engineering hydrology - falmatasaba · preface this teaching material has been compiled from...
TRANSCRIPT
Engineering
Hydrology
Picture Here
Yilma Seleshi (Dr.)
April 2005
PREFACE
This teaching material has been compiled from lecture notes that I have given to the Civil
Engineering undergraduate and postgraduate students in the Department of Civil Engineering,
Addis Ababa University over the last seven years (1997-2004). The teaching material largely
follows the style and content of two standard text books in hydrology. The first text book is the
Prof. Ven Te Chow, David R. Maidment & Larry W. Mays Applied Hydrology and the second
text book is the Prof. Victor Miguel Ponce Engineering Hydrology, Principles and Practices.
Where possible, relevant examples based on Ethiopian rainfall and rivers data are given. Project
type examples taken from the practice in Ethiopia are also given in the Annexes in order to
illustrate the integrated nature of the hydrology study and problem solving is site specific.
The teaching material is broadly composed of 12 chapters dealing with different aspects of
hydrology. The first chapter introduces hydrology and hydrological cycle. The second chapter
discusses rainfall causes and measurements. The third chapter deals with evaporation from lakes
and evapotranspiration. The fourth chapter discusses infiltration as process and its measurements.
Streamflow measurement and hydrographs are discussed in chapter five Hydrology of small and
mid size watersheds are discussed in chapter six and seven respectively. Presentation on river
routing and reservoir routing are given in chapter eight and nine respectively. Hydorlogic
statistics and frequency analysis are given in chapter 11. Finally, chapter 12 briefly discusses the
basics of groundwater hydrology.
Practicing water resources engineers may find the teaching material useful although its main
target is to the undergraduate civil engineering student and to some extent to the postgraduate
students. The Author would very much appreciate receiving comments from students and other
users which improves the content and the style of this teaching material.
ACKNOWLEDGEMENTS
I am grateful to the Addis Ababa University Vice President for Research and Graduate Programs
office for financing the preparation of this teaching material. I would like to express my
appreciation to my wife W/o Hanna Berhanu for her encouragement and support in preparation of
this teaching material.
1 INTRODUCTION ................................................................................................... 1
1.1 Definition of Hydrology ........................................................................................ 1
1.2 Development of Hydrology.................................................................................... 2
1.3 Hydrological Cycle .......................................................................................... 3
1.4 Practice problems .................................................................................................. 8
2 RAINFALL ................................................................................................................. 9
2.1. Causes of rainfall over Ethiopia ....................................................................... 9
2.1.1. General circulation of the global atmosphere ............................................ 9
2.1.2 Formation of the Intertropical Convergence Zone ......................................... 13
2.1.3 The African sector of ITCZ and the associated rainfall ................................ 14
2.1.4 Ethiopian rainfall climate ............................................................................ 16
2.2 The physics of rainfall formation ................................................................... 19
2.2 The physics of rainfall formation ................................................................... 20
2.3 Estimation of Precipitable Water .................................................................... 21
2.4 Measurement of rainfall and optimum number of rain guages ....................... 22
2.5 Estimating missing rainfall data ..................................................................... 24
2.6 Point rainfall analysis and classification of storms ......................................... 29
2.6.1 Point rainfall analysis .................................................................................. 29
2.6.2 Storm temporal pattern classification........................................................... 30
2.7 Estimation of average depth of rainfall over a catchment ............................... 31
2.7.1 Changing point rainfall to area rainfall .......................................................... 31
2.7.2 The arithmetic mean method .................................................................. 32
2.7.3 The Thiessen polygon method ............................................................... 33
2.7.4 The isohyetal method ................................................................................... 33
2.9 Intensity-duration-frequency curves .................................................................... 35
2.10 Double mass analysis ......................................................................................... 37
2.12 Practice Problems .............................................................................................. 46
3. EVAPORATION....................................................................................................... 49
3.1 Definition of some meteorological variables .................................................. 49
3.2.1 Measurements of air and soil temperatures ............................................. 58
3.2.2 Sunshine recorder .................................................................................. 58
3.2.3 Wind speed and direction recorder ........................................................ 60
3.2.4. Dew point temperature measurement ........................................................... 61
3.2.5 Measurement of evaporation and evapotranspiration .................................... 61
3.3 Methods for estimating potential evaporation ................................................. 64
3.3.1 The energy balance method ................................................................... 64
3.3.2. Aerodynamic method ................................................................................... 69
3.3.3. Combined aerodynamic and energy balance method - the combination
method: ................................................................................................................. 71
3.4 Evapotranspiration ......................................................................................... 72
3.5. Analysis of the homogeneity of meteorological data series.................................. 73
3.6 Practice Problems ................................................................................................ 75
4. Infiltration ............................................................................................................... 77
4.1 Factors affecting infiltration .......................................................................... 77
4.2 Measurements of infiltration .......................................................................... 80
4.2.1 Areal measurement ...................................................................................... 80
4.2.2 Point measurement ........................................................................................ 80
4.3 Estimating infiltration rate ................................................................................. 82
4.3.1 Horton infiltration ........................................................................................ 82
4.3.2 The -index method ................................................................................... 85
4.3.3 The Phillip method ....................................................................................... 85
4.3.4* The Green-Ampt method ........................................................................... 86
4.4. Practice Problems .............................................................................................. 93
5. STREAMFLOW MEASUREMENTS and HYDROGRAPH ................................. 94
5.1 Stream gauging site selection ......................................................................... 94
5.1.1 Selection of gauging site ............................................................................. 95
5.2 Stage measurement ............................................................................................. 97
5.2.1 Manual gauge ............................................................................................... 97
5.2.1 Recording gauge .......................................................................................... 98
5.3 Flow velocity measurement and discharge computation .................................... 102
5.4 Dilution gauging .......................................................................................... 107
5.4.1 Sudden Injection method ............................................................................ 107
5.4.2 Continuous and constant rate injection method ........................................... 109
5.5 The slope-area method ...................................................................................... 110
5.6 Orifice formula for bridge opening ............................................................... 114
5.7 Stage discharge relationship - rating curve ........................................................ 117
5.9 Hydrograph separation ...................................................................................... 125
5.10. Practice problems........................................................................................... 128
6. Watershed properties .............................................................................................. 129
6.1 Watershed area ................................................................................................. 129
6.2 Watershed shape .............................................................................................. 130
6.3 Watershed relief ................................................................................................ 131
7 Hydrology of small watersheds ................................................................................ 150
7.1 The rational method ......................................................................................... 150
7.1.1 Determination of tc ............................................................................... 151
7.1.2 Estimation of runoff coefficient C .............................................................. 152
7.1.3 Composite watershed ................................................................................. 153
7.2 Application of the rational method to storm-sewer and culvert size design ... 156
7.2.1 Design of storm sewer ................................................................................ 156
7.3 Practice problems: ........................................................................................... 164
8. Hydrology of midsize watersheds .......................................................................... 166
8.1 The SCS method........................................................................................ 166
8.1.1 SCS Peak discharge and flood hydrograph determination ..................... 168
8.2. The Unit Hydrograph Method ...................................................................... 175
8.2.1 Derivation of unit hydrographs ............................................................. 175
8.3 S- hydrograph ................................................................................................. 181
8.4 Synthetic unit hydrograph ........................................................................... 182
8.5 Instantaneous Geomorphologic Unit Hydrograph............................................... 188
8.6 Practice problems: ............................................................................................ 191
9. River Routing ....................................................................................................... 195
9.1 The Muskingum Method .............................................................................. 196
9.2 Practice problems .............................................................................................. 203
10. Reservoir Routing ................................................................................................. 204
10.1. Level pool or reservoir routing using storage indication or modified pulse
method .................................................................................................................... 205
10.1.2 Reservoir routing with controlled outflow ................................................ 211
10.2 Practice problems........................................................................................... 213
11 Frequency Analysis ............................................................................................... 214
11.1 Concepts of statistics and probability ........................................................... 214
11.1.1 Frequency and probability functions. .................................................... 217
11.3 Statistical parameters ............................................................................... 221
11.4 Fitting data to a probability distribution........................................................ 223
11.5. Common probabilistic models ........................................................................ 226
11.5.1 The Binomial distribution......................................................................... 227
11.5.2 The exponential distribution: .................................................................... 229
11.5.3 Extreme value distribution ...................................................................... 230
11.5.4 Frequency Analysis using Frequency Factor............................................. 235
11.6 Probability plot ........................................................................................ 240
11.7. Testing for outliers .................................................................................. 242
11.8 Practice problems............................................................................................ 244
12. BASICS OF GROUNDWATER HYDROLOGY .............................................. 246
12.1 Introduction ................................................................................................. 246
12.2 Properties of Groundwater ............................................................................... 246
12.3 Groundwater movement ................................................................................... 251
12.3.1 Darcy’s law .......................................................................................... 251
12.3.2 Hydraulic conductivity K ......................................................................... 253
12.3.3 Specific storage and specific yield ............................................................. 256
12.3.4 Transmissivity and storage coefficient. ...................................................... 258
12.4. Determination of hydraulic conductivity in laboratory. ................................... 258
12.5. Flow nets ........................................................................................................ 260
12.6 Unconfined steady state flow: Depuit Assumption .......................................... 264
12.7 Steady-state well hydraulics ............................................................................ 268
List of Tables
Table 0-1: Main synoptic features affecting Ethiopian rainfall (summarized from Babu,
2002). .................................................................................................................... 17
Table 0-2. Recommended minimum densities of meteorological network as applied for
the different physiographic unit of a river basin (WMO, 1994). ............................. 24
Table 0-3: Rain gage chart analysis ........................................................................ 30
Table 0-4. Generalized design criteria - return period- for water-control structures ...... 35
Table 5-1.Recommended minimum density of hydrometric stations .............................. 95
Table 5-2. Values of Co and e in the orifice formula, L = Width of waterway, and W
=unobstructed width of the stream as defined in Figure 5.9: ................................. 114
Figure 1-1a: The hydrological cycle with major components (Ponce, 1989) ..................... 3
Figure 0-1. In general, the amount of solar energy absorbed by the Earth at each latitude
differs from the amount of terrestrial radiation emitted at the same latitude and so
energy has to be transferred from equatorial to polar region (after Burroughs,
1991). .................................................................................................................... 10
Figure 0-2. Hadley circulation: a vigorous upward branch in the tropics, fed by low-level
convergence of moist air flowing over the warm sea and driven by the latent heat
released from the "hot towers" of cumulo-nimbus clouds. The flow from these
cumulus towers can extend to the lower stratosphere, whence there is upper level
divergence as the air streams towards the sub-tropics. Under the action of radiative
cooling to space, this air sinks in the region of the sub-tropical high pressure
systems, thus completing the overall circulation (after Bonnel et al., 1993). ........... 11
Figure 0-3. The position of the ITCZ ( Inter tropical front in some zones) in February and
August (after Barry and Chorley, 1982). ................................................................ 12
Figure 0-4. The meso-scale and synoptic structure of the Intertropical Convergence Zone
(ITCZ), showing a model of the spatial distribution (above) and of the vertical
structure (below) of the convective elements which form the cloud clusters (from
Mason, 1970) ......................................................................................................... 13
Figure 0-5. map of mean annual African rainfall (mm) for approximately 1920 -73
showing the position of the Intertropical Convergence Zone (ITCZ) in July and
January; (b) map of the long-term average winds near the surface in July over and
near Africa (after Folland et al., 1991). ................................................................. 15
Figure 0-6. Schematic diagram of the structure of air masses over western Africa and the
rainfall process in the Sahel region (after Beran and Rodier, 1985). ...................... 16
Figure 0-7. Mean annual isohyetal map of Ethiopia (mm) ............................................ 19
Figure 0-8. Water droplets in clouds are formed by nucleation of vapor on aerosols, then
go through many condensation-evaporation cycle as they circulate in the cloud, until
they aggregate into large enough drops to fall through the cloud base (Chow, 1988).
.............................................................................................................................. 20
Figure 0-9. This diagram illustrates two commonly used non-recording instruments, the
standard BritishMeteorological Office gauge (A) and the U.S. Weather Bureau
standard gauge (B); a Nipher type gauge shield and a wire gauge cylinder which
can be used to assess horizontal interception (D). The construction and principle of
operation of three recording instruments is also shown along with an example of the
chart produced by a tilting-siphon-recording gauge. ............................................. 28
Figure 0-10. Rain gage chart from a rain gage of the reversible, recording type. .... 29
Figure 0-11: Typical rainfall intensity pattern ................................................................ 30
Figure 0-12. Depth-area, or area-reduction, curves. ..................................................... 32
Figure 5-1: Typical streamflow gauging station installed in the Wabi river near Dodola
town upstream of the Melkawakana reservoir (February 2002). ............................. 97
Figure 5-2. The measurement of stage through manual methods and recording
instruments (after Gregory and Walling, 1973) ...................................................... 99
Figure 5-3: A typical chart from vertical float recorder. .......................................... 101
Figure 5-4: Vertical and horizontal axis current meters and wading rod and cable
suspension mounting of the meter body. .............................................................. 102
Figure 5-5: Top: current meter mounted on a measuring rod, (bottom) suspended on a
cable from the bow of a jet-boat. Wide rivers flow (usually greater than 100 m) are
often measured using a boat- the Baro river near Sudan border is the case in
Ethiopia. .............................................................................................................. 103
Figure 5-6. The velocity area technique of discharge measurements: a cable way is used
on large streams for positioning the current meter in the verticals and a special cable
drum can be used to obtain accurate readings of depth and spacing of verticals. The
mean section and mid-section methods are commonly used to compute the discharge
of the individual segments. .................................................................................. 105
Figure 5-7. Dilution gauging: constant rate injection and gulp injection. .................... 108
Figure 5-8. Definition sketch of the orifice formula ..................................................... 116
Figure 5-9. Rating curves in linear (Top) and logarithmic scale of Zarema river near
Zarema, a tributary of Tekeze river (MWR).. ....................................................... 117
Figure 5-10 Daily discharge hydrographs for Wabi Shebele river at Melka Wakana for
the year 1969. ...................................................................................................... 120
Figure 5-11. Separation of sources of streamflow on an idealized hydrograph , (b) sources
of streamflow on a hillslope profile during a dry period, (c) and during a rainfall
event, (d) the extent of a stream network during a dry period, (e) and during a
rainfall event. ....................................................................................................... 123
Figure 5-12. Baseflow separation techniques (Chow et al. 1988). ................................ 126
INTRODUCTION 1 ___________________________________________________________________________
1 INTRODUCTION
1.1 Definition of Hydrology
Hydrology is a multidisciplinary subject that deals with the occurrence,
circulation and distribution of the waters of the Earth. In other words,
hydrology is the study of the location and movement of inland water, both
frozen and liquid, above and below ground. The domain of hydrology
embraces the physical, chemical, and biological reaction of water in natural
and manmade environment (Chow et al, 1988).
Hydrology is applied to major civil engineering projects such as irrigation
schemes, dams and hydroelectric power, and in planning water supply
projects. Hydrological information is essential in:
(1) Estimating reservoir storage capacity that is needed to ensure adequate water
supplies for municipal, irrigation and hydropower needs.
(2) Planning water resources projects the peak discharge and its volume of flood
that have to be adopted in design of irrigation, hydropower, and flood control
projects. If the selected flood is too high, it results in a conservative and
unnecessary costly structures while adoption of a low design flood can result
in the loss of the structure itself and devastating damage to downstream
residence and properties.
(3) Estimating the impact of watershed management on the quantity and quality of
the surface and the groundwater resources.
(4) Planning an integrated water resources development master plan for a basin.
(5) Trans-boundary river water allocation problems, and
(6) Delineation of a probable flood levels to plan a protection of settlements and
projects from flooding or to promote better zoning.
INTRODUCTION 2 ___________________________________________________________________________
1.2 Development of Hydrology
From the very beginning mankind attempted to utilize the precious water
resources of the Earth in a thoughtful way. History tells us that Samaritans and
Egyptians along the Nile Delta, Chinese along the banks of the Hwang - Ho
and Aztecs in South America applied detailed methods for their water
resources management (Shaw, 1994).
The Greek philosophers were the first students of hydrology, with Aristotle
proposing the conversion of moist air into water deep inside mountains as the
source of springs and streams. Homer suggested the idea of an underground
sea as the source of all surface waters.
During the Renaissance, a gradual change occurred from purely philosophical
concepts of hydrology toward observational science. Leonardo da Vinci
(1452-1519) made the first systematic studies of velocity distribution in
streams using a weighted rod held afloat by an inflated bladder. The rod would
be released at a point in the stream, and Leonardo would walk along the bank
marking its progress with an odometer and judging the difference between the
surface and the bottom velocities by the inclination of the rod. Later the
Frenchman, Pierre Perrault (1608-1680), measured surface runoff and found it
to be only a fraction of rainfall.
The year 1850 might be regarded as marking the beginning of the
development of methods in current use in hydrological practice. In 1851,
Mulvaney first described the concept of time of concentration that now forms
the backbone of the rational method of runoff computation, and he also
designed primitive form of rain-gauge that would record time-varying rainfall
intensity during a storm. Five years later Darcy established the basic law of
groundwater motion.
During the following decades, knowledge gradually accumulated: in 1871
Saint Venant derived the equations of one-dimensional surface water flow, in
1891 Manning developed his equation for open channel velocity, in 1908 the
first watershed level measurement of the hydrologic effects of the land-use
change was done, in 1911 Green and Ampt produced their infiltration model,
and in 1925 Streeter and Philps developed the dissolved oxygen sag curve for
rivers.
INTRODUCTION 3 ___________________________________________________________________________ The period from 1930 to 1950 produced a significant step forward in the field
of hydrology as government agencies especially in USA began to develop
their own programs of hydrologic research. Sherman’s 1933 unit hydrograph,
Horton’s 1933 infiltration theory, and Theis’s 1935 nonequilibrium equation
in well hydraulics advanced the state of art significantly. Gumbel 1958
proposed the use of extreme value distribution for frequency analysis of
hydrological data laying a ground for modern statistical hydrology
(Maidement 1993).
1.3 Hydrological Cycle
Water on earth exists in a space called the hydrosphere that extends about 15
km up into the atmosphere and about 1 km down into the lithosphere, the crust
of the earth. Water circulation in the hydrosphere through numerous paths
forms the hydrological cycle. It can be said that the hydrological cycle has no
beginning or end and its main processes occur continuously.
Figure 1-1a: The hydrological cycle with major components (Ponce, 1989)
INTRODUCTION 4 ___________________________________________________________________________
Figure 1.1b. The hydrological cycle with major components (another perspective)
INTRODUCTION 5 ___________________________________________________________________________
Table 1.1: Estimate of the world’s water (Maidement, 1993).
Volume (10 6 km
3 )
Percentage of total
water
Ocean
1370.00
96.50
Groundwater
Fresh
Saline
10.53
12.87
0.76
0.93
Ice sheet and glaciers
24.00
1.65
Lakes:
Fresh
Saline
0.0910
0.0850
0.0070
0.0060
Soil moisture
Biological water
0.0160
0.0012
0.0015
0.0001
Rivers
Marshes
0.0021
0.0110
0.0002
0.0008
Atmospheric vapor
0.0130
0.0010
Table 1.1 gives the relative quantities of the earth’s water contained in each of
the phases of the hydrological cycle. The oceans contain 96.5 % of the earth’s
water, and of the 3.5 % on land, approximately 1% is contained in deep, saline
groundwater or in saline lakes, leaving only 2.5 % of the earth’s water as fresh
water that is 35 million cubic kilometer. Of this fresh water, 68.6% is frozen
into the polar ice caps and a further 30.1 % is contained in shallow aquifers,
leaving only 1.3% of the of the earth’s fresh water mobile in the surface and
atmospheric phases of the hydrological cycle.
The driving force of the circulation is derived from the radiant energy received
from the Sun. The largest atmospheric moisture sources of the earth are
Pacific, Atlantic and Indian oceans. Heating of ocean surface causes
evaporation, the transfer of water from the liquid to the gaseous state, to form
part of the atmosphere, then the water vapor changes back to the liquid again
through the process of condensation to form clouds and, with favorable
atmospheric conditions, precipitation (rain or hail) is produced either to
INTRODUCTION 6 ___________________________________________________________________________ return directly to the ocean storage or to the land surface. Snow may
accumulate in polar regions or on high mountains and consolidate into ice.
In more temperate lands, rainfall may be intercepted by vegetation from
which the intercepted water may return at once to the air by evaporation.
Remaining rainfall reaching the ground may collect to form surface runoff or
it may infiltrate into the ground. The liquid water in the soil then percolates
through the unsaturated layers to reach the water table where the ground soil
stratum becomes saturated, or it is taken up by vegetation from which it may
be transpired back into the atmosphere. The surface runoff and base flow
forming stream-flow or river-flow flows into lakes, swamps, or seas, or
oceans.
An example of marco-hydrological cycle of the Ethiopian part of the Nile
basin is described as follows. Part of rainfall collected over western Ethiopia is
destined to the Mediterranean sea through the Baro, Akobo, Blue Nile and
Atbara rivers. Mediterranean sea is connected to the Atlantic Ocean and to the
Indian Ocean via the Red Sea. Clouds formed from the moisture evaporated
form the Atlantic and Indian Ocean will come back as rainfall over Ethiopian
highlands to complete a micro hydrological cycle.
It appears that the concept of hydrologic cycle is simple, but the phenomenon
is enormously complex and intricate. It is not just one large cycle but rather is
composed of many interrelated cycles of continental, regional, and local
extent. Moreover, although the total volume of water in the global hydrologic
cycle remains essentially constant, the distribution of this continually changing
on continents, in regions, and within local drainage basins.
Under certain well-defined conditions, the response of a watershed to rainfall,
infiltration and evaporation can be estimated if simple assumptions are made.
A watershed is defined as an area of land that drains to a single outlet and is
separated from other watershed by a watershed divide. A water budget
equation connects the elements of the hydrological cycle. For example, for a
given watershed a water budget equation for time step of t is given by
Pt - SRt - Gt- Et - Tt = St (1.1)
Where Pt = rainfall (mm)
SRt = Surface runoff (mm)
Gt = groundwater flow (mm)
INTRODUCTION 7 ___________________________________________________________________________ Et = Evaporation (mm)
Tt = Transpiration (mm)
St = Change in storage (mm)
Note that if t is a year, it may be assumed for preliminary analysis that what is
infiltrated will be shown up in the groundwater flow, and thus the infiltration
term may not be considered in Eq. (1.1) and the change in storage term may be
zero.
Example 1.1 In a given year, a watershed with a drainage area of 215 km2 received
900 mm of rainfall. The average flow rate measured at the outlet of the watershed was
3.1 m3/s. Estimate the amount of water lost due to the combined effects of
evaporation and transpiration. Assume the annual change in storage is zero.
Solution:
The equivalent runoff depth in mm over the watershed is calculated by dividing the
annual volume of runoff by the watershed area.
Runoff depth = (3.1 * 86400 s/day * 365 d/year * 1000mm/m ) / (215 km2 * 106 m2/km2 )
Then using Eq.(1.1): Pt - (Rt + Gt) - Et - Tt = St, we estimate (Et + Tt )
900 mm - (3.1 * 86400 s/day * 365 d/year * 1000mm/m ) / (215 km2 * 10
6
m
2/km
2 ) - (Et + Tt ) = 0.
(Et + Tt ) = 900 - 454.7 = 445.3 mm
445.3 mm of rainfall is consumed annually by evaporation and transpiration over the
whole watershed.
INTRODUCTION 8 ___________________________________________________________________________
1.4 Practice problems
1.1. The following yearly data were collected from a 2000 km2 catchment. Total
precipitation is 620 mm, total combined loss due to evaporation and
evapotranspiration is 350 mm, estimated groundwater outflow is 100 mm,
and mean surface runoff is 150 mm. What is the change in volume of water
(m3) remaining in storage in the catchment at the end of the elapsed year.
1.2. The average annual discharge of the Nile river having basin area of 2.96x106
km2 is 100 x10
9 m
3. Calculate the discharge per unit area in m
3/s/100-km
2.
1.3. A watershed with a drainage area of 450 km2 received 700 mm of
rainfall in a given rainy three months. The average flow rate measured at the
outlet of the watershed over these three months was 15 m3/s. Estimate the
amount of water lost due to the combined effects of evaporation and
transpiration, and groundwater storage.
1.4. What is a hydrological cycle? How does it keep a balance between the water
of the earth and the moisture in the atmosphere?
1.5. List the major water resources projects in your area. What specific
hydrological problems did each project involve?
Rainfall 9
____________________________________________________________
2 RAINFALL
It is common observation that in rainy months dark clouds of large extent often yield
rains of various magnitude and duration. Questions such as sources of rain bearing
clouds, transportation mechanism of these clouds to Ethiopia need to be answered.
In this chapter attempt is made to explain the above questions by describing first the
general global circulation of the atmosphere, followed by the atmospheric circulation
in Africa, and finally, we discuss the physical rainfall producing mechanisms over
Ethiopia.
Note that in this teaching material rainfall has the meaning as precipitation,
although strictly defined rainfall is the liquid form of rainfall excluding hail
and snow. In tropical areas snow-falls has not been observed.
2.1. Causes of rainfall over Ethiopia
To explain causes of rainfall over Ethiopia, first the general circulation at global scale
is described, then African scale atmospheric circulation will be treated. Finally the
circulation that affect the Ethiopian climate under the general setting will be
discussed.
2.1.1. General circulation of the global atmosphere
The weather features we consider here range from local breezes and shower
clouds to the great wave patterns that circle the globe. Formally weather is
defined as the day-to-day state of the atmosphere, consists of short-term
variations of energy and mass exchanges within the atmosphere and between
the earth and the atmosphere. It results from processes that attempt to equalise
differences in the distribution of the net radiant energy from the sun. Acting
over an extended period of time, these exchange of processes accumulate to
become climate. All of these systems are part of the process of the atmosphere
transport of energy. At the largest scale this is a reflection of the fact that the
solar energy absorbed in equatorial region is greater than the outgoing infrared
radiation, whereas in polar regions the reverse applies (Figure 2.1). So, to
achieve a global energy balance the atmosphere and the oceans must transport
energy from the equator to the poles. This is the engine that drives the
principal components of the global weather machine.
Rainfall 10
____________________________________________________________
If the Earth's surface were smooth and of uniform composition, the long-term
mean patterns of wind, temperature, and rainfall will show nothing but zonal
bands but with no longitudinal variation. Moreover, if the Earth did not rotate
energy transport would involve a simple meridional circulation with air rising
at the equator and flowing to the poles, descending, and returning at low level
to the equator. With a rotating Earth the motion involves horizontal vortices
and waves. The distribution of the oceans and continents across the globe
make these motions still more complex, but the broad features retain many of
the features of the simple zonal models (Burroughs, 1991).
Figure 2-1. In general, the amount of solar energy absorbed by the Earth at each
latitude differs from the amount of terrestrial radiation emitted at the same latitude
and so energy has to be transferred from equatorial to polar region (after Burroughs, 1991).
A schematic representation of the zonal wind systems near the Earth's surface
consists of easterly trade winds in the tropics, calm subtropical high pressure
zones, mid-latitude westerlies, and stormy low and high pressure zones close
to the poles. In cross-section these zones can be represented as three
Rainfall 11
____________________________________________________________
circulation systems covering the tropics, the mid-latitudes, and the polar
regions.
The weather in tropical regions is dominated by vertical circulation. Known as
the Hadley cell, after George Hadley, the general motion consists of air rising
near the equator to the heights of up to 20 km and then spreading out 20 -
30N and 20 - 30S latitudes before descending and flowing back toward the
equator. Schematic diagram of the Hadley circulation is shown in Figure 2.2.
Figure 2-2. Hadley circulation: a vigorous upward branch in the tropics, fed by low-
level convergence of moist air flowing over the warm sea and driven by the latent heat released from the "hot towers" of cumulo-nimbus clouds. The flow from these
cumulus towers can extend to the lower stratosphere, whence there is upper level
divergence as the air streams towards the sub-tropics. Under the action of radiative
cooling to space, this air sinks in the region of the sub-tropical high pressure systems, thus completing the overall circulation (after Bonnel et al., 1993).
Rainfall 12
____________________________________________________________
Figure 2-3. The position of the ITCZ ( Inter tropical front in some zones) in February and August (after Barry and Chorley, 1982).
The rising air at the equator is humid, and cools as it rises. This leads to the
formation of touring shower clouds which girdle the Earth and produce heavy
rainfall to the equatorial regions. The precise position of this band of
convective activity, known as the Intertropical Convergence Zone (ITCZ),
varies with the movement of the noon-day Sun throughout the year. It tends to
follow the Sun and the distribution of the oceans and continents in the
equatorial regions.
Figure 2.3 shows the position of the ITCZ in February and August (from Barry
and Chorley, 1982). Note that in view of the discontinuity of convergence in
time and space, the term Inter-tropical Confluence (ITC) is now preferred. In
the following section we discuss in details the characteristics of the ITCZ.
Rainfall 13
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2.1.2 Formation of the Intertropical Convergence Zone
The convection generated by frictional induced convergence in the Trade
Wind boundary layer produces individual cumulus clouds 1 to 10 km in
diameter, which group into meso-scale convective units of some 100 km
across, and that these, in turn, form cloud clusters
Figure 0-4. The meso-scale and synoptic structure of the Intertropical Convergence
Zone (ITCZ), showing a model of the spatial distribution (above) and of the vertical structure (below) of the convective elements which form the cloud clusters (from
Mason, 1970)
Rainfall 14
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100 to 1000 km in diameter (see Fig. 2.4) either along the ITCZ or in the
troughs of lower tropospheric wave disturbances which have wavelengths of
2000 to 3000 km.
Seasonal shifts of the wind field convergence are partly a response to
alternating activity in the subtropical high-pressure cells of the two
hemispheres but, on a shorter time scale, the synoptic activity along the ITC
obscures any simple relationships.
2.1.3 The African sector of ITCZ and the associated rainfall
Most of the rainfall in sub-Saharan Africa is associated with the African sector
of the ITCZ. This ITCZ is associated with the convergence of air streams from
the subtropical highs. Where these flows meet, strong upward motion occurs
which, provided the air contains sufficient moisture, will cause rainfall. The
lifting of the moist air is further enhanced by mountains that are inducing
orograhic rainfall.
Figure 2.5b shows the average convergence of the near surface winds into the
ITCZ in July near and over Africa (Folland et al., 1991). Moist lower and
mid-tropospheric monsoon winds that originate from the tropical Atlantic
enter the region from desert. In the eastern third of the Sahel, the south to
south-west winds may sometimes originate from central Africa or the Indian
Ocean. The ITCZ moves northward in the summer into the southern Sahara
desert in response to increased solar heating and reaches its most northern
position in August (Figure 2.5a).
The position of the ITCZ shown in Figure 2.5b is that at the Earth's surface; a
few kilometres up in the atmosphere the ITCZ is several degrees of latitude
further south. Most of the rainfall associated with the ITCZ generally falls
over a wide band starting several degrees to the south of its surface position,
since here the moist south-west monsoon airflow is sufficiently deep for
rainfall to occur .
Rainfall 15
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Figure 0-5. map of mean annual African rainfall (mm) for approximately 1920 -73
showing the position of the Intertropical Convergence Zone (ITCZ) in July and January; (b) map of the long-term average winds near the surface in July over and
near Africa (after Folland et al., 1991).
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Figure 0-6. Schematic diagram of the structure of air masses over western Africa and the
rainfall process in the Sahel region (after Beran and Rodier, 1985).
2.1.4 Ethiopian rainfall climate
Ethiopia is a mountainous country with about 1.1 million km2 area, and with large
part of the country lying between 1800 and 2400 m above mean sea level. The
highest mountain rises over 4600 m. There are lowland regions (-200 to 500 m) in
the extreme boundaries of the country.
Most of Ethiopia has tropical climate moderated by altitude with a marked wet
season. The eastern lowlands are much dries with a hot semi arid to desert
climate. In the highlands of Ethiopia, temperatures are reasonably warm around
the year but rarely hot.
In Ethiopia in general there are three seasons: the first is the dry season (locally
known as Bega) which prevails from October to January, the second is the small
rainy season (Belg) that runs from February to May and the third is the main rainy
season (Keremt) which prevails from June to September. Rainfall is above 1000
mm a year almost everywhere in the highlands and it rises to as much as 2000 -
3000 mm in the wetter southwestern parts. Annual rainfall decreases when one
Rainfall 17
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moves to the east and north of the country. Night time temperature fall to nearly
or below freezing in mountainous area (> 2500 m). In the northern lowlands,
Danakil depression, the southern lowlands, and Ogaden rainfall is low (< 300
mm/year) and temperatures are high (> 30 C) around the year.
Since Ethiopia is situated in the northeast of Africa it is influenced by the
northeast, to the Southeast and Southwest (west African) monsoons bringing
moisture from the Indian and Atlantic Oceans. In the boreal summer the moisture
gradually penetrates into the countries as the African sector of the Intertropical
Convergence Zone progresses northward. In general, annual rainfall amounts
decreases from the south to the north but the topography as well strongly
influences the rainfall. Ethiopia has one of the largest highland areas in the
tropics; it represents 50% of the land above 2000 m in Africa. The main synoptic
features that affects the Ethiopian rainfall are given Table 2.1.
Table 0-1: Main synoptic features affecting Ethiopian rainfall (summarized from Babu,
2002).
Major synoptic features affecting rainfall over Ethiopia
Season ITCZ (south
Atlantic ocean
effect / El Nino,
SOI )
North Indian
Ocean effect
Low level Jet
and Tropical
easterly Jet
Remarks
June – Sept
(main rainy
season)
ITCZ moves
northwards to Red
Sea. Most of Ethiopia receives
rains
SST condition
influence the
main rains
Active and
moves
northwards
South and
southeastern parts of
Ethiopia do not receive rains.
Feb – May
(small rainy
season)
ITCZ is in south
Ethiopia bringing rains to south and
southwestern
Ethiopia.
Moisture
source for eastern,
southeastern,
and some central
highlands part
of Ethiopia
receives useful rains.
Moves
northwards
As important as the
main rain season for eastern and
northeastern Ethiopia
Oct – Jan
(“dry
season”)
ITCZ is located
further south and brings rain for
extreme south and
southeastern
Ethiopia.
Occasionally
causes some untimely
rainfall in
most part of
Ethiopia
Weak and
migrate southwards
Crop harvesting time
in most of Ethiopia
Rainfall 18
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Rainfall 19
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Figure 0-7. Mean annual isohyetal map of Ethiopia (mm)
Rainfall 20
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2.2 The physics of rainfall formation
The formation of rainfall requires the lifting of an air mass in the atmosphere so
that it cools and some of its moisture condenses. The three mechanisms of air
mass lifting are frontal lifting, where warm air is lifted over cooler air by frontal
passage; orographic lifting, in which an air mass rises to pass over a mountain
range; and convective lifting, where air is drawn upwards by convective action,
such as in the center of a thunderstorm cell.
Figure 0-8. Water droplets in clouds are formed by nucleation of vapor on
aerosols, then go through many condensation-evaporation cycle as they circulate
in the cloud, until they aggregate into large enough drops to fall through the cloud
base (Chow, 1988).
Droplets become
heavy, enough to fall
(0.1 mm)
Droplets
increase in size
by condensation
Many droplets
decrease in size
by evaporation
Some droplets
increase in size by
impact and
aggregation
Droplets form by
nucleation – condensing of
vapor on tiny solid particles
called aerosols (0.001 – 10
m)
Large drops break
up ( 3 – 5 mm)
Rain drops (0.1 – 3 mm)
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Formation of rainfall in clouds is illustrated in Fig. 2.8. As air rises and cools,
water condenses from the vapor to the liquid state. If the temperature is below the
freezing point, then ice crystals are formed instead. Condensation requires a seed
called a condensation nucleus around which the water molecules attach or
nucleate themselves. Particles of dust floating in because the ions electrostatically
attract the polar-bonded water molecules. Ions in the atmosphere include particles
of salt derived from evaporated sea spray, and sulfur and nitrogen compounds
resulting from combustion. The tiny droplets grow by condensation and impact
with their neighbors as they are carried by turbulent air motion, until they become
large enough so that the force of gravity overcomes that of friction and they begin
to fall, further increasing in size as they hit other droplets in the fall path.
However, as the drops falls, water evaporates from its surface and the drop size
diminishes, so the drop may be reduced to the size of an aerosol again and be
carried upwards in the cloud through turbulent action.
2.3 Estimation of Precipitable Water
Precipitable water is the total amount of water vapor in column of air expresses as
the depth of liquid water in mm over the base area of the column. The precipitable
water (w) gives an estimate of maximum possible rainfall under the unreal
assumption of total condensation. It is given by:
(2.1)
Where p is in mb, qv in g/kg and g=9.81 m/s2.
In practice we use the following discrete form of the above equation:
Example 2.1: From a radiosonde (balloon) ascent, the pairs of measurements of pressure
and specific humidity shown in Table below were obtained. The precipitable water in a
column of air up to the 250 mb level is calculated.
p q g
0.1 = (mm) W v
p
p
2
1
(2.2)
dp q g
0.1 = (mm) W v
p
p
2
1
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Pressure (mb) 1005 850 750 700 620 600 500 400 250
Specific
humidity qv
(g/kg)
14.2 12.4 9.5 7.0 6.3 5.6 3.8 1.7 0.2
p 155 100 50 80 20 100 100 150
Mean qv 13.3 10.9 8.2 6.6 5.9 4.7 2.7 0.9
Meanqv . p 2061 1095 4125 532 119 470 275 142
The sum of the qv p = 5107.5, the precipitable water up to the 250 mb level is given by
0.1 /g*5107.5= 52.1 mm.
