Advanced hydrology

Advanced Hydrology
(Web course)
Subhankar Karmakar
Assistant Professor
Centre for Environmental Science and Engineering (CESE)
Indian Institute of Technology Bombay
Powai, Mumbai 400 076
Email: skarmakar@iitb.ac.in
Ph. # +91 22 2576 7857

Hydrologic Cycle
Prof. Subhankar Karmakar
IIT Bombay
Module 1
3 Lectures

The objective of this module is to introduce the
phenomena of weather, different stages of the hydrologic
cycle, hydrologic losses and its measurements.
Module 1

Topics to be covered
 Weather
 Introduction to Hydrology
 Different stages of Hydrology or water cycle
 Hydrologic losses and measurements
 Analytical Methods
Empirical Methods
Module 1

Lecture 1: Weather and hydrologic cycle
Module 1

Weather & Climate
 Weather- “the state of the atmosphere with respect to heat or cold, wetness
or dryness, calm or storm, clearness or cloudiness”.
 Climate – “the average course or condition of the weather at a place usually
over a period of years as exhibited by temperature, wind velocity, and
precipitation”.
(Wikipedia)
Weather refers, generally, to day-to-day temperature and precipitation
activity, whereas climate is the term for the average atmospheric conditions
over longer periods of time.
Module 1Lecture 1

Atmosphere
Troposphere
 Most of the weather occurs.
Stratosphere
19% of the atmosphere’s gases;
 Ozone layer
Mesosphere
Most meteorites burn up here.
Thermosphere
 High energy rays from the sun are
absorbed;
 Hottest layer.
Exosphere
Molecules from atmosphere
escape into space; satellites orbit here.
(http://www.windows.ucar.edu/tour/link=/earth/Atmosphere/layers_activity_print.html) Module 1Lecture 1

Winds and Wind belts
 Exist to circulate heat and
moisture from areas of heating
to areas of cooling
Equator to poles
Low altitudes to high
altitudes
 Three bands of low and high
pressure above and below the
equator (area of low pressure)
Module 1Lecture 1

Cloud Types
Cloud is a visible set of drops of water and fragments of ice suspended in
the atmosphere and located at some altitude above the earth’s surface.
Module 1Lecture 1

Classification of Precipitation events
 Based on the “mechanism” by which air is lifted.
 Frontal lifting:
Warmer air is forced to go above cooler air in equilibrium with a cooler surface.
 Orographic lifting:
Air is forced to go over mountains (and it’s the reason why windward slopes
receive more precipitation).
 Convective Lifting:
Warm air rises from a warm surface and progressively cools down.
 Cyclonic Lifting:
A cyclonic storm is a large, low pressure system that forms when a warm air
mass and a cold air mass collide.
Module 1Lecture 1

Frontal lifting
Module 1Lecture 1

Orographic lifting
Module 1Lecture 1

Convectional lifting
(climateofindia.pbworks.com)
Module 1Lecture 1

Cyclonic lifting
Module 1Lecture 1

Factors affecting Indian climate
Related to Location and Relief Related to Air Pressure and Wind
•Latitude
•Altitude
•Relief
•Distance from Sea
•The Himalayan Mountains
•Distribution of Land & water
•Surface pressure & wind
•Upper air circulation
•Western cyclones
Module 1
Factors affecting Indian climate
Lecture 1

Module 1
Seasons
Cold weather
Hot weather
South west monsoon
Retreating monsoon
Lecture 1

► It extends from December to
February.
► Vertical sun rays shift towards
southern hemisphere.
► North India experiences
intense cold
► Light wind blow makes this
season pleasant in south
India.
► Occasional tropical cyclone
visit eastern coast in this
season.
Tropical Cyclone
Cold Weather Season
Seasons
Module 1Lecture 1

250C
250C
200C
200C
200C
150C
200C
100C`
Temperature-January
Module 1
Seasons
Lecture 1

Pressure-January
1014
HIGH
PRESSURE
Module 1
Seasons
Lecture 1

RAINFALL DUE
TO WESTERN
DISTURBANCES
RAINFALL DUE
TO NORTH EAST
WIND
Winter Rainfall
Module 1
Seasons
Lecture 1

► It extends from March to
May.
► Vertical sun rays shift
towards Northern
hemisphere.
► Temperature rises gradually
from south to north.
► Highest Temperature
experiences in Karnataka in
March, Madhya Pradesh in
April and Rajastan in May. March 300C
April 380C
May 480C
Hot Weather Season
Module 1
Seasons
Lecture 1

Temperature-July
250C
300C
Module 1
Seasons
Lecture 1

Pressure-July
Module 1
Seasons
Lecture 1

LOO
KALBAISAKHI
BARDOLI
CHHEERHA
MANGO
SHOWER
BLOSSOM
SHOWER
Storms in Hot Weather Season
(climateofindia.pbworks.com) Module 1
Seasons
Lecture 1

► It extends from June to
September.
► Intense heating in north
west India creates low
pressure region.
► Low pressure attract the
wind from the surrounding
region.
► After having rains for a few
days sometime monsoon
fails to occur for one or
more weeks is known as
break in the monsoon.
South West Monsoon
LOW PRESSURE
HIGH TEMPERATURE
Module 1
Seasons
Lecture 1

INTER TROPICAL CONVERGENCE ZONE
Arabian
sea Branch
Bay of
Bengal
Branch
Monsoon Wind
Module 1
Seasons
Lecture 1

Onset of SW Monsoon
Module 1
Seasons
Lecture 1

► It extends from October to
November
► Vertical sun rays start shifting
towards Northern
hemisphere.
► Low pressure region shift
from northern parts of India
towards south.
► Owing to the conditions of
high temperature and
humidity, the weather
becomes rather oppressive.
This is commonly known as
the ‘October heat’
LOW PRESSURE
Retreating Monsoon Season
Module 1
Seasons
Lecture 1

Withdrawal of Monsoon
Module 1
Seasons
Lecture 1

> 200cm
100-200cm
50-100 cm
< 50cm
Distribution of Rainfall
(climateofindia.pbworks.com) Module 1
Seasons
Lecture 1

► The variability of rainfall is computed
with the help of the following formula:
C.V.= Standard Deviation/ Mean * 100
► Variability  <25% exist in Western
coasts, Western Ghats, north-eastern
peninsula, eastern plain of the
Ganga, northern-India, Uttaranchal, SW
J & K & HP.
► Variability  >50% found in Western
Rajastan, J & K and interior parts of
Deccan.
► Region with high rainfall has less
variability.
Variability of Rainfall
Module 1
Seasons
Lecture 1

Lecture 2: Weather and hydrologic cycle
(contd.)
Module 1

Hydrology
Hydor + logos (Both are Greek words)
“Hydor” means water and “logos” means study.
Hydrology is a science which deals with the occurrence, circulation and
distribution of water of the earth and earth’s atmosphere.
Hydrological Cycle: It is also known as water cycle. The hydrologic cycle is a
continuous process in which water is evaporated from water surfaces and the
oceans, moves inland as moist air masses, and produces precipitation, if the
correct vertical lifting conditions exist.
Module 1Lecture 1

Hydrologic Cycle
Module 1Lecture 1

Stages of the Hydrologic cycle
 Precipitation
 Infiltration
 Interception
Depression storage
 Run-off
 Evaporation
 Transpiration
 Groundwater
Module 1Lecture 1

Forms of precipitation
 Rain
Water drops that have a diameter of at least 0.5 mm. It can be classified based on
intensity as,
Light rain  up to 2.5 mm/h
Moderate rain 2.5 mm/h to 7.5 mm/h
Heavy rain  > 7.5 mm/h
 Snow
Precipitation in the form of ice crystals which usually combine to form flakes, with
an average density of 0.1 g/cm3.
 Drizzle
Rain-droplets of size less than 0.5 mm and rain intensity of less than 1mm/h is
known as drizzle.
Precipitation
Module 1Lecture 1

Glaze
When rain or drizzle touches ground at 0oC, glaze or freezing rain is
formed.
Sleet
It is frozen raindrops of transparent grains which form when rain falls
through air at subfreezing temperature.
Hail
It is a showery precipitation in the form of irregular pellets or lumps of ice of
size more than 8 mm.
Precipitation
Module 1
Forms of precipitation Contd…
Lecture 1

Rainfall measurement
The instrument used to collect and measure the precipitation is called raingauge.
Types of raingauges:
1) Non-recording : Symon’s gauge
2) Recording
 Tipping-bucket type
 Weighing-bucket type
 Natural-syphon type
Symon’s gauge
Precipitation
Module 1Lecture 1

Recording raingauges
The instrument records the graphical variation of the rainfall, the total
collected quantity in a certain time interval and the intensity of the rainfall
(mm/hour).
It allows continuous measurement of the rainfall.
Precipitation
1. Tipping-bucket type
These buckets are so balanced that when
0.25mm of rain falls into one bucket, it tips
bringing the other bucket in position.
Tipping-bucket type raingauge
Module 1Lecture 1

Precipitation
2. Weighing-bucket type
The catch empties into a bucket mounted
on a weighing scale.
The weight of the bucket and its contents
are recorded on a clock work driven chart.
The instrument gives a plot of cumulative
rainfall against time (mass curve of
rainfall).
Weighing-bucket type raingauge
Module 1Lecture 1

Precipitation
3. Natural Syphon Type (Float Type)
The rainfall collected in the funnel
shaped collector is led into a float
chamber, causing the float to rise.
 As the float rises, a pen attached to
the float through a lever system
records the rainfall on a rotating drum
driven by a clockwork mechanism.
A syphon arrangement empties the
float chamber when the float has
reached a preset maximum level.
Float-type raingauge
Module 1Lecture 1

 Hyetograph
Plot of rainfall intensity against
time, where rainfall intensity is depth of
rainfall per unit time
 Mass curve of rainfall
Plot of accumulated precipitation
against time, plotted in chronological
order.
 Point rainfall
It is also known as station rainfall . It
refers to the rainfall data of a station
Presentation of rainfall data
Rainfall Mass Curve
Precipitation
Module 1Lecture 1

The following methods are used to measure the average precipitation
over an area:
1. Arithmetic Mean Method
2. Thiessen polygon method
3. Isohyetal method
4. Inverse distance weighting
Precipitation
Mean precipitation over an area
1. Arithmetic Mean Method
Simplest method for determining areal average
where, Pi : rainfall at the ith raingauge station
N : total no: of raingauge stations
∑=
=
N
i
iP
N
P
1
1
P1
P2
P3
Module 1Lecture 1

2. Thiessen polygon method
This method assumes that any point in the watershed receives the same
amount of rainfall as that measured at the nearest raingauge station.
Here, rainfall recorded at a gage can be applied to any point at a distance
halfway to the next station in any direction.
Steps:
a) Draw lines joining adjacent gages
b) Draw perpendicular bisectors to the lines created in step a)
c) Extend the lines created in step b) in both directions to form representative
areas for gages
d) Compute representative area for each gage
Module 1
Precipitation
Mean precipitation over an area Contd…
Lecture 1

e) Compute the areal average using the following:
∑=
=
N
i
ii PA
A
P
1
1
mmP 7.20
47
302020151012
=
×+×+×
=
P
1
P
2
P
3
A
1
A
2
A
3
P1 = 10 mm, A1 = 12 Km2
P2 = 20 mm, A2 = 15 Km2
P3 = 30 mm, A3 = 20 km2
3. Isohyetal method
∑=
=
N
i
ii PA
A
P
1
1
mmP .21
50
35102515152055 =
×+×+×+×
=
where, Ai : Area between each pair of adjacent isohyets
Pi : Average precipitation for each pair of
adjacent isohyets
P2
10
20
30
A2=20 , p2 = 15
A4=10 , p3 = 35
P1
P3
A1=5 , p1 = 5
A3=15 , p3 = 25

Steps:
a) Compute distance (di) from ungauged point
to all measurement points.
b) Compute the precipitation at the ungauged
point using the following formula:
N = No: of gauged points
4. Inverse distance weighting (IDW) method
Prediction at a point is more influenced by nearby measurements than that
by distant measurements. The prediction at an ungauged point is inversely
proportional to the distance to the measurement points.
( ) ( )2
21
2
2112 yyxxd −+−=
P1=10
P2= 20
P3=30
d1=25
d2=15
d3=10
p
∑
∑
=
=














=
N
i i
N
i i
i
d
d
P
P
1
2
1
2
1
ˆ mmP 24.25
10
1
15
1
25
1
10
30
15
20
25
10
ˆ
222
222
=
++
++
=
Module 1Lecture 1

 Check for continuity and consistency of rainfall records
 Normal rainfall as standard of comparison
 Normal rainfall: Average value of rainfall at a particular date, month or year
over a specified 30-year period.
Adjustments of precipitation data
Check for Continuity: (Estimation of missing data)
P1, P2, P3,…, Pm  annual precipitation at neighboring M stations 1, 2, 3,…, M
respectively
Px  Missing annual precipitation at station X
N1, N2, N3,…, Nm & Nx normal annual precipitation at all M stations and at X
respectively
Precipitation
Module 1Lecture 1

Check for continuity
1. Arithmetic Average Method:
This method is used when normal annual precipitations at various
stations show variation within 10% w.r.t station X
2. Normal Ratio Method
Used when normal annual precipitations at various stations show
variation >10% w.r.t station X
Precipitation
Module 1
Adjustments of precipitation data Contd…
Lecture 1

Test for consistency of record
Causes of inconsistency in records:
Shifting of raingauge to a new location
 Change in the ecosystem due to calamities
 Occurrence of observational error from a certain date
Relevant when change in trend is >5years
Precipitation
Module 1
Lecture 1

Double Mass Curve Technique
AccumulatedPrecipitationof
StationX,ΣPx
Average accumulated precipitation of
neighbouring stations ΣPav
90
89
88
87
86
85
84
83
82
When each recorded data comes
from the same parent population, they
are consistent.
 Break in the year : 1987
 Correction Ratio : Mc/Ma = c/a
 Pcx = Px*Mc/Ma
Pcx – corrected precipitation at any time period t1 at station X
Px – Original recorded precipitation at time period t1 at station X
Mc – corrected slope of the double mass curve
Ma – original slope of the mass curve
Module 1
Precipitation
Test for consistency of record
Lecture 1

It indicates the areal distribution characteristic of a storm of given duration.
Depth-Area relationship
For a rainfall of given duration, the average depth decreases with the area in
an exponential fashion given by:
where : average depth in cms over an area A km2,
Po : highest amount of rainfall in cm at the storm centre
K, n : constants for a given region
Precipitation
Depth-Area-Duration relationships
)exp(0
n
KAPP −=
P
Module 1Lecture 1

The development of maximum depth-area-duration relationship is known
as DAD analysis.
It is an important aspect of hydro-meteorological study.
Typical DAD curves
(Subramanya, 1994)
Module 1
Precipitation
Depth-Area-Duration relationships Contd…
Lecture 1

 It is necessary to know the rainfall intensities of different durations and different
return periods, in case of many design problems such as runoff
disposal, erosion control, highway construction, culvert design etc.
 The curve that shows the inter-dependency between i (cm/hr), D (hour) and T
(year) is called IDF curve.
 The relation can be expressed in general form as:
( )n
x
aD
Tk
i
+
=
i – Intensity (cm/hr)
D – Duration (hours)
K, x, a, n – are constant for a given
catchment
Intensity-Duration-Frequency (IDF) curves
Precipitation
Module 1Lecture 1

0
2
4
6
8
10
12
14
0 1 2 3 4 5 6
Intensity,cm/hr
Duration, hr
Typical IDF Curve
T = 25 years
T = 50 years
T = 100 years
k = 6.93
x = 0.189
a = 0.5
n = 0.878
Module 1
Precipitation
Intensity-Duration-Frequency (IDF) curves Contd…
Lecture 1

Exercise Problem
• The annual normal rainfall at stations A,B,C and D in a basin are 80.97,
67.59, 76.28 and 92.01cm respectively. In the year 1975, the station D was
inoperative and the stations A,B and C recorded annual precipitations of
91.11, 72.23 and 79.89cm respectively. Estimate the rainfall at station D in
that year.
Precipitation
Module 1Lecture 1

Lecture 3: Hydrologic losses
Module 1

 In engineering hydrology, runoff is the main area of interest. So, evaporation
and transpiration phases are treated as “losses”.
 If precipitation not available for surface runoff is considered as “loss”, then the
following processes are also “losses”:
 Interception
 Depression storage
 Infiltration
 In terms of groundwater, infiltration process is a “gain”.
Hydrologic losses
Module 1Lecture 2

Interception is the part of the rainfall that is intercepted by the earth’s surface
and which subsequently evaporates.
The interception can take place by vegetal cover or depression storage in
puddles and in land formations such as rills and furrows.
Interception can amount up to 15-50% of precipitation, which is a significant part
of the water balance.
Interception
Module 1Lecture 2

 Depression storage is the natural depressions within a catchment area which
store runoff. Generally, after the depression storage is filled, runoff starts.
 A paved surface will not detain as much water as a recently furrowed field.
 The relative importance of depression storage in determining the runoff from a
given storm depends on the amount and intensity of precipitation in the storm.
Depression storage
Module 1Lecture 2

Infiltration
The process by which water on the ground surface enters the soil. The rate of
infiltration is affected by soil characteristics including ease of entry, storage
capacity, and transmission rate through the soil.
The soil texture and structure, vegetation types and cover, water content of the
soli, soil temperature, and rainfall intensity all play a role in controlling
infiltration rate and capacity.
Module 1Lecture 2

Infiltration capacity or amount of infiltration
depends on :
 Soil type
 Surface of entry
 Fluid characteristics.
http://techalive.mtu.edu/meec/module01/images/Infiltration.jpg
Infiltration
Factors affecting infiltration
Module 1Lecture 2

Soil Type : Sand with high porosity will have greater infiltration than clay soil with
low porosity.
Surface of Entry : If soil pores are already filled with water, capacity of the soil to
infiltrate will greatly reduce. Also, if the surface is covered by leaves or impervious
materials like plastic, cement then seepage of water will be blocked.
Fluid Characteristics : Water with high turbidity or suspended solids will face
resistance during infiltration as the pores of the soil may be blocked by the
dissolved solids. Increase in temperature can influence viscosity of water which will
again impact on the movement of water through the surface.
Infiltration
Module 1
Factors affecting infiltration Contd…
Lecture 2

Infiltration
Infiltration capacity :
The maximum rate at which, soil at a given time can absorb water.
f = fc when i ≥ fc
f = when i < fc
where fc = infiltration capacity (cm/hr)
i = intensity of rainfall (cm/hr)
f = rate of infiltration (cm/hr)
Module 1
Infiltration rate
Lecture 2

