, C., Naranyanan, R., Second Edition,  Hydraulic Structures, E&FN Spon, 1996, London UK. 21. Palmieri A., Shah F., Annandale, G.W., Dina A,  Reservoir Conservation  Vol I, The Rescon Approach, The World Bank 2003, Washington D.C. USA. 22. Raghunath H.M.,  Hydrology, Principles, Analysis and Design , New Age International Ltd, 2005, New Dehli, India. 23. Read, H.H., Watson, J., Beginning Geology, George Allen and Unwin Ltd/ Macmillan Education Ltd, 1972, London, UK.. 24. Rowntree K.M., Wadeson R.A, A Hierarchical Geomorphological Model for the Classification of selected South African Rivers , Water Research Commission, Report No 497/1/99, Pretoria, South Africa. 25. Rugumayo A.I., Mwondha I., Musiime T; Sediment Yield Assessment and River Classification with Limited Data Sets ; Proceedings: Sustainable Water Resources Management in the Changing Environment of Monsoon Region, 2004 Colombo, Sri Lanka. 26. Siyam A.M., El Zein S., El Sayed S.M., Mirghani M, Golla S., Saleh A, Klaassen G.,,  Assessment of the Current State of the Nile Basin Reservoir Sedimentation Problems , NBCBNRE, 2005, Khartoum, Sudan 27. Tingasanchali, T. Khan N.M. Combating Reservoir Sedimentation, Proceedings, International Seminar on Managing Water Supply for Growing Demand Bangkok, Thailand, UNESCO 2006, Jakarta, Indonesia. 28. Yang. C.T. Unit Stream Power Equation for Total Load , Journal of Hydrology 1979, No 40 pp 123-138, London, UK. Further Reading rd

1. Dake,J.M.K.., Essentials of Engineering Hydraulics , 3  Edition, Macmillan, 1982,London, UK. 2.  Design of Small Dams , United States Department of the Interior, A Water Resources Technical Publication,1974, Washington, USA. 3. Linsely and Franzini: Water Resources Engineering , McGraw, 1979, London, UK 4. Viessman, W., J.W. Knapp, G.L. Lewis & T.E. Harbaugh (1977).  Introduction to Hydrology, Harper Row, New York Questions 1. The following information is available regarding the relationship between trap efficiency and capacity inflow ratio. Capacity inflow ratio

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Trap efficiency percent

87

93

95

95.5

96

96.5

97

97

97

97.5

Find the probable life of the reservoir with an initial reservoir capacity of 25 million cubic meters, if the average annual flood inflow is 50 million cubic meters and the avera ge annual sediment inflow is 150,000 tons. Assume a specific weight of the sediment equal to 1.2 gm per c.c. The usual life of the reservoir will terminate when 80% of its initial capacity is filled with sediment. 2. Explain the terms: i) Trap Efficiency ii) Capacity Inflow Ratio 3. How can we control reservoir sedimentation?

54

HWRE AR 2010

4. In a flow of these characteristics: Depth, Dm Mean velocity, V  Surface gradient, S

1.5 m 2.0 m/s 0.0015

over a bed characterized by Representative grain size, D 35 Grain specific gravity, s

0.0003 m 2.65

and using the Ackers and White formula, determine the rate of sediment transport expected per metre width of flow? 5. Distinguish between suspended load and bed load. 6. Describe the process of sedimentation in the different stages of a river. 7. What factors that affect the amount of sediment load in a river. 8. Explain the effects of sedimentation in a river. 9 Discuss the performance of three different formula for estimating sediment load. 10.Why should land degradation be addressed in engineering designs?

55