2.4 Measurement of rainfall and optimum number of rain
guages
The basic instrument for rainfall measurement is rain gauge, which samples the
incidence of rainfall at a specific point, through an orifice of known area. Figure
2.9 shows commonly used non-recording and recording instruments. A recording
rain gage records the time it takes for rainfall depth accumulation. Therefore, it
provides not only a measure of rainfall depth but also of rainfall intensity. The
slope of the curve showing accumulated rainfall depth versus time is a measure of
the instantaneous rainfall intensity.
A rainfall gauge site should have some level ground and has ideal shelter. A rain
gauge should be placed from a tree or a building at more than two times of the
height of the tree or the building. It is to be noted that inconsistencies in rainfall
record are often caused by change in the rain gauge site location and
surroundings.
Optimum number of rain gauges. Statistics has been used in determining the
optimum number of rain gauges required to be installed in a given catchment. The
basis behind such statistical calculations is that a certain number of rain gauge
stations are necessary to give average rainfall with a certain percentage of error. If
the allowable error is more, lesser number of gauges would be required. The
optimum number of rain gauges (N) can be obtained using:
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Where: Cv = Coefficient of variation of rainfall based on the existing rain gauge
stations;
E = Allowable percentage error in the estimate of basic mean rainfall
Example: There are four rain gauge stations existing in the catchment of a river. The
average annual rainfall values at these stations are 800, 620, 400 and 540 mm
respectively. (a) Determine the optimum number of rain gauges in the catchment, if it is
desired to limit error in the mean value of rainfall in the catchment to 10%. (b) How
many more gauges will then be required to be installed?
Solution. The mean rainfall is (800 + 620 + 400 + 540) / 4 = 590 mm; the standard
deviation of station annual rainfall is 166.93 mm. The coefficient of variation = mean /
standard deviation = 166.93 / 590 = 28.29
The optimum number of rain gauges is then
Additional number of stations required is 8 - 4 = 4.
WMO recommends a guideline for checking the adequacy of a meteorological
network. Table 2 shows this guideline. It is important to note that to assess the
adequacy of the network of a basin, the number of stations and their spatial
distribution play an important role.
It is important to carefully plan the location of rain gauges in relation to the
location of stream flow gauging stations, in order to perform extension of stream
flow records, flood forecasting, or hydrological analysis. As guideline, rainfall
gauging should be located so that watershed rainfall can be estimated for each
stream gauging station. It is also suggested that gauges are located in the upper,
middle and lower part of a watershed giving good spatial coverage.
2
E
C = N v (2.3)
2
E
C = N v = 004.8
10
29.282
= N
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Table 0-2. Recommended minimum densities of meteorological network as
applied for the different physiographic unit of a river basin (WMO, 1994).
Region Minimum density (km2/ station)
Non-recording Recording
Mountainous region 250 2 500
Interior plains / Flat regions
under difficult conditions
575 5 750
Hilly / undulating 575 5 750
Arid zones* 10 000 100 000
* Arid zone herein is defined as annual rainfall < 450 mm, temperature greater than 27 0C, potential evaporation is 20 times greater than annual rainfall
2.5 Estimating missing rainfall data
Due to the absence of observer or instrumental failure rainfall data record
occasionally are incomplete. In such a case one can estimate the missing data by
using the nearest station rainfall data. If for example rainfall data at day 1 is
missed from station X having mean annual rainfall of Nx and there are three
surrounding stations with mean annual rainfall of N1, N2, and N3 then the missing
data Px can be estimated
Px = (P1 +P2 +P3)/3
(2.3a)
provided N1, N2, and N3 differ within 10% of Nx).
or
provided N1, N2, or N3 differ by more than 10% of Nx . Equation 2.3b is the
method of normal ration.
The second method is a reciprocal weighting factor which takes into account the
distance between the missing data gauge and the other gauges surrounding the
missed gauge. These methods are illustrated in Example 2.2 and 2.3.
Example 2.2 Rainfall station X was inoperative for part of a month during which a
storm occurred. The respective storm totals at three surrounding stations A, B, and C
)N
NP +
N
NP +
N
NP(
3
1 = P
3
x3
2
x2
1
x1x (2.3b)
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were 107, 89, and 122 mm. The normal annual rainfall amounts of station X, A, B, and C
are 978, 1120, 935, and 1200 mm respectively. Estimate the storm rainfall for station X
for its missing month.
Solution Nx = 978 mm, 10% of Nx = 97.8 mm. Thus the maximum permissible
annual rainfall At either of the three stations for taking ordinary mean 978+97.8 = 1075.8
mm. But stations A and C normal annual rainfall are > than 1075.8 so use the normal
ratio method, with that Px is estimated to be 95.3 mm.
Example 2.3.
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Rainfall 27
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Figure 0-9. This diagram illustrates two commonly used non-recording instruments, the standard
BritishMeteorological Office gauge (A) and the U.S. Weather Bureau standard gauge (B); a
Nipher type gauge shield and a wire gauge cylinder which can be used to assess horizontal
interception (D). The construction and principle of operation of three recording instruments is
also shown along with an example of the chart produced by a tilting-siphon-recording gauge.
Rainfall 29
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2.6 Point rainfall analysis and classification of storms
2.6.1 Point rainfall analysis
A typical recording rain gage chart is given in Figure 2.10. The line on the chart is
a cumulative rainfall curve, the slope of the line being proportional to the intensity
of the rainfall. The peak is the point of reversal of the recording gage. An example
of rain gage chart analysis is given in Table 2.3.
Figure 0-10. Rain gage chart from a rain gage of the reversible, recording type.
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Table 0-3: Rain gage chart analysis
Time
(a.m.)
Time
Interval
(min)
Cumulative
time
(min)
Rainfall
during
interval
(mm)
Cumulative
Rainfall
(mm)
Rainfall
Intensity
for Interval
(mm/h)
6:50
7:00 0 10 1 1 6
7:10 10 20 10 11 60
7:15 5 25 11 22 132
7:35 20 45 46 68 138
7:45 10 55 19 87 114
8:25 40 95 31 118 47
9:10 45 40 6 124 8
10:50 100 240 6 130 4
2.6.2 Storm temporal pattern classification
Since no two rainstorms have the same time-intensity relationship, it is often
convenient to group storms with regard to their characteristics as described by
rainfall intensity histogram. Commonly found rainfall intensity patterns such as
uniform intensity, advanced pattern, intermediate pattern, delayed pattern are
shown in Figure 2.11.
Figure 0-11: Typical rainfall intensity pattern
Uniform
Delayed
Advanced
Intermediate
Time
Intensity
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In general, the cold front produces a storm of an advanced type, and the warm
front a uniform or intermediate pattern.
2.7 Estimation of average depth of rainfall over a catchment
Several methods are commonly used for estimating average rainfall over a
watershed. Choice of method requires judgment in consideration of quality and
nature of the data, and the importance, use, and required precision of the result.
Here in three methods of estimating areal average depth of rainfall are discussed.
The first is the arithmetic mean method, the second is the Thiessen polygon
method, and the third is the isohyetal method. Before we discuss these methods,
first we will elaborate techniques for changing point rainfall to area rainfall in
case of only one rain gauge is available in and around the watershed.
2.7.1 Changing point rainfall to area rainfall
When the area of a basin exceeds about 25 km2, rainfall observations at a single
station, even if it is at the center of the catchment, will usually be inadequate for
the design of drainage works. Rainfall records within the catchment and its
immediate surroundings thus must be analyzed to take proper account of the
spatial and temporal variation of rainfall over the basin.
For areas large enough for the average rainfall depth to depart appreciably from
that at a point, one should apply area-reduction factor. Fig. (2.12) provides curves
for calculating areal depths as a percentage of point rainfall values.
Rainfall 32
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Figure 0-12. Depth-area, or area-reduction, curves.
2.7.2 The arithmetic mean method
The central assumption in the arithmetic mean method is that each rainguage has
equal weight and thus the mean depth over a watershed is estimated by:
where: Pj = the station j,
N = the total number of rainguages in and around the watershed.
It is a simple method, and well applicable if the gages are uniformly distributed
over the watershed and individual gage measurements do not vary greatly about
their mean.
N
P = P
jN
j=1
(2.4)
Rainfall 33
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2.7.3 The Thiessen polygon method
The Thiessen polygon method involves assigning relative weights to the gages in
computing the areal average. The assumption in the method is that at any point in
the watershed, the rainfall is the same as that at the nearest gage so the depth
recorded at a given gage is applied out to a distance halfway to the next station in
any direction.
The relative weights of each gage are determined from the corresponding areas of
application in a Thiessen polygon network, the boundaries of the polygons being
formed by the perpendicular bisectors of the lines joining adjacent gages.
(2.5)
Where: Aj = the area of polygon j in the watershed (km2)
Pj = rainfall amount in polygon j (mm)
P = average rainfall (mm)
The disadvantages of the Thiessen method are its inflexibility that is addition of
new station implies construction of new polygon, and it does not directly account
for orographic influences of rainfall.
2.7.4 The isohyetal method
The isohyets are drawn between the gauges over a contour base map taking into
account exposure and orientation of both gauges and the catchment surface. The
rainfall calculation is based on finding the average rainfallPi between each pair
of isohyets, and the area between them in the watershed Aj. Equation (2.5) then
used to estimate the average rainfall over the catchment.
The method is good where there is a dense network of raingages. It is also flexible
and considers orographic effect.
Example 2.4 Rainfall averaging methods: Three raingauges (p1, p2, p3) are installed in a
catchment having an area of 250 km2. Determine the average rainfall using the arithmetic
A = A ,PA A
1 = P J
J
1=j
jj
J
1=j
Rainfall 34
____________________________________________________________
mean, Thiessen polygon, and isohyetal methods if a daily rainfall amounts of p1 = 45
mm, p2 = 48 and p3 = 52 mm were recorded in July 15, 1998..
Solution:
(a) Arithmetic mean method: P = (45 +48 + 52 ) / 3 = 48.3 mm.
(b) The Thiessen polygon method: The area created by the dotted lines and the boundary
of the watershed surrounding each rainfall station is measured. Accordingly for
(c) The isohyetal method:
51 mm
48 mm
45 mm
42 mm
39 mm
36 mm
54 mm
p1 p3
p2
p1 p3
p2
A1 = 94 km2, A2 = 92 km2 and A3 = 64 km2.
Using equation 2.5 we get
P = 1/250 (45*94 + 48*92 + 52*64) = 47.8 mm
p1 =45 p3 = 52
p2 = 48
Areas in between contours beginning from
36 mm line in km2 are:
A1 = 8, A2 =17, A3= 32, A4 =40, A5= 70, A6
= 83. P = 1/250 ((36+39)*0.5* 8+
(39+42)*0.5*17+(42+45)*0.5*32+(45+48)*0.
5*40+(48+51)*0.5*70+(51+54)*0.5*83) =
48.2 mm
Rainfall 35
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2.9 Intensity-duration-frequency curves
Intensity-duration-frequency (IDF) curves (equations) summarize conditional
probability (frequency)of rainfall depths or average intensities. Specifically, IDF
curves are graphical representations of the probability that a certain average
rainfall intensity will occur, given a duration.
IDF curves are used in many hydrological design projects involving urban
drainage, bridge sizing, spillway sizing, etc where there is a need to determine
design storm magnitude (or intensity of rainfall for specified duration) for
required return period.
In establishing IDF curves it is important to know commonly used design
frequencies (return period) for a specific water control structures. Table 2.4 gives
generalized design criteria - return period- for water-control structures.
Table 0-4. Generalized design criteria - return period- for water-control structures
Type of structures Return period (years)
Highway culverts
Low traffic
Intermediate traffic
High traffic
5 -10
10 - 25
50 - 100
Highway bridges
Secondary system
Primary system
10 -50
50 -100
Farm drainage
Culverts
Ditches
5 -50
5 -50
Urban drainage
Storm sewer in small cities
Storm sewer in large cities
2 - 25
25 -50
Airfield
Low traffic
Intermediate traffic
High traffic
5 -10
10 -25
50 -100
Levees
On farms
Around cities
2 -50
50 -200
Dams (small -large) ( 50 - 1000+)
Rainfall 36
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IDF curves can be described mathematically to facilitate calculations. For
example, one can use
(2.6)
Where:
i = the design rainfall intensity (mm/hr)
t = the duration (hr)
c = a coefficient which depends on the exceedence probability
e & f = coefficients which vary with locations
For a given return period, the three constants can be estimated to reproduce i for
three different t's spanning a range of interest. An example of IDF curve for Addis
Ababa is shown in Figure 2.15 .
Figure 2.15: Intensity-duration-frequency curves of Addis Ababa for different return periods
used in the design of the Addis Ababa Ring Road Project.
ft
ci
e
IDF curves of Addis Ababa
0
50
100
150
200
250
300
0 1 2 3 4 5 6
Time (hr)
Rain
fall
(m
m /
hr)
T = 1000
T = 100
T = 50
T = 25
Rainfall 37
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2.10 Double mass analysis
Double mass curve technique is often used to test the consistency of rainfall
record. The procedure is that accumulated rainfall at the gauge station whose
record is in doubt is plotted as ordinate versus the average concurrent
accumulated average rainfall of nearby stations whose rainfall data are reliable.
The procedure is illustrated in Figure 2.16.
Figure 2.16. Double mass analysis technique
Where there is a break point in the graph is noted, the doubt station data may be
adjusted to the previous slope value if the reason for doing so is convincing.
0
100
200
300
400
500
600
0 100 200 300 400 500
Cum
ula
tive a
nnual
valu
es o
f sta
tion z
(m
m)
Cumulative annual value of reference station (mm)
Double-mass anlaysis
Breakpoint
Rainfall 38
____________________________________________________________
Example 2.5.
Rainfall 39
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Rainfall 40
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Rainfall 41
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2.11 Rainfall regime in Ethiopia.
ERA (2002) drainage manual provides six rainfall regime areas (Figure 2.17)
established based on rainfall stations shown in Table 2.4. Also using the statistical
analyses, rainfall intensity-duration curves have been developed for commonly
used design frequencies see “Figures 5-5 through 5-12” of ERA 2002 Manual.
These basic information will be used in estimating peak flood for small and large
watersheds.
Table 2.4 Meteorology Stations with relatively long record period
Meteorological
Region
Station Years of
Record
Meteorological
Region Station Years
of
Record
A1 Axum 18 B Bedele 19
Mekele 35 Gore 45
Maychew 24 Nekempte 27
A2 Gondar 40 Jima 45
Debre Tabor 22 Arba Minch 11
Bahir Dar 35 Sodo 28
Debre Markos 44 Awasa 26
Fitche 25 C Kombolcha 46
Addis Ababa 33 Woldiya 23
A3 Nazareth 40 Sirinka 17
Kulumsa 31 D1 Gode 29*
Robe/Bale 19 Kebri Dihar 38
A4 Metehara 28 D2 Kibre Mengist 24
Dire Dawa 46 Negele 45
Mieso 35 Moyale 18
* max 24 hour rainfall not given Yabelo 34
Years of record through 1997
Rainfall 42
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Figure 2-17 Rainfall Regions (ERA, 2002)
Note: Rainfall data used in the preparation of this figure have been collected from many Ministry
of Water Resources meteorology stations (see Table 2-4). In the course of the preparation of this manual, they have been subjected to statistical techniques. The results indicate that the country
can be divided into the above hydrological regions displaying similar rainfall patterns. The
information is subject to review, and future data may indicate the need for a further refinement in both values and regional boundaries.
Rainfall 43
____________________________________________________________
Intensity-Duration-Frequency
Regions A1 & A4
Figure 5-9
0,0
50,0
100,0
150,0
200,0
250,0
300,0
350,0
400,0
0 10 20 30 40 50 60 70 80 90 100 110 120 130
Duration, min.
Inte
nsit
y, m
m/h
r 2 Year
5 Year
10 Year
25 year
50 Year
100 Year
Rainfall 44
____________________________________________________________
Intensity-Duration-Frequency
Regions A2 & A3
Figure 5-10
0
50
100
150
200
250
300
350
400
0 10 20 30 40 50 60 70 80 90 100 110 120 130
Duration, min.
Inte
ns
ity,
mm
/hr 2 Year
5 Year
10 Year
25 year
50 Year
100 Year
Rainfall 45
____________________________________________________________
Intensity-Duration-Frequency
Bahir Dar & Lake Tana
Figure 5-12
0,0
50,0
100,0
150,0
200,0
250,0
300,0
350,0
400,0
0 10 20 30 40 50 60 70 80 90 100 110 120 130
Duration, min.
Inte
ns
ity,
mm
/hr
2 Year
5 Year
10 Year
25 year
50 Year
100 Year
0.0
50.0
100.0
150.0
200.0
250.0
300.0
350.0
400.0
0 10 20 30 40 50 60 70 80 90 100 110 120 130
Inte
ns
ity, m
m/h
r
Duration, min.
Intensity-Duration-FrequencyRegions B, C & D
Figure 5-11
2 Year
5 Year
10 Year
25 year
50 Year
100 Year
Rainfall 46
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2.12 Practice Problems
1.1 The following rainfall data were recorded at a station for storm of August
24-25, 1981.
1.2 Plot the rainfall hyetograph, compute & plot the cumulative rainfall
hyetograph. Calculate the maximum depth and intensity recorded in 10, 20, and 30
minutes for this storm.
Time (min) 0 5 10 15 20 25 30 35 40
Rainfall (mm) - 2.0 5.0 6.2 5.3 5.2 4.0 3.0 1.0
2.2 The mean and standard deviation of the annual maximum rainfall depths for various
duration recorded at a town are shown below. Determine for each duration, the design
rainfall intensity for return periods of 2, 5, 10, 25, 50, and 100 years. Use the Extreme
Value Type I (Gumbel) distribution. Plot the results as a set of intensity-duration-
frequency curves.
Duration Mean depth (mm) Standard deviation
(mm)
5 min 13 3
10 min 20 6
15 min 26 8
30 min 38 13
1 hr 49 17
2 hr 56 21
3 hr 63 20
1 day 105 63
2.3 The annual precipitation at station X and the average annual rainfall at 8
neighboring stations whose data are reliable are given below. Check the consistency of
the annual rainfall data at station X.
Year
8-stations annual
average rainfall (mm)
Annual rainfall at
station X (mm)
1972 28 35
1973 29 37
1974 31 39 1975 27 35
Rainfall 47
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1976 25 30
1977 21 25
1978 17 20
1979 21 24 1980 26 30
1981 31 31
1982 36 35 1983 39 38
1984 44 40
1985 32 28
1986 30 25 1987 23 21
2.4 The watershed divide line for a given watershed can be approximated by a polygon
whose vertices are located at the following coordinates in km: (5, 5), (-5, 5), (-5, -5), (0, -
10), and (5, -5). The rainfall amounts of a storm were recorded by a number of rain
gauges situated within and nearby the watershed as follows:
Gage number Coordinates (km) Recorded rainfall
(mm)
1 (7, 4) 62
2 (3, 4) 59
3 (-2, 5) 41
4 (-10, 1) 39
5 (-3, -3) 105
6 (-7, -7) 98
7 (2, -3) 60
8 (2, -10) 41
9 (0, 0) 81
Determine the average rainfall on the basin by (a) the arithmetic mean method, (b) the
Thiessen method, and (c) the isohyetal method if the maximum rainfall line is on the
ridge running southwest to northeast through (-3, -3).
Rainfall 48
____________________________________________________________
2.5 The table below gives the annual rainfall amounts of 6 rain gauges installed in a
catchment of 100 km2 . Thiessen polygon area associated for each rain gauge is also
given in km2. It is planned to build a dam at the outlet of the catchment.
Consider the following conditions:
(I) a minimum discharge of 200 l/s should be released from this proposed dam to downstream users throughout the year,
(II) 35 % of the annual rainfall is lost from the catchment through infiltration
and evapotranspiration, and (III) the annual net evaporation loss from 15
km2 surface area of the reservoir is 400mm.
Calculate the annual areal rainfall in the catchment
(b) The annual average net volume of water
Annual rainfall data (mm) and Thiessen polygon area in (km2)
Rain gauge
1
2
3
4
5
6 1999 rainfall (mm)
2052
1915
1868
1723
1640
1510
Thiessen polygon area (km
2)
12
14
25
21
15
13
Evaporation 49
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.
3. EVAPORATION
Evaporation occurs when water is converted into water vapor at the evaporating
surface, the contact between water body and overlapping air. At the evaporative
surface, there is a continuous exchange of liquid water molecule into water vapor
& vice versa.
The two main factors influencing evaporation from an open water surface are the
supply of energy to provide the latent heat of vaporization, and the ability to
transport the vapor away from the evaporative surface. The latent heat of
vaporization lv is the amount of heat absorbed by a unit mass of a substance. Solar
radiation is the main source of heat energy.
The magnitude of annual evaporation is highly dependent on the prevailing
climate in and around the water body. Evaporation has significant impact on
water resources development especially in arid and semi-arid regions.
Evaporation from Lake Nasser in Egypt (arid region) is about 3000 mm/year,
where as evaporation from Lake Koka is about 1500 mm/year, which is half of
that from Lake Nasser.
Evaporation rate is a function of several meteorological and environmental factors
such as net radiation, saturation vapor pressure, actual vapor pressure of air, air
and water surface temperature, wind velocity and atmospheric pressure. Section
3.1 discusses definition and measurements of these variables.
3.1 Definition of some meteorological variables
The atmosphere forms a distinctive, protective layer about 100 km thick around
the Earth. To the hydrologist, the troposphere (the first 11 km) is the most
important layer because it contains 75% of the weight of the atmosphere and
virtually all its moisture. On average, the temperature from ground level to the
tropopause falls steadily with increasing altitude at the rate of 6.5 oC /km. This is
known as the lapse rate.
Evaporation 50
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.
Evaporation at high altitudes is promoted due to low atmospheric pressure as
expressed in the psychrometric constant. The effect is, however, small and in the
calculation procedures, the average value for a location is sufficient. A
simplification of the ideal gas law, assuming 20°C for a standard atmosphere, can
be employed to calculate atmospheric pressure P:
(3.1)
where:
P atmospheric pressure [kPa],
z elevation above sea level [m],
The psychrometric constant, , is given by:
(3.2)
where
= Psychrometric constant [kPa °C-1
],
P = Atmospheric pressure [kPa],
= Latent heat of vaporization, 2.45 [MJ kg-1
],
cp = Specific heat at constant pressure, 1.013 10-3
[MJ kg-1
°C-1
], and
= ratio molecular weight of water vapour / dry air = 0.622.
The specific heat at constant pressure is the amount of energy required to increase
the temperature of a unit mass of air by one degree at constant pressure. Its value
depends on the composition of the air, i.e., on its humidity. For average
atmospheric conditions a value cp = 11.013 10-3
[MJ kg-1
°C-1
] can be used as an
average atmospheric pressure is used for each location.
Air density: air density of moist air (kg/m3)is estimated by a = 3.486 (p/(275 +
T)) where p is the atmospheric pressure in kPa and T is air temperature in degrees
Celsius.
Evaporation 51
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.
Water vapor: the amount of water vapor in the atmosphere is directly related to
the temperature. The water vapor content or humidity of air is usually measured
as a vapor pressure, and the units used is millibar (mb).
Specific humidity: The mass of water vapor per unit mass of moist air is called
specific humidity qv and equals the ratio of the densities of water vapor v and of
moist air a
Vapor pressure: Dalton's law of partial pressures states that the pressure exerted
by a gas (its vapor pressure) is independent of the pressure of other gases; the
vapor pressure e of the water vapor is given by the ideal gas law as
where T is the absolute temperature in K and Rv is the gas constant for water
vapor. If the total pressure exerted by the moist air is p, then p-e is the partial
pressure due to the dry air, and
The gas constant for water vapor is
a
v
v = q (3.3)
TR = e vv (3.4)
vda + = (3.6)
T R = e - p dd (3.5)
0.622
R = R
dv (3.7)
Evaporation 52
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.
where 0.622 is the ratio of the molecular weight of water vapor to the average
molecular weight of dry air.
Combining Eqs.(3.4), (3.5.) and (3.7) we get
The specific humidity qv is approximated by
where TR = p aa
The relationship between the gas constants for moist air and dry air is given by
Saturation vapor pressure es : For a given air temperature, there is a maximum
moisture content the air can hold and the corresponding vapor pressure is called
saturation vapor pressure es . At this vapor pressure, the rates of evaporation and
condensation are equal.
Over a water surface the saturation vapor pressure is related to the air temperature
with equation
Where es is in Pascal (Pa = N/m2) and T is air temperature in degree Celsius.
Due to the non-linearity of the above equation, the mean saturation vapour
pressure for a day, week, decade or month should be computed as the mean
TR )0.622
+ ( = p dv
d
(3.8)
p
e0.622 = q
v (3.9)
J/Kg.K 287 = R ),q0.608 + (1 R = R dvda (3.10)
T + 237.3
T17.27 611 = es exp (3.11)
Evaporation 53
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.
between the saturation vapour pressure at the mean daily maximum and minimum
air temperatures for that period. That is
(3.12)
Using mean air temperature instead of daily minimum and maximum
temperatures results in lower estimates for the mean saturation vapour pressure.
The corresponding vapour pressure deficit (a parameter expressing the
evaporating power of the atmosphere) will also be smaller and the result will be
some underestimation of the reference crop evapotranspiration. Therefore, the
mean saturation vapour pressure should be calculated as the mean between the
saturation vapour pressure at both the daily maximum and minimum air
temperature.
The relative humidity Rh: It is ratio of actual vapor pressure to its saturation
value at a given air temperature T and is given by
Dew-point temperature Td : The dew-point temperature Td is the temperature at
which space becomes saturated when air is cooled under constant pressure and
with constant water-vapor content. It is the temperature having a saturation vapor
pressure es equals to the existing vapor pressure e. Wet bulb thermometer
measures the dew point temperature.
Saturation deficit is the difference between the saturation vapor pressure at air
temperature es and the actual vapor pressure represented by the saturation vapor
pressure at Td which is the amount of water vapor in the air. The saturation deficit
(es - e) represents the further amount of water vapor that the air can hold at the
temperature Ta before becoming saturated.
e
e = R
s
h (3.13)
Evaporation 54
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.
Figure 3.1 Saturated vapor pressure as a function of temperature over water. Point C has
vapor pressure e and temperature T, for which the saturated vapor pressure es. The
temperature at which the air is saturated for vapor pressure e is the dew-point temperature
Td.
Figure 3.1 shows the saturation vapor pressure curve and the Td and T, es and e
relationship. If the barometric pressure is kept constant and the temperature is
reduced, i.e. if the air is cooled at constant barometric pressure, a stage will come
when the air will become saturated with the same amount of vapor.If the cooling
is continued, the vapor will get condensed on the contact surfaces. This
condensation will be in the form of dew if the dew point is > O 0C; and it will be
in the form of frost if the dew point is < 0 0C.
Example 3.1 At a climatic station, air pressure is measured as 100 kPa, air temperature
as 20 0C, and the wet-bulb, or dew-point, temperature as 16
0C. Calculate the
corresponding vapor pressure, relative humidity, specific humidity, and air density.
Solution: The saturated vapor pressure at T = 20 0C is given by
)T + 237.3
T(17.27 611 = es exp
Evaporation 55
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.
= 2339 Pa
and the actual vapor pressure e and the relative humidity are calculated using the dew-
point temperature Td=16 C
The relative humidity is = e/es
= 1819 /2339
= 0.78
= 78%
The air density is calculated from the ideal gas law
)20 + 237.3
20*17.27( 611 = es exp
1819Pa=
)16 + 237.3
16*17.27( 611 = e exp
p
e0.622 = qv
air moist kg
waterof kg0.01133 =
)100000
18190.622( = qv
)T(K)q0.608 + 287(1
p =
v
a
Evaporation 56
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.
= 1.18 kg/m3
Note that the actual vapor pressure can be determined from the difference
between the dry and wet bulb temperatures, the so-called wet bulb depression.
The relationship is expressed by the following equation:
ea = e° (Twet) - g psy (Tdry - Twet)
(3.14)
where
ea = Actual vapour pressure [kPa],
e°(Twet) = Saturation vapor pressure at wet bulb temperature [kPa],
= Psychrometric constant [kPa °C-1
],
Tdry-Twet = Wet bulb depression, with Tdry the dry bulb and Twet the wet bulb
temperature [°C].
The psychrometric constant of the instrument is given by:
gpsy = apsy P (3.15)
where apsy is a coefficient depending on the type of ventilation of the wet bulb
[°C-1
], and P is the atmospheric pressure [kPa]. The coefficient apsy depends
mainly on the design of the psychrometer and rate of ventilation around the wet
bulb. The following values are used:
apsy = 0.000662 for ventilated (Asmann type) psychrometers, with an air
movement of some 5 m/s,
0.000800 for natural ventilated psychrometers (about 1 m/s),
0.001200 for non-ventilated psychrometers installed indoors.
.01133)293*0.608 + 287(1
100000 =
a
Evaporation 57
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.
3.2 Measurements of some meteorological variables
A site for a meteorological station need to be level ground about 10 m by 7 m in
extent covered by short grass and enclosed by open fencing or railings. The site
should not have any steep slopes in the immediate vicinity and should not be
located near trees or buildings. A recommended site plan for the instrument is
shown in Fig. 3.2.
Figure 3.2 Plan of a meteorological station for the northern hemisphere (Shaw,
1994)
Evaporation 58
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.
3.2.1 Measurements of air and soil temperatures
In the ordinary Stevenson Screen, for example, two vertically hung thermometers
are for direct reading of the air temperature (dry bulb) and the reading of the wet
bulb, covered with muslin kept moist by a wick leading from a small reservoir of
distilled water. With these two temperature readings the dew point, vapor pressure
and relative humidity of the air are obtained. Supported horizontally are
maximum and minimum thermometers. The four thermometers are read at 0800
A.M. each day and at this time the maximum and minimum thermometers are
reset. Soil and ground/grass temperature measurements are often taken using soil
and earth/grass thermometers.
The dry and wet bulb temperatures are measured using psychrometers. Most
common are those using two mercury thermometers, one of them having the bulb
covered with a wick saturated with distilled water, and which measures a
temperature lowered due to the evaporative cooling. When they are naturally
ventilated inside a shelter, problems can arise if air flow is not sufficient to
maintain an appropriate evaporation rate and associated cooling. The Assmann
psychrometer has a forced ventilation of the wet bulb and dry bulb thermometers.
The dry and wet bulb temperature can be measured by thermocouples or by
thermistors, the so called thermocouple psychrometers and thermo sound
psychrometers. These psychrometers are used in automatic weather stations and,
when properly maintained and operated, provide very accurate measurements.
3.2.2 Sunshine recorder
A standard Campbell-Stockes sunshine recorder is shown in Fig. 3.3. The
working principle is that the glass sphere focuses the Sun's rays on to a specially
treated calibrated card where they burn a trace. The accumulated lengths of burnt
trace gives a measure of the total length of bright sunshine in hours.
Evaporation 59
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.
Figure 3.3 Sunshine hour recorder Mk. 2 (Campbell - Stokes)
Evaporation 60
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.
3.2.3 Wind speed and direction recorder
A cup anemometer is fixed on a 2 m long pole from the ground and the electrical
recording apparatus is housed conveniently away from the installation. The cup
anemometer can give instantaneous readings of wind velocity (m/s) or provide a
run-of-the-wind a collective distance in km when the counter is read each day
(Figure 3.4).
Figure 3.4 NMSA 1st class meteorological station housed in the cumpus of the
Alemaya University (Photo 2001).
Wind speeds measured at different heights above the soil surface are different.
Surface friction tends to slow down wind passing over it. Wind speed is slowest at
Cup generator anemometer
Evaporation 61
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.
the surface and increases with height. For this reason anemometers are placed at a
chosen standard height, i.e., 10 m in meteorology and 2 or 3 m in
agrometeorology. For the calculation of evapotranspiration, wind speed measured
at 2 m above the surface is required. To adjust wind speed data obtained from
instruments placed at elevations other than the standard height of 2m, a
logarithmic wind speed profile may be used for measurements above a short
grassed surface:
(3.16)
where:
u2 = wind speed at 2 m above ground surface [m/s],
uz =measured wind speed at z m above ground surface [m/s],
z = height of measurement above ground surface [m].
Classes of mean monthly wind speed are (1) les than 1 m/s light wind, (2)
between 1 and 3 m/s light to moderate wind, (3) from 3 to 5 m/s moderate to
strong wind, and (4) above 5 m/s strong wind. Where no wind data are available
within the region, a value of 2 m/s can be used as a temporary estimate. This
value is the average over 2000 weather stations around the globe.
3.2.4. Dew point temperature measurement
Dew point temperature is often measured with a mirror like metallic surface that
is artificially cooled. When dew forms on the surface, its temperature is sensed as
Tdew. Other dew sensor systems use chemical or electric properties of certain
materials that are altered when absorbing water vapour. Instruments for
measuring dew point temperature require careful operation and maintenance and
are seldom available in weather stations. The accuracy of estimation of the actual
vapour pressure from Tdew is generally very high.
3.2.5 Measurement of evaporation and evapotranspiration
Pan evaporation measurement
A practical way to measure evaporation directly is by the use of an evaporation
Evaporation 62
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.
pan. The pan exposes free water surface to the air, and the evaporation rate is
determined by measuring the water loss during one time period, usually one day.
Due to the difference in area of exposure and surrounding meteorological
conditions, evaporation from lakes is less than from the one obtained from the pan
measurement (with annual average multiplying factor about 0.7).
The National Weather Service Class A pan is recommended by the World
Meteorological Organization as a standard instrument for evaporation
measurements. The Class A pan is made of unpainted galvanized iron, has a
diameter of 122 cm and a height of 25.4 cm and is mounted about 15 cm above
the ground on supports which permit free flow of air around and under the pan
(Figure 3.5).
Figure 3.5 Class A evaporation pan
Water loss is determined by daily measurements of water level using a
micrometer hook gage installed in a stilling well set inside the pan. The pan is
initially filled to a height of 20 cm and is refilled when the water level has fallen
below 17.5 cm. Daily evaporation is computed as the difference between two
Evaporation 63
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.
successive observations, corrected to account for any intervening precipitation
measured in a nearby gage. An alternative procedure is to add a measured amount
of water daily to bring the water level in the pan up to a fixed point in the stilling
well. This procedure permits a more accurate measurement of water loss and
assures that the pan has the proper water level at all times.
Little is know about the spatial variability of evaporation. For general purpose and
preliminary evaporation estimates, a density of one station per 5000 km2 appears
to be sufficient.
Indirect measurement of evapotranspiration based on water balances of
watersheds and lakes:
A basic water balance equation, which is applied over a particular time interval,
is given by:
Where:
E = net evapotranspiration loss from the specified volume per unit area
(mm)
P = net precipitation (or irrigation) input to the specified volume per unit area
(mm)
VR = net volume of liquid water entering or leaving the specified volume as
measured inflow or outflow both above and below the surface (m3)
VS = change in liquid water stored within the specified volume (m3)
VL = “leakage,”, i.e., that total volume of liquid water leaving the specified
volume which is not, or cannot be, measured, and which therefore represents an
error in the method (m3)
A = effective area of the sample volume at the land surface (m2)
River runoff is arguably the most accurate hydrologic measurement and is a
valuable, direct determination of the available surface water resource. Careful
gauging can provide stream-flow measurements accurate to about 2 %. Using
carefully selected and well-managed paired watersheds can provide valuable and
convincing evidence of the consequences of land-use change on
evapotranspiration.
AV VVP= E LSR /)(*1000 (3.17)
Evaporation 64
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.
A systematic uncertainty in the evaporation loss deduced from a catchment water
balance arises from the possibility that the unmeasured leakage forms a
significant part of the total water balance.
Lysimeters. A lysimeter is a device in which a volume of soil, typically 0.5 to 2.0
m in diameter, which may be planted with vegetation, is isolated hydrologically
so that leakage VL = 0 in Eq. (3.11). It either permits measurement of drainage VR
or makes it zero and, in the case of a weighing lysimeter, the change in water
storage VS is determined by weight difference. If evapotranspiration from the
lysimeter is to be representative of the surrounding area, it should contain an
undisturbed sample of the soil and vegetation.
3.3 Methods for estimating potential evaporation
Potential Evaporation E0 (mm/day) defined as the quantity of water evaporated
per unit area, per unit time from an idealized extensive free water surface under
existing atmospheric conditions. Potential Evaporation of a given area varies
daily, and is following the variations of the weather.
The three common methods of estimating evaporation will be discussed herein:
the energy balance method, the aerodynamic method, and the combination
method.
3.3.1 The energy balance method
This method is widely used for estimating the amount of evaporation from a large
body of water such as lakes, reservoirs etc.
Consider an evaporation pan of a circular tank containing water, in which the rate
of evaporation is measured by the rate of fall of the water surface (Er = -dh/dt).
Based on the continuity and energy equation, one can derive the energy balance
equation for evaporation as
G) - H - R( l
1 = E sn
wv
r
(3.18)
Evaporation 65
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.