Infiltration
Horton’s Formula: This equation assumes an infinite water supply at the surface
i.e., it assumes saturation conditions at the soil surface.
For measuring the infiltration capacity the following expression are used:
f(t) = fc + (f0 – fc) e–kt for
where k = decay constant ~ T-1
fc = final equilibrium infiltration capacity
f0 = initial infiltration capacity when t = 0
f(t) = infiltration capacity at any time t from start of the rainfall
td = duration of rainfall
Module 1
Infiltration rate Contd…
Lecture 2

f0
ft=fc+(f0-fc)e -kt
fc
f
infiltration
time t
Infiltration
Graphical representation of Horton formula
Measurement of infiltration
1. Flooding type infiltrometer
2. Rainfall simulator
Module 1
Infiltration rate Contd…
Lecture 2

Infiltration
Infiltration indices
The average value of infiltration is called
infiltration index.
Two types of infiltration indices
 φ - index
 w –index
Module 1
Measurement of infiltration
Lecture 2

Infiltration
The indices are mathematically expressed as:
where P=total storm precipitation (cm)
R=total surface runoff (cm)
Ia=Initial losses (cm)
te= elapsed time period (in hours)
The w-index is more accurate than the φ-index because it subtracts initial losses
φ-index=(P-R)/te
w-index=(P-R-Ia)/te
Module 1
Measurement of infiltration Contd...
Infiltration indices
Lecture 2

Example Problem
Infiltration
A 12-hour storm rainfall with the following depths in cm occurred over a basin:
2.0, 2.5, 7.6, 3.8, 10.6, 5.0, 7.0, 10.0, 6.4, 3.8, 1.4 and 1.4. The surface runoff
resulting from the above storm is equivalent to 25.5 cm of depth over the basin.
Determine the average infiltration index (Φ-index) for the basin.
Total rainfall in 12 hours = 61.5 cm
Total runoff in 12 hours = 25.5 cm
Total infiltration in 12 hours = 36 cm
Average infiltration = 3.0 cm/hr
Average rate of infiltration during the central 8 hours
8 Φ +2.0+2.5+1.4+1.4 = 36
Φ = 3.6cm/hr
Module 1Lecture 2

 In this process, water changes from its liquid state to gaseous state.
 Water is transferred from the surface to the atmosphere
through evaporation
Evaporation
Evaporation is directly proportional to :
 Vapor pressure (ew),
 Atmospheric temperature (T),
 Wind speed (W) and
 Heat storage in the water body (A)
Module 1Lecture 2

Evaporation
Vapour pressure: The rate of evaporation is proportional to the difference
between the saturation vapour pressure at the water temperature, ew and the
actual vapour pressure in the air ea.
EL = C (ew - ea)
EL = rate of evaporation (mm/day); C = a constant ; ew and ea are in mm of
mercury;
The above equation is known as Dalton’s law of evaporation. Evaporation takes
place till ew > ea, condensation happen if ew < ea
Module 1
Factors affecting evaporation
Lecture 2

Temperature: The rate of evaporation increase if the water temperature is
increased. The rate of evaporation also increase with the air temperature.
Heat Storage in water body: Deep bodies can store more heat energy than
shallow water bodies. Which causes more evaporation in winter than summer
for deep lakes.
Evaporation
Module 1
Factors affecting evaporation Contd…
Lecture 2

 Soil evaporation: Evaporation from water stored in the pores of the soil i.e., soil
moisture.
 Canopy evaporation: Evaporation from tree canopy.
 Total evaporation from a catchment or an area is the summation of both soil
and canopy evaporation.
Evaporation
Module 1
Types of Evaporation
Lecture 2

Evaporation
The amount of water evaporated from a water surface is estimated by the
following methods:
1. Using evaporimeter data
2. Empirical equations
3. Analytical methods
1. Evaporimeters : Water containing pans which are exposed to the atmosphere
and loss of water by evaporation measured in them in the regular intervals.
a) Class A Evaporation Pan
b) ISI Standard pan
c) Colorado sunken pan
d) USGS Floating pan
Measurement of evaporation
Module 1Lecture 2

Evaporation
Demerits of Evaporation pan:
1. Pan differs in the heat-storing capacity and heat transfer from the sides
and bottom.
Result: reduces the efficiency (sunken pan and floating pan eliminates this
problem)
2. The height of the rim in an evaporation pan affects the wind action over the
surface.
3. The heat-transfer characteristics of the pan material is different from that of
the reservoir.
Module 1
Measurement of evaporation Contd…
1. Evaporimeters
Lecture 2

Evaporation
Pan Coefficient (Cp)
For accurate measurements from evaporation pan a coefficient is introduce, known
as pan coefficient (Cp). Lake evaporation = Cp x pan evaporation
Type of pan Range of Cp Average value Cp
Class A land pan 0.60-0.80 0.70
ISI pan 0.65-1.10 0.80
Colorado sunken pan 0.75-0.86 0.78
USGS Floating pan 0.70-0.82 0.80
Source: Subramanya, 1994
Module 1
Lecture 2

2. Empirical equation
Mayer’s Formula (1915)
EL = Km (ew- ea) (1+ (u9/16))
where EL = Lake evaporation in mm/day;
ew = saturated vapour pressure at the water surface temperature;
ea = actual vapour pressure of over lying air at a specified height;
u9 = monthly mean wind velocity in km/hr at about 9 m above the
ground;
Km= coefficient, 0.36 for large deep waters and 0.50 for small
shallow waters.
Evaporation
Module 1
Lecture 2

A reservoir with a surface area of 250 ha had the following parameters: water temp.
22.5oC, RH = 40%, wind velocity at 9.0 m above the ground = 20 km/hr. Estimate
the volume of the water evaporated from the lake in a week.
Given ew = 20.44, Km =0.36.
Solution:
ea = 0.40 x 20.44 = 8.176 mm Hg; U9 = 20 km/hr;
Substitute the values in Mayer’s Equation .
Now, EL = 9.93 mm/day
For a week it will be 173775 m3.
Evaporation
Example Problem
Module 1Lecture 2

Water Budget method: This is the simplest analytical method.
P + Vis + Vig = Vos + EL + ds + TL
P= daily precipitation;
Vis = daily surface inflow into the lake;
Vig = daily groundwater flow ;
Vos= daily surface outflow from the lake;
Vog= daily seepage outflow;
EL= daily lake evaporation;
ds= increase the lake storage in a day;
TL= daily transportation loss
3. Analytical method
Module 1
Evaporation
Lecture 2

Evapotranspiration
Transpiration + Evaporation
This phenomenon describes transport of water into the atmosphere from
surfaces, including soil (soil evaporation), and vegetation (transpiration).
Hydrologic Budget equation for Evapotranspiration:
P – Rs – Go - Eact = del S
P= precipitation; Rs= Surface runoff; Go= Subsurface outflow; Eact = Actual
evapotranspiration; del S = change in the moisture storage.
Module 1Lecture 2

Highlights in the Module
 Hydrology is a science which deals with the movement, distribution, and quality
of water on Earth including the hydrologic cycle, water resources and
environmental watershed sustainability.
 Stages of the Hydrologic cycle or Water cycle
 Precipitation
 Infiltration
 Interception
 Run-off
 Evaporation
 Transpiration
 Groundwater
Module 1

 Hydrologic Losses : evaporation, transpiration and interception
 Measurement of Precipitation
 Non-Recording Rain gauges: Symons’s gauge
 Recording Rain gauges: tipping bucket type, weighing bucket type and natural
syphon type
 Presentation of Rainfall Data: Mass curve, Hyetograph, Point Rainfall and DAD
curves
 Factors affecting Infiltration: soil characteristics, surface of entry and fluid
characteristics
 Determination of Infiltration rate can be performed using flooding type
infiltrometers and rainfall simulator.
Module 1
Highlights in the Module Contd…

 Factors affecting evaporation : vapour pressure, wind
speed, temperature, atmospheric pressure, presence of soluble salts and heat
storage capacity of lake/reservoir
 Measurement of evaporation: evaporimeters, empirical equations and
analytical methods
 Weather refers, generally, to day-to-day temperature and precipitation
activity, whereas climate is the term for the average atmospheric conditions
over longer periods of time.
 Formation of Precipitation: frontal, convective, cyclonic and orographic
 The four different seasons are: Cold weather, Hot weather, South-West
monsoon and Retreating monsoon
Module 1

IIT Bombay
Philosophy of Mathematical
Models of Watershed Hydrology
Module 2
2 Lectures

Philosophy of mathematical models of
watershed hydrology
Lecture 1

Objectives of this module is to introduce the terms and
concepts in mathematical modelling which will form as a tool
for effective and efficient watershed management through
watershed modelling
Module 2

 Concept of mathematical modeling
 Watershed - Systems Concept
 Classification of Mathematical Models
 Different Components in Mathematical Modelling
Module 2

 A model is a representation of reality in simple form based on
hypotheses and equations:
 There are two types of models
 Conceptual
 Mathematical
Modeling Philosophy
Experiment
Computation
Theory
Module 2

Conceptual Models
Qualitative, usually based on graphs
Represent important system:
components
processes
linkages
Interactions
Conceptual Models can be used:
As an initial step
For hypothesis testing
For mathematical model development
As a framework
For future monitoring, research, and management actions at a site

 Modeling = The use of mathematics as a tool to explain and make predictions of
natural phenomena (Cliff Taubes, 2001)
 Mathematical modelling may involve words, diagrams, mathematical notation
and physical structure
 This aims to gain an understanding of science through the use of mathematical
models on high performance computers
Science
Mathematics
Computer
Science
Module 2
Mathematical Models

Mathematical modeling of watershed can address a wide range of environmental
and water resources problems.
Planning, designing and managing water resources systems involve impact
prediction which requires modelling.
Developing a model is an art which requires knowledge
of the system being modeled, the user’s
objectives, goals and information needs, and some
analytical and programming skills.
(UNESCO, 2005)
Module 2
Mathematical Models Contd…

Mathematical Modeling Process
Working Model
Mathematical
Model
Computational
Model
Results/
Conclusions
Real World Problem
Simplify Represent
Translate
Simulate
Interpret
Module 2

 Mean – average or expected value
 Variance – average of squared deviations from the mean value
 Reliability – Probability (satisfactory state)
 Resilience – Probability (satisfactory state following unsatisfactory state)
 Robustness – adaptability to other than design input conditions
 Vulnerability – expected magnitude or extent of failure when
unsatisfactory state occurs
Consistency- Reliability or uniformity of successive results or events
Module 2
Overall measures of system performance

Watershed - Systems Concept
Input Output
(Eg.
Rainfall, Snow
etc.)
(Eg. Discharge)
http://www.desalresponsegroup.org/alt_watershedmgmt.html
Module 2

The Modeling Process
Model World
Mathematical Model
(Equations)
Real World
Input parameters
Interpret and Test
(Validate) Formulate
Model World
Problem
Model
Results
Mathematical
Analysis
Solutions,
Numericals
Module 2

Model:
A mathematical description of the watershed system.
Model Components:
Variables, parameters, functions, inputs, outputs of the watershed.
Model Solution Algorithm:
A mathematical / computational procedure for performing operations on the
model for getting outputs from inputs of a watershed.
Types of Models
 Descriptive (Simulation)
 Prescriptive (Optimization)
 Deterministic
 Probabilistic or Stochastic
 Static
 Dynamic
 Discrete
 Continuous
 Deductive, inductive, or floating
Basic Concepts
Module 2

Categories of Mathematical Models
Type
Empirical
Based on data analysis
Mechanistic
Mathematical descriptions based on
theory
Time Factor
Static or steady-state
Time-independent
Dynamic
Describe or predict system behavior over
time
Treatment of Data Uncertainty and Variability
Deterministic
Do not address data variability
Stochastic
Address variability/uncertainty
Module 2

Classification of Watershed Models
Based on nature of the algorithms
Empirical
Conceptual
Physically based
Based on nature of input and uncertainty
 Deterministic
 Stochastic
Based on nature of spatial representation
 Lumped
 Distributed
 Black-box
Module 2

 Based on type of storm event
 Single event
 Continuous event
It can also be classified as:
 Physical models
 Hydrologic models of watersheds;
 Scaled models of ships
 Conceptual
Differential equations,
Optimization
 Simulation models
Module 2
Classification of Watershed Models Contd…

Descriptive:
That depicts or describes how things actually work, and answers the
question, "What is this?“
Prescriptive:
suggest what ought to be done (how things should work) according to an
assumption or standard.
Deterministic:
Here, every set of variable states is uniquely determined by parameters in the
model and by sets of previous states of these variables. Therefore, deterministic
models perform the same way for a given set of initial conditions.
Module 2

Probabilistic (stochastic):
In a stochastic model, randomness is present, and variable states are not described
by unique values, but rather by probability distributions.
Static:
A static model does not account for the element of time, while a dynamic model
does.
Dynamic:
Dynamic models typically are represented with difference equations or differential
equations.
Discrete:
A discrete model does not take into account the function of time and usually uses
time-advance methods, while a Continuous model does.
Module 2

Deductive, inductive, or floating: A deductive model is a logical structure based on
a theory. An inductive model arises from empirical findings and generalization from
them. The floating model rests on neither theory nor observation, but is merely the
invocation of expected structure.
Single event model:
Single event model are designed to simulate individual storm events and have no
capabilities for replenishing soil infiltration capacity and other watershed abstraction.
Continuous:
Continuous models typically are represented with f(t) and the changes are reflected
over continuous time intervals.
Module 2

Black Box Models:
These models describe mathematically the relation between rainfall and surface
runoff without describing the physical process by which they are related. e.g. Unit
Hydrograph approach
Lumped models:
These models occupy an intermediate position between the distributed models and
Black Box Models. e.g. Stanford Watershed Model
Distributed Models:
These models are based on complex physical theory, i.e. based on the solution of
unsteady flow equations.
Module 2

Watershed Modelling Terminology
 Input variables
space-time fields of precipitation, temperature, etc.
 Parameters
 Size
 Shape
 Physiography
 Climate
 Hydrogeology
 Socioeconomics
 State variables
space-time fields of soil moisture, etc.
 Drainage
 Land use
 Vegetation
 Geology and Soils
 Hydrology
Module 2

Equations variables
Independent variables
space x
time t
Dependent variables
discharge Q
water level h
All other variables are function of the independent or dependent
variables
Module 2
Watershed Modelling Terminology Contd…

Goals & Objectives
Both goals and objectives are very important to accomplish a project. Goals without
objectives can never be accomplished while objectives without goals will never take
you to where you want to be.
Goals Objectives
Vague, less structured Very concrete, specific and measurable
High level statements that provide
overall context of what the project is
trying to accomplish
Attainable, realistic and low level
statements that describe what the project
will deliver.
Tangible
Intangible
Long term
Short term
Goals
Module 2

Philosophy of mathematical models of
watershed hydrology (contd.)
Lecture 2

time
Precipitation
time
flow
Hydrologic
Model
The goal considered here is to simulate the shape of a hydrograph given a
known input (Eg: rainfall)
Watershed
I. Goal
Module 2
Watershed Modeling Methodology

II. Conceptualization
Source: Wurbs and James, 2002
The hydrologic cycle is a conceptual model that describes the storage and
movement of water between the biosphere, atmosphere, lithosphere, and the
hydrosphere.
Module 2

Note:
For 90 yrs of record,
(2/3) of 90 = 60 yrs for calibration
Remaining (1/3) of 90 = 30 yrs for
validation
III. Model Formulation
Hypothetical data
Goals and Objectives
Conceptualization
Model Formulation
Conceptual
Representation
Calibration &
Verification
Validation
Good
Sensitivity Analysis
Yes
No
Final Model
Module 2

IV. Conceptual Representation
Un measured
Disturbances
Measured
Disturbances
Process State Variable
Eg: velocity, discharge
etc.
( )txc ,
Measured
errors
System Response
Processed Output
( )txc ,0
•Hypothetical data
is considered for
sensitivity analysis
•Field data is not
necessary

Precipitation
Interception
Storage
Surface
Runoff
Groundwater
Storage
Channel
Processes
Interflow
Direct Runoff
Surface
Storage
Baseflow
Percolation
Infiltration
ET
ET
(McCuen, 1989)
ET: evapo-transpiration
Module 2
Flowchart of simple watershed model
IV. Conceptual Representation Contd…

Target
Modelling
Data
Availability
Complexity of
Representation
Guidelines for the Conceptual Model
Eg: Flood event
•Spatial data,
•Time series vs events,
•Surrogate data,
•Heterogeneity of basin characteristics
Issues:
•Catchment scale,
•Accuracy of the analysis,
•Computational aspects
To develop a conceptual watershed model, the following inter-related components
are to be dealt with:
IV. Conceptual Representation Contd…

Calibration is the activity of verifying that a model of a given problem in a specified
domain correctly describes the phenomena that takes place in that domain.
During model calibration, values of various relevant coefficients are adjusted in
order to minimize the differences between model predictions and actual observed
measurements in the field.
Verification is performed to ensure that the model does what it is intended to do.
V. Calibration & Verification
Module 2

Validation is performed using some other dataset (that has not been used as
dataset for calibration)
It is the task of demonstrating that the model is a reasonable representation of the
actual system so that it reproduces system behaviour with enough fidelity to satisfy
analysis objectives.
For most models there are three separate aspects which should be considered
during model validation:
Assumptions
Input parameter values and distributions
Output values and conclusions
VI. Validation
Module 2

VII. Sensitivity Analysis
 Change inputs or parameters, look at the model results
 Sensitivity analysis checks the relationships
Sensitivity
Analysis
Automatic
Trial & Error
•Change input data and re-solve the problem
•Better and better solutions can be
discovered
Module 2

Sensitivity is the rate of change in one factor with respect to change in another
factor.
A modeling tool that can provide a better understanding of the relation between
the model and the physical processes being modeled.
Let the parameters be and system output be ( )txc ,0β
I Model
β1
β2
β3
β4
C1
C2
C3
j
j
i
i
ij
C
C
S
β
β∆
∆
=
Sensitivity of ith output to change in jth parameter:
i = 1, 2, 3; j = 1,2,3,4
VII. Sensitivity Analysis Contd…
Module 2

( ) ( )
( ) ( ) jijjijii
n
j ji
i
n
i ij
j
CCC
n
C
C
n
Here
βββ
β
β
−=∆−=∆
== ∑∑ ==
;
; 11
ModelOutput
Observed Value
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
In this range, model
is not good
A straight line
indicates an
‘excellent’ model
‘A reasonably good model’
Module 2

ModelOutput
Observed Value
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
The model may
become crude if the
system suddenly
changes and the
model does not
incorporate the
relevant changes
occurred.
A ‘Crude Model’
x
x
x
x
x
x
x
x
x x
x
x
x
x
x
Module 2

Mathematical modelling may involve words, diagrams, mathematical notation and
physical structure
Mathematical modeling of watershed can address a wide range of environmental
and water resources problems
Different Components in Modelling are:
1)Goals and Objectives,
2) Conceptualization,
3) Model formulation,
4) Sensitivity Analysis,
5) Conceptual Representation,
6) Calibration & Verification,
7) Validation
Module 2

There are different measures of system performance of models:
 Mean,
Variance,
Reliability,
Resilience,
Robustness,
Vulnerability and
Consistency
Watershed models can be classified based on:
a) Nature of the algorithms,
b) Nature of input and uncertainty,
c) Nature of spatial representation etc.
Module 2

Hydrologic Analysis
IIT Bombay
Module 3
6 Lectures

Module 3
Objective of this module is to introduce the watershed
concepts, rainfall-runoff, hydrograph analysis and unit
hydrograph theory.