If the sensible heat flux Hs (sensible heat loss to surroundings atmosphere to raise
the temperature) and the ground heat flux G are both zero, then an evaporation
rate Er can be calculated as the rate at which all the incoming net radiation is
absorbed by evaporation:
where lv = latent heat of vaporization (J/kg), [lv (kJ/kg) = 2500 - 2.36 T (oC) up
to 40 OC]
w = water density (kg/m3)
Rn = net radiation (W/m2)
Er = rate of evaporation (m/s)
Example 3.2 Calculate by the energy method the evaporation rate from an open water
surface, if the net radiation is 200 W/m2
and the air temperature is 25 C, assuming no
sensible heat or ground heat flux.
Solution: The latent heat of vaporization at 25 C is lv = 2500-2.36*25 =2441 kJ/kg.
Density of water at 25 C is 997 kg/m3
Net radiation estimation
Radiometer or actinometer measures radiant energy received by the ground. For
most studies of evaporation, incident all-wave radiation data are adequate because
wv
nr
l
R = E (3.19)
mm/day 7.10=
mm/day 86400*1000*O1*8.22=
10*8.22=
997*1000*2441
100=
l
R = E
8-
8-
wv
nr
Evaporation 66
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.
the reflectivity of water is nearly constant (average daily values of 5 and 3 %,
short and long wave respectively).
The net radiation Rn is the net input of radiation at the surface at any instant. It is
the difference between the radiation absorbed Ri (1 -) where Ri is the incident
radiation, and that emitted Re.
= albedo, it is the fraction of reflected radiation, for deep water bodies is
about 0.08 because deep water bodies absorb most of the radiation they receive,
and for grass land and a range of agricultural crops = 0.23. In contrast fresh
snow reflects most of the incoming radiation with as high as 0.9, see Table 3.1.
Table 3.1 Plausible values for daily mean short wave radiation reflection coefficient
(Albedo) for broad land cover classes (Maidement, 1993)
Land cover class Short-wave radiation
Reflection coefficient
Open water 0.08
Tall forest 0.11 - 0.16
Tall farm crops (e.g., sugarcane) 0.15 - 0.20
Cereal crops (e.g., wheat) 0.20 - 0.26
Short farm crops (e.g. sugar beet) 0.20 - 0.26
Grass and pasture 0.20 - 0.26
Bare soil 0.10 wet - 0.35 dry
In the absence of measured solar radiation data, the total incoming short-wave
radiation can in most cases be estimated from measured sunshine hours according
to the following empirical relationship:
Where:
n/N = cloudiness fraction
n = bright sunshine hours per day, h
N = total day length, h
R - ) - (1R = R ein (3.20)
oi SN
n = R )61.035.0( (3.21)
Evaporation 67
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.
So = extraterrestrial radiation, MJ m2 day
-1 (Table 3.2)
Table 3.2 Mean solar radiation for cloudless skies, So (MJm -2
day-2
)
Lat.
Deg
Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
0 28.18 29.18 30.02 28.47 26.92 26.25 26.67 27.76 29.60 29.60 28.47 26.80
10 25.25 26.63 29.43 29.60 29.60 29.31 29.43 28.76 29.60 28.05 25.83 24.41
20 21.65 25.00 28.18 30.14 31.40 31.82 31.53 30.14 28.47 25.83 22.48 20.50
30 17.46 21.65 25.96 29.85 32.11 33.20 32.66 30.44 26.67 22.48 18.30 16.04
40 12.27 17.04 22.90 28.34 32.11 33.49 32.66 29.18 23.73 18.42 13.52 10.76
Net long-wave radiation.- There is a significant exchange of radiation energy
between the earth's surface and the atmosphere in the form of radiation at longer
wave lengths, i.e., in the range 3 to 100 m. Both the ground and the atmosphere
emit black-body radiation with a spectrum characteristics of their temperature.
Since the surface is on average warmer than the atmosphere, there is usually a net
loss of energy as thermal radiation from the ground.
The exchange of long-wave radiation Ln between vegetation and soil on the one
hand and atmosphere and clouds on the other, can be represented by the following
radiation law:
Where
Lo = outgoing long-wave radiation (ground to atmosphere), MJm2day
-1
Li = incoming long-wave radiation (atmosphere to ground), MJm2day
-1
f = adjustment for cloud cover
= net emissivity between the atmosphere and the ground
= the Stefan Boltzmann constant = 4.903x10-9 M J m
-2 day
-1 K
-4 = 5.67*10
-8 W/m
2.K
4
T = the absolute air temperature of the evaporating surface in degrees Kelvin (C +273)
The net emissivity can be estimated from
TfLL = L4
oin (3.22)
Evaporation 68
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.
where: a and b = correlation coefficients, a lies in the range 0.34 to 0.44,
and b in the range -0.14 to -0.25.
ed = saturated vapor pressure at dew temperature (kPa)
Adjustment for cloudiness f factor may be estimated from:
Where n/N = ratio of actual to possible hours of sunshine
Note that for general purposes when only sunshine hours, temperature, and
humidity data are available, net radiation (MJ m2 day
-1 ) can be estimated by the
following equation:
Where:
Rn = Net radiation ((MJ m2 day
-1 )
= albedo from Table 3.1
n/N= ratio of actual to possible hours of sunshine
S0 = mean solar radiation from cloudless sky from Table 3.2 (MJ m2 day
-1 )
ed = saturated vapor pressure at dew temperature (kPa)
= the Stefan Boltzmann constant = 4.903x10-9
M J m-2
day--1
K-4
T = the absolute air temperature of the evaporating surface in degrees Kelvin (C +
273)
Rn can be expressed as an equivalent depth of evaporated water in mm by
dividing Rn by w, where w (kg/m3) and (MJ/kg).
eba d (3.23)
N
nf 1.09.0 (3.24)
TeN
nS
N
nR 4
don )14.034.0)(1.09.0()61.035.0)(1( (3.25)
Evaporation 69
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.
3.3.2. Aerodynamic method
Besides the supply of heat energy, the second factor controlling the evaporation rate from
an open water surface is the ability to transport water vapor away from the evaporative
surface. The transport rate is governed by the humidity gradient in the air near the surface
and the wind speed across the surface. The equation for aerodynamic method is
Where:
Ea = Evaporation estimated by aerodynamic method (m/s)
(multiply by [1000 mm/m *86400 s /day] to get in mm/day)
es = saturation vapor pressure at the ambient temperature T (Pa)
ea = ed = actual vapor pressure estimated using dew point temperature Td or by
multiplying es by the relative humidity Rh (Pa)
B = the vapor transfer coefficient (m Pa-1
s-1
)
k = the Von Karman constant = 0.4
u2 = the wind velocity (m/s) measured at height z2 (cm) and z0 is from Table 3.3
a = density of moist air (kg/m3 )
a = density of water (kg/m3 )
p = atmospheric pressure in Pa
])z/z([p
uk0.622=B
where
)e - e( B = E
2
02w
2a
2
aasa
ln
(3.26)
Evaporation 70
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.
Table 3.3 A proximate values of the roughness height of natural surface.
Surface
Roughness height Z0 (cm)
ice, mud flats 0.001
water 0.01-0.06
Grass (up to 10 cm high) 0.1 - 2.0.
Grass (10 -50 cm high) 2 -5
vegetation (1 - 2 m high) 20
trees ( 10-15 m high ) 40-70
Example3.3 Calculate the evaporation rate from open water surface by the aerodynamic
method with air temperature 25C, the relative humidity 40 %, air pressure 101.3 kPa,
and wind speed 3 m/s, all measured at height 2m above the water surface. Assume a
roughness height zo = 0.03 cm.
Solution: The vapor transfer coefficient B is calculated using k = 0.4, a = 1.19 kg/m3
for
air at 25C, and density of water 997 kg/m3,
The evaporation rate is given by
at 25C, es = 3167 Pa and ea = Rh eas = 0.4 * 3167 = 1267 Pa
Ea = 4.54*10 -11
(3167 - 1267) = 8.62*10 -8
m/s
m/(Pa.s)10*4.54 =
])10*(2/3[*997*10*101.3
3*1.19*40.*0.622=
])z/z([p
uk0.622 = B
11-
24-3
2
2
02w
2a
2
ln
ln
)e - eB( = E aasa
Evaporation 71
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.
= 8.62 *10 -8 * (1000 mm/m)(86400 s /day)
Ea =7.45 mm/day
3.3.3. Combined aerodynamic and energy balance method - the combination method:
Evaporation may be computed by the aerodynamic method when energy supply is not
limiting and by the energy balance method when vapor transport is not limiting. But,
normally both of these factors are limiting, so a combination of the two methods is
needed. It is given by:
where:
= the gradient of the saturated vapor pressure curve at air temperature = des/dT,
(Pa/oC) and is
= 66.8 (Pa / C), pscychrometric constant,
Er and Ea = evaporation rate calculated based on energy balance, and aerodynamic
methods respectively (mm/day).
Example 3.5 Use the combination method to calculate the evaporation rate from an open
surface subject to net radiation of 200 W/m2, air temperature 25 C, relative humidity
40%, and wind speed 3 m/s, all recorded at height 2m, and atmospheric pressure 101.3 kPa.
Solution: The evaporation rate corresponding to net radiation of 200 W/m2 is Er = 7.10
mm/day, and for the aerodynamic method is yields Ea = 7.45 mm/day. The combination
method requires values for and .
E +
+ E +
= E ar
(3.27)
)T + (237.3
e 4098 =
2
as (3.28)
CPa/ 67.1 = 10*2441*0.622
10*101.3*1.*1005=
3
3
Evaporation 72
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.
the gradient of the saturated vapor pressure curve at 25oC with eas =32167 Pa for T
25oC
mm/day 7.2 = 7.45*67.1 + 188.7
67.1 + 7.10*
67.1 + 887.7
1887.7=
,E +
+ E +
= E ar
3.4 Evapotranspiration
Evapotranspiration is the combination of evaporation from the soil surface and
transpiration from vegetation. The same factors governing open water evaporation
also govern evapotranspiration, namely energy supply and vapor transport. In
addition, a third factor enters the picture: the supply of moisture at thee
evaporative surface. As the soil dries out, the rate of evapotranspiration drops
below the level it would have maintained in a well watered soil.
The combination method will give good estimate of reference crop
evapotranspiration that is for the rate of evapotranspiration from an extensive
surface of 8 cm to 15 cm tall green grass cover uniform height, actively growing,
completely shading the ground and not short of water.
The potential evapotranspiration of another crop growing under the same
conditions as the reference crop* is calculated by multiplying the reference crop
evapotranspiration Etr by crop coefficient kc, the value of which changes with the
stage of growth of the crop. The actual evapotranspiration Et is found by
multiplying the potential evapotranspiration by a soil coefficient ks ( 0 < ks < 1).
Ekk = E trcst . The values of the crop coefficient KC vary over a range of about (
0.2 < kc < 1.3).
forair J/kg.K 1005 = C 1.00, = K
K ,
Kl0.622
p KC = p
w
h
wv
hp
CPa/ 187.7 = )25 + (237.3
3167*4098 =
)T + (237.3
e
22
s 4098
Evaporation 73
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.
Note that reference crop evapotranspiration Etr can also be estimated by Penman-
Monteith method:
where:
E tr = reference crop evaporation (mm/day)
Rn = net radiation ar crop surface (MJ/m2/d)
G = soil heat flux (MJ/m2/d)
T = average temperature (oC)
U2 = wind speed measured at 2 m height (m/s)
(es - ea) = vapor pressure deficit (kPa)
= slope of vapor pressure curve (kPa/oC) = hygrometric constant (kPa/
oC)
G = 0.4 (T month n mean temperature O
C - T month n-1 mean temperature o
C)
900 = conversion factor
Sometimes Etr is called PET (potential evapotranspiration) although this often
refers to the evapotranspiration of a specific crop. Open water evaporation from
reservoirs may be estimated by multiplying Etr by a factor of 1.2. Estimated
monthly PET over the Tekeze, Awash and Rift Valley, Abay, Dedisa and Dabus,
Wabi Shebele and Genale Dawa, Omo Gibe, and Baro akobo are given in Annex
3.1.
3.5. Analysis of the homogeneity of meteorological data series
Weather data collected at a given weather station during a period of several years
may be not homogeneous, i.e., the data set representing a particular weather
variable may present a sudden change in its mean and variance in relation to the
original values. This phenomenon may occur due to several causes, some of
which are related to changes in instrumentation and observation practices, and
others which relate to modification of the environmental conditions of the site,
such as rapid urbanization or, on the contrary, perhaps development of irrigation
in the area.
Changes relative to data collection may be caused by:
)U0.34 + (1 +
)e - e(U273 + T
900 + G) - R(0.408
= E2
as2n
tr
(3.29)
Evaporation 74
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.
i. change in type of sensor or instrument;
ii. change in the observer and or change in the timing of observations;
iii. "sleeping" data collector;
iv. deterioration of sensors, such as with some types of pyranometers and RH
sensors, or mal-functioning of mechanical parts, such as with a tipping
bucket rain gauge, or by an intermittently broken or snorted wire;
v. aging of bearings on anemometers;
vi. use of incorrect calibration coefficients;
vii. variation in power supply or electronic behavior of instruments;
viii. growth of trees or planting of tall crops or construction of buildings or
fences near a raingauge, anemometer, or evaporation pan;
ix. change in the location of the weather station, or in the types of shelters for
housing temperature and humidity sensors;
x. change in the watering, type or maintenance of vegetation in the vicinity of
the weather station;
xi. significant change in the watering or type of vegetation of the region
surrounding the weather station.
These changes cause observations made prior to the change to belong to a
statistically different population than data collected after the change. It is
therefore necessary to apply appropriate techniques to evaluate whether a given
data set can be considered to be homogeneous and, if not, to introduce the
appropriate corrections. To do so requires the identification of which sub-data
series is to be corrected. To do this requires local information. Crop
evapotranspiration - Guidelines for computing crop water requirements - FAO
Irrigation and drainage paper 56 procedures are used in practice to check
homogeneity of the data (http://www.fao.org/docrep/X0490E).
Evaporation 75
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.
3.6 Practice Problems
3.1 The following mean meteorological data are obtained at an altitude of 2457 m amsl
in the northern part of Ethiopia. Calculate the reference crop evapotranspiration by (a) the
combination method and (b) the Penman Monteith method.
Month
Min Temp. OC
Max. Temp. OC
Humidity
%
Wind at
2m
km/day
Net radiation
MJ/m2/day
Jan 4.9 24.1 56 130 21.5
June 9.2 26.9 38 164 25.8
July 10.7 23.3 54 156 21.7
August 9.8 22.8 72 156 21.6
December min and max temperature are 4.2 and 23 OC. May min and max temperature
are 9.4 and 25.8 OC. Assume p at sea level is 102 kPa for the average temperature of 28
oC.
3.2 Use the combination method to calculate the evaporation rate from an open surface
subject to net radiation of 220 W/m2, air temperature 20 C, relative humidity 65%, and
wind speed 4 m/s, all recorded at height 2m, and atmospheric pressure 102.3 kPa.
3.3 If a dam is constructed in the climatic area described in Problem 1, determine
evaporation loss (million m3) in Jan, June, July and August. Take the average reservoir
area as 150 km2.
3.4 Estimate actual evapotranspiration for cotton in mid season stage in April at Amibara,
Ethiopia. Mean maximum temperature = 36 OC, mean minimum temperature = 22
OC,
men dew point temperature = 9 OC, mean wind speed = 1.5 m/s, mean percentage of
possible sunshine = 94 %, elevation = 300 m, and assume G = 0.0.
3.5 Calculate the daily evaporation rate from an open water surface under the following
climatic condition: incident radiation is 250 W/m2, mean air temperature is 35
oC,
mean relative humidity is 35 %, mean wind speed is 1.5 m/s, mean density of air is 1.0 kg/m
3, air pressure is 100 kPa, all measured at 2 m height. Furthermore, the
roughness height of water is 0.03, the albedo of water is 0.09, the emissivity of water
is 0.97, Stefan Boltzmann constant is 5.67 *10 –8
W/(m2.K
4 ).
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76
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Engineering Hydrology 77
77
4. INFILTRATION
Infiltration is the process of water entry into a soil from rainfall, or irrigation. Soil
water movement (percolation) is the process of water flow from one point to
another point within the soil. Infiltration rate is the rate at which the water
actually infiltrates through the soil during a storm and it must be equal the
infiltration capacities or the rainfall rate, which ever is lesser. Infiltration capacity
the maximum rate at which a soil in any given condition is capable of absorbing
water.
The rate of infiltration is primarily controlled by the rate of soil water movement
below the surface and the soil water movement continues after an infiltration
event, as the infiltrated water is redistributed.
Infiltration and percolation play a key role in surface runoff, groundwater
recharge, evapotranspiration, soil erosion, and transport of chemicals in surface
and subsurface waters.
4.1 Factors affecting infiltration
Infiltration rates vary widely. It is dependent on the condition of the land surface
(cracked, crusted, compacted etc), land vegetation cover, surface soil
characteristics (grain size & gradation), storm characteristics (intensity, duration
& magnitude), surface soil and water temperature, chemical properties of the
water and soil.
surface and soil factors
The surface factors are those affect the movement of water through the air-soil
interface. Cover material protect the soil surface. A bare soil leads to the
formation of a surface crust under the impact of raindrops or other factors, which
breakdown the soil structure and move soil fines into the surface or near-surface
pores. Once formed, a crust impedes infiltration.
Figure 4.1 illustrates that the removal of the surface cover (straw or burlap)
reduces the steady-state infiltration rate from approximately 3 to 4 cm/hr to less
than 1 cm/hr. Figure 4.2 illustrates the difference between crusted, tilled, grass
cover soil on the infiltration curve. The bare tilled soil has higher infiltration than
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Engineering Hydrology 78
78
a crusted soil initially; however, its steady-state rate approaches that of the crusted
soil because a crust is developing. Also, the grass-covered soil has a higher rate
than a crusted soil partially because the grass protects the soil from crusting.
Natural processes such as soil erosion or man-made processes such as tillage, overgrazing
and deforestation can cause change in soil surface configurations.
Figure 4.1 Effect of covered and bare soil on infiltration rates (Maidment, 1993)
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Engineering Hydrology 79
79
Figure 4.2 Effect of surface sealing and crusting on infiltration rates (Maidment, 1993)
The soil properties affecting soil water movement are hydraulic conductivity (a
measure of the soil’s ability to transmit water) and water-retention characteristics
(the ability of the soil to store and release water). These soil water properties are
closely related to soil physical properties.
Soil physical properties include particle size properties and morphological
properties. Particle-size properties are determined from the size distribution of
individual particles in a soil sample. Soil particles smaller than 2 mm are divided
into three soil texture groups: sand, silt, and clay. The morphological properties
having the greatest effect on soil water properties are bulk density, organic matter,
and clay type. These properties are closely related to soil structure and soil surface
area. Bulk density is defined as the ratio of the dry solid weight to the soil bulk
volume. The bulk volume includes the volume of the solids and the pore space.
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Engineering Hydrology 80
80
4.2 Measurements of infiltration
Infiltration is a very complex process, which can vary temporally and spatially.
Selection of measurement techniques and data analysis techniques should
consider these effects, and their spatial dimensions can categorize infiltration
measurement techniques. A brief introduction of infiltration measurement
techniques are described below.
4.2.1 Areal measurement
Areal infiltration estimation is accomplished by analysis of rainfall-runoff data
from a watershed. For a storm with a single runoff peak, the procedure resembles
that of the calculation of a index (see section 4.3.2). The rainfall hyetograph is
integrated to calculate the total rainfall volume. Likewise, the runoff hydrograph
is integrated to calculate the runoff volume. The infiltration volume is obtained by
subtracting runoff volume from rainfall volume. The average infiltration rate is
obtained by dividing infiltration volume by rainfall duration.
4.2.2 Point measurement
Point infiltration measurements are normally made by applying water at a specific site to
a finite area and measuring the intake of the soil. There are four types of infiltrometers:
the ponded-water ring or cylinder type, the sprinkler type, the tension type, and the
furrow type. An infiltrometer should be chosen that replicates the system being
investigated. For example, ring infiltrometers should be used to determine infiltration
rates for inundated soils such as flood irrigation or pond seepage. Sprinkler infiltrometers
should be used where the effect of rainfall on surface conditions influences the
infiltration rate. Tension infiltrometers are used to determine the infiltration rates of soil
matrix in the presence of macropores. Furrow infiltrometers are used when the effect of
flowing water is important, as in furrow irrigation.
Ring or Cylinder Infiltrometers
These infiltrometers are usually metal rings with a diameter of 30 to 100 cm and a
height of 20 cm. The ring is driven into the ground about 5 cm, water is applied
inside the ring with a constant-head device, and intake measurements are recorded
until a constant rate of infiltration is attained. To help eliminate the effect of
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Engineering Hydrology 81
81
lateral spreading use a double-ring infiltrometer, which is a ring infiltrometer with
a second larger ring around it.
Sprinkler infiltrometer - Rain simulator
With the help of rain simulator, water is sprinkled at a uniform rate in excess of
the infiltration capacity, over a certain experimental area. The resultant runoff R is
observed, and from that the infiltration f using f = (P-R)/t. Where P = Rain
sprinkled, R = runoff collected, and t = duration of rainfall.
Example 4.1: A USGS rain-simulator infiltrometer experiment was conducted on
a sandy loam soil. Rainfall was simulated at the rate of 20 cm/hr. The rainfall and
runoff data are given in Table ….
(a) Find and plot the mass-infiltration curve from the experimental data.
(b) Plot an infiltration rate curve.
Table E4.1. Rain-simulator infiltrometer data and infiltration capacity
calculation.
Solution. The measured data are given in Columns 1, 3 and 4. Cumulative
infiltration F is calculated by subtracting the cumulative runoff from the
cumulative rainfall. Infiltration rate is then determined by dividing the F by the
total duration of infiltration. The result is plotted in Figure E4.1.
Rainfall rate 20
Elapsed Time Simulated Measured F f
Time (1)/60 rainfall runoff (3)-(4) (5)/(2)
(min) (hr) (cm) (cm) (cm) (cm/hr)
[1] [2] [3] [4] [5] [6]
0 0.00 0.00 0.00 0.000 11.00
5 0.08 1.67 0.84 0.827 9.92
10 0.17 3.33 1.76 1.573 9.44
15 0.25 5.00 2.76 2.240 8.96
20 0.33 6.67 3.77 2.897 8.69
25 0.42 8.33 4.85 3.483 8.36
30 0.50 10.00 5.93 4.070 8.14
60 1.00 20.00 13.27 6.730 6.73
90 1.50 30.00 21.15 8.850 5.90
120 2.00 40.00 29.33 10.670 5.34
150 2.50 50.00 37.87 12.130 4.85
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Engineering Hydrology 82
82
Figure E4.1. Infiltration rate and cumulative infiltration variation with time.
4.3 Estimating infiltration rate
In the following section four infiltration methods are discussed, that is the Horton
Infiltration, the -index, the Philip infiltration and the Green -Ampt infiltration
equations.
4.3.1 Horton infiltration
In general, for a given constant storm, infiltration rates tend to decrease with time.
The initial infiltration rate is the rate prevailing at the beginning of the storm and
is maximum. Infiltration rates gradually decrease in time and reach a constant
value.
Horton observed the above facts and concluded that infiltration begins at some
rate f o and exponentially decreases until it reaches a constant fc. He proposed the
following infiltration equation where rainfall intensity i greater than fp at all times.
0.00
2.00
4.00
6.00
8.00
10.00
12.00
0.00 0.50 1.00 1.50 2.00 2.50 3.00
Time (hr)
Infi
ltra
tio
n r
ate
(c
m/h
r)
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
Cu
mm
ula
tiv
e In
filt
rati
on
(c
m)
Infiltration rate
Cumulative Infiltration
e )f - f( + f = f -kt
c0cp (4.1)
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Engineering Hydrology 83
83
where:
fp = infiltration capacity in mm/hr at any time t
fo = initial infiltration capacity in mm/hr
fc = final constant infiltration capacity mm/hr at saturation, dependent on soil type
and vegetation
t = time in hour from the beginning of rainfall
k = an exponential decay constant dependent on soil type and vegetation.
Note that infiltration takes place at capacity rates only when the intensity of
rainfall i equals or exceeds fp; that is f =fp when i fp, but when i < fp, f < fp and f
= i.
The cumulative infiltration equation F(t) for the Horton method is found from the
relationship d(F(t)/dt = f(t) = fp and is given by
k
e)f - f(+ tf tF
-kt
c0
c
)1()(
(4.2)
Indicative values for fo, fc, and K are given in Table 4.1.
Table 4.1: estimated values of Horton parameters
Soil / cover complex fo
(mm/hr)
fc
(mm/hr)
K
(1/hr)
Standard agricultural (bare) 280 6 – 220 1.6
Standard agricultural
(vegetated)
900 20 – 290 .8
Peat 325 2 – 29 1.8
Fine sandy clay (bare) 210 2 – 25 2.0
Fine sandy clay (vegetated) 670 10 – 30 1.4
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Engineering Hydrology 84
84
Example 4.2 The infiltration capacities of a given soil at different intervals of time are
measured and values are given in Table E4.1. Find an equation for the infiltration
capacity
Table E4.1.
Time (hr) 0 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00
fp (cm/hr) 10.4 5.6 3.2 2.1 1.5 1.2 1.1 1.0 1.0
Solution The infiltration capacity reaches a constant value equals to fc = 1.0 cm/hr. Now
plotting log 10 (fp - fc) with t on linear scale and estimating slope of the line m = -1/1.31.
Time (hr)
0
5
10
15
0 0.5 1 1.5 2 2.5
time (hr)
f
-4
-2
0
2
4
0 0.5 1 1.5 2 2.5
time (hr)
log
(fp
- fc
)
From this m =-1/1.31= -1/ (k log10e), k = 3.02. Thus the infiltration equation is given by
fp = 1.0 + (10.4 -1.0) e -3.02 t
= 1.0 + 9.4 e -3.02t
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Engineering Hydrology 85
85
4.3.2 The -index method
The -index is the simplest method and is calculated by finding infiltration as a
difference between gross rainfall and observed surface runoff. The -index
method assumes that the loss is uniformly distributed across the rainfall pattern.
Example 4.3 Estimate -index of the catchment having an area 2.26 km2.The observed
runoff caused by the rainfall given in the Table E4.2 is 282 097 m3.
Table E4.2:
Time (hr) Rainfall (mm/hr)
00 to 2 35.6
2 to 5 58.4
5 to 7 27.9
7 to 10 17.8
10 to 12 7.6
Solution: First the rainfall hyetograph is graphed. The runoff depth rd is then calculated
runoff depth = 282097/(2.26*1000*1000) = 125 mm
0.0
10.0
20.0
30.0
40.0
50.0
60.0
70.0
00 to 2 2 to 5 5 to 7 7 to 10 10 to 12
Try -index value of 25 mm/hr, then the first three rainfall will be used in the
calculation. That is: (2*(35.6 –) + 3*(58.4-) + 2*(27.9-)) = 125. This will
give a value of 25 mm/hr. The calculated value crosses the three selected
rainfall intensity signifying that each of these intensities contributes to runoff. The
-index method gives better estimate when losses is calculated after heavy
rainfall and the soil profile is saturated.
4.3.3 The Phillip method
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Engineering Hydrology 86
86
Phillip proposed an equation to estimate cumulative infiltration F(t) by
(4.3)
Where:
S = sorpitivity which is a function of the soil suction potential (representing soil
suction head
K = the hydraulic conductivity of the soil (representing gravity head)
t = time from the beginning of the rainfall.
Noting that f(t) = dF(t)/dt, the Phillip equation for infiltration rate is
(4.4)
Example 4.4 A small tube with a cross-sectional area of 40 cm2 is filled with soil and
laid horizontally. The open end of the tube is saturated, and after 15 minutes, 100 cm3 of
water have infiltrated into the tube. If the saturated hydraulic conductivity of the soil is
0.4 cm/hr, determine how much infiltration would have taken place in 30 minutes if the
soil column had initially been placed upright with its surface saturated.
Solution The cumulative infiltration depth in the horizontal column is F = 100 cm3 / 40
cm2 . For horizontal infiltration, cumulative infiltration is a function of soil suction alone
so that after t = 15 min = 0.25 hr
F(t) = St1/2
, 2.5 = S(0.25) 1/2
S = 5 cm.hr(-1/2
For infiltration down a vertical column, the full equation is used with K = 0.4 cm/hr.
F(t) = St
1/2 + Kt
F(t) =5(0.5)1/2 + 0.4(.5)
=3.74 cm
4.3.4* The Green-Ampt method
The Green-Ampt model is an approximate model utilizing Darcy’s law. The
model is developed with the assumption that water is ponded on the ground
surface. Consider a vertical column of soil of unit horizontal cross-sectional area
KtSttF 5.0)(
KSttf 5.05.0)(
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Engineering Hydrology 87
87
and let a control volume be defined around the wet soil between the surface and
depth L.
ho WATER
y
L
Saturated soil
Wetting front
Dry soil
If the soil was initially of moisture content i throughout its entire depth, the
moisture content will increase from i to n (the porosity) as the wetting front
passes.
The increase in the water stored with in the control volume as a result of infiltration is
L(n- i ) = L for a unit cross-section. By definition the cumulative depth of
water infiltrated into the soil F is given by:
(4.5)
Now from Darcy’s law
(4.6)
And in this case q = -f because q is positive upward while f is positive downward.
Eq.(4.6) can be written as
(4.7)
Here the head at h1 is h0 and the head at the dry soil below the wetting front h2 = -
-L.
(4.8)
)()( inLtF
z
hKq
)(21
21
zz
hhKf
)())(
( 0
L
LK
L
LhKf
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Engineering Hydrology 88
88
h0 is negligible if it is assumed that ponded water becomes surface runoff.
Replacing L by F(t)/() in (4.8), we get
(4.9)
And we know that dF/dt = f, thus we can develop the Green-Ampt equation for
F(t) and this is
(4.10)
and
(4.11)
Equation (4.10) is a non-linear equation in F. It may be solved by the method of
successive substitution in
(4.12)
Given K, t, and , a trial value of F is substituted on the right-hand side (a
good trial value is F = Kt), and a new value of F calculated on the left-hand side,
which is substituted as a trial value on the right-hand side, and so on, until the
calculated values of F converge to a constant.
Note that when the ponded depth ho is not negligible, the value -ho is substituted
for in Eqs. (4.10) and (4.11).
Parameters in the Green-Ampt model
To apply Green-Ampt model the effective hydraulic conductivity K, the wetting
front suction , the porosity n, and the initial moisture i need to be measured or
estimated. These parameters can be determined by fitting to experimental
)(F
FKf
KttF
tF
))(
1ln()(
)1)(
(
tF
Kf
))(
1ln()(
tF
KttF
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Engineering Hydrology 89
89
infiltration data, however, for specific application purposes it is easier to
determine the parameters from readily available data such as soils and land-use
data. Table 4.1 gives average values Green-Ampt parameters.
To incorporate the effects of land cover on infiltration, it is recommended to
divide the area into the following three categories: (1) the area which is bare and
outside the canopy cover, (2) the area which has ground cover, and (3) the bare
area under canopy, and to develop an effective hydraulic conductivity for each
area. Compute the infiltration separately for each area and then sum the three
infiltration amounts weighted to their areal cover to obtain the infiltration for the
area. This method of determining the infiltration assumes that the three areas do
not cascade. If the areas do cascade, this method over predicts infiltration.
Note that for bare ground cover conditions K = Ks/2, for the area which is bare
under canopy the effective hydraulic conductivity can be assumed to be equal to
the saturated hydraulic conductivity Ks of the soil.
The area which has ground cover is assumed to contain macroporosity, and the
effective hydraulic conductivity is equal to the saturated hydraulic conductivity
Ks times a macroporosity factor A. For areas which don not undergo mechanical
disturbance like range land macroporosity factor A is determined from
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Engineering Hydrology 90
90
)032.004.0032.096.0exp( BDCSA
Table 4.1 USDA Soil Texture Green-Ampt Infiltration Parameters (Maidment, 1993)
Soil texture classes
Porosity n
Wetting front soil
suction head
(cm)
Saturated
hydraulic
conductivity
Ks (cm/hr)
Sand 0.437
(0.374-0.500)
4.95
(0.97-25.36)
23.56
Loamy sand 0.437 6.13 5.98
(0.363-506) (1.35-27.94)
Sandy loam 0.453
(0.351-0.555)
11.01
(2.67-45.47)
2.18
Loam 0.463
(0.375-0.551)
8.89
(1.33-59.38)
1.32
Silt loam 0.501
(0.420-0.582)
16.68
(2.92-95.39)
0.68
Sandy clay loam 0.398
(0.332-0.464)
21.85
(4.42-108.0)
0.30
Clay loam 0.464
(0.409-0.519)
20.88
(4.79-91.10)
0.20
Silty clay loam 0.471
(0.418-0.524)
27.30
(5.67-131.5)
0.20
Sandy clay 0.430
(0.370-0.490)
23.90
(4.08-140.2)
0.12
Silty clay 0.479
(0.425-0.533)
29.22
(6.13-139.4)
0.10
Clay 0.475
(0.427-0.523)
31.63
(6.39-156.5)
0.06
(4.13)
And for undisturbed agricultural areas A can be determined from
(4.14)
Where S = Percent sand
C = percent clay
BD = bulk density of the soil (< 2 mm), g/cc, and A > 1.0.
)94.1099.082.2(exp BDSA
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The area which is bare outside canopy is assumed to be crusted and the effective
hydraulic conductivity is equal to the saturated hydraulic conductivity Ks times a
crust factor CRC which is estimated by
(4.15)
where: CRC = crust factor
SC = correction factor for partial saturation of the soil subcrust (see Table 4.2)
= 0.736 – 0.0019 (percent sand)
i = matric potential drop at the crust-subcrust interface, cm (Table 4.2)
= 45.19 – 46.68 (SC)
L = wetting front depth, cm
Table4.2: Mean steady-state matric potential drop i across seals by soil texture
(Maidment 1993)
Soil texture Matric, potential
drop
i (cm)
Reduction factor for sub-
crust conductivity
SC
Sand 2 0.91
Loamy sand 3 0.89
Sandy loam 6 0.86
Loam 7 0.82
Silt loam 10 0.81
Sandy clay loam 5 0.85
Clay loam 8 0.82
Silty clay loam 10 0.76
Sandy clay 6 0.80
Silty clay 11 0.73
Clay 9 0.75
Example 4.4 Compute the infiltration rate f and cumulative infiltration F after one hour
of infiltration into a silt loam soil that initially had an effective saturation of 30 %.
Assume water is ponded to a small but negligible depth on the surface.
Solution:
For a silt loam soil = 16.7 cm, K =0.65 cm/hr, n = 0.501, i = 30% x 0.501
= n - i = 0.501- 30% x 0.501 = 0.35.
= 16.7 x 0.35 = 5.84 cm.
)/(1 L
SCCRC
i
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The cumulative infiltration at t = 1 hour is calculated employing Eq. (4.10), taking a trial
value of F(t) = Kt =0.65 cm.
= 1.27 cm
Substituting F = 1.27 cm in Eq(4.10) we get 1.79 cm and aftera number of iteration F
converges to a constant value of 3.17 cm.
Infiltration rate after one hour is estimated by Eq. (4.11)
= 1.81 cm/hr.
))(
1ln()(
tF
KttF
)68.5
65.01ln(68.51*65.0)1( F
)1)(
(
tF
Kf
)117.3
68.5(63.0 f
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4.4. Practice Problems
4.1 The infiltration rate as a function of time for silt loam are given below. Determine
the best values for the parameters fo, fc, and k for Horton's equation to describe the
infiltration of the silt loam soil at the locality.
Time in hrs 0 0.07 0.16 0.27 0.43 0.67 1.10 2.53
fp (mm/hr) 6.6 5.3 4.3 3.3 2.2 1.3 0.7 0.25
4.2 For clay soil at a given location parameters of Philip's equation were found as S = 45
cm/hr 0.5
, and K = 10 cm/hr. Determine the cumulative infiltration and the infiltration
rate at 0.5 hr increments for a 3-hour period. Plot both as functions of time. Assume
continuously ponded conditions.
4.3. For a sandy loam soil, calculate the infiltration rate (cm/hr) and depth of infiltration
(cm) after one hour if the effective saturation is initially 40 percent, using the Green-
Ampt method. Assume continuously ponded conditions.
4.4. Use the Green-Ampt method to evaluate the infiltration rate and cumulative
infiltration depth of a silty clay soil at 0.1 hour increments up to 6 hours from the
beginning of infiltration. Assume initial effective saturation 20 percent and continuous
ponding.
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5. STREAMFLOW MEASUREMENTS and HYDROGRAPH
Continuous streamflow records are necessary in the design of water supply
systems, in designing hydraulic structures, in the operation of water management
systems, and in estimating sediment or chemical loads of streams. To do these,
systematic records of stage and discharge are essential. This chapter elaborates
common methods practiced in Ethiopia to measure streamflow. The concept of
hydrograph is also discussed.
River stage is the elevation above some arbitrary zero datum of the water surface
at a streamflow gauging station. The datum is sometimes taken as mean sea level
but more often is slightly below the point of zero flow at the gauging station. The
elevation datum is set with reference to at least three permanent reference marks
or benchmarks located in stable ground separate from the recorder structure
following standard surveying work.
It is difficult to make a direct, continuous measurement of discharge in a stream
or river but relatively simple to obtain a continuous record of stage. Thus
measurements of river stage provides the best alternative. Before we discuss some
methods of measuring river stage, first we discuss criteria fort selecting a stream
gauging station.
5.1 Stream gauging site selection
The principles of network design and the proposed use of data should govern the
selection of streams to be gauged. Dense network of gauging station is required
for research works related to runoff estimation, soil erosion estimation, and water
balance calculation at different watershed sizes. Whereas the goal is to construct a
dam to impound water, light network of stream gauging stations is sufficient -
one station at or near the dam site can be adequate. A general-purpose network
must, however, provide the ability to estimate hydrological parameters over a
wide area using for example a regional regression model.
Despite the development of a variety of objectives and statistically based methods
for streamflow and rainfall network design, judgment and experience are still
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indispensable. The WMO Guide to Hydrological Practice recommendations for
network density as a staring point for network design is given in Table 5.1.
Table 5-1.Recommended minimum density of hydrometric stations
Type of region
Range of norms for
minimum network, area,
km2 per station
Range of provisional
norms tolerated in
difficult conditions area,
km2 per station
Flat regions of tropical,
temperate and
Mediterranean zones
1000 - 25000
3000 - 10000
Mountainous regions of
tropical, temperate and
Mediterranean zones
300 -100
1000 - 5000
Small mountainous
islands with very
irregular rainfall, very
dense stream network
140 - 300
Arid zone 5000 - 20000
5.1.1 Selection of gauging site
The selection of a particular site for the gauging station on a given stream should
be guided by the following criteria for an ideal gauge site (WMO 1981):
i. The general course of the stream is straight for about 100 meters upstream
and downstream from the gauging site.
ii. The total flow is confined into the channel at all stages and no flow
bypasses the site as sub-surface flow.
iii. The streambed is not subject to scour and fill and is free of aquatic growth.
iv. Banks are permanent, high enough to contain floods, and are free of brush.
v. Unchanging natural controls are present in the form of a bedrock outcrop
or other stable riffle for low flow, and a channel constriction for high flow,
or a fall or cascade that is un-submerged at all stages to provide a stable
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relation between stage and discharge. If no satisfactory natural low-water
control exists, a suitable site is available for installing an artificial control.
vi. A site is available, just upstream from the control, for housing the stage
recorder where the potential for damage by water-borne debris is minimal
during flood stages; the elevation of the stage recorder itself should be
above any floods likely to occur during the life of the station.
vii. The gauge site is far enough upstream from the confluence with another
stream.
viii. A satisfactory reach for measuring discharge at all stage is available
within reasonable proximity of the gauge site. It is not necessary that low
and high flows be measured at the same cross-section.
ix. The site is readily accessible for ease in installation and operation of the
gauging station.
x. Facilities for telemetry can be made available, if required.
xi. Typical streamflow gauging station installed in the Wabi Shebele river at
upstream fo Melka Wakana Dam is shown in Figure 5.1. In practice rarely
will an ideal site be found for a gauging station and judgement must be
exercised in choosing between possible sites. A gauging site should be
located at a point along the stream where there is a high correlation
between stage and discharge, featuring a one to one correspondence
between stage and discharge. Either section or channel control is
necessary for the rating to be single-valued.
xii. A rapid or fall located immediately downstream of gauging site forces
critical flow through it, providing a section control. In the absence of a
natural section control, an artificial control – for instance, a concrete weir
– can be built to force the rating being single-valued. This type of control
is very stable under low and average flow conditions.
xiii. A long downstream channel of relatively uniform cross-sectional shape,
constant slope, and bottom friction provides a channel control. However, a
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gauging site relying on channel control requires periodic re-calibration to
check its stability. To improve channel control, the gauging site should be
located far from downstream backwater effects caused by reservoirs and
large river confluence.
Figure 5-1: Typical streamflow gauging station installed in the Wabi river
near Dodola town upstream of the Melkawakana reservoir (February 2002).
5.2 Stage measurement
Basically there are two modes of stage measurements. The first is discrete stage
measurements using manual gauges, and the second is continuous stage
measurements using recorders. For the measurement of stage, uncertainties
should not be worse than 10 mm or 0.1 % of the range.
5.2.1 Manual gauge
The simplest way to measure river stage is by means of a staff gage. A staff gauge
is vertically attached to a fixed feature such as a bridge pier or a pile (Figure 5.2).
The scale is positioned so that all possible water levels can be read promptly and
accurately. Another type of manual gauge is the wire gauge. Wire gauge consists
of a reel holding a length of light cable with a weight affixed to the end of the
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cable. The reel is mounted in affixed position – for instance, on a bridge span –
and the water level is measured by unreeling the cable until the weight touches the
water surface. Each revolution of the reel unwinds a specific length of cable,
permitting the calculation of the distance to the water surface. Manual gages are
used where stages do not vary greatly form one measurement to another
measurement. They are impractical in small or flashy streams, where substantial
changes in stages may occur between readings.
5.2.1 Recording gauge
A recording gage measures stages continuously and records them on a strip chart.
The mechanism of a recording gauge is either float actuated or pressure actuated.
In a float actuated recorder, a pen recording the water level on a strip chart is
actuated by a float on the surface of the water. The recorder and float is housed on
suitable enclosure on the top of a stilling well connected to the stream by two
intake pipes (two intake pipes are used incase one of them become clogged)
(Figure 5.2). The stilling well protects the float from debris and ice and dampens
the effect of wave action. This type of gage is commonly used for continuous
measurements of water levels in rivers and lakes.
The pressured actuated recorder or the bubble gage senses the water level by
bubbling a continuous stream of gas (usually CO2) into the water. The bubble
gauge consists of a specially designed servo-manometer, gas-purge system, and
recorder. Nitrogen fed through a tube bubble freely into the stream through an
orifice positioned at a fixed location below the water surface. The pressure in the
tube, equal to that of the piezo-meteric head above the orifice, is transmitted to the
servo-manometer, which converts changes in pressure in the gas-purge system
into pen movements on a strip-chart recorder. Bubble-type water level sensors are
used in applications where a stilling well is either impractical or too expensive
and where the stream carries a heavy sediment load. The Awash river at Awash
town is equipped with pressured actuated recorder.
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Figure 5-2. The measurement of stage through manual methods and recording
instruments (after Gregory and Walling, 1973)
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Figure 5-3: A typical chart from vertical float recorder.
Crest stage gage. This is used to obtain a record of flood crest at sites where
recording gages are not installed. A crest stage gage consists of a wooden staff
gage scale, situated inside a pipe that has small holes for the entry of water. A
small amount of cork is placed in the pipe, floats as the water rises, and adheres to
the staff or scale at the highest water level.
Telemetric gages. Gages with automatic data transmittal capabilities are called
self-reporting gages, or stage sensors. Self-reporting gauges are of the float-
actuated or pressure actuated type. These instruments use telemeters to broadcast
stage measurement in real time, from a stream gauging location to a central site.
This type of gauge is ideally suited for applications where speed of processing is
of utmost important, e.g., for operational hydrology or real-time flood forecasting.
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5.3 Flow velocity measurement and discharge computation
Flow velocity. The velocity of flow in a stream can be measured with a current
meter. Current meters are propeller devices (Figure 5.4) placed in the flow, the
speed with which the propeller rotates being proportional to the flow velocity.
Figure 5-4: Vertical and horizontal axis current meters and wading rod and cable
suspension mounting of the meter body.
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The relation between measured revolution per second of the meter cups N and
water
velocity V is given by
where:
a = the starting velocity or velocity required to overcome mechanical friction.
b = the constant of proportionality, and
Figure 5-5: Top: current meter mounted on a measuring rod, (bottom) suspended on a
cable from the bow of a jet-boat. Wide rivers flow (usually greater than 100 m) are often measured using a boat- the Baro river near Sudan border is the case in Ethiopia.
Initial values of a and b can be found from the calibration tables provided by the
manufacturer. With time the values of a and b are changing and regular
bN + a = V (5.1)
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recalibration is essential. This may be done by towing the current meter through
still water in a tank at a series of known velocities.
The current meter can be hand-held in the flow in a small stream (measurement
by wading), suspended from a bridge or cable way across a large stream, or
lowered from the bow of a boat (Figure 5.5).
Velocity distribution: The flow velocity varies with depth in a stream . Over the
cross-section of an open channel, the velocity distribution depends on the
character of the river banks and of the bed and on the shape of the channel. The
maximum velocities tend to be found just below the water surface and away from
the retarding friction of the banks.
The average velocity occurs say about 0.6 of the depth. It is standard practice to
measure velocity at 0.2 and 0.8 of the depth when the depth is more than 60 cm
and to average the two velocities to determine the average velocity for the vertical
section. For shallow rivers and near the banks on deeper rivers where the depths
are less than 0.6 m, velocity measurements are made at 0.6 of depth of flow.
Discharge computation.
The discharge computation of a stream is calculated from measurements of
velocity and depth. A marked line is stretched across the stream. At regular
intervals along the line, the depth of the water is measured with a graduated rod or
by lowering a weighted line from the surface to the stream bed, and the velocity is
measured using a current meter. The discharge Q at a cross-section of area A is
found by
(5.2)
Where V = streamflow velocity
A = cross sectional area of the flow
dAVQ .
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Figure 5-6. The velocity area technique of discharge measurements: a cable way is used
on large streams for positioning the current meter in the verticals and a special cable
drum can be used to obtain accurate readings of depth and spacing of verticals. The mean
section and mid-section methods are commonly used to compute the discharge of the individual segments.
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in which the integral is approximated by summing the incremental discharges
calculated from
each measurement i, i = 1, 2, ..., n of velocity Vi and depth di.
The measurements represent average values over width wi of the stream.
Example 5.1: Given the following stream gauging data, calculate the discharge.
Vertical No. 1 2 3 4 5 6 7 8 9 10 11
Distance to refernce point (m) 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 65.0
Sounding depth di (m) 0.0 0.5 0.8 1.2 1.5 2.5 3.0 2.0 1.2 2.7 2.9
Velocity at 0.2 di (m/s) 0.0 0.5 0.7 0.9 1.2 1.4 1.7 1.3 0.9 1.7 1.8
Velocity at 0.8 di (m/s) 0.0 0.4 0.6 0.7 0.8 1.1 1.3 1.0 0.7 1.3 1.4
Solution: To use Eq. (5.3) first the average velocity at each sounding depth is calculated,
then the partial width which is constant in this example is calculated (20-15) = 5 m
Vertical No. 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0
Distance to reference point (m) 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 55.0 60.0 65.0
Sounding depth (m) 0.0 0.5 0.8 1.2 1.5 2.5 3.0 2.0 1.2 2.7 2.9
Velocity at 0.2 m (m/s) 0.0 0.5 0.7 0.9 1.2 1.4 1.7 1.3 0.9 1.7 1.8
Velocity at 0.8 m (m/s) 0.0 0.4 0.6 0.7 0.8 1.1 1.3 1.0 0.7 1.3 1.4
Average Velocity (m/s) 0.0 0.5 0.7 0.8 1.0 1.3 1.5 1.2 0.8 1.5 1.6
Partial area Ai (msq) 0.0 2.5 4.0 6.0 7.5 12.5 15.0 10.0 6.0 13.3 4.8
Partial discharge (m3/s) 0.0 1.1 2.6 4.8 7.5 15.6 22.5 11.5 4.8 19.5 7.6
Total Q = 97.58 (m3/s) Total A = 81.61 msq
Average velocity (m/s) = Q/ A = 1.196 m/s
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
1 2 3 4 5 6 7 8 9 10 11
WdV = Q iii
n
=1i
(5.3)
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5.4 Dilution gauging
Dilution gauging method of measuring the discharge in a stream is made by
adding a chemical solution or tracer of known concentration to the flow and then
measuring the dilution of the solution downstream where the chemical is
completely mixed with the stream water. A tracer is a substance that is not
normally present in the stream and that is not likely to be lost by chemical
reaction with other substances. Salt, fluorescein dye, and radioactive materials are
commonly used as tracers. In general there are two methods of dilution gauging,
sudden-injection methods and constant rate injection method. The two methods
are described as follows.
5.4.1 Sudden Injection method
In this method a quantity of tracer volume V1 and concentration C1 is added to the
river by suddenly emptying a flask of tracer solution that is gulp injection. At the
sampling station downstream the entire tracer cloud is monitored to find the
relation between concentration and time. The quantity of tracer or mass of tracer
M is then equal to C1V1. If t1 is the time before the leading edge of the tracer
cloud arrives at the sampling station and t2 is the time after all the tracer has
passed this station the quantity of tracer is given by
Where: C2 = sustained final (equilibrium) concentration of the chemical in the
well mixed flow (mg/l)
C0 = the base value concentration, already present in the river (mg/l)
Using principle of conservation of mass we may estimate the streamflow Q:
C1V1 = dtCCQ
t
t
)( 02
2
1
(5.5)
dtCC
VCQ
t
t
)( 02
11
2
1
(5.6a)
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2
11
TC
CVQ (5.6b)
Figure 5-7. Dilution gauging: constant rate injection and gulp injection.
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5.4.2 Continuous and constant rate injection method
In this method a tracer of known concentration C1 is injected continuously at a
rate q at a sampling station situated downstream of injection point. The mass rate
M at which the tracer enters the test reach is
Assuming that satisfactory mixing of tracer has taken place with the entire flow
across the cross-section with the measured concentration C2 (reaching equilibrium
concentration), we have
201 )( CqQQC qC (5.8)
Solving for Q we get
qC C
C CQ
02
21
(5.9a)
The dilution method is particularly useful for very turbulent flows, which can
provide complete mixing within a relatively short distance. It is also applicable
when the cross section is so rough that alternative methods are unfeasible.
It is to be noted that a highly turbulent and narrow reach is desirable. In this
regard minimum required mixing length can be estimated by
gy
gCCBL
)27.0(13.0 2 (5.9b)
Where: L = mixing length
B = average width of the stream
y = average depth of the stream
C = Chezy’s coefficient of roughness, varying from 15 to 50 for smooth to rough
bed conditions
g = 9.81 m/s2
01 QC qCM (5.7)
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Example 5.2 25 g/l solution of a chemical tracer was discharged into a stream at 0.01 l/s.
At sufficiently far downstream observation point, the chemical was found to reach an
equilibrium concentration of 5 parts per billion. Estimate the stream discharge. The
background concentration of the tracer chemical in stream water may be taken as nil.
Solution. q = 0.01 l/s = 10–5
m3/s, C1 = 25 gm/l = 20 000 mg/l = 20 000 ppm = 20 part
per billion
Q= qCC
CC
02
21
Q = (20 000/0.005)*10-5
Q = 50 m3/s
Example 5.3 A fluorescent tracer with a concentration of 45 gm/l was injected into a
stream at a constant rate of 8 cm3/s. At a downstream section sufficiently far away from
the point of injection, the concentration was found to be 0.008 parts per million. Estimate
the discharge in the stream. The background concentration of the tracer in the stream is
zero.
Solution. using Eq (5.9a) we get
qC C
C CQ
02
21
0008.
)008.45000(10*8 6
Q
Q = 45 m3/s
5.5 The slope-area method
Occasionally, the high stages and swift currents that prevails during floods
increase the risk of accident and bodily harm. Therefore, it is generally not
possible to measure discharge during the passage of a flood. An estimate of peak
discharge can be obtained indirectly by the use of open channel flow formula.
The following guidelines are used in selecting a suitable reach:
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i. High-water marks should be readily recognizable.
ii. The reach should be sufficiently long so that fall can be measured
accurately.
iii. The cross-sectional shape and channel dimensions should be relatively
constant.
iv. The reach should be relatively straight, although a contracting reach is
preferred to an expanding reach, and
v. Bridges, channel bends, waterfalls, and other features causing flow non-
uniformity should be avoided. Note that the accuracy of the slope-area
method improves as the reach length increases.
A suitable reach should satisfy one or more of the following criteria:
i. The ratio of reach length to hydraulic depth should be greater than 75.
ii. The fall should be greater than or equal to 0.15 m, and
iii. the fall should be greater than either of the velocity heads computed at the
upstream and downstream cross sections.
The procedure consists of the following steps:
I. Calculate the conveyance K at the upstream and downstream
sections:
3/21uuu RA
nK (5.10)
3/21ddd RA
nK (5.11)
Where: K = conveyance
A = flow x-sectional area (m2)
R = hydraulic radius (m)
n = reach Manning roughness coefficient
u and d denotes upstream and downstream, respectively
II. Calculate the reach conveyance K , equal to the geometric mean of
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the upstream and downstream conveyances:
2/1)( du KKK (5.12)
III. Calculate the first approximation of the energy slope S:
L
FS (5.13)
Where: F = fall, elevation difference in L;
and L = reach length
IV. Calculate the first approximation to the peak discharge Qi
2/1KSQi (5.14)
V. Calculate the velocity heads:
g
AQh uiu
vu2
)/( 2 (5.15)
g
dAQh uid
vd2
)/( 2 (5.16)
Where: hu and hd are the velocity heads at upstream and downstream sections
respectively,
vu and vd are the velocity heads at upstream and downstream sections
respectively, and
g = gravitational acceleration.
VI. Calculate an updated value of energy slope Si:
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L
hhkFS vdvu
i
)( (5.17)
Where: k = loss coefficient, for expanding flow, i.e., Ad > Au, k = 0.5, for
contracting flow that is Ad < Au, k = 1.0
VII. Calculate an updated value of peak discharge
2/1
ii KSQ (5.18)
VIII Compare the updated value of peak discharge with previous estimate, and
continue the iteration until you close the difference between the newly estimated
peak discharge and the previously estimated peak discharge.
Example 5.3 Use the slope area method to calculate the peak discharge for the following
data: Reach length = 600 m, fall = 0.6 m. Manning n = 0.037.
Upstream flow area = 1550 m2, upstream wetted perimeter = 450 m, upstream
velocity head coefficient = 1.10. Downstream flow area = 1450 m2 , downstream wetted
perimeter = 400 m, downstream velocity head coefficient = 1.12.
Solution: Basic parameters calculations:
Flow area
Wetted Hydraulic Conveyance (m3/s)
Reach conveyance =
93999.14
(m2) perimeter
(m) radius (m)
upstream section 1550 450 3.44 95,545.32 First App. of S = 0.001 downstream section 1450 400 3.63 92,477.97
Peak discharge computation:
Iter. No.
hvu (m)
h vd (m)
Energy slope Peak discharge
(m/m) (m3/s)
1 0.001 2972.515 2 0.20619 0.2399 0.0009438 2887.815 3 0.19461 0.22642 0.000946 2892.639 4 0.19526 0.22718 0.0009468 2892.368 Hence, the final value converges to Q = 2892.3 m3/s
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5.6 Orifice formula for bridge opening
The discharge may be calculated from measurements taken on an existing bridge
over the river by using the orifice formula (TRRL, 1992):
)2/1(2
2/1 ]2
)1()[()2(g
VeDDLDgCQ dudo (5.19)
Where:
Q = Discharge at a section just downstream of the bridge (m3/s)
g = Acceleration due to gravity (9.81 m/s2)
L = Linear waterway, i.e. distance between abutments minus width of piers,
measured perpendicular to the flow (m)
Du = Depth of water immediately upstream of the bridge measured from marks
left by the river in flood (m)
Dd = Depth of water immediately downstream of the bridge measured from marks
on the piers, abutments or wing walls (m)
V = Mean velocity of approach (m/s)
Co and e are coefficients to account for the effect of the structure on flow, as listed
in Table 5.2. Definition sketch of the Orifice formula is shown in Figure 5.8.
Table 5-2. Values of Co and e in the orifice formula, L = Width of waterway, and W
=unobstructed width of the stream as defined in Figure 5.9:
L/W Co e
0.50 0.892 1.050
0.55 0.880 1.030
0.60 0.870 1.000
0.65 0.867 0.975
0.70 0.865 0.925
0.75 0.868 0.860
0.80 0.875 0.720
0.85 0.897 0.510
0.90 0.923 0.285
0.95 0.960 0.125
It is to be noted that whenever possible, flow volumes should be calculated by
both the area-velocity and orifice formula methods, and compared.
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Example 5.4 Calculate the discharge passing through a bridge with a waterway width of
18 m across a stream 30 m wide. In flood the average depth of flow just downstream of
the bridge is 2.0 m and the depth of flow upstream is 2.2 m.
Solution. Discharge at a section just upstream of the bridge, assuming a rectangular
section is A. V = 2.2 x 30 V = 66 V m3/s. Discharge at a section just downstream of the
bridge will be the same and will be given by the orifice formula:
)2/1(2
2/1 ]2
)1()[()2(g
VeDDLDgCQ dudo
L =18, W = 30, L/W = 0.6, thus C0 and e are 0.87 and 1.00 respectively.
)2/1(2
2/1 ]81.9*2
)11()22.2[(2*18)81.9*2(87.0V
Q
Q = 138.6[0.2 + 0.102V2/(2*9.81)]
0.5
Substituting for Q, 66V, we get 66V = 138.6[0.2 + 0.102V2/(2*9.81)]
0.5, V= 1.265
Now Q = 66 V = 66 *1.265 = 83.5 m/s
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Figure 5-8. Definition sketch of the orifice formula
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5.7 Stage discharge relationship - rating curve
The rating curve is constructed by plotting successive measurements of the
discharge and gage height on a graph. Figure 5.9 shows an example of rating
curves in linear and logarithmic scale for Zarema river a tributary of Tekeze river
near the foot of Semen Mountains.
Figure 5-9. Rating curves in linear (Top) and logarithmic scale of Zarema
river near Zarema, a tributary of Tekeze river (MWR)..
Rating curve of Zarema river at Zarema
0
1
2
3
4
5
6
7
0 200 400 600 800 1000 1200 1400 1600
Q (m3/s)
Sta
ge (
m)
Rating curve of Zarema river at Zarema
0.01
0.1
1
10
0.01 0.1 1 10 100 1000 10000
Q (m3/s)
Sta
ge (
m)
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A rating curve must be checked periodically to ensure that the relationship
between Q and H has remained constant. Scouring of the stream bed or deposition
of sediment in the stream can cause the rating curve to change so that the same
recorded gage height produces different Q.
The rating curve is described by a rating equation of the form
where a, b, and H0 are coefficients.
Having paired measured data of (H, Q) the coefficients a and b can be estimated
by taking a trial value of H0 which gives a straight line of the equation:
Or value of H0 adjustment for the low flow, can be estimated with the following
method. Three values of discharge (Q1, Q2, Q3) are selected from known portion
of the curve. One of these should be near the middle of the curve, and the other
value should be near the upper end of the curve. Then the third intermediate value
is estimated by
(5.22)
If H1, H2, and H3 represent the gage heights corresponding to Q1, Q2, and Q3 then
Ho is estimated by
A full rating curve can consist of different rating equation, e.g., one for low flows
and one for high flows, and often for low flows b > 2, for high flows b < 2.
Example5.4 Developed rating curve for the river Zarema near Zarema a tributary of
the Tekeze river having a watershed area of 3259 km2 has the following rating curves:
)H + a(H = Qb
0 (5.20)
)H + (H b + a = Q 0logloglog (5.21)
H2 H + H
H - HH = H
21
2231
03
(5.23)
QQ = Q 312
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119
where H is in m and Q in m3/s. Figure 5.10 shows the graphical form of the above
equations.
Extrapolation of ratings. Extrapolation of ratings is necessary when a water
level is recorded above the highest level and flow gauged level. Without
considering the cross-section geometry and controls, large error may result.
Where the cross section is stable, a simple method is to extend the stage-area and
stage velocity curves and, for given stage values, take the product of velocity and
cross-section area to give discharge values beyond the stage values that have been
gauged.
Stage-area and stage velocity curves can easily produced from the data that are
used for establishing the rating curve of the same river. The stage-area curve then
can be extended above the active channel by using standard land-surveying
methods. Extrapolation of the stage-velocity curve requires understanding of the
high stage control. Where there is channel control and where Manning roughness
is not varying with stage, the Manning equation may be used to estimate the
extrapolated velocity. It is to be noted that an upper bound on velocity is normally
imposed by the Froude number V/(gh)0.5
knowing that the Froude number rarely
exceeds unity in alluvial channels.
Shifting in rating curves. The stage-discharge relationship can vary with time, in
response to degradation, aggradation, or a change in channel shape at the control
section. Shifts in rating curves are best detected from regular gaugings and
become evident when several gaugings deviate from the established curve.
Sediment accumulation or vegetation growth at the control will cause deviation
which increases with time, but a flood can flush away sediment and aquatic weed
and cause a sudden reversal of the rating curve shift.
In gravel-bed rivers a flood may break up the armoring of the surface gravel
material, leading to general degradation until a new armoring layer becomes
established, and rating tend to shift between states of quasi-equilibrium. It may
then be possible to shift the rating curve up or down by the change in the mean
0.43m > H upto flow high for (-0.345) + (H69.997 = Q
0.43m H upto flow low for 0.201) + (H4.241 = Q
1.680
3.057
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bed level, as indicated by plots of stage and bed level versus time. Most of the
rivers in Wollo, such as Mille and Logia exhibit such phenomenon.
In rivers with gentle slopes. discharge for a given stage when the river is rising
may exceed discharge for the same stage when the river is falling (flood
subsiding). In such cases adjustment factors must be applied in calculating
discharge for rising and falling stages.
5.8 Hydrograph
Hydrograph is the graphical representation of the instantaneous rate of discharge
of a stream plotted with respect to time. Annual discharge hydrographs show the
variation of discharge during a year. Annual, monthly and daily discharge
hydrographs for river in the Wabi Shebele basin are shown in Figure 5.10(a) to (c).
Figure 5-10 Daily discharge hydrographs for Wabi Shebele river at Melka Wakana
for the year 1969.
Wabi River at Melkawakana 1969
0.00
20.00
40.00
60.00
80.00
100.00
120.00
140.00
0 50 100 150 200 250 300 350
Date beginning from Jan 1
me
an d
aily f
low
(m
3/s
)
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121
Figure 5.10 (b) Monthly discharge hydrographs for Wabi Shebele river at Melka
Wakana, Imi and Gode
0
100
200
300
400
500
600
700
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
Mo
nth
ly flo
w (M
MC
)
Wabi Shebelle @ Melka Wakana (4 388 sq km, 805 MMC)
Wabi Shebelle @ Imi (91 600 sq km, 2 965 MMC)
Wabi Shebelle @ Gode (127 000 sq km, 3 481 MMC)
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Figure 5.10 (c) Time series of Wabi Shebele annual river flow at Melka Wakana.
Separation of sources of streamflow on idealized hydrograph is shown in Figure 5.12
Wabi at Melka Wakana
400
500
600
700
800
900
1000
1100
1200
1300
1400
1500
1965 1970 1975 1980 1985 1990 1995 2000
Year
An
nu
al fl
ow
in M
illion
Mete
r C
ub
e
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Figure 5-11. Separation of sources of streamflow on an idealized hydrograph , (b)
sources of streamflow on a hillslope profile during a dry period, (c) and during a
rainfall event, (d) the extent of a stream network during a dry period, (e) and during
a rainfall event.
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Perennial streams flow continuously in dry months and have continuous
hydrograph, whereas ephemeral streams yield runoff mostly during rainy months
and their hydrograph are discontinued. If ground water flows toward the stream
during periods of heavy rainfall, the stream is called effluent. If flow is from the
stream to the groundwater system, as in the case of dry seasons, the stream is
called influent. Part of the rivers route may have effluent stream and other part
may be influent stream, change with time.
It consists of a rising limb, crest segment, and falling limb, or recession. The
shape of the rising limb is influenced mainly by the character of the storm. The
point of inflection on the falling side of the hydrograph is commonly assumed to
mark the time at which surface inflow to the channel system ceases. Thereafter,
the recession curve represents withdrawal of water from storage within the basin.
The shape of the recession is largely independent of the characteristics of the
storm causing the rise.
Analytical hydrographs. Analytical expression for streamflow hydrographs such
as gamma function are sometimes used in hydrological studies. It is defined as:
)]/()[()()( pgpttttm
pbpb e
t
tQQ Q= Q
(5.24)
Q = flow rate (m3/s)
Qb = baseflow (m3/s)
Qp = peak flow (m3/s)
For values of tg > tp equation 5.24 exhibits positive skewness.
tp = time to peak (hr)
tg = time -to-centroid (hr)
t = time (hr)
m = tp/(tg-tp)
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Example 5.5 Use Eq. (5.24) to calculate streamflow hydrograph ordinates at hourly
intervals, with the following data: Qb = 10 m3/s, Qp = 85 m
3/s; tp = 2 hr, tg =2.5 h.
Solution Applying Eq. (5.24) we get
Hydrograph
0
20
40
60
80
100
0 1 2 3 4 5 6 7 8 9 10
Time (hr)
Q (
m3/s
)
)]25.2/()2[())25.2/(2()(2
)1085(10 tet
= Q
Values fore example at 1 hr is 44.63 m3/s and at 3.5 hr is 45.02 m3/s. Resulting
hydrograph is plotted above.
Factors affecting the hydrograph shape. The shape of the storm hydrograph is
produced by components from (1) surface runoff, (2) interflow, (3) ground water
or base flow, and (4) channel rainfall. Interflow is that part of rainfall infiltrated
into the soil and move laterally through the upper soil zone until it enters a rills,
or/and small channel or/and stream channel. Interflow may be a large factor in
storms of moderate intensity over watersheds with relatively thin soil covers
overlying rock or hardpan.
The actual shape and timing of the hydrograph is determined largely by the size,
shape, slope, and storage in the watershed and by the intensity and duration of
input rainfall. Geologic features such as the existence of large deep cracks, in the
watershed is also an important factor that determines the hydrograph shape.
5.9 Hydrograph separation
Divisions of a hydrograph into direct runoff or surface runoff and contributed
from the groundwater flow that is the baseflow of the river as a basis for
subsequent analysis is known as hydrograph separation or hydrograph analysis.
Procedures for baseflow separation are usually arbitrary in nature. First it is
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126
necessary to identify the point in receding limb of the measured hydrograph
where direct runoff ends. Generally, this ending point is located in such a way that
the receding time up to that point is about 2 to 4 times the time-to-peak.
Key:
Straight line method
Fixed base method
Variable slope method`
Figure 5-12. Baseflow separation techniques (Chow et al. 1988).
A variety of techniques have been suggested for separating base flow and direct
runoff.
These are (a) the straight line method, (b) the fixed base length method, and the
variable slope method. These methods are illustrated in Figure 5.12.
Recession curves often take the form of exponential decay:
ktoeQtQ /)( (5.25)
Hydrograph
0
20
40
60
80
100
0 1 2 3 4 5 6 7 8 9 10
Time (hr)
Q (
m3/s
)
Log(Q)
N
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Where: Q = baseflow at any time t after the starting time to (m3/s)
Qo = the flow at time to (m3/s)
k = an exponential decay constant or time of storage defined as the time
required for the flow Q to recede to 0.368Qo ( i.e., for t = k, Q = Qoe-1
).
5.9.1 The straight line method
It involves drawing a horizontal line from the point at which surface runoff begins
to the intersection with the recession limb. This is often applicable to ephemeral
streams.
5.9.2 The fixed base method
The surface runoff is assumed to end a fixed time N after the hydrograph peak.
The baseflow before the surface runoff began is projected ahead to the time of the
peak. A straight line is used to connect this projection at the peak to the point on
the recession limb at time N after the peak N can be estimated from
Where: N = time (day)
A = the drainage area in km2
b = 0.8.
This method may be applied for large rivers where their response is in days.
5.9.3 The variable slope method
The base flow curve before the surface runoff began is extrapolated forward to the
time of peak discharge, and the baseflow curve after surface runoff ceases is
extrapolated backward to the time of the point of inflection on the recession limb.
A straight line is used to connect the endpoints of the extrapolated curves.
Ab = N 0.2 (5.25)
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5.10. Practice problems
5.1 Given the following stream gauging data, calculate the discharge and plote the cross
section of the river at this gauging station.
5.2: The following stage versus discharge data were recorded at the streamflow gauging
station. Establish rating curve(s) and their equation (s) for this gauging station.
H (m) 0.11 0.21 0.31 0.33 0.39 0.45 0.95 1.55 2.15 3.95 5.75
Q m3/s 0.119 0.279 0.544 0.612 0.849 1.141 28.80 91.64 180.69 577.60 1140
5.3 Use the slope area method to calculate the peak discharge for the
following data: Reach length = 800 m, fall = 0.8 m. Manning n = 0.004.
Upstream flow area = 1850 m2, upstream wetted perimeter = 550 m, upstream velocity
head coefficient = 1.10.
Downstream flow area = 1650 m2 , downstream wetted perimeter = 500 m,
downstream velocity head coefficient = 1.12.
5.4. Calculate the discharge passing through a bridge with a waterway width of 20 m
across a stream 35 m wide. In flood the average depth of flow just downstream of the
bridge is 2.5 m and the depth of flow upstream is 2.3 m.
Vertical No. 1 2 3 4 5 6 7 8 9 10 11
Distance to reference
point (m)
12.0 22.0 32.0 40.0 48.0 56.0 64.0 74.0 84.0 94.0 104.0
Sounding depth (m) 0.0 1.5 2.8 3.2 4.5 4.8 3.9 3.2 2.2 2.1 1.5
Velocity at 0.2 m (m/s) 0.0 1.5 1.7 1.9 2.2 2.4 2.7 2.3 1.9 1.7 1.8
Velocity at 0.8 m (m/s) 0.0 1.4 1.6 1.7 1.8 2.1 2.3 2.0 1.7 1.3 1.4
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6. WATERSHED PROPERTIES
Surface runoff in watersheds occurs as a progression of the following forms: (1)
over-land flow, (2) rill flow, (3) gully flow, (4) streamflow, and (5) river flow.
Overland flow is runoff that occurs during or immediately after a storm, in the
form of sheet flow over the land surface. Rill flow is runoff that occurs in the
form of small rivulets, primarily by concentration of overland flow. Gully flow is
runoff that has concentrated into depths large enough so that it has the erosive
power to carve its own deep and narrow channel (gully). Streamflow is
concentrated runoff originating in overland flow, rill flow, or gully flow and is
characterized by well-defined channels or streams of sizable depth. Streams carry
their flow into larger streams, which flow into rivers to constitute river flow. It is
to be noted that interflow will contribute to the surface runoff manifested as rill
flow, gully flow, and streamflow.
A watershed can range from as little as 1 ha to hundreds of thousands of square
kilometers. Small watersheds are those where runoff is controlled by overland
flow processes. Large watersheds (river basins) are those where runoff is
controlled by storage processes in the river channels.
The hydrologic characteristics of a watershed are described in terms of the
following properties: (1) area, (2) shape, (3) relief, (4) linear measures, and (5)
drainage patterns.
6.1 Watershed area
Area, or drainage area, or watershed area, or watershed area, is perhaps the most
important watershed property. It determines the potential runoff volume, provided
the storm covers the whole area. The watershed divide is the loci of points
delimiting two adjacent watersheds, i.e., the curve formed from the high points
separating watersheds draining into different outlets. Due to the effect of
subsurface flow (interflow and groundwater flow), the hydrologic watershed
divide may not strictly coincide with the topographic watershed divide. The
hydrologic divide, however, is less tractable than the topographic divide;
therefore, the latter is preferred for practical use.
The topographic divide is delineated on topographic map of suitable scale say 1:
10,000. Overland flow originates at high points and moves towards low points in
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)1.6(n
p cAQ
2L
AK f
5.0
282.0
A
PKc
a direction perpendicular to the contour lines. The area enclosed within the
topographic divide is the watershed area. A planimeter is often used to measure
the watershed area.
In general, the larger the watershed area, the greater the amount of surface runoff
and, consequently, the greater the surface flows. For example, peak flow to a
watershed area may be related by the basic formula
Where: Qp = peak flow
A = watershed area
c and n = parameters to be determined by regional regression
analysis.
6.2 Watershed shape
Watershed shape is the outline described by the horizontal projection of a
watershed. Watersheds vary widely in shape. A quantitative description is
provided by the following formula
(6.2)
Where: Kf = form ratio
A = watershed area
L = watershed length measured along the longest
watercourse.
An alternative description is based on watershed perimeter rather than area. For
this purpose, an equivalent circle is defined as a circle of area equal to that of the
watershed. The compactness ratio is the ratio of the watershed perimeter to that of
the equivalent circle. This leads to
(6.3)
Where; Kc = compactness ratio, P = watershed perimeter
A = watershed area
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6.3 Watershed relief
Relief is the elevation difference between two reference points. Maximum
watershed relief is the elevation difference between the highest point in the
watershed divide and the watershed outlet. The principal watercourse (or main
stream) is the central and the largest watercourse of the watershed and the one
conveying the runoff to the outlet. Relief ratio is the ratio of maximum watershed
relief to the watershed’s longest horizontal straight distance measured in a
direction parallel to that of the principal watercourse. The relief ratio is a measure
of the intensity of the erosion processes active in the watershed (Ponce, 1989).
The overall relief of a watershed is described by hypsometric analysis.
Hypsometric curve is a dimensionless curve showing the variation with elevation
of the watershed sub-area above that elevation. This is illustrated by in Example
6.1. The median elevation of the watershed is obtained from the percent height
corresponding to 50 percent area. The hypsometric curve is used to quantitatively
evaluate a hydrological variable such as precipitation or vegetation cover when
showing a marked tendency to vary with altitude.
Other measure of watershed relief are based on stream and channel
characteristics. The longitudinal profile of a watershed is a plot of elevation
versus horizontal distance. At a given point in the profile, the elevation is usually
a mean value of the channel bed. Between any two points, the channel gradient
(or channel slope) is the ratio of elevation difference to horizontal distance
separating them. Channel gradient vary widely, from as high as 0.10 for very
steep mountain streams to as low as 0.0005 for streams in the plain land /lowland.