 Watershed concepts
 Characteristics of watershed
 Watershed management
 Rainfall-runoff
 Rational Method
 Hydrograph analysis
 Hydrograph relations
 Recession and Base flow separation
 Net storm rainfall and the hydrograph
 Time- Area method
Module 3

 Unit hydrograph theory
 Derivation of UH : Gauged watershed
 S-curve method
 Discrete convolution equation
 Synthetic unit hydrograph
 Snyder’s method
 SCS method
Module 3
(contd..)

Lecture 1: Watershed and rainfall-runoff
relationship
Module 3

Watershed concepts
 The watershed is the basic unit used in
most hydrologic calculations relating to
water balance or computation of
rainfall-runoff
 The watershed boundary (Divide)
defines a contiguous area, such that
the net rainfall or runoff over that area
will contribute to the outlet
 The rainfall that falls outside the
watershed boundary will not contribute
to runoff at the outlet
Watershed diagram
Watershed boundary
Module 3

Watershed concepts Contd…
 Watersheds are characterized in
general, by one main channel and
by tributaries that drain into main
channel at one or more confluence
points
 A “divide” or “drainage divide” is the
line drawn through the highest
elevated points within a watershed
 Divide forms the limits of a single
watershed and the boundary
between two or more watersheds
River
stream
Divide
Sub-catchment or Sub-basin
Module 3

• A water divide is categorized into:-
1. Surface water divide –highest elevation line between basins
(watersheds) that defines the perimeter and sheds water into adjacent
basins, and,
2. Subsurface water divide –which refers to faults, folds, tilted geologic
strata (rock layers), etc., that cause sub-surface flow to move in one
direction or the other.
Surface water divide
Subsurface
water divide
Module 3
Watershed concepts Contd…

 Size: It helps in computing parameters like rainfall
received, retained, amount of runoff etc.
 Shape: Based on the morphological parameters such as geological
structure eg. peer or elongated
 Slope: Reflects the rate of change of elevation with distance along the
main channel and controls the rainfall distribution and movement
 Drainage: Determines the flow characteristics and the erosion behavior
 Soil type: Determines the infiltration rates that can occur for the area
Characteristics of watershed
Module 3

 Land use and land cover: It can affect the overland flow of the
rainwater with the improve in urbanization and increased pavements.
 Main channel and tributary characteristics: It can effect the stream
flow response in various ways such as slope, cross-sectional
area, Manning’s roughness coefficient, presence of obstructions and
channel condition
 Physiography: Lands altitude and physical disposition
 Socio-economics: Depends on the standard of living of the people and
it is important in managing water
Module 3
Characteristics of watershed Contd…

 A watershed management approach is one that considers the watershed as a
whole, rather than separate parts of the watershed in isolation
 Managing the water and other natural resources is an effective and efficient
way to sustain the local economy and environmental health
 Watershed management helps reduce flood damage, decrease the loss of
green space, reduce soil erosion and improve water quality
 Watershed planning brings together the people within the watershed,
regardless of political boundaries, to address a wide array of resource
management issues
Module 3
Watershed Management

 Use an ecological approach that would recover and maintain the biological
diversity, ecological Function, and defining characteristics of natural
ecosystems
 Recognize that humans are part of ecosystems-they shape and are shaped
by the natural systems: the sustainability of ecological and societal systems
are mutually dependent
 Adopt a management approach that recognizes ecosystems and institutions
are characteristically heterogeneous in time and space
 Integrate sustained economic and community activity into the management
of ecosystems
Module 3
Principles for Watershed Management

 Provide for ecosystem governance at appropriate ecological and institutional
scales
 Use adaptive management as the mechanism for achieving both desired
outcomes and new understandings regarding ecosystem conditions
 Integrate the best science available into the decision-making process, while
continuing scientific research to reduce uncertainties
 Implement ecosystem management principles through coordinated
government and non-government plans and activities
 Develop a shared vision of desired human and environmental conditions
Module 3
Principles for Watershed Management Contd…

Lecture 2: Watershed and rainfall-runoff
relationship (contd.)
Module 3

Rainfall-Runoff
 How does runoff occur?
 When rainfall exceeds the infiltration rate at the surface, excess water
begins to accumulate as surface storage in small depressions. As
depression storage begins to fill, overland flow or sheet flow may begin to
occur and this flow is called as “Surface runoff”
 Runoff mainly depends on: Amount of rainfall, soil type, evaporation
capacity and land use
 Amount of rainfall: The runoff is in direct proportion with the rainfall. i.e.
as the rainfall increases, the chance of increase in runoff will also
increases
Module 3

 Soil type: Infiltration rate depends mainly on the soil type. If the soil is having
more void space (porosity), than the infiltration rate will be more causing less
surface runoff (eg. Laterite soil)
 Evaporation capacity: If the evaporation capacity is more, surface runoff will
be reduced
 Components of Runoff
 Overland Flow or Surface Runoff: The water that travels over the ground
surface to a channel. The amount of surface runoff flow may be small since it
may only occur over a permeable soil surface when the rainfall rate exceeds
the local infiltration capacity.
Rainfall-Runoff Contd….
Module 3

 Interflow or Subsurface Storm Flow: The precipitation that infiltrates the soil
surface and move laterally through the upper soil layers until it enters a stream
channel.
 Groundwater Flow or Base Flow: The portion of precipitation that percolates
downward until it reaches the water table. This water accretion may eventually
discharge into the streams if the water table intersects the stream channels of
the basin. However, its contribution to stream flow cannot fluctuate rapidly
because of its very low flow velocity
 Data collection
 The local flood control agencies are responsible for extensive hydrologic
gaging networks within India, and data gathered on an hourly or daily basis
can be plotted for a given watershed to relate rainfall to direct runoff for a
given year.
Module 3
Rainfall-Runoff Contd….

• Travel time for open channel flow (Tt)
Tt = L/V
where L = length of open channel (ft, m)
V = cross-sectional average velocity of flow (ft/s, m/s)
Manning's equation can be used to calculate cross-sectional average velocity
of flow in open channels
where V = cross-sectional average velocity (ft/s, m/s)
kn = 1.486 for English units and kn = 1.0 for SI units
A = cross sectional area of flow (ft2, m2)
n = Manning coefficient of roughness
R = hydraulic radius (ft, m)
S = slope of pipe (ft/ft, m/m)
Module 3
Rational Method
Runoff Measurement Contd….
V = kn / n R2/3 S1/2

Hydraulic radius (R) can be expressed as
R = A/P
where A = cross sectional area of flow (ft2,m2)
P = wetted perimeter (ft, m)
After getting the value of Tt, the time of concentration can be obtained by
Tc = ∑Tt
Rational Method

Values of Runoff coefficients, C (Chow, 1962)
Module 3
Rational Method

Calculation of Tc
• Tc = ∑Tt
where Tt is the travel time i.e. the time it takes for water to travel from
one location to another in a watershed
• Travel time for sheet flow
where,
n = Manning’s roughness coefficient
L = Flow length (meters)
P2 = 2-yr, 24-hr rainfall (in.) and S is the hydraulic grade line or
land surface
Module 3
Rational Method

• Travel time for open channel flow
• Where V is the velocity of flow (in./hr)
• Hence Tt = L/V
• After getting the value of Tt, the time of concentration can be
obtained by
Tc = ∑Tt
Module 3
Rational Method

• Assumptions of rational method
 Steady flow and uniform rainfall rate will produce maximum runoff when all
parts of a watershed are contributing to outflow
 Runoff is assumed to reach a maximum when the rainfall intensity lasts as
long as tc
 Runoff coefficient is assumed constant during a storm event
• Drawbacks of rational method
 The rational method is often used in small urban areas to design drainage
systems and open channels
 For larger watersheds, this process is not suitable since this method is
usually limited to basins less than a few hundred acres in size
Module 3
Rational Method

Lecture 3: Hydrograph analysis
Module 3

Hydrograph analysis
 A hydrograph is a continuous plot of instantaneous discharge v/s time. It
results from a combination of physiographic and meteorological conditions in
a watershed and represents the integrated effects of climate, hydrologic
losses, surface runoff, interflow, and ground water flow
 Detailed analysis of hydrographs is usually important in flood damage
mitigation, flood forecasting, or establishing design flows for structures that
convey floodwaters
 Factors that influence the hydrograph shape and volume
 Meteorological factors
 Physiographic or watershed factors and
 Human factors
Module 3

• Meteorological factors include
 Rainfall intensity and pattern
 Areal distribution or rainfall over the basin and
 Size and duration of the storm event
• Physiographic or watershed factors include
 Size and shape of the drainage area
 Slope of the land surface and main channel
 Channel morphology and drainage type
 Soil types and distribution
 Storage detention in the watershed
• Human factors include the effects of land use and land cover
Hydrograph analysis Contd…

• During the rainfall, hydrologic
losses such as infiltration,
depression storage and detention
storage must be satisfied prior to
the onset of surface runoff
• As the depth of surface detention
increases, overland flow may occur
in portion if a basin
• Water eventually moves into small
rivulets, small channels and finally
the main stream of a watershed
• Some of the water that infiltrates
the soil may move laterally through
upper soil zones (subsurface
stromflow) until it enters a stream
channel
Uniform rainfall
Infiltration
Depression storage
Detention storage
Time (hr)
Runoff(cfs)
Rainfall(in./hr)
Distribution of uniform rainfall

• If the rainfall continues at a constant
intensity for a very long period,
storage is filled at some point and
then an equilibrium discharge can
be reached
• In equilibrium discharge the inflow and
outflow are equal
• The point P indicates the time at
which the entire discharge area
contributes to the flow
• The condition of equilibrium discharge
is seldom observed in nature, except
for very small basins, because of
natural variations in rainfall intensity
and duration
Rainfall
Equilibrium
discharge
Runoff(cfs)
Rainfall(in./hr)
Time (hr)
P
Equilibrium hydrograph
Module 3

Hydrograph relations
• The typical hydrograph is
characterized by a
1. Rising limb
2. Crest
3. Recession curve
• The inflation point on the falling limb
is often assumed to be the point
where direct runoff ends
Net rainfall = Vol. DRO
Crest
Falling limb
Inflation point
Recession
Direct runoff
(DRO)
Recession
Rising
limb
Pn
Q
Time
Base flow (BF)
Hydrograph relations
Module 3

Recession and Base flow separation
• In this the hydrograph is divided into
two parts
1. Direct runoff (DRO) and
2. Base flow (BF)
• DRO include some interflow whereas
BF is considered to be mostly from
contributing ground water
• Recession curve method is used to
separate DRO from BF and can by an
exponential depletion equation
qt = qo ·e-kt where
qt = discharge at a later time t
qo = specified initial discharge
k = recession constant
C
D
B
A
Q
Time
N=bA0.2
Base flow separation
Module 3

• There are three types of baseflow separation techniques
1. Straight line method
2. Fixed base method
3. Constant slope method
1. Straight line method
• Assume baseflow constant regardless of stream height (discharge)
• Draw a horizontal line segment (A-D) from beginning of runoff to intersection
with recession curve
2. Constant slope method
• connect inflection point on receding limb of storm hydrograph to beginning of
storm hydrograph
• Assumes flow from aquifers began prior to start of current storm, arbitrarily
sets it to inflection point
• Draw a line connecting the point (A-C) connecting a point N time periods
after the peak.
Module 3
Baseflow Separation Methods

3. Fixed Base Method
• Assume baseflow decreases while stream flow increases (i.e. to peak of
storm hydrograph)
• Draw line segment (A –B) from baseflow recession to a point directly below
the hydrograph peak
• Draw line segment (B-C) connecting a point N time periods after the peak
where
N = time in days where DRO is terminated, A= Discharge area in km2,
b= coefficient, taken as 0.827
Module 3
Baseflow Separation Methods Contd…

 The distribution of gross rainfall can be given by the continuity equation as
Gross rainfall = depression storage+ evaporation+ infiltration+
surface runoff
 In case, where depression storage is small and evaporation can be
neglected, we can compute rainfall excess which equals to direct runoff,
DRO, by
Rainfall excess (Pn) = DRO = gross rainfall – (infiltration+
depression storage)
Module 3
Rainfall excess

• The simpler method to determine
rainfall excess include
1. Horton infiltration method
2. Ø index method
• Note:- In this, the initial loss is
included for depression storage
Rainfallandinfiltration
Depression storage
Net storm rainfall
Ø index
Horton infiltration
Time
Infiltration loss curves
Module 3
Rainfall excess Contd…

• Horton infiltration method
Horton method estimates infiltration with an exponential-type equation that
slowly declines in time as rainfall continues and is given by
f= fc + (fo – fc) e-kt ( when rainfall intensity i>f)
where
f = infiltration capacity (in./hr)
fo = initial infiltration capacity (in./hr)
fc = final infiltration capacity (in./hr)
k = empirical constant (hr-1)
• Ø index method
It is the simplest method and is calculated by finding the loss difference
between gross precipitation and observed surface runoff measured as a hydrograph
Module 3
Rainfall excess Contd…

• Rainfall of magnitude 3.8 cm and 2.8 cm occurring on two consecutive 4-h
durations on a catchment area 27km2 produced the following hydrograph of
flow at the outlet of the catchment. Estimate the rainfall excess and φ-index
Time from
start of
rainfall (h) -6 0 6 12 18 24 30 36 42 48 54 60 66
Observed
flow
(m3/s) 6 5 13 26 21 16 12 9 7 5 5 4.5 4.5
Example Problem-1
Module 3
Baseflow separation:
Using Simple straight line method,
N = 0.83 A0.2 = 0.83 (27)0.2
= 1.6 days = 38.5 h
So the baseflow starts at 0th h and ends at the point (12+38.5)h

Hydrograph
6 5
13
26
21
16
12
9
7
5 5 4.5 4.5
0
5
10
15
20
25
30
-10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
Discharge(m3/s)
Time (hr)
Module 3
 50.5 h ( say 48 h approx.)
Constant baseflow of 5m3/s
Example Problem-1 Contd…

Time (h) FH Ordinates(m3/s) DRH Ordinates (m3/s)
-6 6 1
0 5 0
6 13 8
12 26 21
18 21 16
24 16 11
30 12 7
36 9 4
42 7 2
48 5 0
54 5 0
60 4.5 0
66 4.5 0
DRH ordinates are obtained from subtracting the corresponding FH with the base flow i.e. 5 m3/s
Module 3

Hydrograph
6
5
13
26
21
16
12
9
7
5 5 4.5 4.5
0
5
10
15
20
25
30
-10 -5 0 5 10 15 20 25 30 35 40 45 50 55 60 65 70
Discharge(m3/s)
Time (hr)
Area of Direct
runoff hydrograph
Module 3

Area of DRH = (6*60*60)[1/2 (8)+1/2 (8+21)+
1/2 (21+16)+ 1/2 (16+11)+
1/2 (11+7)+ 1/2 (7+4)+ 1/2 (4+2)+ 1/2 (2)]
= 1.4904 * 106m3 (total direct runoff due to storm)
Run-off depth = Runoff volume/catchment area
= 1.4904 * 106/27* 106
= 0.0552m = 5.52 cm = rainfall excess
Total rainfall = 3.8 +2.8 = 6.6cm
Duration = 8h
φ-index = (P-R)/t = (6.6-5.52)/8 = 0.135cm/h
Module 3

A storm over a catchment of area 5.0 km2 had a duration of 14hours. The mass
curve of rainfall of the storm is as follows:
If the φ-index of the catchment is 0.4cm/h, determine the effective rainfall
hyetograph and the volume of direct runoff from the catchment due to the
storm.
Time from
start of
storm (h) 0 2 4 6 8 10 12 14
Accumulat
ed rainfall
(cm) 0 0.6 2.8 5.2 6.6 7.5 9.2 9.6
Module 3
Example Problem-2

Time from
start of
storm(h)
Time
interval ∆t
Accumulated
rainfall in ∆t
(cm)
Depth of
rainfall in
∆t (cm)
φ ∆t
(cm) ER (cm)
Intensity
of ER
(cm/h)
0 _ 0 _ _ _ _
2 2 0.6 0.6 0.8 0 0
4 2 2.8 2.2 0.8 1.4 0.7
6 2 5.2 2.4 0.8 1.6 0.8
8 2 6.7 1.5 0.8 0.7 0.35
10 2 7.5 0.8 0.8 0 0
12 2 9.2 1.7 0.8 0.9 0.45
14 2 9.6 0.4 0.8 0 0
Module 3
• Total effective rainfall = Direct runoff due to storm = area of ER hyetograph
= (0.7+0.8+0.35+0.45)*2 = 4.6 cm
• Volume of direct runoff = (4.6/100) * 5.0*(1000)2
= 230000m3

Run-off Measurement
• This method assumes that the outflow hydrograph results from pure
translation of direct runoff to the outlet, at an uniform velocity, ignoring any
storage effect in the watershed
• The relation ship is defined by dividing a watershed into subareas with
distinct runoff translation times to the outlet
• The subareas are delineated with isochrones of equal translation time
numbered upstream from the outlet
• In a uniform rainfall intensity distribution over the watershed, water first flows
from areas immediately adjacent to the outlet, and the percentage of total
area contributing increases progressively in time
• The surface runoff from area A1 reaches the outlet first followed by
contributions from A2, A3 and A4,
Module 3
Time- Area method

2A
1A
3A
4A
Isochrone of
Equal time to outlet
hr5hr10hr15
jiin ARARARQ 1211 ...+++= −
2R
1R
3R
Time, t
Rainfall
2A
1A
3A
4A
0 5 10 15 20
Time, t
Area
Outlet
Module 3
Run-off Measurement Contd…
Time- Area method

where
Qn = hydrograph ordinate at time n (cfs)
Ri = excess rainfall ordinate at time i (cfs)
Aj = time –area histogram ordinate at time j (ft2)
Limitation of time area method
• This method is limited because of the difficulty of constructing isochronal
lines and the hydrograph must be further adjusted to represent storage
effects in the watershed
Module 3
Time- Area method
Run-off Measurement Contd…