The channel gradient can be obtained in different ways. The first method uses the
max. and min. elevations, the second method is defined as the constant slope that
makes the shaded area above the longitudinal profile equal to the shaded area
below the longitudinal profile. Typical watershed relief types in the upper ,middle and
lower watersheds of the Wabi Shebele basin is given in Figure 1.,
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132
Upper watershed, Enemora river Near Ticho February 2002, (elevation > 3000 meter
above sea level)
Middle course area – Galeti near Hirna, February 2002, (1800 m above sea level)
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133
Lower watershed- Wabi Shebele at Gode Februray 2002, (elevation about 400 meter
above sea level)
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134
Example 6.1: The following data have been obtained by planimetering 135 km2
watershed.
Plot the hypsometric curve for the above watershed
Solution:
ELEVATION
(M)
SUBAREA ABOVE INDICATED
ELEVATION (KM2)
1010 135
1020 85
1030 65
1040 30
1050 12
1060 4
1070 0
Elevation, E, (m) Catch. Area, A, (km2) A/135 *100 Y = (E -Emin)/(Emax-Emin) *100
1010 135 100 0
1020 85 63 20
1030 65 48 40
1040 30 22 60
1050 12 9 80
1060 4 3 100
Y
Ai / Ac
0
20
40
60
80
100
0 20 40 60 80 100
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6.4 Stream networks
The watershed length (or hydraulic length) is the length measured along the
principal watercourse (Fig. 6.1a). The principal watercourse is the central and
largest watercourse of the watershed and the one conveying runoff to the outlet.
The length to the watershed centroid is the length measured along the principal
watercourse from the watershed outlet to a point located closest to the watershed
centroid (Fig. 6.1a).
Figure 6.2 Typical recognizable drainage patter from aerial photograph.
Figure 6.1. Top: Linear measures of a watershed, (Bottom) Concept of stream order.
.
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136
Figure 6.2 Typical recognizable drainage patter from aerial photograph
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137
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138
(6.1)
(6
(6.2)
(6.3)
Table 6.1. Indicative values for bifurcation ratio, length ratio and drainage area order
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139
(6.4)
(6.5)
(6.6)
(6.7)
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140
(6.8)
(6.9)
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141
(6.10)
(6.11)
(6.12)
(6.13)
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142
(6.14)
(6.15)
(6.16)
(6.17)
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143
(6.18)
(6.20)
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144
(6.21)
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145
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146
(6.22)
(6.23)
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147
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148
6.5 Practice Problems.
6.1 Based on the following topographic map, delineate the watershed and
calculate form ratio and compactness ratio.
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149
Example watershed in Wabi Shebele basin - catchment area divide line of Jilbo
river and Birhamo irrigation area (done based on 1: 10,000 Topo-map provided by
Ethiopian Mapping Authority.
Hydrology of Small Watersheds
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150
7 Hydrology of small watersheds
Watershed area plays an important role in setting methodology for analyzing
surface runoff, groundwater flow, evaporation and other hydrological variables. In
practice watershed are divided into small, medium and large watersheds
depending on their watershed area. This chapter discusses salient hydrologic
features of small watersheds.
The following characteristics describe a small watershed:
(1) rainfall can be assumed to be uniformly distributed in time and space,
(2) storm duration usually exceeds concentration time,
(3) runoff is primarily by overland flow, and
(4) channel storage processes are negligible.
The upper limit of small watershed is difficult to define because of the natural
variability such as watershed slope and vegetation cover varies from watershed to
watershed and within watershed. In practice, either the concentration time of less
than 1 hr or watershed area less than 2.5 - 10 km2
has been used to define the
upper limit of a small watershed. The rational method is the most widely used
method of estimation runoff from small watersheds.
7.1 The rational method
The rational method is widely used around the world for flood estimation on
small rural watersheds and is the most widely used method for urban drainage
design. It is generally considered to be an approximate deterministic model
representing the flood peak that results from a given rainfall, with the runoff
coefficient being the ratio of the peak rate of runoff to the rainfall intensity over a
given watershed area.
The rational method takes into account the following hydrological
characteristics or processes: (1) rainfall intensity, (2) rainfall duration, (3) rainfall
frequency, (4) watershed area, (5) hydrologic abstraction, and (6) runoff
concentration.
Hydrology of Small Watersheds
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151
The rational method does not take into account directly the following
characteristics or processes:
(1) spatial or temporal variations in either total or effective rainfall, and
(2) concentration time much greater than rainfall duration.
The rational method formula is
(7.1)
Where:
Qp = the peak discharge for the required return period (m3/s)
C = the runoff coefficient (Table 7.1)
I = the rainfall intensity for a required return period of duration equal to
critical storm duration (mm/hr)
A = the drainage watershed area (km2)
The average rainfall intensity I has a duration equal to the critical storm duration,
normally taken as the time of concentration tc. For design, I is estimated from the
rainfall intensity-duration-frequency data for the location, with its frequency the
same as that selected for the design flood. Time of concentration is an idealized
concept and is defined as the time taken for a drop of water falling on the most
remote point of a drainage basin to reach the outlet, where remoteness relates to
time of travel rather than distance. In other words, it is the time after
commencement of rainfall excess when all portion of the drainage basin are
contributing simultaneously to flow at the outlet.
7.1.1 Determination of tc
For urban areas, values of tc are normally calculated as length divided by velocity
determined by hydraulic formulas. For rural drainage basins, tc is generally
estimated by means of an empirical formula such as Kirpich’s equation:
(7.2)
Where:
L = the length of channel from divide to outlet (km)
S = the average channel slope (m/m)
tc.= the time of concentration (min)
CIAQp 278.0
385.077.0976.3 SLtc
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7.1.2 Estimation of runoff coefficient C
Estimating the value of the runoff coefficient is the greatest difficulty and the
major source of uncertainty in application of the rational method. The coefficient
must account for all the factors affecting the relation of peak flow to average
rainfall intensity other than area and response time. A better estimate would be
obtained from measurements of runoff volume at the outlet of the case watershed
and rainfall volume over the watershed. Indicative runoff coefficients for urban
areas and for rural areas are given in Table 7.1(a) and 7.1(b).
Table 7.1(a) Average runoff coefficients for urban areas: 5-year and 10-year design
frequency (Maidment, 1993).
Description of area Runoff coefficient
Business
Downtown areas
Neighborhood areas
0.70 to 0.95
0.50 to 0.70
Residential
Single-family areas
Multiple units, detached
Multiple units, attached
0.30 to 0.50
0.40 to 0.60
0.60 to 0.75
Residential (suburban) 0.25 to 0.40
Apartment –dwelling area 0.50 to 0.70
Industrial
Light areas
Heavy areas
0.50 to 0.80
0.60 to 0.90
Parks, cemeteries 0.10 to 0.25
Playgrounds 0.10 to 0.25
Railroad yard areas 0.20 to 0.40
Unimproved areas 0.10 to 0.30
Characteristics of surface Runoff coefficient
Streets: Asphaltic
Concrete
Brick
0.70 to 0.95
0.80 to 0.95
0.70 to 0.85
Drives and walks 0.70 to 0.85
Roofs 0.75 to 0.95
Lawns, sandy soil
Flat (2 %)
Average (2 to 7 %)
Steep (7%)
0.05 to 0.10
0.10 to 0.15
0.15 to 0.20
Lawns, heavy soil: Flat (2 %)
Average (2 to 7 %)
Steep (7%)
0.13 to 0.17
0.18 to 0.22
0.25 to 0.35
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Table 7.1(b) Average runoff coefficients for rural areas (Schwab, et al., 1993)
Topography and
Vegetation
Soil Texture
Open Sandy Loam Clay and Silt Loam Tight Clay
Woodland
Flat
Rolling
Hilly
0.10
0.25
0.30
0.30
0.35
0.50
0.40
0.50
0.60
Pasture
Flat
Rolling
Hilly
0.10
0.16
0.22
0.30
0.36
0.42
0.40
0.55
0.60
Cultivated land
Flat
Rolling
Hilly
0.30
0.40
0.52
0.50
0.60
0.72
0.60
0.70
0.82
Note: Flat (0 – 5 % slope); rolling (5 –10 % slope), hilly (10 – 30 % slope)
7.1.3 Composite watershed
A composite watershed is one that drains two or more adjacent watersheds of
widely differing characteristics in time of concentration and runoff coefficients.
To apply the rational method to composite watershed, as in Figure 7.1, several
rainfall duration are chosen, ranging from the smallest time of concentration to
the largest adjacent watershed time of concentration.
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A
Figure 7.1 Composite watersheds having different time of concentration. The
objective is to find the peak flow at S.
The calculation proceeds by trial and error, with each trial associated with
each rainfall duration. To calculate the partial contribution from the watershed,
which has the largest time of concentration, an assumption must be made
regarding the rate at which the flow is concentrated at the watershed outlet. The
rainfall duration that gives the highest combined peak flow (A plus B) is taken as
the design rainfall duration.
Example 7.1 A watershed has a runoff coefficient of 0.20, area 150 ha with the general
slope of 0.001 and maximum length of travel of overland flow of 1.25 km. Information
on the storm of 50 years return period is given as follows:
Duration (min) 15 30 45 60 80
Rainfall (mm) 40 60 75 100 120
Estimate the peak flow to be drained by a culvert for a 50-year storm.
Solution. First the time of concentration tc is calculated using Eq. (7.2)
tc = 67.5 min
Maximum depth of rainfall for 67.5 min duration is obtained by interpolation from the
given rainfall duration and return period.
= 100 + [(120-100)/20]*7.5
385.077.0976.3 SLtc
385.077.0 )001.0()25.1(976.3 ct
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= 107.5 mm
Intensity of rainfall is then calculated (mm/hr):
= (107.5 / 67.5) * 60 = 95.5 mm/hr
Finally the peak flow to be drained by a culvert for a 50-year storm is calculated by
Q = 0.278 CIA
Q = 0.278* 0.20*95.5*1.50
Q = 7.96 m3/s
Example 7.2 Calculate the peak discharge by the rational method for a 1.5 km2
composite watershed shown in Figure 7.1 with the following characteristics:
Subarea A
Heavy industrial
Subarea B
Residential
(suburban)
Area km2 0.6 0.9
Runoff coefficient 0.6 0.3
Time of concentration (min) 40 100
Assume a return period T = 10 y and the following IDF function:
Where:
I = rainfall intensity (mm/hr)
T = return period in years
tr = rainfall duration in minutes
Solution:
To compute the contribution of sub-area B, assume that the flow concentrates linearly at
the outlet, i.e., each equal increment of time causes an equal increment of area
contributing to the flow at the outlet.
First choose rainfall duration between 40 and 100 minutes at 10 min. intervals. For each
rainfall duration, rainfall intensity is calculated using
67.0
22.0
)20(
1100
rt
TI
67.0
22.0
)20(
1100
rt
TI
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The Assumption of linear concentration for sub-area B leads to the following:
Rainfal duration Rainfall intensity Contributing area of B Peak flow at C
(min) (mm/hr) (km2) (m3/s)
40 117.50 0.12 12.94
50 105.97 0.24 12.73
60 96.90 0.36 12.61
70 89.55 0.48 12.55
80 83.44 0.60 12.53
90 78.28 0.72 12.54
100 73.85 0.84 12.56
For the above calculation it is clear that the peak flood is 12.94 m3/s corresponding to 40
min. of time of concentration.
7.2 Application of the rational method to storm-sewer and
culvert size design
7.2.1 Design of storm sewer
The design of storm sewer systems is a direct application of the principles from
both hydrology and hydraulics. IDF curves are used to specify rainfall intensities.
Watershed characteristics are used to estimate the volume and flow rate of the
runoff from the rainfall. Flow equation are used to calculate pipe and channel size
necessary to convey the calculated rate of flow.
Determination of storm sewer flow rates. A storm sewer is typically designed for
a specific return period of storm, usually 10 or 25 years. The duration D used in
the determination of the rainfall intensity is equal to the time of concentration of
the contributing watersheds. Storm sewer design regulations usually specify a
minimum time of concentration, and if the watershed time of concentration is less
than the specified minimum, then use the specified minimum time of
concentration. In cases where a storm sewer inlet has upstream piping, the
maximum of the watershed time of concentration or the accumulated upstream
travel time is used.
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Hydraulic grade line calculation. Once flow rates have been calculated throughout
the system, the hydraulic grades are then determined. The flow in the sewer is
assumed to be open channel flow. Within storm sewer pipe systems the slope of
the hydraulic grade line (HGL) can be calculated using Manning’s Equation:
(7.3)
Where:
V = velocity in the sewer (m/s)
R = hydraulic radius of the sewer (m)
S = Slope of the hydraulic grade line taken parallel to the sewer slope
(m/m)
n = Manning roughness, commonly used design values are:
n = 0.013 for verified sewer pipe, and
n = 0.015 for concrete pipe
With a known slope of the HGL, it is possible to calculate upstream hydraulic
grade (HG) with a known downstream grade.
HG upstream = HG downstream + SL (7.4)
The term SL is also the headloss in the pipe section. Headloss at junctions, inlets,
or manholes can be calculated using the equation:
(7.5)
Where:
HL = headloss (m)
K = headloss coefficient dependent on geometry
V = maximum velocity influent to junction (m/s)
g = gravitational constant
c
c
c
tupstreamdAccumulate
trequiredMinimum
tWatershed
D max
2/13/21SR
nV
g
VKH L
2
2
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By summing headlosses in upstream direction the hydraulic grade at any
point within the system can be calculated.
Example 7.3 A storm sewer system must be designed to convey storm water from three
watersheds to a pond with a free discharge tailewater condition. The following watershed
information was collected:
Watershed 1 Watershed 2 Watershed 3
Area (ha) 0.5059 0.2711 0.3804
C 0.78 0.75 0.85
tc (min) 10.0 10.0 10.
Ground elevation (m) 29.61 29.22 28.35
The inlet for watershed 1 is 90 m from the inlet for watershed 3. The inlet for watershed 2
is 75 m from the inlet for watershed 3. The inlet for watershed 3 is 13.5 m upstream of
the outlet pond. The IDF curve for the region for 25-yr return period is estimated by
Where:
I = rainfall intensity (mm/hr)
D = rainfall duration (min)
The following constraints must be met in the design of the system:
Minimum Maximum
Velocity (m/s) 0.75 4.5
Cover (m) 0.9 3.0
Assume that the pipe roughness (Manning’s n) is 0.02 for all pipes. The ground elevation
at the outlet is 27.93 m.
Solution:
First the system is schematized:
Watershed 3
Watershed 1
Watershed 2
8.0)12(
1600
DI
outlet 75 m
90 m
135 m
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Catch 1 catch. 2 catch. 3
area (ha) 0.51 0.27 0.38
C 0.78 0.75 0.85
tc (min) 10.00 10.00 11.08
I (mm/hr) 134.95 134.95 129.88
Q (m3/s) 0.15 0.08 0.12
Capacity full
for 450 mm pipe (m3/s) 0.2192724 Q actual/Qfull 0.674613
Using the calculated flow rates and knowing the ground elevations at inlet 1, inlet 2, and
inlet 3 we can try pipe sizes and invert elevations for the piping from inlet 1 to inlet 3,
from inlet 2 to inlet 3. This pipe size should minimize cover while still meeting the cover
constraints. The slope of the pipe can match the slope of the ground, therefore the slope
can be calculated
An 18 in (450 mm) pipe from inlet 1 to inlet 3 has a normal depth of 265 mm and flow
velocity of 1.5 m/s at this depth. A 15 in (375 mm) pipe from inlet 2 to inlet 3 has a
normal depth of 207 mm and flow velocity of 1.15 m/s at this depth. The pipe invert
should be placed so that the cover constraint is met.
Upstream invert pipe 1 = ground elevation – ground cover – pipe diameter
= 29.61 – 0.90 – 0.45 = 28.260 m
Downstream invert pipe 1 = 28.35 – 0.90 – 0.45 = 27.000 m
Upstream invert pipe 2 = 29.22 – 0.90 – 0..375 = 27.945 m
Downstream invert pipe 2 = 28.35 – 0.90 – 0.375 =27.075 m
Time of flow in each pipe must be calculated – knowing the pipe length and the velocity
we can determine this time.
0116.075
35.2822.29
014.090
35.2861.29
32
31
S
S
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Time pipe 1 = L/ velocity of flow = 90 m / (1.467 m/s) = 61.1 sec = 1.02 min.
Time pipe 2 = L/ velocity of flow = 75 m / (1.158 m/s) = 64.76 sec = 1.08 min.
The time of concentration to use for inlet 3 is therefore 10 min + 1.08 min. or 11.08 min.
This duration gives an intensity of 130 mm/hr and a flow rate of 0.1556 m3/s. The total
discharge through pipe 3 will be then the sum of the inlet discharges and the influent
discharge.
Q pipe 3 = 0.12 + 0.15 +0.08 = 0.35 m3/s
If the slope for pipe 3 follows the ground slope then S pipe 3 = 0.0031. A 30 in (750 mm) at
this slope and discharge will have a normal depth of 516 mm and a velocity of 1.01 m/s.
The pipe invert should be placed so that the cover constraint is met.
Upstream invert pipe 3 = ground elevation – ground cover – pipe diameter
= 28.35 – 0.90 – 0.75 = 26.70 m
Downstream invert pipe 3 = 27.93 – 0.90 – 0.75 = 26.28 m
Assuming there is no backwater condition at the outlet and that flow will exist at the
normal depth with backwater or drawdown, the hydraulic grade in the system can be
calculated. Junction losses at inlet 3 can be estimated using the junction headloss
equation. The maximum influent velocity is 1.467 m/s. Using an assumed K = 0.5, the
head loss is
HG outlet = invert elevation + normal depth = 26.28 + 0.516 = 26.841 m
HG outflow inlet 3 = invert elevation + normal depth = 26.70 + 0.516 = 27.2161 m
HG inflow inlet 3 = 27.2161 + 0.0548 = 27.2708 m
HG outflow pipe 2 = max. (27.2708, 27.075 + 0.265) = 27.34 m
HG outflow pipe 1 = max. (27.2708, 27.000 + 0.207) = 27.27 m
mmg
VKHL 8.54
81.9*2
467.15.0
2
22
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The outflow of pipe 1 has a mild backwater condition (27.2708 –27.207 = 0.0638 m).
Since the backwater is not severe we will assume that flow reaches normal depth before
reaching the upstream invert.
HG inflow pipe 1 = 28.260 + 0.207 = 28.467 m
HG inflow pipe 2 = 27.945 + 0.265 = 28.210 m
This pipe system will meet then all the required constraints.
Example 7.4 A typical plan for design of a small storm-sewer project is shown in
Figure 7.2. Table 7.2 shows a summary of the computations illustrating the application of
the rational method to determine design flows based on the following data:
i. Runoff coefficients
(a) Residential area: C = 0.3
(b) Business area: C = 0.6
ii. Areal weighting of runoff coefficients where required
iii. Selected design frequency = 5 years
iv. Intensity-Duration- Frequency curve
v. Inlet time = 20 min.
vi. Manning n in sewer = 0.013
vii. Free outfall to river at elevation = 80 m
viii. A drop of 3 cm across each manhole where no change in pipe size occurs (to
account in head losses). When a change in pipe size occurs, set the elevation of
0.8 of pipe depths equal, and provide corresponding fall in manhole invert.
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Figure 7.2.
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163
Excel File:
Hydrology of Small Watersheds
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164
7.3 Practice problems:
7.1. Using the plan given in Example 7.4 for design of a small storm-sewer project and
the following data:
(1) Runoff coefficients
(a) Residential area: C = 0.4
(b) Business area: C = 0.7
(c) Areal weighting of runoff coefficients where required
(2) Selected design frequency = 5 years
(3) Intensity-Duration- Frequency equation:
Where:
I = rainfall intensity (mm/hr)
D = rainfall duration (min)
(4) Inlet time = 25 min.
(5) Manning n in sewer = 0.013
(6) Free outfall to river at elevation = 80 m
(7) A drop of 3 cm across each manhole where no changes in pipe size occurs (to account
in head losses). When a change in pipe size occurs, set the elevation of 0.8 of pipe depths
equal, and provide corresponding fall in manhole invert.
Design the storm water sewer system.
7.2. Determine a 10-year peak flow at a storm water inlet from a 40 ha area in rolling
terrain. An inlet time of 20 min may be assumed. IDF curve of Problem 7.1 may be used.
Land use is as follows.
Land use Area (ha) Runoff coefficient
Single-family residential 30 0.40
Commercial 3 0.60
Park 7 0.15
75.0)10(
1500
DI
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Hydrology of Midsize Watersheds
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166
8. Hydrology of midsize watersheds
The following characteristics describe a midsize watershed: (1) rainfall intensity
varies within the storm duration, (2) rainfall can be assumed to be uniformly
distributed in space, (3) runoff is by overland flow and stream channel flow, and
(4) channel storage processes is negligible.
For midsize watersheds, runoff response is primarily a function of the
characteristics of the storm hyetograph, with concentration time playing a
secondary role. Watershed values ranging from few 100 km2 to 5000 km
2 may be
considered as midsize watersheds. The lower limit however could go up to 50 ha
depending on the design guideline followed for a specific purpose.
Commonly used hydrological techniques for estimating flood hydrograph from
midsize watershed are the Soil Conservation Service (SCS) and the unit
hydrograph methods.
8.1 The SCS method
The SCS method is widely used for estimating floods on small to medium-sized
ungaged drainage basins around the world (Graphical presentation is given in
Figure 8.1) . The method was developed based on 24-hr rainfall runoff data in
USA. In its derivation it is assumed that no runoff occurs until rainfall equals an
initial abstraction (that is losses before runoff begins) Ia, and also satisfies
cumulative infiltration F (the actual retention before runoff begins) or water
retained in the drainage basin, excluding Ia. The potential retention (the potential
retention before runoff begins ) S is the value that (F + Ia) would reach in a very
long storm.
If Pe is the effective storm rainfall equal to (P - Ia), and rd = depth of runoff
the basic assumption in the method is
(8.1) e
d
P
r
S
F
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Figure 8-1: SCS Relation between Direct Runoff, Curve Number and Precipitation
Eq. (8.1) states that the ratio of actual retention to potential retention is equal to
the ratio of actual runoff to potential runoff. The empirical relation Ia = 0.2S was
Hydrology of Midsize Watersheds
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168
adopted as the best approximation from observed data, and so Pe = (P - 0.2S). For
convenience and to standardize application of SCS method, the potential retention
is expressed in the form of a dimensionless runoff curve number CN.
8.1.1 SCS Peak discharge and flood hydrograph determination
The peak discharge in the SCS method is derived from the triangular
approximation to the hydrograph shown in Figure 8.2 resulting from rainfall
excess of duration D.
Rainfall excess
La
Peak flow
Direct runoff
qp
D 1.67 Tp
Tp
2.67 Tp
The lag La of the peak flow, time from the centroid of rainfall excess to the peak
of the hydrograph, is assumed to be 0.6tc. Then the time of rise Tp to the peak of
the hydrograph is
cp tD= T 6.05.0 (8.2)
The base length of the hydrograph is assumed to be 2.67Tp. Then from a
triangular hydrograph assumption (excess rainfall depth = runoff depth) the peak
discharge can then be estimated from
Figure 8.2: SCS triangular hydrograph
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169
c
dp
tD
Ar = q
6.05.0
208.0
(8.3)
Where:
qp = peak discharge (m3/s)
rd = the excess rainfall depth (mm) determined from Eq. (8.4)
A = watershed area (km2)
tc = time of concentration (hr)
D = duration of excess rainfall (hr)
The depth of runoff resulting from a required return period rainfall depth of
duration corresponding to the time of concentration tc is estimated by
0.8S + P
)0.2S - (P = r
2
d (8.4)
where:
rd = depth of runoff equal to depth of excess rainfall (mm)
S = the potential retention (mm)
P = design rainfall amount of duration tc corresponding to T years return period (mm)
and S (mm) is estimated using
)1 - CN
100254( =S (8.5)
To estimate the time of concentration tc the Kirpich formula (Chapter 7) may be
used, that is
GL 3.97 = t-0.3850.77
c
where:
L = the length of the river from the divide to the outlet (km)
G = the average river slope (m/m)
tc = time of concentration (min)
The explicit consideration of the various factors that are thought to affect flood
runoff makes the method attractive. Designers however may have uncertainties in
choosing the CN and in determining the method for tc. It is found that assumed
antecedent moisture condition had major effect and that results were better for
bare soil or sparse vegetation than for dense vegetation. Therefore care is required
in its application, and there is a need for checking of the method against observed
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170
flood data for the region of interest or with other methods. Table 8.1 & Table 8.2
provide experimental values of CN for different land use or crop, treatment
practice, hydrological soil group and antecedent moisture conditions. The use of
SCS method is illustrated by Example 8.1.
Example 8.1 A certain watershed experienced 12.7 cm heavy storm in a single day. The
watershed is covered by pasture with medium grazing, and 32 % of B soils and 68 % of C
soils. This event has been preceded by 6.35 cm of rainfall in the last 5 days. Following
the SCS methodology, determine the direct runoff for the 12.7 cm rainfall event.
Solution. From Table 8.1, for pasture range fair hydrologic condition for B soil the CN =
68 and for C soil the CN = 79. The weighted curve number for the AMC II is
CN = 0.32*68 + 0.68*79 = 76
AMC III is taken because for the last 5 days there was substantial rainfall. The CN for
the AMC III is
II
IIIII
CN
CN = CN
0057.043.0
8876*0057.043.0
76
= CN III
Then S is calculated using
)1 - CN
100254( = S
= 254(100/88 - 1) = 35 mm
The direct runoff depth is
0.8S + P
)0.2S - (P = r
2
d
= mm0.8 +
)0.2 - (
2
9335*127
35*127
The direct runoff produced by the 127 mm heavy storm is thus 93 mm. It is 73 % of the
total rainfall. If this rainfall would have occurred on the AMC I - dry condition then
5876*013.03.2
76
= CN I
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)1 - CN
100254( = S
= 254(100/58 - 1) = 183 mm
0.8S + P
)0.2S - (P = r
2
d = mm0.8 +
)0.2 - (
2
29183*127
183*127
which is 29/127=23 % of the total rainfall. The AMC III and AMC I gave results of
dramatic difference.
Table 8.1: Runoff curve numbers for hydrological soil-cover complexes for antecedent
rainfall condition II and Ia = 0.2S. For Conditions I and III see Table 8.2 (Maidment, 1993)
Land use or
Crop
Treatment or
practice
Hydrologic
condition
Hydrologic soil group
A B C D
Fallow Straight row - 77 86 91 94
Row crops Straight row Poor 72 81 88 91
Straight row Good 67 78 85 89
Contoured Poor 70 97 84 88
Contoured Good 65 75 82 86
Terraced Poor 66 74 80 82
Terraced Good 62 71 78 81
Small grain Straight row Poor 65 76 84 88
Straight row Good 63 75 83 87
Contoured Poor 63 74 82 85
Contoured Good 61 73 81 84
Terraced Poor 61 72 79 82
Terraced Good 59 70 78 81
Close-seeded
legumes or rotation
meadow
Straight row Poor 66 77 85 89
Straight row Good 58 72 81 85
Contoured Poor 64 75 83 85
Contoured Good 55 69 78 83
Terraced Poor 63 73 80 83
Terraced Good 51 67 76 80
Pasture range Poor 68 79 86 89
Fair 49 68 79 84
Good 39 61 74 80
Contoured Poor 47 67 81 88
Contoured Fair 25 59 75 83
Contoured Good 6 35 70 79
Meadow (permanent) Good 30 58 71 78
Wood (farm woodlots) Poor 45 66 77 83
Fair 36 60 73 79
Good 25 55 70 77
Farmsteads --- 59 74 82 86
74 84 90 92
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Soil Group Description Final infiltration
rate (mm/hr)
A Lowest runoff potential. Include deep sands with
very little silt and clay, also deep loess and
aggregated silt.
8-12
B Moderately low runoff potential: mostly sandy soils
less deep than A and loess less deep and aggregated
than A, but the group as a whole has above-average
infiltration after thorough wetting.
4-8
C Moderately high runoff potential: comprises shallow
soils and soils containing considerable clay and
colloids, though less than those of group D. Clay
loams, Shallow sandy loam, The group has below
average infiltration after pre-saturation
1-4
D Highest runoff potential. Includes mostly clays of
high swelling soil, heavy plastic clays, but the group
also includes some shallow soils with nearly
impermeable sub-horizons near the surface
0-1
Table 8.2 Antecedent rainfall conditions and curve numbers (for Ia = 0.2S)
Curve number for
Condition II
Factor to convert curve number for condition II to
Condition I Condition III
10 0.40 2.22
20 0.45 1.85
30 0.50 1.67
40 0.55 1.50
50 0.62 1.40
60 0.67 1.30
70 0.73 1.21
80 0.79 1.14
90 0.87 1.07
100 1.00 1.00
5-day antecedent rainfall (mm)
Condition
General description
Dormant Growing
Season Season
I Optimum soil condition from about
lower plastic limit to wilting point
< 13 < 36
II Average value for annual floods 13 – 28 36 – 53
III Heavy rainfall or light rainfall and
low temperatures within 5 days prior
to the given storm
> 28 > 53
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Note that the dry and wet antecedent moisture conditions, AMC I and AMC III may be
calculated from
II
IIIII
II
III
CN
CN = CN
CN
CN = CN
0057.043.0,
013.03.2
(8.7)
or estimated using the coefficients given in Table 8.1
Example 8.2 Determine (a) the design peak runoff rate, for a 50-year return period
storm from a 120 km2 watershed having IDF curve (I in mm/hr, T in years and tc in
minutes) given by
78.0
18.0
)20(
500
ct
T = I
and with the following characteristics:
Subarea
(km2)
Topography
Slope (%)
Soil group Land use, treatment, and
hydrological condition
75 10-35 C Row crop, contoured, good
45 20-45 B Woodland, good
The maximum length of flow is 15 km and the difference in elevation along this path is
450 m.
Solution.
First we estimate the time of concentration:
SL 3.97 = t-0.3850.77
c
)15000/450(15-0.3850.77
c 3.97 = t
= 123 min.
for the return period of 50 years and tc = 123 min, the design intensity of rainfall is
estimated by
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78.0
18.0
)20(
500
ct
T = I
78.0
18.0
)20123(
50*500
= I
= 22 mm/hr
The average curve number CN for the watershed is
2120
451
120
75CNCN = CN
From Table 8.1 and 8.2 CNI for the soil group C and Row crop, contoured, good
condition is 82, CNII for the soil group B and wood land and good condition is 55, and
CN = 72. Estimating s with
)1 - CN
100254( = S
)1 - 100
254( = S72
= 98.7 mm
The net rainfall estimated from
0.8S + P
)0.2S - (P = r
2
d
7.98*44
7.98*2*22
0.8 +
)0.2 - ( = r
2
d
= 4.8 mm
The peak discharge then is
c
dp
tD
Ar = q
6.05.0
208.0
2*6.02*5.0
8.4*120*208.0
= q p
= 54 m3/s
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8.2. THE UNIT HYDROGRAPH METHOD
A unit hydrograph is a specific type of hydrograph in that it represents the effects
of the physical characteristics of the basin on the completely defined, standardized
input rainfall excess. The essence of the unit hydrograph is that since the physical
characteristics of the watershed shape, size, slope etc are relatively constant over
few years, one might expect considerable similarity in the shape of hydrographs
resulting from similar rainfall characteristics. It is called a unit hydrograph
because, for convenience, the runoff volume under the hydrograph is commonly
adjusted to 1 cm or 1 mm equivalent depth over the watershed. Formally we
define the unit hydrograph of a drainage basin (watershed) as a hydrograph of
direct runoff resulting from one centimeter of rainfall excess of a specified
duration generated uniformly over the watershed area at a uniform rate.
An important feature of the unit hydrograph is its specified time period. This is
the duration of rainfall excess that produce the unit hydrograph, and its duration
must be included in the name of the unit hydrograph. For a given basin a 1-hr unit
hydrograph will be different from the 3-hr unit hydrograph. A 1-hr unit
hydrograph is produced by 1 mm of rainfall excess falling over the basin in 1 hr at
a rate of 1 mm/hr, and a 3-hr unit hydrograph by 1 mm of rainfall excess
occurring uniformly during 1 3-hr period, at a rate of 1/3 mm/hr.
8.2.1 Derivation of unit hydrographs
The unit hydrograph is best derived from the hydrograph of a storm of reasonably
uniform intensity, duration of desired length, and a relatively large runoff volume.
In this section we discuss two methods of deriving a unit hydrograph from
observed rainfall and the resulting hydrograph of a given watershed.
Method one: simple proportioning of the direct runoff
The steps are:
1. Separate the base flow from direct runoff,
2. Determine the volume of direct runoff, and
3. Divide the ordinate of the direct runoff hydrograph by observed
runoff
depth.
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Example 8.3 Given the following hydrograph of a given watershed having drainage area
of 104 km2 derive the unit hydrograph for the watershed.
Solution: The direct runoff ordinates are obtained by subtracting the base flow from the
total streamflow, that Col. [3]- Col. [4]. Calculate the direct runoff depth
= 0.1170 m
= 11.70 cm
Then the unit hydrograph ordinate Col. [6] is obtained by dividing Col. [5] by 11.70 cm.
Note that it is informative to indicate the full unit of the unit hydrograph in this case
m3/s/cm.
Date Hour Total flow
(m3/s)
Base
flow (m
3/s)
Direct
stream flow qi
(m3/s)
Unit
hydrograph ordinate (m
3/s
/cm) [3] [4] [5] [6]
16-Feb 6000 11 8 3 0.0
8000 170 8 162 13.9
1000 260 6 254 21.7
1200 266 6 260 22.3
1400 226 8 218 18.7
1600 188 9 179 15.3
1800 157 11 146 12.5
2000 130 12 118 10.1
2200 108 14 94 8.0
2400 91 16 75 6.4
17-Feb 2000 76 17 59 5.1
4000 64 19 45 3.9
6000 54 21 33 2.8
8000 46 22 24 2.1
1000 38 24 14 1.2
1200 32 26 6 0.5
1400 27 27 0 0.0
1690
A
qt
r
n
i
i
d
1
610*104
1690*3600*2dr
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Method 2: Unit hydrograph derivation by convolution method
The process by which the design storm is combined with the transfer function
(that is the unit hydrograph) to produce the direct runoff hydrograph is called
convolution. Analytically speaking, convolution is referred to as theory of linear
superposition. Conceptually, it is a process of multiplication, translation with time
and addition.
The basic assumptions inherent in the convolution method are:
a. The excess rainfall has a constant intensity within the effective
duration. This implies that the storm selected for analysis should
be of short duration, since these will most likely produce an intense
and nearly constant excess rainfall rate, yielding a well-defined
single peaked hydrograph of short time base
b. The excess rainfall is uniformly distributed throughout the whole
drainage area. In this case drainage area should not be too large
(about 30 km2, max) and for large watershed, the area should be
subdivided and each sub-area analyzed for storms covering the
whole sub-area.
c. The base time of the direct runoff hydrograph resulting from an
excess rainfall of given duration is constant.
d. The ordinates of all direct runoff's of a common base time are
directly proportional to the total amount of direct runoff
represented by each hydrograph.
e. For a given watershed, the hydrograph resulting from a given
excess rainfall reflects the unchanging characteristics of the
watershed. So unit hydrographs are applicable incase of channel
and watershed physical conditions such as afforestation /
deforestation/ widening of channel remain unchanged and
watersheds do not have appreciable storage. For the changed
condition a new unit hydrograph should be produced.
The discrete convolution equation. The discrete convolution equation is given by
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178
Where: Qn = direct runoff (m3/s),
Rm = excess rainfall (mm),
U n-m-1 = the unit hydrograph ordinate (m3/s/mm),
M = the number of pulses of excess rainfall, and
N = the number of pulses of direct runoff in the storm
considered.
n = 1, 2, ..., N
m = 1, 2, ..., M
Note that the time interval used in defining the excess rainfall hyetograph
ordinates must be the same as that for which the unit hydrograph was specified.
Total number of discharge ordinate N derived from M excess rainfall pulses is
given by M + 1 + the number of unit hydrograph ordinates V. That is N =
M+1+V
Suppose that there are M pulses of excess rainfall and N pulses of direct runoff in
the storm considered, then N equations can be written for Qn, n = 1, 2, ..., N, in
terms of N - M +1 unknown values of the unit hydrograph. If Qn and Rm are
given and Un-m+1 is required, the set of equations is over determined, because
there are more equations N than unknowns N-M+1. Thus, unique solution is not
possible find.
Example 8.4: In a storm, the rainfall excess of 0.5 cm, 0.7 cm, 0.0 cm and 0.8 cm
occurred in four successive hours. The storm hydrograph due to this storm has the hourly
ordinates (m3/s)as given below: 0.5, 44.5, 110.5, 85.5, 102.8, 94.0, 38.4, 18.6, 10.9, 5.3,
2.9, 0.5. If there is a constant base flow of 0.5 m3/s, find the hourly ordinates of the unit
hydrograph.
Solution: The direct runoff ordinates Qn (m3/s) are 0.0, 44.0, 110.0, 85.0, 102.3,
93.5, 37.9, 18.1, 10.4, 4.8, 2.4, 0.0. The depth of effective rainfall are R1 = 0.5
cm, R2 = 0.7 cm, R3 = 0.0, and R4 = 0.8 cm.
Using equation (8.9)
UR = Q 1+m-nm
Mn
1=m
n
(8.9)
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The number of unit hydrographs ordinates to be found are = N-M+1 = 12-4+1= 9
Calculation fro
Q1 = R1U1 = 0.5 x U1= 0 m3/s, U1 = 0.0 m
3/s/cm
For the second time interval,
Q2 = R2U1 + R1U2 = 0.7 x 0.0 + 0.5 x U2 = 44.0 m3/s, U2 = 88.0 m3/s/cm
For the third time interval
Q3 = R3U1 + R2U2+ R1U3 =0.0 *0.05 + 0.70 *88.0 + 0.5 *U3 = 111.0 m3/s.
U3 = 96.8 m3/s/cm
Similarly the remaining ordinates are found U4 =34.5 m3/s/cm d as U5 =15.5
m3/s/cm, U6 =10.4 m
3/s/cm U7 =-33.0 m
3/s/cm U8 =0.0 m
3/s/cm U9 =0.0 m
3/s/cm.
Unit hydrograph application. Once the unit hydrograph has been determined it
can be applied to find the direct runoff and stream flow hydrographs using Eq.
(8.9).
where Qn = direct runoff (to be calculated),
Rm = excess rainfall (observed or design rainfall excess),
U n-m-1 = the unit hydrograph ordinate (already known),
M = the number of pulses of excess rainfall, and
N = the number of pulses of direct runoff in the storm
considered.
n = 1, 2, ..., N
m = 1, 2, ..., M
Example8.5 Calculate the streamflow hydrograph for a storm with rainfall excess of
nearly 0 cm in the first half hour, 4 cm in the second half-hour and 1 cm the third half-
hour. Use the half hour unit hydrograph ordinate given in column [3]. Asume the
UR = Q 1+m-nm
Mn
1=m
n
UR = Q 1+m-nm
Mn
1=m
n
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180
baseflow is constant at 20 m3/s throughout the flood. Calculate also the watershed area
from which this Unit hydrograph is derived.
Solution The calculation of the direct runoff hydrograph by convolution is shown in the
Table below.
The time interval is in t = 0.5 h intervals. For the first time interval, n =1 in the
equation
Q1 = R1U1 = 0.0 x 12.5 = 0 m3/s.
For the second time interval,
Q2 = R2U1 + R1U2 = 4.00 x 12.5 +0.0 x 16.25 = 50 m3/s.
For the third time interval
Q3 = R3U1 + R2U2+ R1U3 =1.00 *12.5 +4.00 *16.25 + 0.0 *48.12 = 77.5 m3/s.
and so on. In tabular form it is easily calculated by simply shifting one time step of the
resulting hydrograph from individual excess rainfall. see columns [4] , [5] and [6].
[1] [2] [3] [4] [5] [6] [7] [8]
Time Excess
Rainfall (cm)
UH m3/s/cm hydrographs Base flow stream
flow
(o.5 hr) 0 cm ER 4cm ER 1cm ER (m3/s) (m3/s)
0*[3] 4*[3] 1*[3] [6]+[7]
1 0 12.50 0.00 20.00 20.00
2 4 16.25 0.00 50.00 20.00 70.00
3 1 48.12 0.00 65.00 12.50 20.00 97.50
4 58.07 0.00 192.48 16.25 20.00 228.73
5 14.02 0.00 232.28 48.12 20.00 300.40
6 19.16 0.00 56.08 58.07 20.00 134.15
7 0.00 76.64 14.02 20.00 110.66
8 0.00 0.00 19.16 20.00 39.16
20.00 20.00
The peak flow resulting from the storm was 300 m3/s and occurred at 2 hours. The
watershed area is calculated from the principle that the volume of the direct runoff under
the unit hydrograph is 1 cm in our case 1 cm, and it is 30 km2.
UR = Q 1+m-nm
Mn
1=m
n
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8.3 S- hydrograph
The S-hydrograph method allows the conversion of an X- hour unit hydrograph
into a Y-hour unit hydrograph, regardless of the ratio between X and Y. The
procedure consists of the following steps:
i. Determine the X hour S-hydrograph. Accumulating the unit hydrograph
ordinates at intervals equal to X, thus deriving the X-hour S-hydrograph.
ii. Lag the X -hour S-hydrograph by a time interval equal to Y hours.
iii. Subtract ordinates of these two S-hydrographs.
iv. Multiply the resulting hydrograph ordinates by X/Y to obtain the Y-hour
unit hydrograph.
v. The volume under X-hour and Y-hour unit hydrograph is the same. If Tb is
the time base of the X-hour unit hydrograph, the time base of the Y-hour
unit hydrograph is Tb - X + Y.
Example 8.6 Derive a 3-hr UH from 2-hr UH. The 2-hr UH is given in Columns
[1] and [2]
[1] [2] [3] [4] [5] [6]
Time (hr) 2-h 2-h lagged 3-hr UH
UH
(m3/s.cm)
SH
(m3/s.cm)
3h
(m3/s.cm)
[3]-[4]
(m3/s.cm)
[5]*2/3
(m3/s.cm)
0 0 0 0 0
1 50 50 50 33
2 150 150 150 100
3 300 350 0 350 233
4 600 750 50 700 467
5 750 1100 150 950 633
6 650 1400 350 1050 700
7 550 1650 750 900 600
8 450 1850 1100 750 500
9 350 2000 1400 600 400
10 250 2100 1650 450 300
11 150 2150 1850 300 200
12 50 2150 2000 150 100
13 0 2150 2100 50 33
14 0 2150 2150 0 0
sum = 4300 sum = 4300
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One-hour period ordinates of the 2 hr UH is read from the graph of the 2 hr UH Column
[2]. In calculating 2-h SH direct cumulative are not used, only sequential 2 hr cumulative
is considered.
8.4 Synthetic unit hydrograph
Where there are no streamflow data to develop unit hydrograph of a given
watershed by the foregoing methods, synthetic unit hydrograph method is utilized.
Synthetic unit hydrograph procedures are used to develop unit hydrographs for
other locations on the stream in the same watershed or for nearby watersheds of
similar physio-hydro-climatological characteristics. This section describes
Snyder’s synthetic unit hydrograph which is widely used in practice for un-
gauged watersheds.
Snyder found synthetic relations for some characteristics of a standard unit
hydrograph based on a study made mainly in the watersheds of the Appalachians
highlands, USA ranging from 30 to 30 000 km2. Five characteristics of a required
unit hydrograph for a given excess rainfall duration may be calculated:
i. the peak discharge per unit of watershed area qpR,
ii. the basin lag tpr (time difference between the centroid of the excess rainfall
hyetograph and the unit hydrograph peak),
iii. the base time tb, and
iv. the widths W (in time units) of the unit hydrograph at 50% and 75% of the
peak discharge.
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183
Figure 8.1 Snyder synthetic standard (a) and required (b) unit hydrographs.
Snyder defined a standard unit hydrograph as one whose rainfall duration tr is
related to the basin lag tp by
( 8.10)
For a standard unit hydrograph Snyder found the followings:
(Step 1) The basin lag is
(8.11)
Where:
tp = basin lag in hours
L = the length of the main stream in km from outlet to the
upstream divide
Lc = the distance in km from the outlet to a point on the stream nearest the
centroid of the watershed area.
C1 = 0.75
Ct = a coefficient derived from gauged watershed in the same region (0.3 to 0.6).
rp tt 5.5
`)( 3.0
1 ctp LLCCt
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184
Step (2) The peak discharge per unit drainage area in m3/s/cm/km
2 of the
standard unit hydrograph is
(8.12)
Where: C2 = 2.75
Cp = varies from 0.56 to 0.69
To compute Ct and Cp for a gauged watershed, the values of L and Lc are
measured from the basin map. From a derived unit hydrograph of the watershed
we obtain its effective duration tR in hours, its basin lag tpR in hours, and its peak
discharge per unit drainage area, qpR in m3/s/km
2.
If tpR = 5.5tR then tR = tr , tpR = tp and qpR = qp, and Ct and Cp are computed from
Eq. (8.11) and (8.12).
If tpR is quite different from 5.5tR, the standard basin lag is estimated by
(8.13)
Solve simultaneously for tr and tp, .then compute values of Ct and Cp with
qpR = qp and tpR = tp
Step (3) the relationship between, qp and the peak discharge per unit area qpR of
the required unit hydrograph is
(8.14)
(Step 4)
p
p
pt
CCq
2
rp
RrpRp
tt
tttt
5.5
4
pR
pp
pRt
tqq
pR
bq
t56.5
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185
(8.15)
(Step 5)
(8.16)
Where Cw = 1.22 for 75% of width
= 2.14 for 50% of width
Example 8.7 From its basin map of a given watershed, the following quantities are
measured: L = 150 km, Lc = 75 km, and watershed area = 3500 km2. From the unit
hydrograph derived for the watershed, the following are determined: tR = 12 hr, tpR = 34
hr, and qpR = 157.5 m3/s/cm. Determine the coefficients Ct and Cp.
Solution.
From the given data, 5.5tR = 66 hr, which is quite different from 34 hr. using Eq. (8.13)
And solving the above simultaneous equations we get: tr = 5.9 hr and tp = 32.5 hr.
Ct is calculated from
Ct = 2.65
The peak discharge per unit area for 1 cm depth of unit hydrograph is = 157.5/
3500 = 0.045 m3/s/cm/km
2. The coefficient Cp is calculated using
08.1 pRwqCW
rp
rp
tt
tt
5.5
4
1234
`)( 3.0
1 ctp LLCCt
`)75*150(75.05.32 3.0
tC
p
p
pt
CCq
2
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186
Example 8.8 Compute the six-hour synthetic unit hydrograph of a watershed having a
drainage area of 2500 km2 with L = 100 km and Lc = 50 km. This watershed has similar
pysio-climatic characteristics to the watershed of Example 8.7.
Solution.
The values of Ct and Cp determined above can also be used for this watershed.
For the six-hour unit hydrograph, tR = 6 hr, then tpR is calculated
Then qp in (m3/s/cm/km
2)
and
56.0
34
75.2045.0
p
p
C
C
`)( 3.0
1 ctp LLCCt
hrt
t
p
p
5.25
`)50*100(64.2*75.0 3.0
hrt
tp
r 64.45.5
5.25
5.5
hrt
t
tttt
pR
pR
RrppR
8.25
4
664.45.25
4
0604.05.25
56.0*75.2pq
0597.08.25/255*0604.0 pR
pp
pRt
tqq
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187
the peak discharge is qpr*A = 0.0597 * 2500 = 149.2 m3/s/cm.
The unit hydrograph coordinate then can be constructed based on the parameters that
derived above. The depth of the runoff under the unit hydrograph should be a unit.
hrqW
hrqW
pR
pR
9.440597.0*14.214.2
6.250597.0*22.122.1
08.108.1
50
08.108.1
75
hrq
tpR
b 930597.0
56.556.5
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188
8.5 Instantaneous Geomorphologic Unit Hydrograph
(8.17)
(8.19)
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189
(8.20)
(8.21)
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190
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191
8.6 Practice problems:
8.1 Determine (a) the design peak runoff rate, for a 50-year return period storm from
a 120 km2 watershed having IDF curve (I in mm/hr, T in years and tc in minutes) given by
78.0
18.0
)20(
500
ct
T = I
and with the following characteristics:
Subarea
(km2)
Topography
Slope (%)
Soil group Land use, treatment, and
hydrological condition
75 10-35 C Row crop, contoured, good 45 20-45 B Woodland, good
The maximum length of flow is 15 km and the difference in elevation along this path is
450 m.
8.2. A small rural watershed near Addis Ababa experienced 98 mm heavy storm in a
single day of early July. The watershed is covered by poor close-seed legumes 56%, poor pasture with grazing 25%, good fallow lands and 10 % and good wood lots 9%.
Hydrological soil groups respectively are B, A, B and C. Determine the direct runoff
from the 98 mm rainfall event
(a) by considering AMC I prevailed,
(b) by considering AMC III prevailed, and
(c) by considering AMC II prevailed.
8.3. A 15-minute unit hydrograph ordinates of a watershed are given in Table 1A.
The index of the watershed is 40 mm/hr.
(a) Determine the peak-flow that would result from a storm lasting 1.5 hours given in
Table 1B.
(b) Calculate the watershed area.
Table 1A: The unit hydrograph ordinates.
Time (min)
0 25 50 75 100 125 150 175
U(t)
m3/s/mm
0 85 145 234 126 78 48 0
Table 1B: rainfall data
Time (hr) 0.0 0.5 1.0 1.5
Accumulated
rainfall (mm)
0.0 25 56 95
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192
8.4 In July 10, 2000 the following rainfall and the resulting streamflow were recorded at a gauging site of the watershed having drainage area of 195 km2 .
Time (hr)
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
Accumulated
rainfall (mm)
20
45
78
Streamflow
(m3/s)
70
90
180
360
600
560
340
140
100
The base flow is 20 m3/sec.
In August 20, 2002 the following heavy rainfall were recorded over the watershed:
Time (hr)
1
2
3
Rainfall (mm)
19
36
40
(A) Estimate the peak flood caused by August 20, 2002 storm.
(B) If flooding at the gaging site occurs when the river stage is greater than 4 m, did the
August 20, 2002 rainfall cause flooding?
The gauging site has a rating equation defined by
where: Q in m3/sec, and H in m.
8.5 A watershed has a runoff coefficient of 0.4, area 350 ha. with general slope of 0.1 %
and maximum length of travel of overland flow of 2.5 km. Information on storm of 30
years return period is given as follows:
Duration (min) 15 30 45 60 80
Rainfall (mm) 40 60 75 100 120
Determine the size of the culvert that safely evacuates the 30-year flood from the watershed. Take an allowable velocity of 1.5 m/s through the culvert.
8.6. The following rainfall-runoff data were measured in an allowable velocity of 1.5 m/s through the culvert.
Rainfall P (cm) Runoff Qd (cm)
16.2 13.6
12.5 11.4 8.2 5.3
9.4 6.2
12.9 10.4
)0.2 + (H 80.0 = Q1.6
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193
Assuming that the data encompasses a wide range of antecedent moisture conditions, estimate the AMC II runoff curve number.
8.7. The following rainfall distribution was observed during a 6-hr storm
Time (hr) 0 - 2 2 - 4 4 - 6
Intensity (mm/hr) 10 15 12
The runoff curve number is CN = 76, calculate the ф index.
8.8 Given the following flood hydrograph and effective storm pattern, calculate the unit hydrograph ordinates of the watershed by the forward convolution equation.
Time (hr) 0 1 2 3 4 5 6 7 8 9 10 11 12
Flow (m3/s) 0 5 18 46 74 93 91 73 47 23 9 2 0
Time (hr) 0 1 2 3 4 5 6
Effective rainfall
(cm/hr) 0.5 0.8 1 0.7 0.5 0.2
Physiographic data on two nearly hydrological homogenous watersheds (1) and (2) are given below. Based on observed flood and the corresponding storm, a 3-hr unit
hydrograph was developed for watershed (1) and its peak discharge is 65 m3/s, the time
to peak from the beginning of the excess rainfall is 12 hr. Develop a unit hydrograph for
watershed (2).
Physiographic
characteristics
Watershed (1) Watershed (2)
L (Km) 40 60 Lca (km) 20 30
A (km2) 200 450
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River Routing
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195
9. River Routing
Flow routing is a mathematical procedure for predicting the changing magnitude,
speed, and shape of a flood wave as a function of time at one or more points along
a watercourse (waterway or channel). The watercourse may be a river, reservoir,
canal, drainage ditch, or storm sewer. The inflow hydrograph can result from
design rainfall, reservoir release (spillway, gate, and turbine release and / or dam
failures), and landslide into reservoirs.
River routing uses mathematical relations to calculate outflow from a river
channel given inflow, lateral contributions, and channel characteristics are known.
Channel reach refers to a specific length of river channel possessing certain
translation and storage properties.
The terms river routing and flood routing are often used interchangeably. This is
attributed to the fact that most stream channel-routing applications are in flood
flow analysis, flood control design or flood forecasting.
Two general approaches to river routing are recognized: (1) hydrologic and (2)
hydraulic. As will be discussed in Chapter 10, in the case of reservoir routing,
hydrologic river routing is based on the storage concept. Hydraulic river routing is
based on the principles of mass and momentum conservation. There are three
types of hydraulic routing techniques: (1) kinematics wave, (2) diffusion wave,
and (3) dynamic wave. The dynamic wave is the most complete model of
unsteady open channel flow, kinematics and diffusion waves are convenient and
practical approximations to the dynamic wave.
Hybrid model, possessing essential properties of the hydrologic routing and
hydraulic routing methods are being developed. The Muskingum-Cung method is
an example of a hybrid model.
This chapter discusses a commonly used hydrologic routing method such as the
Muskingum method.
River Routing
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196
9.1 The Muskingum Method
The Muskingum method of flood routing was developed in the 1930s in
connection with the design of flood protection schemes in the Muskingum River
Basin, Ohio, USA. It is the most widely used method of hydrologic river routing,
with numerous applications throughout the world.
For a river channel reach where the water surface cannot be assumed horizontal,
in case of flood flow, the stored volume becomes a function of the stages at both
ends of the reach, and not at the downstream (outflow) end only.
In a typical reach, the different components of storage may be defined for a given
instant in time as shown in Figure 9.1.
I Q
Wedge storage
Prism storage
Fig. 9.1: River reach storages.
The continuity equation holds at any given time: dS/dt = I(t) - Q(t) where the total
storage S is the sum of prism storage and wedge storage. The prism storage Sp is
taken to be a direct function of the storage at the downstream end of the reach and
the prism storage is the function of the outflow Sp = f1(Q). The wedge storage Sw
exists because the inflow, I, differs from outflow Q and so may be assumed to be
a function of the difference between inflow and outflow Sw = f2(I - Q).
The total storage may be represented by:
S = f1(Q) + f2 (I - Q) ( 9.1)
with due regard paid to the sign of the f2 term.
Assuming that in Eq. (9.1) f1(Q) and f2 (I - Q) could be both a linear functions, i.e.
f1(Q) = KQ and f2 (I - Q) = b (I - Q), we have
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S = bI + (K-b)Q = K [(b/K) I + (1 -b/k)Q] (9.2)
and writing X = b/K, we get
S = K [XI + ( 1 - X) Q] (9.3)
X is a dimensionless weighting factor indicating the relative importance of I and
Q in determining the storage in the reach and also the length of the reach. The
value of X has limits of zero and 0.5, with typical values in the range 0.2 to 0.4.
Most streams have X values between 0.1 and 0.3. K has the dimension of time.
Note that when X = 0, there is no wedge storage and hence no backwater, this is
the case for a level-pool reservoir. The parameter K is the time of travel of the
flood wave through the channel reach and should have the same time unit as t ,
the time interval of the inflow hydrograph.
The routing equation for the Muskingum method is derived by combining Eq.
(9.1) and Eq. (9.3) and the result is given in Eq. (9.4). The condition is that C1 + C
2 + C3 = 1, and often the range for t is K/3 t . K.
Determination of K. If observed inflow and outflow hydrograph are available for
a river reach, the values of K and X can be determined. Assuming various values
of X and using known values of the inflow and outflow, successive values of the
numerator and denominator of the following expression can be computed:
t + X) - 2K(1
t - X) - 2K(1 = C
t + X) - 2K(1
2KX + t = C
t + X) - 2K(1
2KX - t = C
where
QC + IC + IC = Q
3
2
1
j3j21+j11+j
(9.4)
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The computed values of the numerator and denominator are plotted for each time
interval, with the numerator on the vertical axis and the denominator on the
horizontal axis. This usually produces a graph in the form of a loop. The value of
X that produces a loop closest to a single line is taken to be the correct value for
the reach, and K is equal to the slope of the line. Finally K may be computed form
the average value determined from Eq. (9.5) for the correct values of X. Note that
since K is the time required for the incremental flood wave to traverse the reach,
its value may also be estimated as the observed time of travel of peak flow
through the reach.
With K = t and X = 0.5, flow conditions are such that the outflow hydrograph
retains the same shape as the inflow hydrograph, but it is translated downstream a
time equal to K. For X = 0, Muskingum routing reduces to linear reservoir
routing.
As a guide K and X are chosen such that
One rule of thumb used in practice is that the ratio t/K be approximately 1 and X
be in the range 0 to 0.5 together with Eq. (9.6).
)Q- QX)( - (1 + )I - IX(
)]Q + Q( - )I + It[(0.5 = K
j1+jj1+j
j1+jj1+j (9.5)
5.01
XandXK
t0.5 X (9.6)
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Example 9.1: Given the inflow and outflow hydrograph in Table E9.1 from Cols.1 to 3.
Estimate the routing parameters K and X.
Solution: Channel storage is calculated by :
)]Q + Q( - )I + It[(0.5 S= Sj1+jj1+jjj 1
Several values of X are tried, within the range 0.0 to 0.5, for example, 0.1, 0.2, and 0.3.
For each trial value of X, the weighted flow (XIj + (1-X)Q j) are calculated, as shown in
Table E9.1. Calculating the slope of the storage vs. weighted outflow curve then solves
the value of K. In this case the value of K is 2 days for X =0.1. It is to be noted that there
is greater storage during the falling stage than during a rising stage of a flood for a given
discharge.
Table E9.1. Derivation of X in the Muskingum method, example 9.1
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Time Inflow Outflow Storage Weighted flow (m3/s)
(days) (m3/s) (m3/s) (m3/s).d X = 0.1 X=0.2 X = 0.3
0 352 352 0 0 0 0
1 587 383 102 403 424 444
2 1353 571 595 650 728 806
3 2725 1090 1803 1254 1417 1581
4 4408 2021 3814 2259 2498 2737
5 5987 3265 6369 3537 3809 4081
6 6704 4542 8812 4758 4974 5190
7 6951 5514 10611 5658 5801 5945
8 6839 6124 11687 6196 6267 6339
9 6207 6353 11972 6338 6323 6309
10 5346 6177 11483 6094 6011 5928
11 4560 5713 10491 5598 5482 5367
12 3861 5121 9285 4995 4869 4743
13 3007 4462 7928 4316 4171 4025
14 2358 3745 6507 3606 3467 3328
15 1779 3066 5170 2937 2809 2680
16 1405 2458 4000 2352 2247 2142
17 1123 1963 3054 1879 1795 1711
18 952 1576 2322 1513 1451 1389
19 730 1276 1737 1221 1167 1112
20 605 829 1352 807 784 762
21 514 1022 986 971 920 870
22 422 680 603 654 628 603
23 352 559 371 538 517 497
24 352 469 209 457 445 434
25 352 418 118 411 405 398
X = 0.1 X = 0.2 X = 0.3
y = 1.8386x
R2 = 0.9896
0
2000
4000
6000
8000
10000
12000
14000
0 2000 4000 6000 8000
0
2000
4000
6000
8000
10000
12000
14000
0 2000 4000 6000 8000
0
2000
4000
6000
8000
10000
12000
14000
0 2000 4000 6000 8000
Where estimate of inflow and outflow hydrograph is not readily available
standard practice is to assume a value of 0.2 for X, with a smaller value for
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channel systems with large floodplains and larger values, near 0.4, for natural
channels. The following relationships for estimating K and X:
Where:
L = the reach length (m)
Vo = the average velocity (m/s)
yo = the full flow depth (m)
So = the slope of the channel bottom (m/m)
F = the Froud number V0 / [gyo] 2
Example 9.2: The flood hydrograph ordinates tabulated in Table E9.2 arrived at location
A (Figure E9.2) on 20 July 1999 at 1:00 p.m. Determine the peak flow and arrival time
(calendar day) of this flood at downstream location B. Muskingum coefficients of the
reach from A to B are: X = 0.35 and K = 1.2 days. The initial outflow at B was 10 m3/s.
A B
A B
Figure E9.2. Schematics of a river reach (From A to B)
Table E9.2: Flood hydrograph ordinates at Section A-A.
Time (hr)
1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0
Inflow
(m3/s)
15 80 150 180 200 140 125 75 45 25
Solution:
The Muskingum coefficients are determined for X = 0.35 and K = 1.2 hr. The values are
C1 = 0.0625, C 2 = 0.7188, and C3 = 0.2188. Then the outflow is calculated in Table E 9.2,
oV
L K
6.0 (9.10)
LS
yFX
0
0
2
9
413.05.0
(9.6)
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and plotted in Figure E9.2. The arrival time of the peak flood at section B-B is at 20 July
1999 at 6:00 p.m.
Table E9.2 Calculation of Outflow hydrograph from inflow hydrograph
1 2 3 4 5 6
Routing Inflow Outflow
period j I Q
(hr) (m3/s) C1 I j+1 C2 I j C3 Q j (m3/s)
Initial condition 10 10
1 15 0.94 7.19 2.19 10.31
2 80 5.00 10.78 2.26 18.04
3 150 9.38 57.50 3.95 70.82
4 180 11.25 107.81 15.49 134.55
5 200 12.50 129.38 29.43 171.31
6 140 8.75 143.75 37.47 189.97
7 125 7.81 100.63 41.56 149.99
8 75 4.69 89.84 32.81 127.34
9 45 2.81 53.91 27.86 84.57
10 25 1.56 32.34 18.50 52.41
11 25 1.56 17.97 11.46 31.00
12 25 1.56 17.97 6.78 26.31
13 25 1.56 17.97 5.76 25.29
14 25 1.56 17.97 5.53 25.06
15 25 1.56 17.97 5.48 25.01
16 25 1.56 17.97 5.47 25.00
Figure E9.2. Inflow and outflow hydrograph.
0
50
100
150
200
250
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16Time (hr)
Q (
m3/s
)
Inflow hydrograph
Outflow hydrograph
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9.2 Practice problems
9.1. Given the following inflow and outflow hydrograph for a given reach, determine K
and X.
Time (hr) 1 2 3 4 5 6 7 8 9 10 11
Inflow (m3/s) 45 85 160 185 200 170 145 135 98 45 75
Outflow (m3/s) 45 50 75 88 145 160 175 145 78 56 40
9.2. Given the following inflow hydrograph to a certain stream channel reach,
calculate the outflow hydrograph by the Muskingum method.
Time (hr) 1 2 3 4 5 6 7 8 9 10 11
Inflow (m3/s) 35 75 88 125 185 140 120 98 78 35 30
and K = 1 hr, X = 0.2 and t = 1 hr.
9.3. A storm event occurred on a given catchment that produced a rainfall pattern of 5
cm/hr for the first 10 min, 10 cm/hr in the second 10 min, and 5 cm/hr in the next 10 min.
The catchment is divided into three sub-catchments.
A 1 C
B 2
The unit hydrograph of the three sub-catchments are given in the following table. Sub
basins A and B had a loss rate of 2.5 cm/hr for the first 10 min and 1.0 cm/hr thereafter.
Sub-basin C had a loss rate of 1.0 cm/hr for the first 10 min and 0 cm/hr thereafter.
Determine the resulting flood at point 2 given the Muskingum coefficients K = 20 min
and X = 0.2.
10 - minutes unit hydrographs
Sub-catchment A Sub-catchment B Sub-catchment C
Time (min) Q
(m3/s/cm)
Time (min) Q
(m3/s/cm)
Time (min) Q (m3/s/cm)
0 0 0 0 0 0
10 5 10 5 10 16
20 10 20 12 20 33
30 15 30 16 30 50
40 20 40 21 40 33
50 25 50 26 50 16
60 20 60 19 60 0
70 15 70 18
80 10 80 10
90 5 90 5
100 0 100 0
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10. Reservoir Routing
Flood routing refers to the process of calculating the passage of a flood
hydrograph through a system. It is a procedure to determine the time and
magnitude of flow at a point on a downstream water course from known or
assumed hydrograph at one or more points upstream. If the system is reservoir
through which the flood is routed the term storage routing or reservoir routing is
used.
Reservoir routing method is used:
i. For flood forecasting in the lower parts of a river basin after passing
through reservoir, the case of Awash river downstream of Koka dam,
ii. For sizing spillways and determining dam / cofferdam height
iii. For conducting river basin watershed studies for watersheds where one or
more storage facilities exist. Specifically, for watersheds in which existing
reservoir are located, a reservoir routing is necessary to evaluate
watershed plans such as location of water supply structures, and regional
flood control measures.
Note that in order to develop an operational flood routing procedures for a major
river system, detailed knowledge of the main stream and the various feeder
channels is necessary.
In comparison to other hydrological problems, storage routing is relatively
complex. There are a number of variables involved, including (1) the input
(upstream) hydrograph; (2) the output (downstream) hydrograph; (3) the stage-
storage volume relationship measured from the site; (4) the energy loss (weir and
orifice) coefficients; (5) physical characteristics (e.g., weir length, diameter of the
riser pipe, length of the discharge pipe, etc.) of the outlet facility; (6) the storage
volume versus time relationship; (7) the depth-discharge relationship; and (8) the
target peak discharge allowed from the reservoir. The problem is further
complicated in that the inflow and outflow hydrographs can be from either storms
that have occurred (i.e., actually measured events) or design storm values.
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205
10.1. Level pool or reservoir routing using storage indication or
modified pulse method
Level pool routing is the procedure for calculating the outflow hydrograph from a
reservoir with a horizontal water surface, given its inflow hydrograph and storage
outflow characteristics.
For a hydrological system, input I(t), output Q(t), and storage S(t) are related by
the continuity equation
The time horizon is broken into intervals of duration t, indexed by j, that is t = 0,
t, 2t, ..., j t, (j+1)t, ..., and the continuity equation is integrated over each time
interval. For the j-th time interval:
(10.2)
The inflow values at the beginning and end of the j-th time interval are Ij and I j+1,
and the corresponding values of the outflow are Q j and Q j+1. If the variation of
the inflow and outflow over the interval is approximately linear (for t small), the
change in storage over the interval Sj+1 - Sj can be found by rewriting the above
equation as
In order to solve the above equation let us separate group first the known (the
right quantity from the equality) and the unknown (the left one) variables in the
following equation:
Q(t) - I(t) = dt
dS (10.1)
t2
Q + Q - t
2
I + I = S- S
1+jj1+jj
j1+j (10.3)
Q(t)dt - I(t)dt = dS
t1)+(j
tj
t1)+(j
tj
S
S
1+j
j
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206
The procedure then is first established storage-outflow relationship: 2S/t +Q and
Q based on the existing storage-water elevation and outflow-water elevation data,
physical characteristics of the reservoir.
The value of t is taken as the time interval of the inflow hydrograph. For a given
value of water surface elevation, the value of storage S and discharge Q are
determined, then the value of 2S/t +Q is calculated and plotted.
In routing the inflow through time interval j, all terms on the right side of Eq.
(10.4) are known, and so the values of 2S j+1 /t + Q j+1 can be computed. The
corresponding value of Q j+1 can be determined by linear interpolation of tabular
values.
To set up the data required for the next time interval, the value of 2Sj+1 / t - Q j+1
is calculated by
The computation is then repeated for subsequent routing periods.
Depth (stage) storage relationships for a given contour lines can be computed as follows.
The area within contour lines of the site can be planimetered, with the storage in any
depth increment Δh equal to the product of the average area and the depth increment.
Thus the storage increment ΔS is given by:
)Q - t
S2( + )I + I( = Q +
t
S2j
j
1+jj1+j
1+j
(10.4)
Q2 - )Q + t
S2( = )Q -
t
S2(
1+j1+j
1+j
1+j
1+j
(10.5)
hAAS ii )(*5.0 1 (10.6)
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207
Figure 10.1 Development of the storage-outflow function for level pool routing on the
basis of storage-elevation and elevation-outflow curves.
0.0
100.0
200.0
300.0
400.0
500.0
600.0
700.0
800.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
H (m)
Q( m
3 /s)
0.0
1000.0
2000.0
3000.0
4000.0
5000.0
6000.0
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
H (m)
S(m
3)
0.0
100.0
200.0
300.0
400.0
500.0
600.0
700.0
800.0
0.0 500.0 1000.0 1500.0 2000.0 2500.0
2S/dt+Q (m3/s)
Q(
m3 /s
)
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208
Example 10.1 . The design of an emergency spillway calls for a broad-crested weir of
width L = 10.0 m; rating coefficient Cd = 1.7; and exponent n = 1.5. The spillway crest is
at elevation 1070. Above this level, the reservoir walls can be considered to be vertical,
with a surface area pf 100 ha. The dam crest is at elevation 1076 m. Base flow is 17 m3/s,
and initially the reservoir level is at elevation 1071 m. Route the design hydrograph given
in Table E10.2 through the reservoir. What is the maximum pool elevation reached?
Solution. The calculation of the storage indication function above the spillway crest
elevation are shown in Table E10.1a. Outflow is calculated based on the Q = Cd L H n =
1.7 *10* H 1.5
.
The routing is summarized in Table E10.1b. The inflow hydrograph is given in Column 2
and 3. Columns 5 and 6 give calculated vale of 2Sj / t - Q j and 2Sj+1 / t + Q j+1. The
initial outflow is 17 m3/s; the initial storage indication value (when the reservoir water
level is 1 m above the spillway crest)
)Q - t
S2( + )I + I( = Q +
t
S2j
j
1+jj1+j
1+j
for j = 0 )Q - t
S2( + )I + I( = Q +
t
S20
0101
1
is = ( 17 +17 ) + 2*100000/3600 - 17 = 572.56 m3/s, with corresponding outflow of Q1 =
17 m3/s. For the next iteration for j = 1, we calculate Q2 - )Q +
t
S2( = )Q -
t
S2(
11
1
1
1
value based on the last estimated value, that is 2S1 / t - Q1 = 572.56 - 2*17 = 538.56
m3/s. Then
)Q - t
S2( + )I + I( = Q +
t
S2j
j
1+jj1+j
1+j
for j = 1 is )Q - t
S2( + )I + I( = Q +
t
S21
1212
2
= (17+20) + (538.56) = 575.56 m3/s.
The corresponding Q2 is then obtained from Table E10.1a as 17.4 m3/s. The recursive
procedure continues until iteration continues until the outflow has substantially reached
the base flow condition. The maximum pool elevation (MPE) occurs at maximum spill of
72.5 m3/s. It can be calculated from the ogee spillway equation and H = 2.63 m depth,
and the MPE is 1070.0 + 2.63 = 1072.63 m.
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209
Table E10.1a Storage discharge relationship.
[1] [2] [3] [4] [5]
Elevation
Head above
spillway
crest
Q
(outflow)
m3/s
S Storage
(m3)
2Sj+1/dt
+Qj+1
(m3/
s)
(m) (1070 m)
1070.0 0.0 0.0 0 0
1071.0 1.0 17.0 1000000 573
1072.0 2.0 48.1 2000000 1159
1073.0 3.0 88.3 3000000 1755
1074.0 4.0 136.0 4000000 2358
1075.0 5.0 190.1 5000000 2968
1076.0 6.0 249.8 6000000 3583
1077.0 7.0 314.8 7000000 4204
1078.0 8.0 384.7 8000000 4829
0
20
40
60
80
100
120
140
160
0 5 10 15 20 25
Time (hr)
Q (
m3/s
)
Inflow design Hydrograph
Outflow hydrograph over the spillway
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210
Table E10.1b: Design hydrograph and reservoir routing calculation
Time Index Time
Design Inflow
hydrograph Ij +I j+1 2Sj/dt - Qj 2Sj+1/dt +Qj+1 Qj+1
j (hr) (m3/s) (m3/s) (m3/s) (m3/s) (m3/s)
1 0 17 34 538.6 572.6 17.0
2 1 20 37 538.6 575.6 17.4
3 2 50 70 540.7 610.7 19.0
4 3 100 150 572.7 722.7 24.3
5 4 130 230 674.1 904.1 33.7
6 5 150 280 836.7 1116.7 45.9
7 6 140 290 1024.8 1314.8 58.3
8 7 110 250 1198.2 1448.2 67.2
9 8 90 200 1313.9 1513.9 71.7
10 9 70 160 1370.5 1530.5 72.8
11 10 50 120 1384.9 1504.9 71.0
12 11 30 80 1362.8 1442.8 66.8
13 12 20 50 1309.2 1359.2 61.2
14 13 17 37 1236.8 1273.8 55.7
15 14 17 34 1162.4 1196.4 50.8
16 15 17 34 1094.8 1128.8 46.7
17 16 17 34 1035.5 1069.5 43.1
18 17 17 34 983.3 1017.3 40.1
19 18 17 34 937.2 971.2 37.4
20 19 17 34 896.3 930.3 35.2
21 20 17 34 860.0 894.0 33.2
22 21 17 34 827.6 861.6 31.4
23 22 17 34 798.8 832.8 29.9
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211
10.1.2 Reservoir routing with controlled outflow
Most large reservoirs have some type of outflow control, wherein the amount of
outflow is regulated by gated spillways. In this case, both hydraulic conditions
and operational rules determine the prescribed outflow. Operational rules take
into accounts the various use of water.
The differential equation of storage can be used to route flows through reservoirs with
controlled outflow. In general, the outflow can be either (1) uncontrolled, (2) controlled
(gated), or (3) a combination of controlled and uncontrolled. The discretized equation,
including controlled outflow, is
in which Qr is the mean regulated outflow during the time interval t. With Qr
known, the solution proceeds in the same way as with the uncontrolled out flow
case.
In the case where the entire outflow is controlled, Eq. (10.6) reduces to
By which the storage volume can be updated based on average inflows and mean
regulated outflow.
Example 10.2 Discharge from a reservoir is over a spillway with discharge
characteristic:
Q = 120 H 1.5
Where: Q in m3/s and H is the head over the spillway (m).