• Find the storm hydrograph for the following data using time area method.
Given rainfall excess ordinate at time is 0.5 in./hr
A B C D
Area (ac) 100 200 300 100
Time to gage G (hr) 1 2 3 4
A
B
C
D
G
Module 3
Time area histogram method uses
Qn = RiA1 + Ri-2A2 +…….+ RiAj
For n = 5, i = 5, and j = 5
Q5 = R5A1 + R4A2+ R3A3 + R2A4
(0.5 in./ hr) (100 ac) + (0.5 in./hr) (200 ac) + (0.5 in./hr)
(300ac) + (0.5 in./hr) (100)
Q5 = 350 ac-in./hr
Note that 1 ac-in./hr ≈ 1 cfs, hence
Q5 = 350 cfs
Example Problem

Example Problem Contd…
Tim
e
(hr)
Hydrograp
h
Ordinate
(R1:Rn)
Basi
n
No.
Time
to
gage
Basin area
A1:An (ac)
R1:An R2:An R2:An R2:An R2:An Storm
hydrograph
0 0
1 0.5 A 1 100 * 50 50
2 0.5 B 2 200 100 50 +150
3 0.5 C 3 300 150 100 50 300
4 0.5 D 4 400 50 150 100 50 350
5 50 150 100 50 350
6 50 150 100 300
7 50 150 200
8 50 50
9 0
Excel spreadsheet calculation
* =(R1*A1) = (0.5*100) and + = (adding the columns from 6 to 10) Module 3

0
50
100
150
200
250
300
350
400
0 1 2 3 4 5 6 7 8 9 10
Contribution of
each sub area
A
A A A
B B B
C C
D
Time (hr)
Q(CFS)
Module 3

Lecture 4: Introduction to unit hydrograph
Module 3

Unit hydrograph (UH)
• The unit hydrograph is the unit pulse response function of a linear hydrologic
system.
• First proposed by Sherman (1932), the unit hydrograph (originally named
unit-graph) of a watershed is defined as a direct runoff hydrograph (DRH)
resulting from 1 in (usually taken as 1 cm in SI units) of excess rainfall
generated uniformly over the drainage area at a constant rate for an effective
duration.
• Sherman originally used the word “unit” to denote a unit of time. But since
that time it has often been interpreted as a unit depth of excess rainfall.
• Sherman classified runoff into surface runoff and groundwater runoff and
defined the unit hydrograph for use only with surface runoff.
Module 3

The unit hydrograph is a simple linear model that can be used to derive the
hydrograph resulting from any amount of excess rainfall. The following basic
assumptions are inherent in this model;
1. Rainfall excess of equal duration are assumed to produce hydrographs
with equivalent time bases regardless of the intensity of the rain
2. Direct runoff ordinates for a storm of given duration are assumed directly
proportional to rainfall excess volumes.
3. The time distribution of direct runoff is assumed independent of
antecedent precipitation
4. Rainfall distribution is assumed to be the same for all storms of equal
duration, both spatially and temporally
Unit hydrograph Contd….
Module 3

Terminologies
1. Duration of effective rainfall : the time
from start to finish of effective rainfall
2. Lag time (L or tp): the time from the
center of mass of rainfall excess to the
peak of the hydrograph
3. Time of rise (TR): the time from the start
of rainfall excess to the peak of the
hydrograph
4. Time base (Tb): the total duration of the
DRO hydrograph
Base flow
Direct runoff
Inflection
point
TR
tp
Effective rainfall/excess rainfall
Q(cfs)
Module 3
Derivation of UH : Gauged watershed

1. Storms should be selected with a simple structure with relatively uniform spatial
and temporal distributions
2. Watershed sizes should generally fall between 1.0 and 100 mi2 in modern
watershed analysis
3. Direct runoff should range 0.5 to 2 in.
4. Duration of rainfall excess D should be approximately 25% to 30% of lag time tp
5. A number of storms of similar duration should be analyzed to obtain an average
UH for that duration
6. Step 5 should be repeated for several rainfall of different durations
Module 3
Unit hydrograph
Rules to be observed in developing UH from gaged watersheds

1. Analyze the hydrograph and separate base flow
2. Measure the total volume of DRO under the hydrograph and convert time to
inches (mm) over the watershed
3. Convert total rainfall to rainfall excess through infiltration methods, such that
rainfall excess = DRO, and evaluate duration D of the rainfall excess that
produced the DRO hydrograph
4. Divide the ordinates of the DRO hydrograph by the volume in inches (mm)
and plot these results as the UH for the basin. Time base Tb is assumed
constant for storms of equal duration and thus it will not change
5. Check the volume of the UH to make sure it is 1.0 in.(1.0mm), and graphically
adjust ordinates as required
Module 3
Unit hydrograph
Essential steps for developing UH from single storm hydrograph

Obtain a Unit Hydrograph for a basin of 315 km2 of area using the rainfall and
stream flow data tabulated below.
Time (hr) Observed hydrograph(m3/s)
0 100
1 100
2 300
3 700
4 1000
5 800
6 600
7 400
8 300
9 200
10 100
11 100
Time
(hr)
Gross PPT
(GRH) (cm/h)
0-1 0.5
1-2 2.5
2-3 2.5
3-4 0.5
Stream flow data Rainfall data
Module 3
Unit hydrograph
Example Problem

• Empirical unit hydrograph derivation separates the base flow from the observed
stream flow hydrograph in order to obtain the direct runoff hydrograph (DRH). For
this example, use the horizontal line method to separate the base flow. From
observation of the hydrograph data, the stream flow at the start of the rising limb
of the hydrograph is 100 m3/s
• Compute the volume of direct runoff. This volume must be equal to the volume of
the effective rainfall hyetograph (ERH)
VDRH = (200+600+900+700+500+300+200+100) m3/s (3600) s = 12'600,000 m3
• Express VDRH in equivalent units of depth:
VDRH in equivalent units of depth = VDRH/Abasin = 12'600,000 m3/(315000000
m2) = 0.04 m = 4 cm
Module 3
Unit hydrograph

Obtain a Unit Hydrograph by normalizing the DRH. Normalizing implies dividing the
ordinates of the DRH by the VDRH in equivalent units of depth
Time (hr) Observed
hydrograph(m3/s)
Direct Runoff
Hydrograph
(DRH) (m3/s)
Unit Hydrograph
(m3/s/cm)
0 100 0 0
1 100 0 0
2 300 200 50
3 700 600 150
4 1000 900 225
5 800 700 175
6 600 500 125
7 400 300 75
8 300 200 50
9 200 100 25
10 100 0 0
11 100 0 0
Module 3

Module 3
Unit hydrograph
0
200
400
600
800
1000
1200
0 2 4 6 8 10 12
Q(m3/s)
Time (hr)
Observed hydrograph
Unit hydrograph
DRH

• Determine the duration D of the ERH associated with the UH obtained in 4.
In order to do this:
1. Determine the volume of losses, VLosses which is equal to the difference
between the volume of gross rainfall, VGRH, and the volume of the
direct runoff hydrograph, VDRH .
VLosses = VGRH - VDRH = (0.5 + 2.5 + 2.5 +0.5) cm/h 1 h - 4 cm = 2 cm
2. Compute the f-index equal to the ratio of the volume of losses to the
rainfall duration, tr. Thus,
ø-index = VLosses/tr = 2 cm / 4 h = 0.5 cm/h
3. Determine the ERH by subtracting the infiltration (e.g., ø-index) from
the GRH:
Module 3
Unit hydrograph

Time (hr) Effective
precipitation (ERH)
(cm/hr)
0-1 0
1-2 2
2-3 2
3-4 0
As observed in the table, the duration of the effective rainfall hyetograph is 2 hours.
Thus, D = 2 hours, and the Unit Hydrograph obtained above is a 2-hour Unit
Hydrograph.
Module 3
Unit hydrograph

Lecture 5: Derivation of S-curve and
discrete convolution equations
Module 3

• It is the hydrograph of direct surface discharge that would result from a
continuous succession of unit storms producing 1cm(in.)in tr –hr
• If the time base of the unit hydrograph is Tb hr, it reaches constant outflow
(Qe) at T hr, since 1 cm of net rain on the catchment is being supplied and
removed every tr hour and only T/tr unit graphs are necessary to produce an
S-curve and develop constant outflow given by,
Qe = (2.78·A) / tr
where
Qe = constant outflow (cumec)
tr = duration of the unit graph (hr)
A = area of the basin (km2 or acres)
Module 3
Unit hydrograph
S – Curve method

Unit hydrograph in
Succession produce
Constant outflow
Qe cumec
Time t (hr)
tr
I
Lagged s-curve
Lagged by tr-hr
S-curve
hydrograph
To obtain tr-hr UG
multiply the S-curve
difference by tr/tr
I
Constant flow Qe (Cumec)
Successive unit storms of Pnet = 1 cm
DischargeQ(Cumec)Intensity(cm/hr)
Lagged
Changing the duration of UG by S-curve technique
Module 3
Unit hydrograph
S – Curve method Contd…

• Convert the following 2-hr UH to a 3-hr UH using the S-curve method
Time (hr) 2-hr UH ordinate (cfs)
0 0
1 75
2 250
3 300
4 275
5 200
6 100
7 75
8 50
9 25
10 0
Module 3
Example Problem
Unit hydrograph
Solution
Make a spreadsheet with the 2-hr
UH ordinates, then copy them in
the next column lagged by D=2
hours. Keep adding columns until
the row sums are fairly constant.
The sums are the ordinates of your
S-curve

Module 3
Unit hydrograph
Time
(hr)
2-hr
UH
2-HR lagged UH’s Sum
0 0 0
1 75 75
2 250 0 250
3 300 75 375
4 275 250 0 525
5 200 300 75 575
6 100 275 250 0 625
7 75 200 300 75 650
8 50 100 275 250 0 675
9 25 75 200 300 75 675
10 0 50 100 275 250 0 675
11 25 75 200 300 75 675

0
100
200
300
400
500
600
700
800
0 2 4 6 8 10 12 14
Q(cfs)
Time (hr)
S-curve
2 hr UH
Lagged by 2 hr
Draw your S-curve, as shown in figure below
Make a spreadsheet with the 2-hr UH ordinates, then copy them in the next column
lagged by D=2 hours. Keep adding columns until the row sums are fairly constant.
The sums are the ordinates of your S-curve.
Module 3
Unit hydrograph

Time (hr) S-curve
ordinate
S-curve
lagged 3hr
Difference 3-HR UH
ordinate
0 0 0 0
1 75 75 50
2 250 250 166.7
3 375 0 375 250
4 525 75 450 300
5 575 250 352 216
6 625 375 250 166.7
7 650 525 125 83.3
8 675 575 100 66.7
9 675 625 50 33.3
10 675 650 25 16.7
11 675 675 0 0
Unit hydrograph

Find the one hour unit hydrograph using the excess rainfall hyetograph and
direct runoff hydrograph given in the table
Time (1hr) Excess Rainfall (in) Direct Runoff (cfs)
1 1.06 428
2 1.93 1923
3 1.81 5297
4 9131
5 10625
6 7834
7 3921
8 1846
9 1402
10 830
11 313
Module 3
Example Problem
Unit hydrograph

Discrete Convolution Equation
∑ − +
=
=
m*
n m n m 1
m 1
Q P U m* = min(n,M)
Where Qn = Direct runoff
Pm = Excess rainfall
Un-m+1 = Unit hydrograph ordinates
Suppose that there are M pulses of excess rainfall.
If N pulses of direct runoff are considered, then N equations can be written Qn in
terms of N-M+1unknown values of unit hydrograph ordinates, where n=
1, 2, …,N.

P1 P2 P3
Input Pn
U1
U2 U3 U4 U5
Un-m+1
n-m+1
Unit pulse response applied to
P1
Unit pulse response applied to
P2
n-m+1
Un-m+1
Output Qn
Output ∑ − +
=
=
m*
n m n m 1
m 1
Q P U
Combination of 3 rainfall
UH

The set of equations for discrete time convolution
∑ − +
=
=
m*
n m n m 1
m 1
Q P U
n = 1, 2,…,N
=1 1 1Q PU
= +2 2 1 1 2Q P U PU
= + +3 3 1 2 2 1 3Q P U P U PU
−= + + +M M 1 M 1 2 1 MQ P U P U ..... PU
+ +=+ + + +M 1 M 2 2 M 1 M 1Q 0 P U ..... P U PU
− − − − += + + + + + + +N 1 M N M M 1 N M 1Q 0 0 ..... 0 0 ..... P U P U
− − += + + + + + + +N M 1 N M 1Q 0 0 ..... 0 0 ..... 0 P U

Solution
• The ERH and DRH in table have M=3 and N=11 pulses respectively.
• Hence, the number of pulses in the unit hydrograph is N-M+1=11-3+1=9.
• Substituting the ordinates of the ERH and DRH into the equations in table
yields a set of 11 simultaneous equations
Module 3
Unit hydrograph
− −
= = =2 2 1
1
1
Q P U 1,928 1.93x404
U 1,079 cfs/in
P 1.06
Similarly calculate for remaining ordinates and the final UH is tabulated below
n 1 2 3 4 5 6 7 8 9
Un (cfs/in) 404 1,079 2,343 2,506 1,460 453 381 274 173

Lecture 6: Synthetic unit hydrograph
Module 3

Synthetic Unit Hydrograph
• In India, only a small number of streams are gauged (i.e., stream flows due
to single and multiple storms, are measured)
• There are many drainage basins (catchments) for which no stream flow
records are available and unit hydrographs may be required for such basins
• In such cases, hydrographs may be synthesized directly from other
catchments, which are hydrologically and meteorologically homogeneous,
or indirectly from other catchments through the application of empirical
relationship
• Methods for synthesizing hydrographs for ungauged areas have been
developed from time to time by Bernard, Clark, McCarthy and Snyder. The
best known approach is due to Snyder (1938)
Module 3

• Snyder (1938) was the to develop a synthetic UH based on a study of
watersheds in the Appalachian Highlands. In basins ranging from 10 –
10,000 mi.2
Snyder relations are
tp = Ct(LLC)0.3
where
tp= basin lag (hr)
L= length of the main stream from the outlet to the divide (mi)
Lc = length along the main stream to a point nearest the watershed
centroid (mi)
Ct= Coefficient usually ranging from 1.8 to 2.2
Module 3
Snyder’s method
Synthetic unit hydrograph

Qp = 640 CpA/tp
where
Qp = peak discharge of the UH (cfs)
A = Drainage area (mi2)
Cp = storage coefficient ranging from 0.4 to
0.8, where larger values of cp are associated
with smaller values of Ct
Tb = 3+tp/8
where Tb is the time base of hydrograph
Note: For small watershed the above eq.
should be replaced by multiplying tp by the
value varies from 3-5
• The above 3 equations define points for a
UH produced by an excess rainfall of
duration D= tp/5.5 Snyder’s hydrograph parameter
Snyder’s method Contd…
Module 3

Use Snyder’s method to develop a UH for the area of 100mi2 described below.
Sketch the appropriate shape. What duration rainfall does this correspond to?
Ct = 1.8, L= 18mi,
Cp = 0.6, Lc= 10mi
Calculate tp
tp = Ct(LLC)0.3
= 1.8(18·10) 0.3 hr,
= 8.6 hr
Module 3
Example Problem
Calculate Qp
Qp= 640(cp)(A)/tp
= 640(0.6)(100)/8.6
= 4465 cfs
Since this is a small watershed,
Tb ≈4tp = 4(8.6)
= 34.4 hr
Duration of rainfall
D= tp/5.5 hr
= 8.6/5.5 hr
= 1.6 hr

0
1000
2000
3000
4000
5000
0 5 10 15 20 25 30 35 40
Q(cfs)
Time (hr)
Qp
W75
W50 Area drawn to represent 1
in. of runoff over the
watershed
W75 = 440(QP/A)-1.08
W50 = 770(QP/A)-1.08
(widths are distributed 1/3 before Qp
and 2/3 after)
Module 3

• Unit = 1 inch of runoff (not rainfall) in 1
hour
• Can be scaled to other depths and times
• Based on unit hydrographs from many
watersheds
• The earliest method assumed a
hydrograph as a simple triangle, with
rainfall duration D, time of rise TR (hr),
time of fall B. and peak flow Qp (cfs).
tp
Qp
TR B
SCS triangular UH
Module 3
SCS (Soil Conservation Service) Unit Hydrograph

• The volume of direct runoff is
or
where B is given by
Therefore runoff eq. becomes, for 1 in. of rainfall excess,
=
BT
vol
Q
R
p
+
=
2
RTB 67.1=
R
p
T
vol
Q
75.0
=
R
p
T
A
Q
484
=
where
A= area of basin (sq mi)
TR = time of rise (hr)
Module 3
R
p
T
A
Q
)008.1()640(75.0
=
22
BQTQ
Vol
pRp
+=
SCS Unit Hydrograph Contd…

• Time of rise TR is given by
where
D= rainfall duration (hr)
tp= lag time from centroid of rainfall to QP
Lag time is given by
where
L= length to divide (ft)
Y= average watershed slope (in present)
CN= curve number for various soil/land use
Module 3
pR t
D
T +=
2
0.5
0.7
0.8
L
19000y
9
CN
1000






−
=pt

Runoff curve number for different land use (source: Woo-Sung et al.,1998)
Module 3

Use the SCS method to develop a UH for the area of 10 mi2 described below.
Use rainfall duration of D = 2 hr
Ct = 1.8, L= 5mi,
Cp = 0.6, Lc= 2mi
The watershed consist CN = 78 and the average slope in the watershed is
100 ft/mi. Sketch the resulting SCS triangular hydrograph .
Module 3
Example Problem
Solution
Find tp by the eq.
Convert L= 5mi, or (5*5280 ft/mi) = 26400 ft.
Slope is 100 ft/mi, so y = (100ft/mi) (1mi/5280 ft)(100%) = 1.9%
Substituting these values in eq. of tp, we get tp = 3.36 hr
0.5
0.7
0.8
L
19000y
9
CN
1000






−
=pt

Find TR using eq.
Given rainfall duration is 2 hr, TR = 4.36 hr, the rise of the hydrograph
Then find Qp using the eq, given A= 10 mi2
. Hence Qp = 1.110 cfs
Module 3
R
p
T
A
Q
484
=
pR t
D
T +=
2
To complete the graph, it is also necessary to know the time of fall B. The
volume is known to be 1 in. of direct runoff over the watershed.
So, Vol. = (10mi2) (5280ft/mi)2 (ac/43560ft2) (1 in.) = 6400 ac-in
Hence from eq.
B = 7.17 hr
22
BQTQ
Vol
pRp
+=