The reservoir surface area is 12.5 km2
at spillway crest level and increases linearly by 2
km2 per meter rise of water level above crest level.
r
1+jj1+jjj1+jQ
2
Q + Q -
2
I + I =
t
S- S
(10.6)
r
1+jj
j1+j tQ - 2
I + It S= S (10.7)
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212
The design storm inflow, assumed to start with the reservoir just full, is given by a
triangular hydrograph, base length 40 hr and a peak flow of 450 m3/s occurring after 18
hours after the start of flow. Estimate the peak outflow over the spillway and its time of
occurrence to start of inflow (adapted from Shaw, 1994).
Solution. A level water surface in the reservoir is assumed. Temporary storage above the
crest level is given by:
)()5.12(*10)0.25.12( 3
0
26
0
mHHdhh Adh= S
HH
Outflow over the crest are given by Q = 120H1.5
Taking the time interval of the inflow hydrograph 2 hr = 7200 s, we have
)Q - t
S2( + )I + I( = Q +
t
S2j
j
1+jj1+j
1+j
5.1
1
2
11
6 1207200/)5.12(*10*2
jjj1+j
1+jHHH = Q +
t
S2
5.1
1
2
1
1
1 1207.27734722
jjj1+j
1+jHHH = Q +
t
S2
and
5.1
1
2
11 1207.27734722
jjj1+j
1+jHHH = Q
t
S2
Now we need to derive the values of the above two functions.
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213
10.2 Practice problems
10.1. Design the emergency spillway width (rectangular cross section) for the following
dam, reservoir, and flood conditions: dam crest elevation = 483 m; emergency spillway
crest elevation = 475 m, coefficient of spillway rating = 1.7; exponent of spillway rating
= 1.5. Elevation-storage relation:
Elevation (m) 475 477 479 481 483
Storage (hm3) 5.1 5.3 5.6 6.4 7.6
Inflow hydrograph to reservoir:
Time (hr) 0 1 2 3 4 5 6 7 8 9 10 11 12 13
Inflow
(m3/s)
0 10 30 50 60 150 250 350 280 210 190 170 130 100
Time (hr) 14 15 16 17 18 19 20 21 22 23 24
Inflow
(m3/s)
90 75 50 40 30 15 10 5 2 1 0
Assume design freeboard = 3 m and initial reservoir pool level at spillway crest.
10.2 Solve Example 10.1 if the inflow hydrograph is changed to:
Time (min)
0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150
Inflow
(m3/s)
0.
0
2.7 4.4 6.9 8.7 9.4 15.1 19.6 20.9 10.7 8.6 5.5 4.4 2.1 1.5 0.
Frequancy analysis
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214
11 Frequency Analysis
Extreme rainfall events and the resulting floods can take thousands of live and
cause billions of dollars in damage. Flood plain management and design of flood
control works, reservoirs, bridges, and other investigations need to reflect the
likelihood or probability of such events. Hydrological studies also need to address
the impact of unusually low rainfalls causing low stream flows which affects for
example water quality and water supply.
The term frequency analysis refers to the techniques whose objective is to analyze
the occurrence of hydrologic variable within statistical framework, by using
measured data and basing predictions on statistical laws.
Frequency analyses try to answer the following problems:
(1) Given n years of daily streamflow record for stream X, what is the
maximum (or minimum) flow Q that is likely to recur with a
frequency of once in T years on average?
(2) What is the return period associated with a maximum (or
minimum) flow Q. In more general term, the preceding questions
can be stated as follows: given n years of streamflow data for
stream X and L years design life of a certain structure, what is the
probability P of a discharge Q being exceeded at least once during
the design life L?.
Frequency analysis is made using appropriate probability distribution function to
the random variable under consideration. The next section briefly summarizes
probability distribution functions commonly used in hydrology.
11.1 Concepts of statistics and probability
Hydrological processes evolve in space and time in a manner that is partly
predictable, or deterministic and partly random, and such processes are called
stochastic processes. In this chapter, pure random processes are discussed using
statistical parameters and functions.
Frequancy analysis
______________________________________________________________________________
215
Let X is a random variable that is described by a probability distribution function
and represent for example annual rainfall amount at a specified location. Let a set
of observations x1, x2, …, xn of this random variable sample be drawn from a
hypothetical infinite population possessing constant statistical properties that is
stationarity (having no significant trend and variation in variance).
Defining sample space as a set of all possible samples that could be drawn from
the population, and an event as a subset of the sample space; the probability of an
event A, P(A), is the chance that it will occur when an observation of the random
variable is made.
If a sample of n observation has nA values in the range of event A, then the
relative frequency of A is
and the P(A) is given by
The basic three probability laws are:
(1) Total probability law: If the sample space is completely divided into m
non overlapping areas or events A1, A2, …, Am then
P(A1) + P(A2) + …+ Am = P() = 1
(11.3)
(2) Complementarity: It follows that if A is the complement of A, that is A
=
- A, then
P(A) = 1 – P(A) (11.4)
(3) Conditional probability:
)1.11(/ nnf As
)2.11(/lim)( nnAP An
)5.11()(
)()/(
AP
BAPABP
Frequancy analysis
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216
The above equation being read as P(B/A) the conditional probability that event B
will occur given that event A has already occurred is P(A B) the joint
probability that events A and B will both occur divided by P(A) the probability of
event A occurrence.
Example 11.1. The values of annual rainfall at Addis Ababa from 1900 to 1990 are
given in Table E11.1. Plot the time series and find the probability that the
annual rainfall R in any year is less than 1000 mm, greater than 1400 mm
and between 1000 and 1400.
Solution. The annual rainfall R at Addis Ababa over 90 years from 1900 to 1989 is
plotted in Figure E11.1. We see that there was extreme rainfalls in years 1947 (1939 mm)
and in year 1962 (903 mm).
Table E11.1: Annual rainfall amounts (mm) at Addis Ababa, Ethiopia, 1900 to 1989.
year 1900 1910 1920 1930 1940 1950 1960 1970 1980
0 1164 1270 1077 1460 937 946 1009 1423 1255
1 1241 1244 1041 1023 1105 935 1365 1175 1175
2 986 1162 1560 976 1154 1101 904 938 1209
3 1433 1175 1282 1181 1055 922 1015 1274 1192
4 1112 1439 1200 1027 1083 1199 1275 1192 1128
5 1101 1901 1179 1283 1006 1277 963 930 1190
6 1545 1729 1595 1419 1362 1025 1225 1124 1234
7 1047 1590 1271 1099 1939 1318 1167 1473 1212
8 1133 962 1343 1054 1313 1311 1102 1045 1203
9 1265 992 1245 1134 1354 1028 1328 1262 1324
There are n = 1989 – 1900 + 1 = 90 data points. Let A be the event R < 1000 mm, B is
the event R > 1400 mm. The number of events falling in these ranges are nA = 12, nB =
13.
So the P(A) 12 / 90 = 0.133
P(B) 13 / 90 = 0.144, and
P( 1000 < R < 1400) = 1 – P(A) – P(B)
= 1- 0.133 – 0.144 = 0.723 = 65/90.
Frequancy analysis
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217
11.1.1 Frequency and probability functions.
Relative frequency function fs (x) is given by
The number of observations ni in interval I, covering the range [xi – x, xi].
Equation 11.6 is an estimate of P(xi – x, X xi), that is the probability that the
random variable X will lie in the interval [xi – x, xi].
Frequency histogram is used to display the distribution of frequencies over
selected intervals. For example, the frequency histogram for the rainfall at Addis
Ababa with x = 50 mm is given in Figure E11.2.
Time series of Addis Ababa rainfall (1900 - 1989)
500
700
900
1100
1300
1500
1700
1900
2100
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000
year
annua
l ra
infa
ll a
mo
unts
(m
m)
)6.11(/)( nnxf iis
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218
Figure E11.1: Frequency histogram of the Addis Ababa annual rainfall (1900 - 1989)
Cumulative frequency function Fs(x) is given by
This is an estimate of P(X xi), that is the cumulative probability of xi.
Probability density function us defined as:
and the probability distribution function is defined as
and
Note that the relative frequency, cumulative frequency and probability
distribution functions are all dimensionless function varying over the range [0, 1].
0
5
10
15
20
25
30Absolute
frequency
< 949 950-
999
1000-
1049
1050-
1199
1200-
1349
1350-
1499
1500-
1649
1650-
1799
>
1800
Rainfall interval (mm)
)7.11()()(1
j
i
j
sis xfxF
)8.11()(
lim)(
0x
xfxf s
xn
)9.11()(
lim)(
0x
xFxF
xn
)10.11()(
)(dx
xdFxf
Frequancy analysis
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219
However the probability density function f(x) has a dimension [x]–1
and varies
over the range [0, ) and has the property of:
a b x
Figure 11.2: A probability density function
One of the best-known probability density functions is that forming the familiar
bell-shaped curve for the normal distribution:
Where mean and standard deviation are the parameters of the normal
distribution.
Defining standardized normal variable z as
Then
)11.11(1)(
dxxf
f(x)
0)()(
)()()()(
d
d
b
a
dxxfaXP
aFbFdxxfbXaP
)12.11(]2
)(exp[
2
1)(
2
2
xxf
xz
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220
The standard normal probability distribution function is then
Example 11.2. The annual mean flows of a certain stream have been found to be
normally distributed with mean 90 m3/s and standard deviation 30 m
3/s.
Calculate the probability that a flow larger than 150 m3/s will occur.
Solution. Let X be the random variable describing annual mean flow of the river given
above. The standardized variable
z value for flow equal to 150 m3/s is (150-90)/30 = 2.00
x f(x)
-4.0 0.0001338
-3.5 0.0008727
-3.0 0.0044318
-2.5 0.0175283
-2.0 0.0539910
-1.5 0.1295176
-1.0 0.2419707
-0.5 0.3520653
0.0 0.3989423
0.5 0.3520653
1.0 0.2419707
1.5 0.1295176
2.0 0.0539910
2.5 0.0175283
3.0 0.0044318
3.5 0.0008727
4.0 0.0001338
normal density function
0.0000
0.1000
0.2000
0.3000
0.4000
0.5000
-6.00 -4.00 -2.00 0.00 2.00 4.00 6.00x
f(x)
)13.11(]2
exp[2
1)(
2
zz
zf
)14.11(]2
exp[2
1)(
2
z
duu
xF
30
90
xxz
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221
The required probability is that P(X > 150 m3/s) = P(z > 2.0)
It is known that P(z > 2.0) = 1- P(z < 2.0) = 1- F(2). = 1 – (0.5 + 0.4772) = 0.0228.
So the probability that a flow larger than 50 m3/s will occur is 0.0228.
11.3 Statistical parameters
The objective of statistics is to extract the essential information from a set of data.
Statistical parameters are characteristics of a population, such as and . A
statistical parameter is the expected value E of some function of a random
variable.
The sample mean is calculated from
The variability of data is measured by the variance 2 and is defined by:
The sample variance s2 is estimated by
Coefficient of variation CV is defined by
)15.11()()(
dxxxfXmeantheXE
x
)16.11(1
1
n
i
ixn
x
)18.11()(1
1 2
1
2 xxn
sn
i
i
)17.11()()())(( 222
dxxfxXE
Frequancy analysis
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222
And sample CV is estimated by s /
The symmetry of a distribution about the mean is measured by the coefficient of
skewness :
Sample estimate
Example 11.3: Calculate the sample mean, sample standard deviation, and
sample coefficient of skewness of the Addis Ababa rainfall
given in Example 11.1.
Solution: Sample mean is calculated from Eq. (11.16)
Sample standard deviation is
Sample skewness is
)20.11()()(])[( 3
3
3
dxxfxXE
)19.11(
CV
)21.11()2)(1(
)(
3
1
3
snn
xxn
Cs
n
i
i
mmxxn
xi
i
n
i
i 120690
11 90
11
mmxxxn
si
i
n
i
i 203)1206(190
1)(
1
1 290
1
2
1
2
x
Frequancy analysis
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223
11.4 Fitting data to a probability distribution
As discussed in the previous section, a probability distribution is a function
representing the probability of occurrences of a random variable. By fitting a
distribution to a set of hydrologic data, a great deal of the probabilistic
information in the sample can be compactly summarized in the function and its
associated parameters. Two methods can be used for fitting a probability
distribution: the first is the method of moment and the second is the method of
maximum likelihood.
The method of moment. The principle in the method of moment is to equate the
moments of the probability density function about the origin to the corresponding
moments of the sample data.
f(x)dx
x = moment arm
xi = moment arm
204.1203)290)(190(
)1206(90
)2)(1(
)(
3
90
1
3
3
1
3
i
i
n
i
i x
snn
xxn
Cs
f(x)
fs(x)=1/n
Frequancy analysis
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224
Example11.4: The exponential distribution can be used to describe various kinds of
hydrologic data, such as the interval times between rainfall events. The
probability density function is given by
Determine the relationship between the parameter and the first moment about the origin
.
Solution:
The method of maximum likelihood. The central principle in the method of
maximum likelihood is that the best value of a parameter of a probability should
be that value which maximizes the likelihood or joint probability of occurrence of
the observed sample.
Let a sample of independent and identically distributed observations x1, x2, …, xn
of interval dx be taken. P(X= xi) = f(xi) = the value of the probability density for
X = xi if f(xi), and the probability that the random variable will occur in the
interval including xi is f(xi)dx. Since it is assumed that the observations are
independent, the joint probability of occurrence is simply the product of the
probability of the observations, thus the joint probability of occurrence is
(11.22)
The likelihood function L is given by
(11.23)
Or ln(L) is
(11.24)
xexf )(
xexXE x
1
)(0
n
i
n
i dxxf1
)]([
n
i ixfL1
)(
])(ln[)ln(1
n
i
ixfL
Frequancy analysis
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225
Example 11.5: The following data are the observed times between rainfall events at a
given location. Assuming that the inter-arrival time of rainfall events follows an
exponential distribution; determine the parameter for this process by the method of
maximum likelihood. The time between rainfall events (days) are: 2.2, 1.5, 0.6, 3.4, 2.1,
1.3, 0.8, 0.5, 4.0, and 2.5.
Solution:
The log-likelihood function is
The maximum value of ln(L) occurs when
Thus
Testing goodness of fit. By comparing the theoretical and sample values of the
relative frequency or the accumulative frequency function, one can test the
goodness of fit of a probability distribution. In the case of the relative frequency
function, the 2 test is used. The sample value of the relative frequency of interval
i is calculated using Eq. (11.6)
and the theoretical value is estimated from
]ln[])(ln[)ln(11
n
i
xn
i
iiexfL
n
i
ii
n
i
xnx11
ln)(ln
0)(ln
L
)/1(89.1
11
0)(ln
1
hrx
xnL n
i
i
nnxf iis /)(
Frequancy analysis
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226
The 2 test statistic is given by
(11.25)
Where m = the number of intervals.
To describe the 2 test, the
2 probability distribution must be defined. A
2
distribution with v degrees of freedom is the distribution for the sum of squares of
v independent normal random variables zi; this sum is the random variable
(11.26)
The 2 distribution function is tabulated in Annex 1. In the
2 test, v = m - p- 1,
where m is the number of intervals as before, and p is the number of parameters
used in fitting the proposed distribution. A confidence level is chosen for the test;
it is often expressed as 1 - , where is termed the significant level. A typical
value for the confidence level is 95 percent. The null hypothesis for the test is that
the proposed probability distribution fits the data adequately. This hypothesis is
rejected (i.e., the fit is deemed inadequate) if value of in Eq. (11.25) is larger
than a limiting value,
as determined from the 2 distribution with v degrees of freedom as the value
having cumulative probability 1 - .
11.5. Common probabilistic models
Many discrete probability mass functions and continuos probability density
functions are used in Hydrology. The most common are the binomial,
exponential, normal, gamma (Pearson Type 3), log-normal, log-gamma (log-
)()()( 1 iii xFxFxp
2
c
m
i i
iisc
xp
xpxfn
1
22
)(
)]()([
v
i
ivz
1
22
2
c
21, v
Frequancy analysis
______________________________________________________________________________
227
Pearson Type III) and Gumbel (extreme value type I). A description of some
commonly used probability distribution in hydrology is given below.
11.5.1 The Binomial distribution
It is common to examine a sequence of independent events for which the outcome
of each can be either a success or a failure; e.g., either the T-yr flood occurs or it
does not. Such a sequence consists of Bernoulli trials, independent trials for which
the probability of success at each trial is a constant p. The binomial distribution
answers the question, what is the probability of exactly x successes in n Bernoulli
trials?
The probability that there will be x successes followed by n-x failures is just the
product of the probability of the n independent events: px
(1-p) n-x
. But this
represents just one possible sequence for x successes and n-x failures; all possible
sequences must be considered, including those in which the successes do not
occur consecutively. The number of possible ways (combinations) of choosing x
events out of n possible events is given by the binomial coefficient
(11.27)
Thus, the desired probability is the product of the probability of any one sequence
and the number of ways in which such a sequence can occur is
(11.28)
Where x = 0, 1, 2, 3, …, n.
The mean and the variance of x are given by
(11.29)
(1130)
)!(!
!
xnx
nn
x
xnx
n
x
ppxP
)1()(
)1()(
)(
pnpxVar
npxE
Frequancy analysis
______________________________________________________________________________
228
The skewness is
(11.31)
Example 11.6: Consider the 50-yr flood, that is a flood having a return period of 50
years, T = 50 years, and then the probability of exceedence is given by P( the flood > x
value) = p = 1/T = 0.02.
(a) What is the probability that at least one 50-yr flood occur during the 30-yr
life time of a flood control project?
(b) What is the probability that the 100-yr flood will not occur in 10-yr? In 100
yr?
(c) In general what is the probability of having no floods greater than the T-yr
flood during a sequence of T yr?
Solution: (a) The probability of occurrence in any one year (event) is p = 1/T . The
probability (at least one occurrence in n events) is called the risk. Thus the risk is the sum
of the probabilities of 1 flood, 2 floods, 3 floods, …, n floods occurring during the n-yr
period. In other words, risk is 1- probability of no occurrence in n yr [1-P(0)].
Risk = 1 – P(0)
= 1- (1-p)n
= 1 – (1 – 1/T)n
Reliability = 1 - Risk
For the problem at hand, p = 1/T = 1/ 50 = 0.02
Risk = 1 – (1 – 1/T)n
= 1- (1 – 0.02)30
= 0.455
(b) Here p = 1/100 = 0.01, for n =10 yr, P(x =0) = 0.92. For n =100, P(x=0) =0.37.
(c) P (x =0) = (1 – 1/T)T
as T gets larger, P (x=0) approaches 1/e = 0.368. The Risk of
flooding in T –yrs is then 1- 0.368 = 2/3.
5.0)1(
21
pnp
pCs
0300
30
0
)1()1()0(
ppppP xnx
n
x
Frequancy analysis
______________________________________________________________________________
229
Example 11.7: A cofferdam has been built to protect homes in a floodplain until a major
channel project can be completed. The cofferdam was built for the 20 –yr
flood event. The channel project will require 3-yr to complete. What are
the probabilities that:
(a) The cofferdam will not be overtopped during the 3 yr (the
reliability)?
(b) The cofferdam will be overtopped in any one year?
(c) The cofferdam will be overtopped exactly once in 3 yr?
(d) The cofferdam will be overtopped at least once in 3 yr (the risk)
(e) The cofferdam will be overtopped only in the third year?
Solution:
(a) Reliability = (1 – 1/T)n
= (1 – 1/20)3 = 0.86
(b) Prob = 1 /T = 0.05
(c)
(d) Risk = 1 – Reliability = 0.14
(e) Prob =(1-p)(1-p)p =0.95 2. (0.05) = 0.045
11.5.2 The exponential distribution:
Consider a process of random arrivals such that the arrivals (events) are
independent, the process is stationary, and it is not possible to have more than one
arrival at an instant in time. If the random variable t represents the inter-arrival
time (time between events), it is found to be exponentially distributed with
probability density function of
(11.32)
The mean and the variance of t:
(11.33)
(1134)
The skewness is 2.
135.0)1()1(,05.0 131
3
1
ppxPpfor
.0,)( tetf t
2/1)(
/1)(
tVar
tE
.0,1)( tetF t
Frequancy analysis
______________________________________________________________________________
230
(11.35)
Example 11.8. During the course of a year, about 110 independent storm events occur at
a given location, and their average duration is 5.3 hours. Ignoring
seasonal variations, a year of 8760 hours, calculate the storm average
inter-arrival time. What is the probability that at least 4 days = 96 hr
elapse between storms? What is the probability that the separation
between two storms will be less than or equal to 12 hrs.
Solution: the average inter-event time is estimated by subtracting the total rainfall
periods from the total hours (rainy and non-rainy) and dividing by the number of rainfall
events.
The inter-event time = (8760 –110*5.3)/110 = 74.3 hr. and = 1/74.3 = 0.0135.
The probability that at least 4 days = 96 hr elapse between storms is Prob (t 96) = 1 – F
(96) = e(-0.0135*96)
.
The probability that the separation between two storms will be less than or equal to 12 hrs
is Prob (t < 12) = F (12) = 1- e(-0.0135*12)
. = 0.15
11.5.3 Extreme value distribution
Many time interests exist in extreme events such as the maximum peak discharge
of a stream or minimum daily flows. The extreme value of a set of random
variables is also a random variable. The probability distribution of this extreme
value random variable will in general depend on the sample size and the parent
distribution from which the sample was obtained.
The study of extreme hydrological events involves the selection of a sequence of
the largest or smallest observations from sets of data. For example, the study of
peak flows uses just the largest flow recorded each year at a gauging station - for
30 years of data only 30 points are selected.
Gumbel distribution: Extreme Value Type I
The extreme Value Type I (EVI) probability distribution is
(11.36)
xux
xF ))exp(exp()(
Frequancy analysis
______________________________________________________________________________
231
The parameters are estimated by:
(11.37)
A reduced variate y can be defined as:
(11.38)
Substituting the reduced variate into Eq.(11.37) yields
F(x) = exp(-exp(-y)) (11.39)
Solving for y:
(11.40)
The return period and the cumulative probability function is related by
P(X xT) = 1/T
= 1- P(X < xT )
= 1- F(xT)
Also
F(xT) = (T-1)/T
And finally we get y in terms of return period for EVI distribution:
(11.41)
5772.0,6
xus
uxy
)(
1lnln
xFy
1lnln
T
TyT
Frequancy analysis
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232
and related to xT by
xT = u + yT (11.42)
Heavy storm are most commonly modeled by the EVI distribution.
Example 11.9. Annual maximum values of 10-minutes-duration rainfall at Chicago,
Illinois, from 1913 to 1947 are given in Table E11.9. Develop a model
for storm rainfall frequency analysis using Extreme value Type I
distribution and calculate the 5-, 10-, and 50 – year return period
maximum values of 10-minute rainfall at Chicago.
Table E11.2. Annual maximum 10-minutes rainfall (mm) at Chicago, Illinois,
1913-1947
Year 10-min R (mm) Year 10-min R (mm)
1913 12 1930 8 1914 17 1931 24
1915 9 1932 24
1916 15 1933 20 1917 10 1934 16
1918 12 1935 18
1919 19 1936 28
1920 13 1937 16 1921 19 1938 13
1922 14 1939 16
1923 20 1940 9 1924 17 1941 18
1925 17 1942 14
1926 17 1943 23
1927 15 1944 17 1928 22 1945 17
1929 12 1946 16
1947 15
Solution: The sample moments calculated from the data in Table E11.2. The mean is
16.5 mm and the standard deviation is 4.5 mm.
The parameters of the EVI distribution is then:
4.145.3*5772.05.165772.0
5.35.466
xu
s
Frequancy analysis
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233
The probability model is
To determine the values of xT for various of return period T, it is convenient to use the
reduced variate yT given by Eq. (11.41). For T = 5 years
And Eq. (11.42) yields
xT = u +yT
=14.4 + 3.5 x 1.500
= 19.6 mm.
By the same method, the 10-, and 50- year values are estimated to be 22.4 mm and 28.2
mm respectively.
It may be noted from the data in Table E11.2 that the 50-year return period rainfall was
equaled once in the 35 years data (in 1936), and that the 10-year return period rainfall
was exceeded four times during this period.
Extreme Value Type III (Weibull) distribution – Low flow analysis:
Weibull distribution has found greatest use in hydrology as the distribution of low
stream flows. It is defined as:
(11.43)
The cumulative Weibull, F(x), is given by
(11.44)
The mean and the variance of the distribution are:
.0,,0),))/()((exp()()()( 1 xxxxf
)))/()((exp(1)( xxF
xx
xF
xux
xF
0))5.3
4.14exp(exp()(
))exp(exp()(
500.115
5lnln
1lnln
T
TyT
Frequancy analysis
______________________________________________________________________________
234
(11.45)
(11.46)
Where is a displacement parameter to create 0 as the lower bound of the
parameter x.
The estimate of the parameters is done using:
(11.47)
(11.48)
Where A() and B() is taken from Table 11.2.
Example 11.10: The minimum annual daily discharge on a stream are found to have an
average of 4.6 m3/s, a standard deviation of 1.8 m
3/s and a coefficient of
skew of 1.4. Evaluate the probability of the annual mean flow being less
than 3.69 m3/s.
Solution: Weibull distribution is used here for low flow analysis
Using estimated coefficient of skewness value = 1.4 , then the corresponding
parameters are read from Table 11.2.
1/ = 0.79, A() = 0.098, B() = 1.36
estimates of = 1.266,
)/11()()( XE
)]/11()/21([)()( 22 XVar
)( A
)( B
Frequancy analysis
______________________________________________________________________________
235
Table 11.2. Values of A() and B()
(Skewness) 1/ A() B()
-1.000 0.02 0.446 40.005
-0.971 0.03 0.444 26.987
-0.917 0.04 0.442 20.481
-0.867 0.05 0.439 16.576
-0.638 0.10 0.425 8.737
-0.254 0.20 0.389 4.755
0.069 0.30 0.346 3.370
0.359 0.40 0.297 2.634
0.631 0.50 0.246 2.159
0.896 0.60 0.193 1.815
1.160 0.70 0.142 1.549
1.430 0.80 0.092 1.334
1.708 0.90 0.044 1.154
2.000 1.00 0.000 1.000
2.309 1.10 -0.040 0.867
2.640 1.20 -0.077 0.752
2.996 1.30 -0.109 0.652
3.382 1.40 -0.136 0.563
3.802 1.50 -0.160 0.486
4.262 1.60 -0.180 0.418
4.767 1.70 -0.196 0.359
5.323 1.80 -0.208 0.308
5.938 1.90 -0.217 0.263
6.619 2.00 -0.224 0.224
7.374 2.10 -0.227 0.190
8.214 2.20 -0.229 0.161
Prob (X 3.7) = F(3.7) is give by
= 0.368
11.5.4 Frequency Analysis using Frequency Factor
8.4)098.0(8.16.4)( A
4.2)36.1(8.18.4)( B
)))4.28.4/()4.27.3((exp(1)7.3( 266.1F
Frequancy analysis
______________________________________________________________________________
236
Calculating the magnitude of extreme events by the method outlined in Example
11.9 requires that the probability distribution function be invertible, that is, for a
value for T or [F(xT) = T/(T-1)], the corresponding value of xT can be determined.
Some probability distribution functions are not readily invertible, including the
Normal and Pearson Type III distributions, and an alternative method of
calculating the magnitudes of extreme events is required for these distributions.
The magnitude of xT of a hydrological event may be expressed as:
(11.49)
which may be approximated by
(11.50)
in the event that the variable analyzed is y = log x, then the same method is
applied to the statistics for the logarithms of the data, using
(11.
51)
and the required value of xT is found by taking antilog of yT.
The Frequency factor for Normal Distribution
The frequency factor can be expressed from Eq. (11.50) as
(11.52)
This is the same as the standard normal variable z defined in this chapter.
The value of z corresponding to an exceedence probability of p = 1/T can be
calculated by finding the value of an intermediate variable w:
(11.53)
TT Kx
sKxx TT
sKyy TT
T
T
xK
)5.00(1
ln
5.0
2
p
pwT
Frequancy analysis
______________________________________________________________________________
237
then calculating z using the approximation
(11.54)
When p > 0.5, 1-p is substituted for p in Eq. (11.53) and the value of z computed
by Eq.(11.54).
The Frequency factor for Extreme value distribution Type I (EVI)
For the EVI distribution the frequency factor is given by
(11.55)
Extreme value distribution Type II (EVII):
For the Extreme value distribution Type II (EVII) the logarithm of the variate
follows the EVI distribution. For this case Eq.(11.51) is used to calculate yT, using
the value of KT from Eq.(11.55).
Log-Pearson Type III distribution.
Log-Pearson Type III distribution the first step is to take the logarithms of the
hydrologic data, y = log x. Then the mean y, the standard deviation sy and
coefficient of skewness Cs are calculated for the logarithms of the data. The
frequency factor depends on the return period T and the coefficient of skewness
Cs. When Cs= 0, the frequency factor is equal to the standard normal variable z.
When Cs 0, KT is approximated by
(11.56)
where k = Cs/6.
The value of z for a given return period can be calculated using Eq.(11.53) &
(11.54).
32
2
001308.0189269.0432788.11
010328.0802853.0515517.2
www
wwwz
1lnln5772.0
6
T
TKT
5432232
3
1)1()6(
3
1)1( kzkkzkzzkzzKT
Frequancy analysis
______________________________________________________________________________
238
Example11.11.The annual maximum daily discharge measured on the Beressa river at
Debere Birhan gauging site are given in Table E11.3. The Beresa river is
a tributary of Jemma River which is lying in Abay basin and has
watershed area of 220 km2 . Calculate the 5- and 50- year return period
annual maximum discharge of the Beressa river at Deberebirhan using
lognormal, EVI, and log-Pearson Type III distributions.
Table E11.3. Maximum daily discharge of the Beresa River (m3/s)
Year Q (m3/s) Year Q (m3/s)
1961 60.4 1980 84.4
1962 59.5 1981 missed
1963 82.5 1982 180.0
1964 90.0 1983 107.1
1965 32.8 1984 66.8
1966 75.0 1985 92.0
1967 58.0 1986 89.4
1968 112.5 1987 17.9
1969 151.4 1988 67.7
1970 80.7 1989 37.4
1971 144.0 1990 53.5
1972 63.1 1991 56.1
1973 81.3 1992 54.5
1974 163.0 1993 56.6
1975 83.7 1994 252.2
1976 140.0 1995 148.7
1977 58.0 1996 126.0
1978 74.5 1997 91.9
1979 101.0
Solution: Let X be the maximum annual discharge, then the mean x = 91.48 m3/s , the
standard deviation sx = 46.89 m3/s, and coefficient of skewness Cs = 1.4. For the log 10
data Y = Log(X), then the mean y = 1.91 m3/s , the standard deviation sy = 0.22 m
3/s,
and coefficient of skewness Cs = -0.3971.
Lognormal distribution.. The frequency factor can be obtained from Eq.(11.55). For T =
50 years, KT = 2.054
Then
x50 = 10 2.36
=229 m3/s
Similarly for T = 5 years, KT = 0.842.
sKyy TT
sm
y
/36.2
22.0*054.291.1
3
50
sKyy TT
Frequancy analysis
______________________________________________________________________________
239
Then
x5 = 10 2.095
=124 m3/s
EVI distribution. The frequency factor can be obtained from Eq.(11.54). For T = 50
years, KT = 2.592
Then
Similarly for T = 5 years, KT = 0.719.
Then
Log-Pearson Type III distribution. For Cs = -0.3971, the value of K50 is obtained using
Eq. (11.56), K50 1.834,
x50 = 10 2.313
=205 m3/s
Similarly for T = 5 years, KT = 0.855, and x5 = 125 m3/s.
In summary:
Return period
5 years 50 years
Lognormal 124 229
EVI 125 213
Log-Pearson Type III 125 205
In this example the values of Beressa flood at Deberebirhan estimated by the
lognormal, EVI and log-pearson Type III distribution are very close to each other.
sm
y
/095.2
22.0*842.091.1
3
5
sm
x
/213
89.46*592.248.91
3
50
sKxx TT
sKxx TT
sm
x
/125
89.46*719.048.91
3
0
sKyy TT
313.2
22.0*834.191.150
y
Frequancy analysis
______________________________________________________________________________
240
Commonly accepted practice first the data has to be fitted to candidate
distributions and select the model that describes the observed data very well and
apply the selected model in estimating the required flood of a given return period.
11.6 Probability plot
As a check that a probability distribution fit a set of hydrological data, the data
may be plotted on specially designed probability paper, or using a plotting scale
that linearizes the distribution function. The plotted data are then fitted with a
straight line for interpolation purposes.
Plotting position refers to the probability value assigned to each piece of data to
be plotted. Numerous methods have been proposed for the determination of
plotting positions. Most plotting position formulas are represented by :
(11.57)
Where m = is the rank 1 for the maximum, and n is for the minimum value
n = the number of data points used in the analysis.
b = 0.5 Hazen’s plotting position
b = 0.3 Chegodayev’s plotting position
b = 3/8 Blom’s plotting position
b = 1.3 Tukey’s plotting position
b = 0.44 Gringorten’s plotting position
Example 11.12. Considering that the probability distribution of the maximum flow of
the Berassa river used in Example 11.11 follows the Gumbel distribution,
plot the values on Gumbel paper.
Solution: The data is ranked first col [4], and the plotting position method is chosen, b =
0.4 Gringorten’s plotting position, and the return period is calculated for the data Col
[5]. Then the reduced variate yT for the Gumbel distribution is calculated for the T
associated
[1] [2] [3] [4] [5] [6] [7] [8] [9]
Year Flow Rank Flow Empirical Emprical
CDF
Gumbel distribution
T = (n+1-2a)/
(m-a) 1-1/T Reduced Predicted Observed
Tbn
bmxXP m
1
21)(
Frequancy analysis
______________________________________________________________________________
241
variate
y
flow
Flow
1961 60.4 1 252.2 60.3 0.983 4.091 220.1 252.2
1962 59.5 2 180.0 22.6 0.956 3.097 183.7 180.0
1963 82.5 3 163.0 13.9 0.928 2.597 165.4 163.0
1964 90.0 4 151.4 10.1 0.901 2.256 153.0 151.4
1965 32.8 5 148.7 7.9 0.873 1.996 143.4 148.7
1966 75.0 6 144.0 6.5 0.845 1.783 135.7 144.0
1967 58.0 7 140.0 5.5 0.818 1.603 129.1 140.0
1968 112.5 8 126.0 4.8 0.790 1.445 123.3 126.0
1969 151.4 9 112.5 4.2 0.762 1.305 118.2 112.5
1970 80.7 10 107.1 3.8 0.735 1.177 113.5 107.1
1971 144.0 11 101.0 3.4 0.707 1.060 109.2 101.0
1972 63.1 12 92.0 3.1 0.680 0.951 105.2 92.0
1973 81.3 13 91.9 2.9 0.652 0.849 101.5 91.9
1974 163.0 14 90.0 2.7 0.624 0.753 97.9 90.0
1975 83.7 15 89.4 2.5 0.597 0.661 94.6 89.4
1976 140.0 16 84.4 2.3 0.569 0.573 91.4 84.4
1977 58.0 17 83.7 2.2 0.541 0.489 88.3 83.7
1978 74.5 18 82.5 2.1 0.514 0.407 85.3 82.5
1979 101.0 19 81.3 1.9 0.486 0.327 82.4 81.3
1980 84.4 20 80.7 1.8 0.459 0.249 79.5 80.7
1982 180.0 21 75.0 1.8 0.431 0.172 76.7 75.0
1983 107.1 22 74.5 1.7 0.403 0.096 73.9 74.5
1984 66.8 23 67.7 1.6 0.376 0.021 71.2 67.7
1985 92.0 24 66.8 1.5 0.348 -0.054 68.4 66.8
1986 89.4 25 63.1 1.5 0.320 -0.129 65.7 63.1
1987 17.9 26 60.4 1.4 0.293 -0.206 62.9 60.4
1988 67.7 27 59.5 1.4 0.265 -0.283 60.0 59.5
1989 37.4 28 58.0 1.3 0.238 -0.363 57.1 58.0
1990 53.5 29 58.0 1.3 0.210 -0.445 54.1 58.0
1991 56.1 30 56.6 1.2 0.182 -0.532 50.9 56.6
1992 54.5 31 56.1 1.2 0.155 -0.624 47.6 56.1
1993 56.6 32 54.5 1.1 0.127 -0.724 43.9 54.5
1994 252.2 33 53.5 1.1 0.099 -0.836 39.8 53.5
1995 148.7 34 37.4 1.1 0.072 -0.968 35.0 37.4
1996 126.0 35 32.8 1.0 0.044 -1.138 28.8 32.8
1997 91.9 36 17.9 1.0 0.017 -1.411 18.8 17.9
with the plotting position and the flow data col [7]. The Gumbel predicted flow is done in
Col. [8] using Eq.(11.42). Then plot the predicted col [8] and observed col [9] flows on
the Gumbel scale with x- axis being col [7]. It is seen that the data fits well the Gumbel
distribution except at the extreme value.
Frequancy analysis
______________________________________________________________________________
242
11.7. Testing for outliers
Outliers are data points that depart significantly from the trend of the remaining
data. The retention and deletion of these outliers significantly affect the
magnitude of the statistical parameters computed from the data, especially small
sample size.
Water Resources Council (1981) recommends that if the station skew is greater
than +0.4, tests for high outliers are considered first; if the station skew is less
than –0.4, test for low outliers are considered first. Where the station skew is
between 0.4, test for both high and low outliers should be applied before
eliminating any outliers from the date set.
The following frequency equation can be used to detect high outliers:
(11.58)
Where; yH = the high (+) / low (-) outlier threshold in log units
Kn = values are Given in Table 11.3 for one sided test that detect
outlier at the 10-percent level of significance in normally
distributed data.
If the logarithms of the values in a sample are greater / less than yH in the above
ynH sKyy
0.0
50.0
100.0
150.0
200.0
250.0
300.0
-2.000 -1.500 -1.000 -0.500 0.000 0.500 1.000 1.500 2.000 2.500 3.000 3.500 4.000 4.500
Peak Q
(m
3/s
)
Gumbel reduced variate y
Gumbe Frequency Plot of Beresa River peak Flow
Frequancy analysis
______________________________________________________________________________
243
equation, then they are considered high / low outlier.
Table 11.3 Outlier test Kn value
Sample size n
Kn Sample size n
Kn Sample size n
Kn Sample size n
Kn
10 2.036 24 2.467 38 2.661 60 2.837
11 2.088 25 2.486 39 2.671 65 2.866
12 2.134 26 2.502 40 2.682 70 2.893 13 2.175 27 2.519 41 2.692 75 2.917
14 2.213 28 2.534 42 2.700 80 2.940
15 2.247 29 2.549 43 2.710 85 2.961 16 2.279 30 2.563 44 2.719 90 2.981
17 2.309 31 2.577 45 2.727 95 3.000
18 2.335 32 2.591 46 2.736 100 3.017
19 2.361 33 2.604 47 2.744 110 3.049 20 2.385 34 2.616 48 2.753 120 3.078
21 2.408 35 2.628 49 2.760 130 3.104
22 2.429 36 2.639 50 2.768 140 3.129 23 2.448 37 2.650 55 2.804
Example 11.13. Check the data given in Example 11.11 for outliers?
Solution: The mean and standard deviation of log transformed peak flow with
sample size n = 36 are 1.9089 m3/s and 0.2217 m
3/s respectively. For n = 36 the
value of Kn = 2.639.
= 2.485 & 1.31933
Corresponding to Q = 305 m3/s and 21 m
3/s.
The maximum value is 257 m3/s and the minimum is 17 m
3/s. It is seen that low
outlier is found but it is very near to the boundary of 21 m3/s. So the data may be
acceptable in a sense that no outlier is found. However, one should check the
reason behind the low outlier, by comparing to the rainfall in the rainy months of
June, July, and August of the year 1987.
ynH sKyy
2217.0*639.29.1 Hy
Frequancy analysis
______________________________________________________________________________
244
11.8 Practice problems
11.1 The values of annual rainfall at Addis Ababa from 1900 to 1990 are given in
Table Find the mean, standard deviation, coefficient of variation, and skewness
for two period: (a) for data from 1900 – 1945, and 1946 –1990. Fit the data using
normal distribution and check its goodness of fit over the two periods indicated.
Plot the data normal probability paper to check its fitness.
11.2 Fit the data of the peak flood of the Beresa river (given in Example 11.11) using
log-Pearson Type III distribution. Plot is in log-Pearson paper
11.3 The record of annual peak discharge at a stream gaging station is as follows:
year 1961 1962 1963 1964 1965 1966 1967 1968 1969
Q
(m3/s)
45.3 27.5 16.9 41.1 31.2 19.9 22.7 59.0 35.4
Determine using the lognormal distribution:
(a) The probability that an annual flood peak of 42.5 m3/s will not be exceeded.
(b) The return period of the dischrge of 42.5 m3/s
(c) The magnitude of a 20-year flood
Frequancy analysis
______________________________________________________________________________
245
Groundwater 246.
_________________________________________________________________
12. BASICS OF GROUNDWATER HYDROLOGY
Our study of hydrology up to this point has concentrated on various aspects of
surface water processes, including rainfall, runoff, infiltration, evaporation,
channel flow, and storage in reservoirs. An engineering hydrologist also must be
able to address processes of groundwater hydrology and flow. This chapter
presents a concise treatment of groundwater topics following Bedeint & Huber
(1992), including properties of groundwater, groundwater flow and well
hydraulics. This represents a minimum coverage for the practicing hydrologist or
engineering student.
Groundwater hydrology is of great importance because of the use of aquifer
systems for water supply. Properties of the porous media and subsurface geology
govern both the rate and direction of groundwater flow in any aquifer system. The
injection or accidental spill of waste into an aquifer or the pumping of the aquifer
for water supply may alter the natural hydraulic flow patterns.
12.2 Properties of Groundwater
Groundwater can be characterized according to vertical distribution. Figure 12.1
indicates the divisions of subsurface the collection of groundwater data and
evaluation of these data in terms water. The soil water zone, which extends from
the ground surface down through the major root zone, varies with soil type and
vegetation. The amount of water present in the soil water zone depends primarily
on recent exposure to rainfall and infiltration. Hygroscopic water remains
adsorbed to the surface of soil grains, while gravitational water drains through the
soil under the influence of gravity. Capillary water is held by surface tension
forces just above the water table, which is defined as the level to which water rise
in a well drilled into the saturated zone.
The vadose zone extends from the lower edge of the soil water zone to the upper
limit of the capillary zone. Thickness may vary from zero for high water table
condition to more than 100 m in arid regions. Vadose zone water is held in place
12.1 Introduction
Groundwater 247.
_________________________________________________________________
by hygrospcopic and capillary forces, and infiltration water passes downward
toward the water table as gravitational flow, subject to retardation by these forces.
Figure 12.1: Vertical zones of subsurface water
The capillary zone, or fringe, extends from the water table up to the limit of
capillary rise, which varies inversely with pore size of the soil and directely with
the surface tension. Capillary rise can range from 2.5 cm for fine gravel to more
than 200 cm for silt. Just above the water table almost all pores contain capillary
water, and the water content decreases with height depending on the type of soil,
A typical soil moisture curve is shown in Fig. 12.2.
In the zone of saturation, which occurs beneath the water table, the porosity is a
direct measure of the water contained per unit volume, expressed as the ratio of
the volume of voids to the total volume, Only a portion of the water can be
removed from the saturated zone by drainage or by pumping from a well.
Groundwater 248.
_________________________________________________________________
Figure 12.2: Typical soil moisture relationship
A vertical hole dug into the earth penetrating an aquifer is referred to as a well
Wells are used for pumping, recharge, disposal, and water level observation.
Often the portion of the well hole that is open to the aquifer is screened to prevent
aquifer material from entering the well.
An aquifer can be defined as a formation that contains sufficient saturated
permeable material to yield significant quantities of water to wells and springs.
Aquifer are generally areally extensive and may be overlain or underlain by a
confining bed, defined as a relatively impermeable material. An aquiclude is a
relatively impermeable confining unit, such as clay, and an aquitard is a poorly
permeable stratum, such as a sandy clay unit, that may leak water to adjacent
aquifers.
Aquifers can be characterized by the porosity of a rock or soil, expressed as the
ratio of the volume of voids Vv to the total bulk volume (bulk volume includes
volume of solids plus pore space) V:
(12.1)
Where m is the density of the grains (normally assumed as 2.65 g/cm3) and b is
the bulk density, defined as the oven-dried mass of the sample divided by its
original volume. In soils containing a large percentage of clays (> 10 %) the clay
mineralogy or clay type has significant effect on soil water properties. For
m
bv
V
Vn
1
Groundwater 249.
_________________________________________________________________
example expandable clay such as montmorillonite has a significantly lower
hydraulic conductivity and high water retention than non-expandable clays such
as kaolinite. Table 12.1 shows a range of porosities for a number of aquifer
materials.
Table 12.1: Representative values of Porosity (Maidement, 1996)
Sediment or rock type Porosity
Clays 0.40 – 0.60
Silts 0.35 – 0.50
Fine sands 0.20 – 0.45
Coarse sands 0.15 – 0.35
Shales (near-surface, weathered) 0.30 – 0.50
Shales (at depth) 0.01 – 0.10
Limestones 0.05 – 0.35
Bedded salt 0.001 – 0.005
Unfractured igneous rocks 0.0001 – 0.01
Fractured igneous rocks 0.01 – 0.10
Basalts 0.01 – 0.25
Unconsolidated geologic material are normally classified according to their size
and distribution. Soil classification based on particle size is shown in Figure 12.3
based on different current classification schemes. Partcile sizes are measured my
mechanically sieving grains larger than 0.05 mm and measuring rate of settlement
for smaller particles in suspension.
Figure 12.3: Triangular representation of soil textures for describing various
combinations of sand, silt and clay.
Groundwater 250.
_________________________________________________________________
The texture of a soil is defined by the relative proportion of sand, silt and clay
present in the particles size analysis and can be expressed most easily on a
triangle of soil texture (Fig. 12.3):
Most aquifers can be considered as underground storage reservoirs that receive
recharge from rainfall or from an artificial source. Water flows out of aquifer due
to gravity or to pumping from wells. Aquifers may be classified as unconfined or
as confined, depending on the existence of a water table. A leaky confined aquifer
represents a stratum that allows water to flow through the confining zone.
.
Figure 12.4: Schematic cross section illustrating unconfined and confined aquifer
A perched water table is an example where unconfined water bodies sites on top of a clay
lens, separated from the main aquifer; a vertical cross-section illustrating unconfined and
confined aquifers. A confined aquifer is one in which water table exists and rises and falls
with changes in volume of water stored.
Confined aquifers, called artesian aquifers, occur where groundwater is confined
by a relatively impermeable stratum, or confining layer, and water is under
pressure greater than atmospheric. If a well penetrates such an aquifer, the water
level will rise above the bottom of the confining unit. If the water level rises
above the land surface, a flowing well or spring results and is referred as an
artesian well or spring.
A recharge area supplies water to confined aquifer, and such an aquifer can
convey water from the recharge area to locations of natural or artificial discharge.
The peizometric surface (or potentiometric surface) of a confined aquifer is the
Groundwater 251.
_________________________________________________________________
hydrostatic level of water in the aquifer, defined by the water level that occurs in a
lined penetrating well.
Contour maps and profiles can be prepared of the water table for an unconfined
aquifer or the piezometric surface for confined aquifer called equipotential lines.
Once determined from a series of wells in an aquifer, orthogonal lines can be
drawn to indicate the general direction of groundwater flow, in the direction of
decreasing head.
12.3 Groundwater movement
12.3.1 Darcy’s law
The movement of groundwater is well established by hydraulic principles in 1856
by Henri Darcy, who investigated the flow of water through beds of permeable
sand. Darcy discovered one of the most important laws of hydrology – that the
flow rate through porous media is proportional to the head loss and inversely
proportional to the length of the flow path.
Figure 12.5 shows that the experimental setup for determining head-loss through
a sand column, with piezometers located a distance L apart.
Figure 12.5: Head loss through a sand column
Groundwater 252.
_________________________________________________________________
Total energy of this system can be expressed by the Bernoulli equation
(12.2)
Where: p = pressure of water (N/m2)
= specific weight of water (N/m3)
v = velocity of flow in the medium (m/s)
z = elevation (m)
h1 = head loss (m)
Because velocities are very small in porous media, velocity heads may be
neglected, allowing head loss to be expressed:
(12.3)
Darcy related flow rate to head loss and length of column through a
proportionality constant K, the hydraulic conductivity, a measure of the
permeability of the porous media and is given by:
(12.4)
The negative sign indicates that flow of water is in the direction of decreasing
head.
The Darcy velocity that results from Eq.(12.4) is an average velocity through the
entire cross section of the column. The actual flow is limited to the pore space
only, so that the seepage velocity Vs is equal to the Darcy velocity divided by
porosity:
(11.5)
12
2
221
2
11
22hz
g
vpz
g
vp
2
21
11 z
pz
ph
dL
dhK
A
QV
n
V
nA
QVs
Groundwater 253.
_________________________________________________________________
It should be pointed out that Darcy’s law applies to laminar flow in porous media,
and experiments indicate that Darcy’s law is valid for Reynolds number less than
1 and perhaps as high as 10. This represents an upper limit to the validity of
Darcy’s law, which turns out to be applicable in most groundwater systems.
Deviations can occur near pumped wells and in fractured aquifers systems with
large openings.
12.3.2 Hydraulic conductivity K
The hydraulic conductivity is a measure of the ability of a fluid to move through
the interconnected void spaces of sediment or rock. Hydraulic conductivity is a
function of both the medium and the fluid. To separate the effect of the medium
from those of the fluid, the permeability (k) is defined in the following expression:
(12.6)
Where = the dynamic viscosity of the fluid (kg/(m.s) or Pa.s)
= the fluid density (kg/m3)
k = intrinsic permeability (m2)
Table 12.2 gives representative values of hydraulic conductivity and permeability
for different sediment and rock types. The table identifies a commonly observed
range in hydraulic conductivity for each of the sediment and rock types. It does
not identify bounding values, nor does it differentiate between laboratory and in
situ measurements.
Table 12.2: representative values of hydraulic conductivity and permeability (Maidment,
1996)
Sediment or rock type Hydraulic conductivity
(m/day)
Permeability
(m2)
Clays 10-7 – 10
-3 10
-19 – 10
-15
Silts 10-4 – 10
0 10
-16 – 10
-12
Fine to coarse sand 10-2 – 10
3 10
-14 – 10
-9
Gravels 10 2 – 10
5 10
-10 – 10
-7
Shales (matrix) 10-8 – 10
-4 10
-20 – 10
-16
Shales (fractured and weathered) 10-4 – 10
-0 10
-16 – 10
-12
Sandstones (well cemented) 10-5 – 10
-2 10
-17 – 10
-14
Sandstones (friable) 10-3 – 10
-0 10
-15 – 10
-12
gkK
Groundwater 254.
_________________________________________________________________
Salt 10-10
– 10-8
10-22
– 10-20
Anhydrite 10-7 – 10
-6 10
-19 – 10
-18
Unfractured igneous and metamorphic rock 10-9 – 10
-5 10
-21 – 10
-17
Fractured igneous and metamorphic rock 10-5 – 10
-1 10
-17 – 10
-13
Rock type
Cenozoic flood basalts:
Dense, unfractured 10-6 – 10
-3 10
-18 – 10
-15
Vesicular 10-4 – 10
-3 10
-16 – 10
-15
Interbeds 10-3 – 10
3 10
-15 – 10
-9
Quaternary basalts:
Vesicular 10 1 – 10
3 10
-11 – 10
-9
Tuffs:
Densely welded (matrix) < 10-6 < 10
-18
Densely welded (fractured) 10-6 – 10
1 10
-18 – 10
-11
Nonwelded 10-3 – 10
-2 10
-15 – 10
-14
A hydrologic unit is homogeneous if its hydraulic properties are the same at
every location. If the hydraulic conductivity of a sediment or rock is independent
of the direction of flow, the porous medium is isotropic. In this case fluid flow is
the same direction as the hydraulic gradient. If the value of hydraulic conductivity
of a sediment or rock is dependent of the direction of flow, the porous medium is
anisotropic. In an anisotropic medium, flow lines are not aligned with the
direction of the hydraulic gradient but instead are rotated toward the direction of
higher hydraulic conductivity. However, because of spatial and temporal
variability in the geologic processes that create and modify rocks and sediments,
no unit is truly homogeneous. Given the normal variation of hydraulic
conductivity measurements, question arises as to how best to average a set of
measurements to estimate a single value for the effective hydraulic conductivity
(Ke) when using a homogeneous medium approximation. For steady-state flow,
with a spatially uniform hydraulic conductivity, the following averaging rules
apply:
1. Perfectly stratifies medium, n layers, with layer thickness di and
hydaulic conductivity Ki:
When the flow is parallel to layering (arithmetic mean) horizontal
hydraulic conductivity
(12.7)
n
i i
i
ii
n
i
Ae
K
d
Kd
KK
1
1
Groundwater 255.
_________________________________________________________________
Flow perpendicular to layering (harmonic mean) vertical hydraulic
conductivity:
(12.8)
In typical field situation in alluvial deposits, it is found that the hydraulic
conductivity in the vertical direction to be less than the value in the horizontal
direction. The ratios of horizontal hydraulic conductivity to vertical hydraulic
conductivity fall in the range of 2 to 10 for alluvium, with values upto 100 where
clay layers exists.
2. Heterogeneous medium, nonstratified, m measurements
Two-dimensional models (geometric mean)
(12.9)
Three-dimensional models
(12.10)
Where 2
y is the variance of the natural logarithms of the hydraulic conductivity
measurements.
If the hydraulic gradient is not constant, as in radial flow towards a pumping well,
no general averaging rules are available.
Variation of fluid density and viscosity with temperature can be estimated using:
n
i i
i
i
n
i
He
K
d
d
KK
1
1
m
mGe KKKKK /1
21 )....(
)6/1( 2
yGe KK
2
0
6
0
4 )(1056.2)(1017.31 TTxTTxow
Groundwater 256.
_________________________________________________________________
(12.11)
(12.12)
Where: 0 = density of water at To = 20 0C which is 998.2 kg/m
3.
= dynamic viscosity (Pa. s) at To = 20 0C, = 1.005 x 10
-3 Pa.s.
If fluid density is a function of total dissolved solids (TDS), then
(12.13)
Where: C = concentration of TDS (g/L)
C0 = TDS in fresh water corresponding to 0 (g/L)
c = 7.14 x 10 –4 L/g within the salinity range from fresh water to
seawater.
12.3.3 Specific storage and specific yield
Specific storage
When fluid pressures decline within a porous medium, two responses occur:
(1) the fluid volume expands under the lower value of fluid pressure, and
(2) the pore space decreases as an additional fraction of the overburden
pressure must be carried by the solid matrix.
The magnitude of the first response is controlled by the compressibility of water
(), the magnitude of the second by the compressibility of the porous medium ().
The compressibility of a porous medium is defined as the ratio of the change in
volume (-V/V) to the change in effective stress (e), where e = - p. The
compressibility is the inverse of the bulk modulus.
15.133
37.248
5 101039.2 Tx
)(1 00 CCc
Groundwater 257.
_________________________________________________________________
It is convenient to have a single parameter that characterizes the volume of water
released from a porous medium with a decline in fluid pressure. Specific storage
[L-1
] is defined as the volume of water that a unit volume of porous medium
releases from storage per unit change in hydraulic head. Specific storage is
calculated as:
(12.14)
Representative values of compressibility and specific storage are given in Table
12.3.
Table 12.3: Representative values of compressibility and specific storage.
Specific yield
Unconfined aquifers respond to groundwater withdrawals differently than
confined aquifers. Confined aquifers yield water to wells by the mechanisms of
fluid volume expansion and compaction of pore volume. By definition a confined
aquifer remains saturated. In addition to their elastic storage properties,
unconfined aquifers yield water by desaturation of the pore spaces as the water
table declines. Water released from unconfined aquifer greatly exceeds that of a
confined aquifer for equal water level declines. To reflect this difference, the
storativity of an unconfined aquifer is termed the specific yield, Sy, defined as the
volume of water released from a unit area of the aquifer for a unit decline in the
water table (Table 12.4). The specific yield is a significant fraction of the
effective porosity.
Table 12.4 gives representative values of the specific yield.
Sediment or rock type Specific yield, %
Sediment or Rock Type Compressibility
(Pa-1
)
Specific storage
Ss (m-1
)
Unconsolidated clays 10-6 – 10
-8 10
-2 – 10
-4
Unconsolidated sands 10-7 – 10
-9 10
-3 – 10
-5
Unconsolidated gravels 10-8 – 10
-10 10
-4 – 10
-6
Compacted sediment 10-9 – 10
-11 10
-5 – 10
-7
Igneous and metamorphic rocks 10-9 – 10
-11 10
-5 – 10
-7
Water 4.4x 10 -10
) ngSs
Groundwater 258.
_________________________________________________________________
Clays 1 – 5
Silts 10 – 20
Fine sands 10 – 30
Sands and gravels 20 – 30
Sandstones 5 – 20
Shale 0.5 – 5
Limestone 0.5 – 20
One consequence of groundwater withdrawals in geologic settings that contain
compressible sediments is subsidence of the land surface (e.g., the Central Valley
of California and Mexico City). Most of the consolidation occurs not in the
aquifers but in the interstratified silts and clays that slowly releases water from
storage with the decline in hydraulic head. While subsidence can be arrested by
halting the decline in hydraulic head, because of inelastic nature of many clays,
subsidence is largely irreversible.
12.3.4 Transmissivity and storage coefficient.
In the case of horizontal flow through a layer of thickness b, two derived
parameters are commonly used. Transmissivity, T, [L2T
-1] is the product of
hydraulic conductivity and the layer thickness and storage coefficient S (or
storativity) is the product of specific storage and thickness.
(12.15)
(12.16)
It should be emphasized that S and T are specifically defined for two-dimensional,
horizontal analysis, having the aquifer thickness included within their values.
12.4. Determination of hydraulic conductivity in laboratory.
Hydraulic conductivity in saturated zones can be determined by a number of
techniques in the laboratory as well as in the field. Constant head and falling head
permeameters are used in the laboratory for measuring K. In the field, pump tests,
KbT
bSS s
Groundwater 259.
_________________________________________________________________
slug tests, and tracer tests are available for determination of K. These tests are
described in more detail in the well hydraulics section.
A permeameter (Figure 12.8) is used in the laboratory to measure K by
maintaining flow through a small column of material and measuring flow rate and
head loss. For a constant head permeameter, Darcy’s law can be directly applied
to find K, where V is the volume flowing in time t through a sample of area A,
length L, and with constant head h:
(12.17)
Figure 12.8: Permeameter for measuring hydraulic conductivity of geologic samples. (a)
constant head. (b) Falling head.
Ath
VLK
Groundwater 260.
_________________________________________________________________
The falling head permeameter test consists of measuring the rate of fall of the
water level in the tube or column and noting:
(12.18)
(12.19)
After equating and integrating,
(12.20)
Where L, r, and rc are shown in Figure 12.8 and t is the time interval for water to
fall from h1 to h2.
12.5. Flow nets
Darcy’s law was originally derived in one dimension. Many groundwater
problems, however, are really two- or three-dimensional. Flow rate and direction
need to be determined using flow nets which consists of a set of streamlines and
equipotential lines and boundary condition in two dimension.
Equipotential lines are prepared based on observed water levels in wells
penetrating an isotropic aquifer. Flow lines are then drawn orthogonally to
indicate the direction of flow. For the flow net of Fig 12.9, the hydraulic gradient i
is given by
(12.21)
and constant flow q per unit thickness between two adjacent flow lines is
dt
dhrQ 2
dl
dhKrQ c
2
2
1
2
2
lnh
h
tr
LrK
c
ds
dhi
dmds
dhKq
Groundwater 261.
_________________________________________________________________
Figure 12.9: Typical flow net
If we assume ds = dm for a square net, then for n squares between two flow lines
over which total head is divided (h = H/n) and for m divided flow channels:
(12.22)
Where: K = hydraulic conductivity of the aquifer
m = the number of flow channels .
n = the num,ber of squares over the direction of flow
H = total head loss in direction of flow.
n
KmHmqQ
Groundwater 262.
_________________________________________________________________
Since no flow can cross an impermeable boundary, streamlines must parallel it.
Also, streamlines are usually horizontal through high K material and vertical
through low K material because of refraction of lines across a boundary between
different K media following (see Figure 12.10):
(12.23)
Figure 12.10: Refraction of flow lines across a boundary between media of different
hydraulic conductivities.
Flow nets for seepage through a layered system may not be orthogonal and are
shown in Figure 12.1. Most of the flows is in the lower layer which has high K
(Fig 12.11(a).
2
1
2
1
tan
tan
K
K
Groundwater 263.
_________________________________________________________________
Figure 12.11: Flow nets fort seepage from one side of a channel through two different
anisotropic two-layer systems. (a) Ku/KL = 1/50. (b) Ku/KL = 50. The anisotropic ratio for
all layers is Kx / Kz = 10.
Groundwater 264.
_________________________________________________________________
12.6 Unconfined steady state flow: Depuit Assumption
For the case of unconfined groundwater flow, Dupuit developed a theory that
allows for a simple solution based on several important assumptions:
1. The water table or free surface is only slightly inclined.
2. Streamlines may be considered horizontal and equipotential lines
vertical
3. Slopes of the surface and hydraulic gradient are equal
Figure 12.12 shows the graphical example of Dupuit’s assumptions for essentially
one-dimensional flow. The free-surface from x = 0 to x = L can be derived by
considering Darcy’s law and the governing one-dimensional equation.
Figure 12.12: Steady flow in an unconfined aquifer between two water bodies with
vertical boundaries.
Example 12.1: Derive the equation of one-dimensional flow in an unconfined aquifer
with recharge rate of W, using the Dupuit assumptions and Figure E12.1.
Groundwater 265.
_________________________________________________________________
Figure E12.1.
Solution
a) Designate recharge intensity as W, it can be seen that
From Darcy’s law for one-dimensional flow, the flow per unit width is
Substituting the second equation into the first yields
or
Wdx
dq
dx
dhKhq
Wdx
hdK
2
22
2
K
W
dx
hd 22
22
Groundwater 266.
_________________________________________________________________
Integrating gives:
Where a and b are constants. The boundary condition h = ho at x =0 gives
b = ho2
and the boundary condition h = hL at x = L gives
Finally we get the Dupuit Parabola formulae:
Differentiation of the parabolic equation gives
But Darcy’s law gives
and
baxK
Wxh
22
K
WL
L
hha L
)( 2
0
2
)()( 2
0
2
2
0 xLK
Wxx
L
hhhh L
)2()(
22
0
2
xLK
W
L
hh
dx
dhh L
K
q
dx
dhh
)2()(
22
0
2
xLK
W
L
hh
K
q L
Groundwater 267.
_________________________________________________________________
simplifying
Example 12.2: Two rivers located 1000 m apart fully penetrate an aquifer (Fig. E12.1).
The aquifer has a K value of 0.5 m/day. The region receives an average
rainfall of 15 cm/year and evaporation is about 10 cm/yr. Assume that
the water elevation in River 1 is 20 m and the water elevation in River 2
is 18 m. Using the equation derived in part (a) determine the location
and height of the water divide. (b) What is the daily discharge per m
width into each river?
Solution
The water divide is located at hmax, that is at x = d, where q = 0. Using L =1000, K = 0.5
m/day, h0 – 20 m, hL – 18 m, W = (15-10) = 5 cm/yr = 1.469x10-4 m/day:
At x = d, h = hmax
)2
()(2
22
0
LxWhh
L
Kq L
md
xxxd
hhLW
KLd
LdWhh
L
K
LxWhh
L
Kq
L
L
L
2.361
)1820(10369.110002
5.0
2
1000
)(22
)2
()(2
0
)2
()(2
22
4
22
0
22
0
22
0
.9.20
)2.3611000(5.0
2.361410369.12.361
1000
)2018(20
)()(
22
2
max
2
0
22
0max
m
xxh
xLK
Wxx
L
hhhh L
Groundwater 268.
_________________________________________________________________
(b) For discharge into River 1, set x = 0 m:
The negative sign indicates that flow is in the opposite direction from the x-direction into
the River 1. Similarly for discharge into River 2, set x = L = 1000 m, and q = 0.08745
m2/day into River 2.
12.7 Steady-state well hydraulics
Steady radial flow to a well – confined aquifer
The drawdown curve or cone of depression varies with distance from a pumping
well in a confined aquifer (Fig. 12.13). The flow is assumed two-dimensional for
a completely penetrating well in a homogenous, isotropic aquifer of unlimited
extent. For horizontal flow, the above assumptions apply and Q at any r equals,
from Darcy’s law:
(12.24)
for steady radial flow to a well. Integrating after separation of variables, with h =
hw at r =rw at the well yields:
(12.25)
)2
()(2
22
0
LxWhh
L
Kq L
daymq
xx
q
/0495.0
)2/10000(10369.1)1820(10002
5.0
2
422
dr
dhrbKQ 2
)/ln(2
w
w
rr
hhKbQ
Groundwater 269.
_________________________________________________________________
Figure 12.13: Radial flow to a well penetrating an extensive confined aquifer For steady
radial flow to a well.
Near the well the relationship holds and can be rearranged to yield an estimate for
transimissivity T:
(12.26)
by observing heads h1 and h2 at two adjacent observation wells located at r1 and
r2, respectively, from the pumping well. In practice, it is often necessary to use
unsteady-state analyses because of the long time required to reach steady state.
Steady radial flow to a well – unconfined aquifer
1
2
12
ln)(2 r
r
hh
QKbT
Groundwater 270.
_________________________________________________________________
Applying Darcy’s law for radial flow in an unconfined, homogenous, isotropic,
and horizontal aquifer and using Dupuit’s assumptions (Figure 12.4)
(12.27)
Figure 12.14: Radial flow to a well penetrating an extensive unconfined aquifer For
steady radial flow to a well.
Integrating using the boundary condition specified in Fig. 12.14:
(12.28)
dr
dhrKhQ 2
1
2
2
1
2
2
lnr
r
hhKQ
Groundwater 271.
_________________________________________________________________
Groundwater 272.
_________________________________________________________________
Groundwater 273.
_________________________________________________________________
Annex 1. Monthly Potential Evapotranspiration and Altitude relationship over
Ethiopia
Groundwater 274.
_________________________________________________________________
Table 1. Tekeze and Mereb basins monthly potential evapotraspiration (PET) and altitude
relationships. Y = PET (mm/day), X = altitude (m).
Equation R2 t -value
Coeff. Cons.
January Y = - 0.0007 X + 5.3209 0.71 24.71 -6.21
February Y = - 0.0009 X + 6.2027 0.75 24.74 -6.40
March Y = - 0.0009 X + 6.8341 0.70 23.96 -5.77
April Y = - 0.0009 X + 7.319 0.68 22.68 -5.39
May Y = - 0.0007 X + 6.8455 0.64 23.72 -4.39
June Y = - 0.0007 X + 6.618 0.62 21.87 -4.27
July Y = - 0.8656 Ln(X) + 10.681 0.63 19.34 -4.29
August Y = - 0.5021 Ln(X) + 7.3998 0.65 21.92 -3.14
September Y = - 0.0005 X + 5.1666 0.60 23.87 -3.89
October Y = - 0.0007 X + 6.1164 0.76 24.89 -6.06
November Y = - 0.0008 X + 5.6512 0.89 38.61 -10.83
December Y = - 0.0008 X + 5.204 0.90 39.30 -10.86
Table 2. Awash and Rift Valley basins monthly potential evapotraspiration (PET) and
altitude relationships. Y = PET (mm/day), X = altitude (m).
Equation R2 t -value
Coeff. Cons.
January Y = - 0.0006 X + 4.8328 0.71 28.06 -5.48
February Y = - 0.0005 X + 5.1702 0.70 33.79 -5.75
March Y = - 0.0006 X + 5.5959 0.72 33.81 -6.46
April Y = - 0.0008 X + 5.7231 0.76 32.08 -7.58
May Y = 35.054 X - 0.286
0.73 23.09 -6.20
June Y = - 0.0023 X + 8.1067 0.72 13.95 -7.01
July Y = - 0.002 X + 6.8319 0.75 16.46 -7.90
August Y = - 0.0014 X + 5.8871 0.84 23.77 -10.14
September Y = - 0.0012 X + 5.7409 0.80 23.29 -8.85
October Y = 28.698 X - 0.2662 0.76 22.62 -5.65
November Y = 14.348 X - 0.1728 0.75 30.86 -5.42
December Y = - 0.0005 X + 4.8254 0.70 28.05 -5.53
Groundwater 275.
_________________________________________________________________
Table 3. Abay basin except Dedisa and Dabus watersheds monthly potential
evapotraspiration (PET) and altitude relationships. Y = PET (mm/day), X =
altitude (m).
Equation R2 t -value
Coeff. Cons.
January Y = - 0.0007 X + 5.4389 0.64 14.88 -4.72
February Y = - 0.0008 X + 6.1292 0.56 12.73 -3.87
March Y = - 0.0010 X + 6.9370 0.61 12.18 -4.14
April Y = - 0.0012 X + 7.4885 0.69 13.45 -5.16
May Y = - 0.0008 X + 6.3138 0.69 15.60 -4.70
June Y = - 0.0006 X + 5.1495 0.76 16.57 -4.75
July Y = - 0.0006 X + 3.7716 0.79 18.91 -5.09
August Y = - 0.0001 X + 2.8461 0.70 37.73 -2.32
September Y = - 0.0006 X + 4.6550 0.59 13.57 -3.83
October Y = - 0.0008 X + 5.7701 0.72 15.30 -4.87
November Y = - 0.0009 X + 5.9514 0.61 10.60 -3.72
December Y = - 0.0006 X + 5.0342 0.77 15.09 -4.31
Table 4. Dedisa and Dabus watersheds (tributary of Abay basin) monthly potential
evapotraspiration (PET) and altitude relationships. Y = PET (mm/day), X =
altitude (m).
Equation R2 t -value
Coeff. Cons.
January Y = - 0.0005 X + 4.9403 0.45 12.23 -2.22
February Y = - 0.0006 X + 5.6164 0.54 15.39 -3.07
March Y = - 0.0006 X + 5.6967
0.82 30.32 -6.02
April Y = - 0.0008 X + 6.2272 0.48 11.39 -2.55
May Y = - 0.0006 X + 4.7427 0.55 13.91 -2.94
June Y = - 0.0008 X + 4.5708 0.62 11.14 -3.35
July Y = - 0.0006 X + 3.8977 0.76 18.14 -5.05
August Y = - 0.0002 X + 2.9166 0.81 46.07 -4.61
September Y = - 0.0003 X + 3.9142 0.66 22.58 -3.43
October Y = - 0.0005 X + 4.5854 0.59 16.72 -3.16
November Y = - 0.0007 X + 4.9180 0.58 12.33 -3.13
December Y = - 0.0006 X + 4.9761 0.68 17.08 -3.56
Groundwater 276.
_________________________________________________________________
Table 5. Wabi Shebelle and Geneale Dawa basins monthly potential evapotraspiration
(PET) and altitude relationships. Y = PET (mm/day), X = altitude (m).
Equation R2 t - value
Coeff. Cons.
January Y = 20.324 X -0.2219
0.71 -6.17 23.14
February Y = 22.152 X -0.2261
0.75 -6.25 23.18
March Y = 25.49 X -0.2400
0.84 -9.08 31.42
April Y = -0.0007 X + 5.3951 0.77 -8.62 32.31
May Y = -0.0007 X + 5.3468 0.79 -8.82 30.67
June Y = -0.0009 X + 5.5797 0.77 -8.30 24.09
July Y = 48.563 X -0.3599
0.83 -11.49 28.74
August Y = 51.061 X -0.3622
0.82 -9.91 24.93
September Y = 53.177 X -0.3621
0.87 -10.17 25.64
October Y = -0.0007 X + 5.1307 0.80 -9.54 30.86
November Y = -0.0006 X + 5.081 0.74 -7.68 28.46
December Y = -0.0006 X + 5.054 0.74 -5.14 20.54
Table 6. Omo Gibe basin monthly potential evapotraspiration (PET) and altitude
relationships. Y = PET (mm/day), X = altitude (m).
Equation R2 t - value
Coeff. Cons.
January Y = - 0.0010 X + 5.4424 0.88 19.24 -4.78
February Y = - 0.0008 X + 5.3333 0.71 20.94 -5.70
March Y = - 0.0009 X + 5.5393 0.73 21.40 -6.32
April Y = - 0.0007 X + 5.0331 0.69 21.65 -5.53
May Y = - 0.0007 X + 4.5584
0.65 17.39 -4.93
June Y = - 0.0007 X + 4.1211 0.52 12.45 -3.73
July Y = - 0.0006 X + 3.6115 0.62 14.53 -4.79
August Y = - 0.0005 X + 3.2901 0.56 15.18 -4.21
September Y = - 0.0006 X + 3.9205 0.58 14.78 -4.03
October Y = - 0.0005 X + 4.1586 0.56 19.96 -4.67
November Y = - 0.0006 X + 4.4332 0.58 18.92 -4.38
December Y = - 0.0006 X + 4.5068 0.62 18.71 -4.78
Groundwater 277.
_________________________________________________________________
Table 7. Baro Akobo basin monthly potential evapotraspiration (PET) and altitude
relationships. Y = PET (mm/day), X = altitude (m).
Equation R2 t -value
Coeff. Cons.
January Y = - 0.0010 X + 5.4424 0.88 35.79 -9.63
FebruarY Y = - 0.0009 X + 5.9371 0.80 33.68 -7.15
March Y = - 0.0006 X + 5.5812 0.80 43.64 -6.83
April Y = - 0.0006 X + 5.7733 0.67 30.85 -5.09
MaY Y = - 0.0005 X + 4.7500 0.71 37.81 -5.69
June Y = - 0.0004 X + 4.1620 0.65 35.01 -4.95
JulY Y = - 0.0005 X + 3.9859 0.82 40.60 -7.66
August Y = - 0.0002 X + 3.4565 0.48 39.53 -3.05
September Y = - 0.0005 X + 4.3136 0.83 46.16 -8.08
October Y = - 0.0006 X + 4.7104 0.88 55.32 -9.79
November Y = - 0.0008 X + 4.9346 0.94 59.99 -14.35
December Y = - 0.0009 X + 5.0132 0.89 41.74 -10.46
Groundwater 278.
_________________________________________________________________
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Shaw, E. (1994). Hydrology In Practice. Chapman & Hall.
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