0
200
400
600
800
1000
1200
0 2 4 6 8 10 12 14
Q(cfs)
Time (hr)
Qp= 1110 (cfs)
TR=4.36 (hr) B=7.17 (hr)
Module 3

Exercise problems
1. The stream flows due to three successive storms of 2.9, 4.9 and 3.9 cm of 6
hours duration each on a basin are given below. The area of the basin is 118.8
km2 . Assuming a constant base flow of 20 cumec, derive a 6-hour unit
hydrograph for the basin. An average storm loss of 0.15 cm/hr can be assumed
(Hint :- Use UH convolution method)
Time (hr) 0 3 6 9 12 15 18 21 24 27 30 33
Flow
(cumec)
20 50 92 140 199 202 204 144 84 45 29 20
Module 3

2. The ordinates of a 4-hour unit hydrograph for a particular basin are given
below. Derive the ordinates of (i) the S-curve hydrograph, and (ii) the 2-hour
unit hydrograph, and plot them, area of the basin is 630 km2
Time (hr) Discharge (cumec)
0 0
2 25
4 100
6 160
8 190
10 170
12 110
Time (hr) Discharge (cumec)
14 70
16 30
18 20
20 6
22 1.5
24 0
Module 3
Exercise problems Contd…

3. The following are the ordinates of the 9-hour unit hydrograph for the entire
catchment of the river Damodar up to Tenughat dam site: and the catchment
characteristics are , A = 4480 km2, L = 318 km, Lca = 198 km. Derive a 3-hour
unit hydrograph for the catchment area of river Damodar up to the head of
Tenughat reservoir, given the catchment characteristics as, A = 3780km2, L =
284 km, Lca = 184km. Use Snyder’s approach with necessary modifications for
the shape of the hydrograph.
Time (hr) 0 9 18 27 36 45 54 63 72 81 90
Flow
(cumec)
0 69 1000 210 118 74 46 26 13 4 0
Module 3
Exercise problems Contd…

 This module presents the concept of Rainfall-Runoff analysis, or the
conversion of precipitation to runoff or streamflow, which is a central
problem of engineering hydrology.
 Gross rainfall must be adjusted for losses to infiltration, evaporation and
depression storage to obtain rainfall excess, which equals Direct Runoff
(DRO).
 The concept of the Unit hydrograph allows for the conversion of rainfall
excess into a basin hydrograph, through lagging procedure called
hydrograph convolution.
 The concept of synthetic hydrograph allows the construction of hydrograph,
where no streamflow data are available for the particular catchment.
Module 3

Hydrologic Analysis
(Contd.)
IIT Bombay
Module 4
3 Lectures

Module 4
Objective of this module is to learn linear-kinematic
wave models and overland flow models

 Kinematic wave modeling
 Continuity equation
 Momentum equation
 Saint Venant equation
 Kinematic overland flow modeling
 Kinematic channel modeling
Module 4

Lecture 1: Kinematic wave method
Module 4

Kinematic wave method
• This method assumes that the weight or gravity force of flowing
water is simply balanced by the resistive forces of bed friction
• This method can be used to derive overland flow hydrographs,
which can be added to produce collector or channel hydrographs
and eventually, as stream or channel hydrograph
• This method is the combination of continuity equation and a
simplified form of St. Venant equations
(Note:- The complete description of St. Venant equations is provided
in Module-6)
Module 4

Q
S0
Ѳ
Z
V2/2g
h
dx
Ff
x
dx
x
Q
Q
∂
∂
+
Datum
Continuity Equation
y
FH
Fg
Energy line
Kinematic modeling methods
Module 4

The general equation of continuity,
Inflow-Outflow = rate of change of storage
Inflow =
Outflow =
Storage change = tx
t
A
∆∆
∂
∂
where,
q= rate of lateral inflow per unit length of
channel
A = cross- sectional area
Continuity Equation Contd…
Module 4
txqt
x
x
Q
Q ∆∆+∆•




 ∆
•
∂
∂
−
2
t
x
x
Q
Q ∆•




 ∆
•
∂
∂
−
2

The equation of continuity becomes, after dividing by ∆x and ∆t,
• For unit width b of channel with v= average velocity, the continuity equation
can be written as
q
x
Q
t
A
=
∂
∂
+
∂
∂
Module 4
b
q
t
y
x
y
v
x
v
y =
∂
∂
+
∂
∂
+
∂
∂
Continuity Equation Contd…

Momentum equation
It is based on Newton’s second law and that is, Net force = rate of change of
momentum
The following are the three main external forces are acting on area A
Hydrostatic : FH =
Gravitational : Fg=
Frictional : Ff=
= specific weight of water (ρg)
y= distance from the water surface to
the centroid of the pressure prism
Sf= friction slope, obtained by solving for
the slope in a uniform flow
equation, (manning’s equation)
So= Bed slope
γ
Module 4
( ) x
x
yA
∆
∂
∂
−γ
AS xγ− ∆0
f
AS xγ− ∆

• The rate of change of momentum is expressed from Newton’s second law
as
where the total derivative of v W.R.T t can be expressed
( )mv
dt
d
F =
x
v
v
t
v
dt
dv
∂
∂
+
∂
∂
=
………..4.1
………..4.2
Module 4
Momentum Equation Contd…

• Equating Eq. 4.1 to the sum of the three external forces results in
= g(So-Sf)
• For negligible lateral inflow and a wide channel, the Eq. 4.3 can be
rearranged to yield
Sf = So
………..4.3
………..4.4
Saint Venant equation
Module 4
( )
A
vq
x
yA
A
g
x
v
v
t
v
+
∂
∂
+
∂
∂
+
∂
∂
tg
v
xg
vv
x
y
∂
∂
−
∂
∂
−
∂
∂
−
1
Momentum Equation Contd…

• In developing the general unsteady flow equation it is assumed that the flow is
one-dimensional (variation of flow depth and velocity are considered to vary
only in the longitudinal X- direction of the channel
• The velocity is constant and the water surface is horizontal across any section
perpendicular to the longitudinal flow axis
• All flows are gradually varied with hydrostatic pressure such that all the vertical
accelerations within the water column cab be neglected
• The longitudinal axis of the flow channel can be approximated by a straight
line, therefore, no lateral secondary circulations occur
Assumptions of Saint Venant equations
Module 4

• The slope of the channel bottom is small (less than 1:10)
• The channel boundaries may be treated as fixed non-eroding and non-
aggarading
• Resistance to flow may be described by empirical resistance equations such
as the manning or Chezy equations
• The flow is incompressible and homogeneous in density
Module 4
Assumptions of Saint Venant equations Contd…

Forms of momentum Equation
Module 4
Type of flow Momentum equation
Kinematic wave ( study
uniform)
Sf = So
Diffusion (non inertia) model Sf = So
Steady no-uniform Sf = So
Unsteady non-uniform Sf = So
x
y
∂
∂
−
Dynamic
wave
x
v
g
v
x
y
∂
∂






−
∂
∂
−
t
v
g
1
x
v
g
v
x
y
∂
∂






−
∂
∂






−
∂
∂
−

Possible types of open channel flow
Module 4

Kinematic wave Dynamic wave
It is defined as the study of motion
exclusive of the influences of mass
and force
In this the influences of mass and force
are included
When the inertial and pressure forces
are not important to the movement of
wave then the kinematic waves
governs the flow
When inertial and pressure forces are
important then dynamic waves govern
the moment of long waves in shallow
water (large flood wave in a wide river)
Force of this nature will remain
approximately uniform all along the
channel (Steady and uniform flow)
Flows of this nature will be unsteady
and non-uniform along the length of
the channel
Froude No. < 2 Froude No. > 2
Difference between kinematic and dynamic wave
Kinematic modelling methods

Fr =
Froude number
gd
v
Where
V= velocity of flow
g= acceleration due to gravity
d= hydraulic depth of water
Wave celerity (C) gdc =
1. Flows with Froude numbers greater than one are classified as supercritical
flows
2. Froude number greater than two tend to be unstable, that are classified as
dynamic wave
3. Froude number less then 2 are classified as kinematic wave
Module 4

Visualization of dynamic and kinematic waves
Module 4

Lecture 2: Kinematic overland flow
routing
Module 4

• For the conditions of kinematic flow, and with no appreciable backwater effect,
the discharge can be described as a function of area only, for all x and t;
Q= α · Am
where,
Q= discharge in cfs
A= cross-sectional area
α , m = kinematic wave routing parameters
Kinematic overland flow routing
………..4.5
Module 4

• Henderson (1966) normalized momentum Eq. 4.4 in the form of
Governing equations
………..4.6
Less than one, than the equation will represent
Kinematic flow
where Qo=flow under uniform
condition
Hence, for the kinematic flow condition,
Q≈Qo ………..4.7
Kinematic routing methods
Module 4
2
1
11
1 













+
∂
∂
+
∂
∂
+
∂
∂
−=
gy
qv
tg
v
xg
vv
x
y
S
QQ
o
o

• Woolhiser and Liggett (1967) analyzed characteristics of the rising overland
flow hydrograph and found that the dynamic terms can generally be
neglected if,
or
where,
L= length of the plane
Fr= Froude number
y= depth at the end of the plane
S0= slope
k= dimensionless kinematic flow number
………..4.8
Module 4
102
≥=
yFr
LS
k o
102
≥=
v
LgS
k o
Governing equations Contd…

Q* is the dimensionless flow v/s t* (dimensionless time) for varies values of k in
Eq. 8. It can be seen that for k≤10, large errors in calculation of Q* result by
deleting dynamic terms from the momentum Eq. for overland flow
Effect of kinematic
wave number k on the
rising hydrograph
Module 4

• The momentum Eq. for an overland flow segment on a wide plane with
shallow flows can be derived from Eq. 4.5 and manning's Eq. for overland
flow
• Rewriting the Eq. 4.9 in terms of flow per unit width for an overland flow qo,
we have
………..4.9
= conveyance factor
mo= 5/3 from manning’s Eq.
So= Average overland flow slope
yo= mean depth of overland flow
………..4.10
Module 4
3/5
yS
n
k
q o
m
=
om
ooo yq α= o
m
o S
n
k
=α

Estimates of Manning’s roughness coefficients for overland flow
Module 4

• The continuity Eq. is
Finally, by substituting Eq. 4.11 in Eq. 4.9, we have
 Eq. 4.10 and Eq.4.12 form the complete kinematic wave equation for
overland flow
where,
i= rate of gross rainfall (ft/s)
f= infiltration rate
qo= flow per unit width ( cfs/ft)
yo= mean depth of overland
flow
………..4.11
………..4.12
Module 4
fi
x
q
t
y oo
−=
∂
∂
+
∂
∂
fi
x
y
ym
t
y om
ooo
o o
−=
∂
∂
+
∂
∂ −1
α

Lecture 3: Kinematic channel modeling
Module 4

 Representative of collectors or stream channels
 Triangular
 Rectangular
 Trapezoidal
 Circular
 These are completely characterized by slope, length, cross-sectional
dimensions, shape and Manning’s n value.
Kinematic channel modeling
Module 4

Basic channel shapes and their variations
Module 4

• The basic forms of the equations are similar to the overland flow Eq. (Eqs.4 .10
and 4.12). For stream channels or collectors,
Equations of kinematic channel modeling
………..4.13
……….4.14
where,
Ac= cross sectional flow area (ft2)
Qc= discharge
qo= overland inflow per unit length (cfs/ft)
αc, mc= kinematic wave parameter for the particular channel
Module 4
o
cc
q
x
Q
t
A
=
∂
∂
+
∂
∂
cm
ccc AQ α=

shape αc mc
Triangular 4/3
Square 4/3
Rectangular 5/3
Trapezoidal Variable, function of A and W
Circular 5/4
Kinematic channel parameters
Module 4
3/1
2
1
94.0






+ z
z
n
s
n
s72.0
( )3/249.1 −
W
n
s
( )6/1804.0
cD
n
s

• Determine αc and mc for the case of a triangular prismatic channel
1
1
ZZ
yc
Example Problem
Module 4

Solution
and yc = channel depth
Wetted perimeter =
hydraulic radius =
Substituting these into manning’s Eq. given by
2
cc ZyAArea ==
c
c
P
A
R =
Module 4
2
12 zyP cc +=
3/2
3/5
49.1
c
c
c
P
A
s
n
Q =

From Eq.14, . Therefore,
and mc= 4/3
Module 4
( )
( ) 3/123/2
3/103/5
159.1
49.1
Zy
yZ
s
n
Q
c
c
c
+
=
( ) 3/42
3/1
2
1
94.0
cc Zy
Z
Z
s
n
Q 





+
=
( ) 3/4
3/1
2
1
94.0
cc A
Z
Z
s
n
Q 





+
=
cm
ccc AQ α=
3/1
2
1
94.0






+
=
Z
Z
s
n
cα

Highlights in the module
 This module presents the concept of kinetic wave model which assumes the
that the weight or gravity force of flowing water is simply balanced by the
resistive forces of bed friction
 The brief introduction to St. Venant equations is provided in this module,
whereas, the complete part of this is covered in module-6.
Module 4

Flood Routing
IIT Bombay
Module 5
4 Lectures

The objective of this module is to introduce the concepts and
methods of lumped and distributed flood routing along with
an insight into Muskingum method.
Module 5

 Lumped flow routing
 Level pool method
 Kinematic wave/Channel routing
 Muskingum method
 Distributed Flow routing
 Diffusion wave routing
 Muskingum-Cunge method
 Dynamic wave routing
Module 5

Lecture 1: Introduction to flood routing
Module 5

Flood Routing
“Flood routing is a technique of determining the flood
hydrograph at a section of a river by utilizing the data of
flood flow at one or more upstream sections.”
( Subramanya, 1984)
Module 5

Applications of Flood Routing
Flood:
 Flood Forecasting
 Flood Protection
 Flood Warning
Design:
Water conveyance (Spillway) systems
Protective measures
Hydro-system operation
Water Dynamics:
Ungauged rivers
Peak flow estimation
River-aquifer interaction
For accounting changes in flow hydrograph as a flood wave passes downstream
Module 5

Types of flood routing
 Lumped/hydrologic
 Flow f(time)
 Continuity equation and Flow/Storage relationship
 Distributed/hydraulic
 Flow  f(space, time)
 Continuity and Momentum equations
Module 5

Flow Routing Analysis
It is a procedure to determine the flow hydrograph at a point on a watershed from a
known hydrograph upstream.
Upstream
hydrograph
Inflow)( =tI
Inflow
Q Transfer
Function
Outflow)( =tQ
Downstream hydrograph
OutflowQ
Module 5

Flow Routing Analysis Contd…
As flood wave travels downstream, it undergoes
Peak attenuation
Translation
Q
tTp
Qp
Q
tTp
Qp
Q
tTp
Qp
Module 5

Flood Routing Methods
Lumped / Hydrologic flow routing:
Flow is calculated as a function of time alone at a particular location.
Hydrologic routing methods employ essentially the equation of continuity and
flow/storage relationship
Distributed / Hydraulic routing:
Flow is calculated as a function of space and time throughout the system
Hydraulic methods use continuity and momentum equation along with the
equation of motion of unsteady flow (St. Venant equations).
Module 5

Hydrologic routing
1. Level pool method (Modified Puls)
 Storage is nonlinear function of Q
 Reservoir routing
2. Muskingum method
 Storage is linear function of I and Q
 Channel routing
3. Series of reservoir models
 Storage is linear function of Q and its time derivatives
Module 5

Continuity equation for hydrologic routing
Flood hydrograph through a reservoir or a channel reach is a gradually varied
unsteady flow. If we consider some hydrologic system with input I(t), output Q(t), and
storage S(t), then the equation of continuity in hydrologic routing methods is the
following:
Change in storage
Change in time
Module 5

Rate change of flow storage can be also represented by this following equation:
Even if the inflow hydrograph, I(t) is known, this equation cannot be solved directly
to obtain the outflow hydrograph, Q(t), because both Q and S are unknown. A
second relation, the storage function is needed to relate S, I, and Q. The particular
form of the storage equation depends on the system: a reservoir or a river reach.
Change in storage
Change in time
Contd..
Module 5
Continuity equation for hydrologic routing

Lecture 2: Level pool routing and modified
Pul’s method
Module 5

Hydrologic flow routing
When a reservoir has a horizontal water surface elevation, the storage function is a
function of its water surface elevation or depth in the pool. The outflow is also a
function of the water surface elevation, or head on the outlet works.
S= f(O)
where S= storage and O= Outflow
Module 5
1. Level Pool Routing

1. Level Pool Routing Contd..
I= inflow
Q= outflow
S =storage
t=time
Q
S
Module 5

The peak outflow occurs when the outflow hydrograph intersects the inflow
hydrograph.
1. Level Pool Routing Contd…
Maximum storage occurs when
As the horizontal water surface is assumed in the reservoir, the reservoir
storage routing is known as Level Pool Routing. The outflow from a reservoir is a
function of the reservoir elevation only. The storage in the reservoir is also a
function of the reservoir elevation.
Module 5

In a small time interval the difference between the total inflow and outflow in a
reach is equal to the change in storage( ) in that reach
Where = average inflow in time t, = average outflow in time t. If suffixes 1 and
2 denote the beginning and end of the time interval t then the above equation
becomes
Module 5

The time interval should be sufficiently short so that the inflow and out flow
hydrographs can be assumed to be straight lines in that time interval.
Module 5

In reservoir routing, the following data are known:
(i) Elevation vs Storage
(ii) Elevation vs Outflow discharge and hence storage vs outflow discharge
(iii) Inflow hydrograph, and
(iv) Initial values of inflow, outflow O, and storage S at time t = 0.
Module 5

A variety of methods are available for routing of floods through a reservoir. All of
them use the general equation but in various rearranged manners.
Pul’s Method: This is a semi-graphical method.
All the terms on the left hand side are known and hence right hand side at the end of
the time step . Since S = f(h) and O = f(h), the right hand side is a function of
elevation h for a chosen time interval .Graphs can be prepared for h vs O, h vs S
and h vs . .
Module 5
Modified Pul’s Method

STEPS
For practical use, this semi-graphical is very convenient:
1. From the known storage-elevation and discharge-elevation data, prepare a
curve of vs elevation. Here is any chosen interval, approximately
20 to 40% of the time rise of the inflow hydrograph.
2. On the same plot prepare a curve of outflow discharge vs elevation.
3. The storage, elevation and outflow discharge at the starting of routing are
known. For the first time interval , , and are known and
hence the term is determined form the Pul’s method.
4. The water –surface elevation corresponding to is found by using the
plot of step (1). The outflow discharge Q2 at the end of the time step is
found from the plot of step (2).
Module 5
Level Pool Routing
Modified Pul’s Method Contd…

STEPS
4. The water –surface elevation corresponding to is found by using
the plot of step (1). The outflow discharge Q2 at the end of the time step
is found from the plot of step (2).
5. Deducting from gives for the beginning of
the next step.
6. The procedure is repeated till the entire inflow hydrograph is routed.
Module 5
Modified Pul’s Method Contd…
Level Pool Routing

Lecture 3: Channel routing methods
Module 5

2. Channel Routing
In very long channels the entire flood wave also travels a considerable distance
resulting in a time redistribution and time of translation as well. Thus, in a
river, the redistribution due to storage effects modifies the shape, while the
translation changes its position in time. In reservoir, the storage is a unique
function of the outflow discharge S = f(O).
Storage in the channel is a function of both outflow and inflow discharges and
hence a different routing method is needed. The water surface in a channel
reach is not only parallel to the channel bottom but also varies with time.
Module 5

The total volume in storage for a channel reach having a flood wave can be
considered as prism storage + wedge storage.
Prism storage: The volume that would exist if uniform flow occurred at the
downstream depth i.e. the volume formed by an imaginary plane parallel to the
channel bottom drawn at the outflow section water surface.
Wedge storage: It is the wedge like volume formed between the actual water
surface profile and the top surface of the prism storage. At a fixed depth at a
downstream section of a river reach the prism storage is constant while the wedge
storage changes from a positive value at an advancing flood to a negative value
during a receding flood.
Module 5
2. Channel Routing Contd…

Prism Storage: It is the volume
that would exits if uniform flow
occurred at the downstream
depth, i.e. the volume formed by
an imaginary plane parallel to the
channel bottom drawn at the
outflow section water surface.
Module 5

Wedge storage : It is the wedge-like volume formed between the actual water
surface profile and the top surface of the prism storage.
Module 5

At a fixed depth at a downstream section of a river reach, the prism storage is constant
while , the wedge storage changes from a positive value for advancing flood to a
negative value during a receding flood.
Total storage in the channel reach can be expressed as :
where k and x are coefficients and m= a constant exponent . It has been found that
m varies from 0.6 for rectangular channels to a value of about 1.0 for natural
channels, Q = outflow
Module 5

Assuming that the cross sectional area of the flood flow section is directly
proportional to the discharge at the section, the volume of prism storage is equal
to KQ where K is a proportionality coefficient, and the volume of the wedge
storage is equal to KX(I- Q), where X is a weighing factor having the range 0 < X
< 0.5. The total storage is therefore the sum of two components
)( QIKXKQS −+=
It is known as Muskingum storage equation representing a linear model for
routing flow in streams.
Module 5
Muskingum Method
Channel routing

QQ
QI −
I
Q
II
IQ −
I Q
Advancing
Flood
Wave
I > Q
Receding
Flood
Wave
Q > I
KQS =Prism
)(Wedge QIKXS −=
K is a proportionality coefficient,
X is a weighing factor on inflow versus
outflow (0 ≤ X ≤ 0.5)
X = 0.0 - 0.3  Natural stream
)( QIKXKQS −+=
])1([ QXXIKS −+=
Module 5
Muskingum Method Contd…
Channel routing

])1([ QXXIKS −+=
The value of X depends on the shape of the modeled wedge storage. It is zero for
reservoir type storage (zero wedge storage or level pool case S = KQ) and 0.5
for a full wedge. In natural streams mean value of X is near 0.2. The parameter K
is the time of travel of the flood wave through the channel reaches also
known as storage time constant and has the dimensions of time.
Module 5
Channel routing

From the Muskingum storage equation, the values of storage at time j and
j+1 can be written as
and
So, change in storage over time interval ∆t is,
From the continuity equation the storage for the same time interval ∆t is,
Module 5
Channel routing

Equating these two equations,
Collecting similar terms and simplifying
This is the Muskingum’s routing equation for channels
Module 5
Channel routing

Muskingum’s routing equation for channels:
where
For best results, the routing interval ∆t should be so chosen that K>∆t>2KX.
If ∆t<2KX, the coefficient C1 will be negative. Generally negative values of
coefficients are avoided by choosing appropriate values of ∆t.
Module 5
Channel routing

To use Muskingum equation to route a given inflow hydrograph through a
channel reach:
 K , X and Oj should be known.
Procedure:
(i)knowing K and X, select an appropriate value of t
(ii) calculate C1, C2, and C3
(iii) starting from the initial conditions known inflow, outflow calculate the outflow for
the next time step.
(iv) Repeat the calculations for the entire inflow hydrograph.
Module 5
Channel routing

Lecture 4: Hydraulic routing
Module 5

Hydraulic/Distributed flow routing
 Flow is calculated as a function of space and time throughout the system
 Hydraulic methods use continuity and momentum equation along with the
 St. Venant Equations (Refer to Module 6 for more details)
 Kinematic wave routing
 Muskingum-Cunge method
Module 5

It is the relationship between the Muskingum method and the Saint-Venant equations.
Inflow-Outflow Equation:
The constants C0, C1 and C2 are functions of wave celerity, c.
Q discharge and y  depth of flow
Muskingum-Cunge method
Diffusion wave routing
Module 5
,
dy
dA
dy
dQ
dA
dQ
c ==
t
OIIO 210 CCC ttttt ++= ∆+∆+

where,
Q0 = Reference discharge,
S0 = Reach Slope,
QB = Baseflow
Qp = Peak flow taken from the inflow hydrograph
Module 5
Muskingum-Cunge method Contd…






∆
−=
xSTc
Q
X
***
1
2
1
0
0
( )BpB QQQQ −+= 50.00

Dynamic Wave Routing
Flow in natural channels is unsteady, non-uniform with junctions, tributaries,
variable cross-sections, variable resistances, variable depths, etc. The complete
St.Venant equation represents the dynamic wave routing. (Refer to Module 6 for more
details)
Valley
storage
Prism
storage
Wedge
storage
Non-conservative
form of continuity
equation
Module 5




∂
∂
+



∂
∂
+
∂
∂
=
x
y
V
x
V
y
t
y
0

Dynamic Wave Routing Contd…
Momentum equation considering all relevant forces acting on the system:
Local
acceleration
term
Convective
acceleration
term
Pressure
force term
Friction
force term
Module 5
0)(
11 2
=−−
∂
∂
+





∂
∂
+
∂
∂
fo
SSg
x
y
g
A
Q
xAt
Q
A
Gravity
force term

Example Problem
Given:
Inflow hydrograph
K = 2.3 hr, X = 0.15, ∆t = 1 hour, Initial Q = 90 cfs
Find:
Outflow hydrograph using Muskingum routing method
Module 5
5927.0
1)15.01(3.2*2
1)15.01(*3.2*2
)1(2
)1(2
3442.0
1)15.01(3.2*2
15.0*3.2*21
)1(2
2
0631.0
1)15.01(3.2*2
15.0*3.2*21
)1(2
2
3
2
1
=
+−
−−
=
∆+−
∆−−
=
=
+−
+
=
∆+−
+∆
=
=
+−
−
=
∆+−
−∆
=
tXK
tXK
C
tXK
KXt
C
tXK
KXt
C
Period Inflow
(hr) (cfs)
1 93
2 137
3 208
4 320
5 442
6 546
7 630
8 678
9 691
10 675
11 634
12 571
13 477
14 390
15 329
16 247
17 184
18 134
19 108
20 90

jjjj QCICICQ 32111 ++= ++
C1 = 0.0631, C2 = 0.3442, C3 = 0.5927
Period Inflow C1Ij+1 C2Ij C3Qj Outflow
(hr) (cfs) (cfs)
1 93 0 0 0 90
2 137 9 32 53.343 94.343
3 208 13 47 55.9171 115.9171
4 320 20 72 68.70406 160.7041
5 442 28 110 95.2493 233.2493
6 546 34 152 138.2469 324.2469
7 630 40 188 192.1811 420.1811
8 678 43 217 249.0413 509.0413
9 691 44 233 301.7088 578.7088
10 675 43 238 343.0007 624.0007
11 634 40 232 369.8452 641.8452
12 571 36 218 380.4217 634.4217
13 477 30 197 376.0217 603.0217
14 390 25 164 357.411 546.411
15 329 21 134 323.8578 478.8578
16 247 16 113 283.819 412.819
17 184 12 85 244.6778 341.6778
18 134 8 63 202.5124 273.5124
19 108 7 46 162.1108 215.1108
20 90 6 37 127.4962 170.4962
0
100
200
300
400
500
600
700
800
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Discharge(cfs)
Time (hr)
Inflow Outflow

Exercise Problem
An inflow hydrograph is measured for a cross section of a stream. Compute the
outflow hydrograph at a point five miles downstream using the Muskinghum
method . Assuming K = 12hr, x=0.15, and outflow equals inflow initially. Plot the
inflow and outflow hydrograph.
Time Inflow (cfs)
9:00A.M. 50
3:00P.M 75
9:00 P.M. 150
3:00A.M. 450
9:00A.M. 1000
3:00P.M. 840
9:00P.M. 750
3:00A.M. 600
9:00A.M. 300
3:00P.M. 100
9:00P.M. 50
Module 5

 Flood routing is a technique of determining the flood hydrograph at a section
of a river by utilizing the data of flood flow at one or more upstream sections
 As a flood wave travels downstream, it undergoes:
 Peak attenuation
 Translation
 Types of flood routing
 Lumped/hydrologic
 Distributed/hydraulic
 Lumped / Hydrologic flow routing
 Flow is calculated as a function of time alone at a particular location.
 Equation of continuity and flow/storage relationship
Module 5

 Hydrologic flow routing methods
 Level pool method (Modified Puls)
 Channel routingMuskingum method
 Series of reservoir models
 Distributed / Hydraulic routing
 Flow is calculated as a function of space and time throughout the system
 Hydraulic methods use continuity and momentum equation along with the
Module 5

 Distributed / Hydraulic routing methods
 Muskingum Cunge method
 Complete solution to St.Venant equations
Module 5

IIT Bombay
Equations Governing Hydrologic and
Hydraulic Routing
Module 6
5 Lectures

Objectives of this module is to understand the physical
phenomena behind the Reynolds transport theorem and
Saint Venant equations.
Module 6

 Reynolds Transport Theorem
 Control Volume Concept
 Open Channel Flow
 Saint Venant Equations
 Continuity Equation
 Momentum Equation
 Energy Equation
Module 6

Lecture 1: Reynolds transport theorem and
open channel flow
Module 6

Fluids Problems-Approaches
1. Experimental Analysis
2. Differential Analysis
3. Control Volume Analysis
 Looks at specific regions, rather than specific masses
Module 6

Reynolds' transport theorem (Leibniz-Reynolds' transport theorem) is a 3-D
generalization of the Leibniz integral rule. The theorem is named after Osborne
Reynolds (1842–1912).
 Control volume: A definite volume specified in space. Matter in a control volume
can change with time as matter enters and leaves its control surface.
 Extensive properties (B) : Properties depend on the mass contained in a fluid,
 Intensive properties (β) : Properties do not depend on the mass.
Osborne Reynolds
Reynolds Transport Theorem
dB
or dB dm
dm
β β= =
mass (m)  momentum 
P.E. (mgh)  K.E. (1/2 mv2) 
1
)(
==
dm
md
β )( vm
gh=β
Module 6
v
dm
vmd
==
)(
β
2
2
1 v=β

This theorem applies to any transportable property,
including mass, momentum and energy.
Module 6
Reynolds Transport Theorem:
The total rate of change of any extensive property B (=βdm = βρd∀) of a system
occupying a control volume C.V. at time ‘t’ is equal to the sum of:
a) the temporal rate of change of B within the C.V.
b) the net flux of B through the control surface C.S. that surrounds the C.V.
∫∫∫∫∫ +∀
∂
∂
=
....
.
sc
rel
vc
dAVd
tdt
dB
βρβρ
Reynolds Transport Theorem Contd…

 Reynolds Transport Theorem can be applied to a control volume of finite size
 No flow details within the control volume is required
 Flow details at the control surfaces is required
We will use Reynolds Transport Theorem to solve
many practical fluids problems
Here, control volume is the sum
of I & II
fluid particles at time ‘t’
o fluid particles at time ‘t+∆t’
Module 6
Reynolds Transport Theorem Contd…

I
II III
- Control volume position occupied by fluid at time ‘t’
- Position occupied by fluid at time ‘t+∆t’
- Position occupied by fluid at time ‘t’, but not at ‘t+∆t’
- Position occupied by fluid at both the time ‘t’ and ‘t+∆t’
- Position occupied by fluid at time ‘t+∆t’, but not at ‘t’
I
II
III
Fixed frame in space
(upper and lower boundaries are impervious)
Finally, control volume at time ‘t’  I&II and at time ‘t+∆t’ II&III
Let us consider the following system,
Module 6

1st term 2nd term
1st term is equivalent to the change in extensive property stored in control
volume (as ∆t 0)
Module 6
( ) ( )[ ]
( )
( ) ( )[ ] ( ) ( )[ ]{ }tIttIIItIIttII
t
sys
tIIIttIIIII
t
sys
BBBB
tdt
dB
eqFrom
dincontainedfluidtheofmassdm
volumeelementald
d
dm
where
ddmdBHere
BBBB
tdt
dB
Lt
Lt
−−−
∆
=
∀=
=∀
∀
=
∀==
+−+
∆
=
∆+∆+
→∆
∆+
→∆
1
)1.6.(
,
)1.6...(
1
0
0
ρ
ρββ

Extensive property in the control volume,
Here, area vector is always normally outward to the surface
dA

control surface

V
θ
∆l
Module 6
( )
dAl
ldA
ttimeindAthroughgpasfluidofVolume
vectorvelocityv
dAmagnitudeofvectorAreaAd
.cos
cos
""""sin
θ
θ
∆−=
∆−=
∆
=
=

Module 6
( )
( )
( )
)(
)(
:
cos
sin
..
surfacecontrolthroughpropertyextensiveofoutflowNet
volumecontroltheinpropertiesextensiveofchangeofRate
fluidofpropertyextensiveofchangeofrateTotal
dt
dB
changeShapepropertiesextensiveinChangeB
dt
d
onlypropertiesextensiveinChangeB
t
Note
ldA
surfacecontrolthethroughgpasfluidofvolumeTotal
sys
sc
+=
=⇒
+⇒
⇒
∂
∂
∆−= ∫∫ θ

:sAssumption
tconsisi tan)( ρ
Module 6
( )
)(inf)()()(,
),()(.
0.
0,
.var
)2.6......(0.
0
,,.
....
....
..
....
velowtIandveoutflowtQstorageS
wheretItQdAVandSd
dAVd
dt
d
d
dt
d
flowstatesteadyFor
flowunsteadydensityiableofequationtheisThis
dAVd
dt
d
dt
massd
dt
dB
massofonconservatioflawperasNowmassbetoBConsider
scvc
scvc
vc
scvc
sys
−=+==
−==∀∴
=+∀
=∀−
=+∀∴
==
∫∫∫∫∫
∫∫∫∫∫
∫∫∫
∫∫∫∫∫
ρ
ρρ

Reynolds transport
equation becomes,
Unsteady State
Steady State
Module 6
.
,int
,1
,
)(
)()(
)()(
)()(,0
.,0)()(
0)()(.
1
1
1
1
)1()1(1
jjj
jjj
j
j
jjjj
tj
tj
tj
tj
S
S
tandttimethebetweensystemthefromOutflowQ
tandttimethebetweensystemtheoInflowI
jtimeatstorageS
jtimeatstorageSwhere
QISS
dttQdttIdS
dttQdttIdS
tQtIthen
dt
dS
provedhencetQtI
dt
dS
or
tItQ
dt
dS
ei
j
j
−
−
−
−
∆
∆−
∆
∆−
=
=
−=
=
−=−
−=∴
−=∴
==∴
=−=
=−+
∫∫∫
−
flowsteadyandtconsisii tan)( ρ
∑∫ ⋅+∀=
CSCV
sys
d
dt
d
dt
dB
AV

βρβρ
∑ ⋅=
CS
sys
dt
dB
AV

βρ

Module 6
,
)()(
),4.6(
)5.6......(),3.(
)4.6......()(
,2
)3.6......()(
,1
221102
221101
1
1101
2212
1101
Thus
QIQISSor
QIIQSS
inSofvaluengSubstituti
IQSSEqFrom
QISS
j
QISS
jIf
−+−+=
−=−+−
+−=
−=−
=
−=−
=
∑=
−+=
j
t
ttj
QISS
1
0
)(

 An open channel is a waterway, canal or conduit in which a liquid flows
with a free surface.
 In most applications, the liquid is water and the air above the flow is
usually at rest and at standard atmospheric pressure.
udel.edu/~inamdar/EGTE215/Open_channel.pdf
Open channel flow
Module 6

Pipe flow Open channel flow
Flow driven by Pressure work Gravity(i.e. Potential Energy)
Flow cross-section Known (Fixed by
geometry)
Varies based on the depth of flow
Characteristic flow
parameters
Velocity deduced from
continuity equation
Flow depth and velocity deduced by
solving simultaneously the
continuity and momentum
equations
Specific boundary
conditions
Atmospheric pressure at the water
surface
Pipe flow vs. Open channel flow
(Source:www.uq.edu.au/~e2hchans/reprints/b32_chap01.pdf)

Different flow conditions in an open channel
Section 1 – rapidly varying flow
Section 2 – gradually varying flow
Section 3 – hydraulic jump
Section 4 – weir and waterfall
Section 5 – gradually varying
Section 6 – hydraulic drop due to
change in channel slope
Module 6

Open Channel Flow
Unsteady Steady
Varied Uniform Varied
Gradually
Rapidly
Gradually
Rapidly
(NPTEL, ComputationalHydraulics)
Module 6

Different Types of flow in an open channel
1. Steady Uniform flow
2. Steady Gradually-varied flow
3. Steady Rapidly-varied flow
4. Unsteady flow
Module 6

Case (1) – Steady uniform flow:
 Steady flow is where there is no change with time, ∂/∂t = 0.
 Distant from control structures, gravity and friction are in balance, and if the
cross-section is constant, the flow is uniform, ∂/∂x = 0
Case (2) – Steady gradually-varied flow:
 Gravity and friction are in balance here too, but when a control is introduced
which imposes a water level at a certain point, the height of the surface varies
along the channel for some distance.
i.e. ∂/∂t = 0, ∂/∂x ‡ 0.
Case (3) – Steady rapidly-varied flow:
 Here depth change is rapid.
Case (4) – Unsteady flow:
 Here conditions vary with time and position as a wave traverses the waterway.
Module 6

Types of Open Channel Flows
Module 6

Hydraulic Routing
Hydraulic
routing
Momentum
Equation
Physics of water
movement
Hydraulic routing is intended to describe the dynamics of the water or flood wave
movement more accurately
Hydrological routing
Continuity equation + f (storage, outflow, and possibly inflow) relationships 
assumed, empirical, or analytical in nature. Eg: stage-discharge relationship.
Module 6

Dynamic Routing- Advantages
 Higher degree of accuracy when modeling flood situations because it
includes parameters that other methods neglect.
 Relies less on previous flood data and more on the physical properties of
the storm. This is extremely important when record rainfalls occur or other
extreme events.
 Provides more hydraulic information about the event, which can be used to
determine the transportation of sediment along the waterway.
Module 6

Lecture 2: Navier-Stokes and Saint Venant
equations
Module 6

Navier-Stokes Equations
 St.Venant equations are derived from Navier-Stokes Equations for shallow
water flow conditions.
 The Navier-Stokes Equations are a general model which can be used to model
water flows in many applications.
 A general flood wave for 1-D situation can be described by the Saint-Venant
equations.
Claude-Louis Navier Sir George Gabriel Stokes
Module 6

Navier-Stokes Equations Contd…
 It consists of 4 nonlinear PDE of mixed hyperbolic-parabolic type describing
the fluid hydrodynamics in 3D.
 Expression of F=ma for a fluid in a differential volume
 The acceleration vector contains local and convective acceleration terms
where i: x, y, z
ui: u, v, w
uj: u, v, w
Module 6
( )
( )
( )8.6
7.6
6.6
z
w
w
y
w
v
x
w
u
t
w
a
z
v
w
y
v
v
x
v
u
t
v
a
z
u
w
y
u
v
x
u
u
t
u
a
z
y
x
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=
j
i
j
i
i
x
u
u
t
u
a
∂
∂
+
∂
∂
=

 The force vector is broken into a surface force and a body force per unit volume.
 The body force vector is due only to gravity while the pressure forces and the
viscous shear stresses make up the surface forces(i.e. per unit mass).
Module 6
)11.6(
1
)10.6(
1
)9.6(
1






∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−+=






∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−+=






∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−+=
zyxz
p
gf
zyxy
p
gf
zyxx
p
gf
zzyzxz
zz
zyyyxy
yy
zxyxxx
xx
τττ
ρ
τττ
ρ
τττ
ρ

 The stresses are related to fluid element displacements by invoking the
Stokes viscosity law for an incompressible fluid.
( )
( )
( )15.6
14.6
)13.6(
12.62,2,2






∂
∂
+
∂
∂
==






∂
∂
+
∂
∂
==






∂
∂
+
∂
∂
==
∂
∂
=
∂
∂
=
∂
∂
=
y
w
z
v
z
u
x
w
x
v
y
u
x
w
x
v
x
u
zyyz
zxxz
yxxy
zzyyxx
µττ
µττ
µττ
µτµτµτ
Module 6

 Substituting eqs. 6.12-6.15 into eqs. 6.9-6.11, we get,
Module 6
notationEinstein
xx
u
x
p
gf
z
w
y
w
x
w
z
p
gf
z
v
y
v
x
v
y
p
gf
z
u
y
u
x
u
x
p
gf
jj
i
i
ii
zz
yy
xx
∂∂
∂
+
∂
∂
−=






∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=






∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=






∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
−=
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
)18.6(
1
)17.6(
1
)16.6(
1
ν
ρ
ν
ρ
ν
ρ
ν
ρ

 The equation of continuity for an incompressible fluid
 The three N-S momentum equations can be written in compact form as
Module 6
)19.6(
1 2
i
jj
i
ij
i
j
i
g
xx
u
x
p
x
u
u
t
u
+
∂∂
∂
+
∂
∂−
=
∂
∂
+
∂
∂
ν
ρ
)20.6(0
0
=
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂
i
i
x
u
z
w
y
v
x
u

The Saint Venant Equations were formulated in the 19th
century by two mathematicians, de Saint Venant and
Bousinnesque.
The solution of the St. Venant equations is known as dynamic
routing, which is generally the standard to which other methods
are measured or compared.
Jean Claude Saint-Venant
Joseph Valentin Boussinesq
Continuity equation:
Momentum equation:
Q-Discharge through the channel
A-Area of cross-section of flow
y- Depth of flow
S0-Channel bottom slope
Sf- Friction slope
0=
∂
∂
+
∂
∂
t
A
x
Q
0)(
11 2
=−−
∂
∂
+





∂
∂
+
∂
∂
fo
SSg
x
y
g
A
Q
xAt
Q
A
Saint Venant Equations

Assumptions of St. Venant Equations
• Flow is one-dimensional
• Hydrostatic pressure prevails and vertical accelerations are negligible
• Streamline curvature is small.
• Bottom slope of the channel is small.
• Manning’s and Chezy’s equation are used to describe resistance effects
• The fluid is incompressible
• Channel boundaries are considered fixed and therefore not susceptible to
erosion or deposition.
1D gradually varied unsteady flow in an open channel is given by St. Venant
equations:
 Continuity Equation ( based on Conservation of Mass)
 Momentum Equation ( based on Conservation of Momentum)
Module 6

In the diagrams given,
Q = inflow to the control volume
q = lateral inflow
= Rate of change of flow
with distance
= Outflow from the C.V.
= Change in mass
1-D Open channel flow
dx
x
Q
Q
∂
∂
+
t
Adx
∂
∂ )(ρ
x
Q
∂
∂
Plan View
Elevation View
Module 6

St. Venant equations
Conservation of Mass
In any control volume consisting of the fluid (water) under consideration, the net
change of mass in the control volume due to inflow and outflow is equal to the net
rate of change of mass in the control volume
Continuity equation:
0=
∂
∂
+
∂
∂
t
A
x
Q Q-Discharge through the channel
A-Area of cross-section of flow
Module 6

Q = AV = volume water discharge [L3/T]
ρQ = Mass water discharge = ρAV [M/T]
∂/∂t(Mass in control volume) = Net mass inflow rate (assuming q=0)
Continuity Equation-Derivation
Module 6
( ) ( )
( ) ( )
0
sec
arg,;0
0.
=





∂
∂
+
∂
∂
⇒
−
==





∂
∂
+
∂
∂
∆⇒
=∆
∂
∂
+∆
∂
∂
∆
∂
∂
−=−=∆
∂
∂
∆+
x
Q
t
A
tioncrossthethrough
edischQAVHere
x
AV
t
A
x
x
x
AV
x
t
A
ei
x
x
AV
AVAVx
t
A
xxx
ρ
ρ
ρ
ρρρ
ρ

In 1-D open channel flow continuity
equation becomes,
0
)(
=
∂
∂
+
∂
∂
t
y
x
Vy
Non-conservation form
(velocity is dependent variable)
0=−
∂
∂
+
∂
∂
q
t
A
x
Q
Conservation form
Module 6
0=
∂
∂
+
∂
∂
+
∂
∂
t
y
x
V
y
x
y
V

Example Problem
Calculate the inlet velocity Vin from the diagram shown.
Module 6
)0025.0(1*2)0025.0(101.0*1.0
)(
0
2
gVx
AVAV
dt
dh
A
AVAVhA
dt
d
d
dt
d
in
outoutinintank
outoutinintank
CSCV
+−=
+−=
+−=
⋅+∀=
−
∑∫
ρρρ
ρρ AV

smVin
/47.4=

( )ss
mvF ∆=∑
Momentum
In mechanics, as per Newton’s 2nd Law:
Net force = time rate of change of momentum
Sum of forces in
the s direction
Change in momentum
in the s direction
mass
Velocity in the s
direction

Momentum Equation
 The change in momentum of a body of water in a flowing channel is
equal to the resultant of all the external forces acting on that body.
Sum of forces
on the C.V.
Momentum stored
within the C.V
Momentum flow
across the C. S.
Module 6
∫∫∫∫∫∑ +∀=
....
.
scvc
dAVVdV
dt
d
F ρρ

∫∫∫∫∫∑ +∀=
....
.
scvc
dAVVdV
dt
d
F ρρ
This law states that the rate of change of momentum in the control volume is equal
to the net forces acting on the control volume
Since the water under consideration is moving, it is acted upon by external forces
which will lead to the Newton’s second law
Sum of forces on
the C.V.
Momentum stored
within the C.V
Momentum flow
across the C. S.
Module 6
0)(
11 2
=−−
∂
∂
+





∂
∂
+
∂
∂
fo
SSg
x
y
g
A
Q
xAt
Q
A
Conservation of Momentum

Applications of different forms of momentum equation
 Kinematic wave: when gravity forces and friction forces balance each other
(steep slope channels with no back water effects)
 Diffusion wave: when pressure forces are important in addition to gravity and
frictional forces
 Dynamic wave: when both inertial and pressure forces are important and
backwater effects are not negligible (mild slope channels with downstream
control)
Module 6

The three most common approximations or simplifications are:
Kinematic
Diffusion
Quasi-steady models
Approximations to the full dynamic equations
Kinematic wave routing:
 Assumes that the motion of the hydrograph along the channel is controlled by
gravity and friction forces. Therefore, uniform flow is assumed to take place. Then
momentum equation becomes a wave equation:
where Q is the discharge, t the time, x the distance along the channel, and c the
celerity of the wave (speed).
 A kinematic wave travels downstream with speed c without experiencing any
attenuation or change in shape. Therefore, diffusion is absent.
0=
∂
∂
+
∂
∂
x
Q
c
t
Q

 The diffusion wave approximation includes the pressure differential term
but still considers the inertial terms negligible; this constitutes an
improvement over the kinematic wave approximation.
 The pressure differential term allows for diffusion (attenuation) of the
flood wave and the inclusion of a downstream boundary condition which
can account for backwater effects.
 This is appropriate for most natural, slow-rising flood waves but may lead
to problems for flash flood or dam break waves
x
y
SSf
∂
∂
−= 0
Module 6

 It incorporates the convective acceleration term but not the local
acceleration term, as indicated below:
 In channel routing calculations, the convective acceleration term and
local acceleration term are opposite in sign and thus tend to negate each
other. If only one term is used, an error results which is greater in
magnitude than the error created if both terms were excluded
(Brunner, 1992).
 Therefore, the quasi-steady approximation is not used in channel routing.
Module 6
Quasi-Steady Dynamic Wave Routing
)()(0
xg
VV
x
y
SSf
∂
∂
−
∂
∂
−=

Lecture 3: Derivation of momentum
equation
Module 6

(a) (b)
Diagrammatic representation of (a) Isovels and (b) Velocity profile in an open
channel flow
60
Isovels
100
80
40
Velocity Distribution in an Open Channel Flow
60
Module 6

γh
h
γ(h-anh/g)
Gradually Varied Flow
Convex surface
an : acceleration component
acting normal to the streamlines
For horizontal surface
For curved surface
γanh/g)
Convex Curvi-linear flow
Hydrostatic pressure distribution in a curved surface is,
y
an
h
Module 6

γh
h
Concave surface
For horizontal surface
For curved surface
γanh/g)
h
Concave Curvi-linear flow
an
y
Gradually Varied Flow (Contd..)
Module 6

I, Q
y1 y2
t
Note: y=y(x,t), Depth of flow varies with distance and time
A = A (x,y) , Area of flow
Information can be known
about:
1) Critical inflow hydrograph
2) How much area may be
flooded for critical cases
For mass (m)  1
)(
==
dm
md
β
From Reynold’s transport theorem, we have
Module 6
Gradually Varied Flow (Contd..)

Module 6
( )
,var
0
,,
.
flowunsteadydensityiableofequationtheisThis
dt
massd
dt
dB
massofonconservatioflawperasNow
massbetoBConsider
sys
==
tconsisand tanρ
txoffunctionareiablestheallthatindicates
t
dAVd
t scvc
&var
0.
....
∂
∂
=+∀
∂
∂
⇒ ∫∫∫∫∫ ρρ
0.
....
=+∀∴ ∫∫∫∫∫ scvc
dAVd
dt
d
ρρ

Control volume
(A.dx)
-Longitudinal Section of the Channel
Module 6
( ) )21.6(..
..
0 ∫∫∫∫ ++
∂
∂
=
outletinlet
dAVdAV
t
dxA
or ρρ
ρ

y
η
dη
1 2
dx
q/2
Component in x-
direction
q/2
-Top view of the open channel section
-Cross-sectional view of a
compound channel
Total contribution is q.dx
Module 6

x
Q
t
A
q
∂
∂
+
∂
∂
=
Basic equation or Conservative form of
continuity equation (Applicable for kinematic &
non-prismatic channels)
Module 6
( ) ( )
( )
( )
or
x
Q
q
t
A
dxbysidesbothDividing
dx
x
Q
qdx
t
dxA
dx
x
Q
qdx
t
dxA
dx
x
Q
QqdxQ
t
dxA
aswrittenrebecanEquation






∂
∂
−−
∂
∂
=












∂
∂
−−
∂
∂
=






∂
∂
+−
∂
∂
=












∂
∂
+++−
∂
∂
=
−
0
,''
,
..
..
..
0
,)21.6(
ρ
ρ
ρ
ρρ
ρ
ρρ
ρ

dη
η
dy
y
y- η
B’
Module 6
( )
( ) ( )
0,
,;,
.
,
=
∂
∂




∂
∂






∂
∂
+



∂
∂
+



∂
∂






∂
∂
=
==
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
=
x
A
channelprismaticFor
x
y
y
A
V
x
V
A
t
y
y
A
qor
txyytyfAHere
x
A
V
x
V
A
t
A
qor
x
VA
t
A
qNow

Module 6
( ) ( )




∂
∂
+



∂
∂
+
∂
∂
=
==




∂
∂
+



∂
∂
+



∂
∂
=
==
=⇒=




∂
∂
++



∂
∂
+



∂
∂
=
≅
∂
∂
=
x
y
V
x
V
D
t
y
or
channeltheofwidthunitBqgAssu
x
y
BV
x
V
DB
t
y
Bqor
txyytyfAHere
DBABADdepthhydraulicNow
channelprismaticfor
x
y
BV
x
V
A
t
y
Bqor
B
y
A
So
dyBdA
0
)(1'&0,min
'.'.'
,;,
'.'/)(,
,'0'
',
'

Valley
storage
Prism
storage
Wedge
storage
Non-conservative
form of continuity
equation
Now, from Reynolds transport theorem:
From conservation law of momentum
(Newton’s 2nd law of motion)
∑= F
dt
dB
momentum )( vm
Module 6
or
yByB
B
ADei
lengthchannelthethroughoutuniformsameisdepthIf
=== '/'.
'
.,.
,)(




∂
∂
+



∂
∂
+
∂
∂
=
x
y
V
x
V
y
t
y
0
∫∫∫∫∫ +∀
∂
∂
=
....
.
scvc
dAVd
tdt
dB
βρβρ
v
dm
vmd
==
)(
β

Unsteady Non-uniform
flow
Steady Non-uniform
flow
0∑ =FFor steady uniform flow,
Module 6
∫∫∫∫∫∑ +∀
∂
∂
=∴
....
.
scvc
dAVVdV
t
F ρρ

Forces acting on the C.V.
-Plan View
-Elevation View
Module 6

Fg = Gravity force due to weight of water in the C.V.
Ff = friction force due to shear stress along the bottom and sides of the C.V.
Fe = contraction/expansion force due to abrupt changes in the channel cross-
section
Fw = wind shear force due to frictional resistance of wind at the water surface
Fp = unbalanced pressure forces due to hydrostatic forces on the left and
right hand side of the C.V. and pressure force exerted by banks
Forces acting on the C.V. (Contd..)
Module 6

γ.η γ.(y-η)
a) Hydrostatic Force
dη
η
dy
y
y- η
B’
b
Area of the Elemental strip=b.d η
Force=area*hydrostatic pressure
or dF=(b.d η)*γ.(y-η)
Module 6
( ) ( )
y
F b.d y d
η
η γ η η
=
= −∫1
0

Module 6
( )
( ) ( )
p
y
p
F
&F F dx
x
&F F F
F F
F (F dx ) dx
x x
F b.d y dx
x
and y y( x,t ) (u sin g Leibnitz rule)
η
η γ η
=
∂ 
= +  
∂ 
= −
∂ ∂   
=− + =−   
∂ ∂   
∂ 
⇒ =− − ∂ 
=
∫
1
2 1
1 2
1 1
1 1
0
1 2
1
F dx
x
F
F 





∂
∂
+ 1
1
a) Hydrostatic Force Contd…
( ) ( )
{ } ( )
b( t )
a( t )
b( t )
a( t )
[ General rule :
F t x,t dx
d d
F( t ) x,t dx
dt dt
φ
φ
=
 
=   
∫
∫

Module 6
( )
( ) ( )
( ) ( )
( )
( )
( )
( ) ( )
b( t )
a( t )
y
p
y
d x,t db(t ) da(t )
dx b(t ),t . a(t ),t .
dt dt dt
Therefore,
d d
F b y d b y y b y
x dx dx
(b)
b y y d dx
x x
y
b
η
η
η
η
φ
φ φ
η
γ η η γ γ
γ η η η
γ
=
=
=
=
       
= + −             
   ∂ 
=− − − − + −    ∂     
∂ ∂  
=− − + −  ∂ ∂  
∂
= −
∂
∫
∫
∫
0
0
0
0
( )
( )
y
y y
b
y d dx
x x x
y b
b d y d dx
x x
η
γ η η
γ η γ η η
 ∂ ∂  
− + −   ∂ ∂   
∂ ∂ 
= + − ∂ ∂ 
∫
∫ ∫
0
0 0
( )
( ) dxd
x
b
ydxA
x
y
dxd
x
b
ydxdb
x
y
y
yy




∂
∂
−−



∂
∂
−=




∂
∂
−−



∂
∂
−=
∫
∫∫
ηηγγ
ηηγηγ
0
00

b) Boundary Reaction
Module 6
( )
.
.
.
,
)2()2(sec
)1()1(sec.
alsoondistributipressurechydrostatitheinChange
dxd
x
b
dx
x
db
dx
x
A
areaelementalinChange
tionatareaElementalAdx
x
A
tionatAdbareaElementary
element
element
element
element
∴
∂
∂
=





∂
∂
=





∂
∂
=
∴
−=+





∂
∂
−==
η
η
η
γy
y
y-η
η
γ(y-η)

dη
dη
b1
q/2
q/2
b
dx
Module 6
( )
( )
y
y
(bd ) y
In similar way,
(bd )dx. . y ,
x
additional hydrostatic force
for change in width of the channel.
η γ η
η γ η
−
∂
−
∂
∫
∫
0
0
b) Boundary Reaction Contd…

W
Module 6
c) Body Force
θ
w.sinθ
θ
γ
sin.
..,
,tan)(
WF
dxAWWeight
tconsbeAareaLet
g =∴
=∴

Module 6
( )
depthmeanhydraulicorradiushydraulicRwhere
SRS
P
A
or
balancedentirelyareforcestwoThesesAssumption
stressshearperimeterwettedPwhereSdxAdxP
issidesandbottomchannelforforceShear
SdxAFor
SandofvaluessmallveryFor
g
=
=





=
===
=
=≅
,....
:
,.....
,
...
tantansin,
000
000
0
0
γγτ
τγτ
γ
θθθθ

c) Body Force Contd…
.,
, 0
hereflowuniformunsteadyconsidertoneedweHowever
SSflowuniformsteadyFor f
=

d) Frictional Force
e) Wind Force
Module 6
( )
ff
f
f
SdxAFor
SRdxP
dxPforceShear
SR
...
....
..
..
0
0
γ
γ
τ
γτ
−=
−=
−=∴
=
f
rrf
w
ww
w
W
VVC
channeltheoflengththethroughout
uniformisitandBiswidthtopIfdxBwindforceF
bestressshearwindLet
ρ
ρ
τ
τ
τ
−=−=
==
2
...
)
(..

.,
.,
veissoveisVveisVIf
veissoveisVveisVIf
wrr
wrr
+−⇒−
−+⇒+
τ
τ
ffw
f
f
WdxBWdxBF
factorstressshearwindW
tcoefficienstressshearwindC
...)(. ρρ −=−=∴
=
=
f) Eddy Losses:
( )
2
2
2
2
22
2
2
)(






∂
∂
=
∂
∂
=






∂
∂
=






∂
∂
∝
⇒
A
Q
xg
K
v
xg
K
S
g
v
x
KSor
g
v
x
S
SlinegradientenergytheofSlopelossEddy
ee
e
ee
e
e
ω
Vω
Vω.cos ω
V(=Q/A)
Module 6

,,
.)(
exp
...:sin
12
equationmomentumNow
casestheallinvealwaysisKthatEnsure
VVObviously
tcoefficienansionorncontractioKHere
SdxAFlosseddythegcauforceDrag
e
e
ee
+
>>
=
−= γ
AdVVdV
dt
d
F
sccv
.....
.
ρρ ∫∫∫∫∫∑ +∀=
as v
dm
vmd
==
)(
β
( ) AdVVAdVVdxAV
dt
d
F
outletinlet
......... ρρρ ∫∫∫∫∑ ++=
Module 6

Module 6
( )
( )dxqVQVdxqQand
directionvealongvelocityofcomponentbetoVConsider
factorcorrectionmomentumHere
AdVVAdVVdxAV
dt
d
F
x
x
outletinlet
.......,.
)(
)1(.........
11
1
ρβρβ
β
ρρρ
==
+
=
++= ∫∫∫∫∑
)..(1
QVρβ
( ) )....(1
dxQV
x
QV ρρβ
∂
∂
+
Forces acting on the C.V. (Contd..)

Body
force
Friction
force
Contraction/
Expansion
force
Wind
force
Hydrostatic
force
Boundary
Reaction
Module 6
( ) ( )( ) ( )
( ) ( )
:exp,
........
..............
),1(,
1
111
forcetotaltheoutfindtoansionstheallcombiningNow
dxQV
x
dxqVdxAV
t
dxQV
x
QVdxqVQVdxAV
dt
d
equationNow
x
x












∂
∂
+−
∂
∂
=






∂
∂
+++−
ρρβρ
ρρβρβρβρ
( )
( ) dxd
x
b
y
dxd
x
b
yAdx
x
y
BdxWgAdxSgAdxSgAdxS
y
y
fef
ηηγ
ηηγγρρρρ
∂
∂
−+
∂
∂
−−
∂
∂
−−−−
∫
∫
0
0
0

Module 6
( ) ( )
( )
( )
( ) 





∂
∂
+−





∂
∂
=
∂
∂
−−






∂
∂
+−
∂
∂
=
∂
∂
−−
==












∂
∂
+−
∂
∂
=




∂
∂
+−
∂
∂
=












∂
∂
+−
∂
∂
=




∂
∂
−−−−
A
Q
xgAgA
qV
t
Q
gAx
y
SSor
gAbySHRSHLdividingNow
A
Q
x
qV
t
Q
x
y
gASSgA
SWgAssu
A
Q
x
qV
t
Q
QV
x
qV
t
Q
dxQV
x
qVAV
t
A
x
y
gBWgASgASgASor
dxbySHRSHLdividingNow
x
f
xf
ef
x
x
x
fef
2
11
0
2
110
2
11
11
1
0
.1
:""..&..,
.
,0&0min
.
..
...
:""..&..,
ββ
ββ
ββ
ββ
β
ρ

Module 6
( )
,""..&..
0
11,10
:)(
0
2
1
gbySHRSHLgMultiplyin
SS
x
y
A
Q
xgAt
Q
gA
andqIf
channelsprismaticforequationmomentumofformonConservati
f
=−−
∂
∂
−





∂
∂
+





∂
∂
== β
Local
acceleration
term
Convective
acceleration
term
Pressure
force term
Gravity
force
term
Friction
force term
0)(
11 2
=−−
∂
∂
+





∂
∂
+
∂
∂
fo
SSg
x
y
g
A
Q
xAt
Q
A

Module 6
( )
0)(.2.
0,,
0)(.2.
0)(
.
11
..2
1
..
1
0)(
.1.1
.,
0)(
11
:
2
tan
tan
2
2
tan
2
22
2
=−−
∂
∂
+





∂
∂
∂
∂
+



∂
∂
+





∂
∂
∂
∂
+
∂
∂
=
∂
∂
=−−
∂
∂
+





∂
∂
∂
∂
+
∂
∂
+



∂
∂
+





∂
∂
∂
∂
+
∂
∂
=−−
∂
∂
+






∂
∂
∂
∂
+



∂
∂
+



∂
∂
+



∂
∂
+
∂
∂
=−−
∂
∂
+





∂
∂
+
∂
∂
=
=−−
∂
∂
+





∂
∂
+
∂
∂
=
=
=
fo
tconsy
fo
tconsy
fo
tconsy
fo
fo
SSg
x
y
g
x
y
y
A
A
V
x
V
V
t
y
y
A
A
V
t
V
or
x
A
channelprismaticforNow
SSg
x
y
g
x
y
y
A
x
A
A
V
x
V
V
t
y
y
A
A
V
t
V
or
SSg
x
y
g
x
y
y
A
V
Ax
A
V
Ax
V
VA
At
A
V
t
V
A
A
or
SSg
x
y
g
A
VA
xAt
VA
A
or
VAQTaking
SSg
x
y
g
A
Q
xAt
Q
A
EquationMomentumofformonConservati

Kinematic Wave
Diffusion Wave
Dynamic Wave
Module 6
0)(
0)(..
2
2
=−−
∂
∂
+



∂
∂
+



∂
∂
+



∂
∂
+



∂
∂
+
∂
∂
=−−
∂
∂
+





∂
∂
∂
∂
+



∂
∂
+



∂
∂
+



∂
∂
+
∂
∂
fo
fo
SSg
x
y
g
x
y
y
V
x
V
V
x
V
V
t
y
y
V
t
V
or
SSg
x
y
g
x
y
y
A
A
V
x
V
V
x
V
V
t
y
B
A
V
t
V
or
0)(.. =−−
∂
∂
+



∂
∂
+



∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
fo
SSg
x
y
g
x
V
V
x
y
V
x
V
y
t
y
y
V
t
V
Continuity Equation for
Non-Conservation form(=0)
0)( =−−
∂
∂
+
∂
∂
+
∂
∂
fo
SSg
x
y
g
x
V
V
t
V

Module 6
gA
qV
SS
x
y
x
V
g
V
t
V
g
gbySHRSHLDividing
A
qV
SSg
x
y
g
x
V
V
t
V
flowernalisthereifBut
x
fo
x
fo
1
1
)(
1
,""..&..
)(
,int
β
β
=−−
∂
∂
+
∂
∂
+
∂
∂
=−−
∂
∂
+
∂
∂
+
∂
∂
fo SS
x
y
x
V
g
V
t
V
g
=+
∂
∂
−
∂
∂
−
∂
∂
−
1
Steady, uniform flow
Steady, non-uniform flow
Unsteady, non-uniform flow

)24.6()(
1
)23.6()(
)22.6(0)(
)(
)0(
)0(
1
1
1
1
gA
qV
SS
x
y
x
V
g
V
t
V
g
gA
qV
SS
x
y
x
V
g
V
gA
qV
SSor
gA
qV
SS
flowuniformfortermonacceleraticonvective
x
V
flowsteadyfortermonacceleratilocal
t
V
Here
x
fo
x
fo
x
fo
x
fo
β
β
β
β
=−−
∂
∂
+
∂
∂
+
∂
∂
=−−
∂
∂
+
∂
∂
=+−
=−−
=⇒
∂
∂
=⇒
∂
∂
,flowuniformsteadyFor −
,flowuniformnonsteadyFor −
,flowuniformnonunsteadyFor −
Module 6

Lecture 4: Derivation of momentum equation
(contd.)
Module 6

2D Saint Venant Equations
 Obtained from Reynolds Navier-Stokes equations by depth-
averaging.
 Suitable for flow over a dyke, through the breach, over the floodplain.
 Assumptions: hydrostatic pressure distribution, small channel slope
Module 6
.
)()()(
.
)()()(
.0
)()(
3
1
22
2
2
3
1
22
2
2
eqmomentumy
h
vu
vgn
y
z
gh
y
h
gh
y
hv
x
huv
t
hv
eqmomentumx
h
vu
ugn
x
z
gh
x
h
gh
y
huv
x
hu
t
hu
eqcontinuity
y
hv
x
hu
t
h
b
b
+
−
∂
∂
−=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
+
−
∂
∂
−=
∂
∂
+
∂
∂
+
∂
∂
+
∂
∂
=
∂
∂
+
∂
∂
+
∂
∂

Solutions to
St. Venant equations
Method of
characteristics
Finite Difference
methods
Explicit Implicit
Finite Element
Methods
Solutions to St. Venant equations
Further reading - Note: NPTEL (Computational Hydraulics),
http://www.efm.leeds.ac.uk/CIVE/CIVE3400/stvenant.pdf.
Module 6

Solutions to St. Venant equations Contd…
• Analytical
– Solved by integrating partial differential equations
– Applicable to only a few special simple cases of kinematic waves
• Numerical
– Finite difference approximation
– Calculations are performed on a grid placed over the (x, t) plane
– Flow and water surface elevation are obtained for incremental time
and distances along the channel
Module 6

x-t plane for finite differences calculations
Finite Difference Scheme (FDS)
Upstreamboundary
Downstreamboundary
Time,t
Distance, x
Module 6

∆x ∆x
∆t h0, Q0, t0 h1, Q1, t0 h2, Q2, t0
h0, Q0, t1 h1, Q1, t1 h2, Q2, t2
∆x ∆x
∆t
i, ji-1, j i+1, j
i-1, j+1 i-1, j+1 i+1, j+1
x-t plane
Cross-sectional view
in x-t plane
Module 6
Finite Difference Scheme (FDS) Contd…

Module 6
Explicit Implicit
Temporal derivative
Spatial derivative
Temporal derivative
Spatial derivative
j j
i i
u uu
t t
+
−∂
≈
∂ ∆
1
j j
i i
u uu
x x
+ −
−∂
≈
∂ ∆
1 1
2
t
uuuu
t
u j
i
j
i
j
i
j
i
∆
−−+
≈
∂
∂ +
+
+
+
2
1
1
1
1
x
uu
x
uu
x
u j
i
j
i
j
i
j
i
∆
−
−+
∆
−
≈
∂
∂ +
++
+ 1
11
1
)1( θθ

Time step j+1
Time step j
Time step j-1
Space
Time
Reach
Spatial derivative is written using terms on known time line ‘j’
Explicit FDS
Module 6

Time step j+1/2
Time step j
Time
Time step j+1
i i+1i-1
Space
h1
h3
h5
h7
Q
Q
Q
Center
point
Spatial and temporal derivatives use unknown time lines for computation ‘j+1’
Implicit FDS
Module 6
2
4
6

v2
v1
P1
P2
W
θ
Wsinθ
Rf
L
fs
RPWPF −−+=∑ 21
sinθ
Momentum Analysis in an Open Channel
For a constant mass and a per unit width consideration in a rectangular channel,
( ) ( )12 vvqmvs −=∆ ρ
Module 6

Here,
 Rf is the frictional resistance.
 P1 and P2 are pressure forces per unit width given by:
 Combining terms we get:
 Considering a short section so that Rf is negligible and the channel slope is
small so that sinθ ≅ 0, the equation can be written as:
2
y
P
2
γ
=
)(sin
22
12
2
2
2
1
vvqRW
yy
f
−=−+− ρθ
γγ
2
2
2
1
2
1
qv
2
y
qv
2
y
ρ+
γ
=ρ+
γ
Momentum Analysis in an Open Channel Contd…
Module 6

Specific force plus momentum curve
• i.e, M
g
qv
2
y
g
qv
2
y 2
2
21
2
1
=+=+
M is the specific force plus
momentum and is constant for
both y1 & y2.
There are two possible depths
for a given M called sequent
depths. The depth associated
with the minimum M is yc.
MMc
y
yc
M
y1
y2
q1
q2
y = yc
Module 6

 M is called the momentum function
or the specific force plus
momentum.
 For a constant q, M can be plotted
against depth to create a curve
similar to the specific energy curve.
 Under steady conditions, M is
constant from point to point along a
channel reach.
 When q = q1 = q2 (at steady
conditions when depth and velocity
remain constant at a point)
v
2
v
1
y1
y2
1
2
M
g
qv
2
y
g
qv
2
y 2
2
21
2
1
=+=+
Module 6

Lecture 5: Energy equation and numerical
problems
Module 6

Energy in Gradually Varied
Open channel flow
Module 6
 In a closed conduit there can be a pressure gradient that drives the flow.
 An open channel has atmospheric pressure at the surface.
 The HGL (Hydraulic Gradient Line) is thus the same as the fluid surface.

Energy equation applied to open channel
p1 and p2 : pressure forces per unit width at sections 1 & 2 respectively
v1 and v2 : velocity of flow at sections 1 & 2 respectively
and : energy coefficients
z1 and z2: elevations of channel bottom at sections 1 & 2 respectively w.r.t any
datum
hL: energy head loss of flow through the channel
1
α 2
α
L
hz
g
vp
z
g
vp
+++=++ 2
2
2
2
2
1
2
1
1
1
22
α
γ
α
γ
Module 6

Simplifications made to the Energy Equation:
1. Assume turbulent flow (α = 1).
2. Assume the slope is zero locally, so that z1 = z2
3. Write pressure in terms of depth (y = p / γ).
4. Assume friction is negligible (hL = 0).
i.e.
g
v
y
g
v
y
22
2
2
2
2
1
1
+=+
21
EE =
Module 6
Energy equation applied to open channel Contd…

Specific Energy Equation
Module 6
Energy at a particular point in the channel = Potential Energy + Kinetic Energy
where y is the depth of flow, v is the velocity, Q is the discharge, A is the cross-
sectional flow area and E is the specific energy i.e energy w.r.t channel bottom.
2
2
2
2
2
gA
QyE
or
g
vyE
+=
+=

Total energy
 When energy is measured with respect to another fixed datum , it’s
called Total Energy
where z is the height of the channel bottom from the datum
 Pressure head (y) is the ratio of pressure and the specific weight of
water
 Elevation head or the datum head (z) is the height of the section
under consideration above a datum
 Velocity head (v2/2g) is due to the average velocity of flow in that
vertical section
Module 6
g
vzyE
2
2
++=

Example Problem
Module 6
Channel width (rectangular) = 2m, Depth = 1m, Q = 3.0 m3/s, Height above
datum = 2m. Compute specific and total energy
Ans: A = b*y = 2.0*1.0 = 2 m2
Specific energy =
Total energy =
Datum height + specific energy = 2.0 + 1.20 = 3.20 m
2
2
2gA
QyE +=
2
2
2*81.9*2
31+=E

Specific Energy Diagram
Module 6
The specific energy can be plotted graphically as a function of depth of
flow : E = Es + Ek
where
)(
2 2
2
energyKinetic
gA
QEk
=
)( energyStaticyEs
=
2
2
2gA
QyE +=

Specific Energy Diagram Contd…
Module 6

As the depth of flow increases,
the static energy increases and
the kinetic energy decreases,
The total energy curve
approaches the static energy
curve for high depths and the
kinetic energy curve for small
depths
ks
EyEy 1∞⇒∞
Specific Energy Diagram Contd…
Module 6
As discharge (Q) increases, the specific energy curves move to the upper right
portion of the graph

Thus, for flat slope (+ other assumptions…) we can graph y against E:
(Recall for given flow, E1 = E2 )
Curve for
different, higher Q.
For given Q and E, usually have 2 allowed depths:
Subcritical and supercritical flow.
Module 6

The specific energy is minimum (Emin) for a particular  critical depth –
Depth
Froude’s number = 1.0. & velocity = Vc.
Emin  only energy value with a singular depth!
Depths < critical depths – supercritical flow (Calm, tranquil flow)
Froude Number > 1.0. V > Vc.
Depths > critical depths – subcritical flow (Rapid flow, “whitewater”)
Froude Number < 1.0. V < Vc.
Example: Flow past a sluice gate
Module 6

Critical Depth and Froude Number
It can easily be shown that at ,
Module 6
At the turning point (the left-most point of
the blue curve), there is just one value of
y(E).
This point can be found from

The Froude number can be defined as:
(Recall that the Reynolds number is the ratio of acceleration to viscous forces).
The Froude number is the ratio of acceleration to gravity
Perhaps more illustrative is the fact that surface (gravity) waves move at a speed
of
Flows with Fr < 1 move slower than gravity waves.
Flows with Fr > 1 move faster than gravity waves.
Flows with Fr = 1 move at the same speed as gravity waves.
Module 6
Critical Depth and Froude Number Contd…

Flows sometimes switch from supercritical to subcritical:
(The switch depends on upstream and downstream velocities)
Gravity waves: If you throw a rock into the water, the entire circular wave will
travel downstream in supercritical flow.
In subcritical flow, the part of the wave trying to travel upstream will in fact move
upstream (against the flow of the current).
Module 6
Critical Depth and Froude Number Contd…

Example Problem
Will the flow over a bump be supercritical or subcritical?
As it turns out:
Left = subcritical
Right = supercritical
Using the Bernoulli equation for frictionless, steady, incompressible flow along a
streamline:
or
Left
Left
Right
Right
Module 6

Apply Bernoulli equation along free surface streamline (p=0):
For a channel of rectangular cross-section,
Module 6

Substitute Q = V z b into Bernoulli equation:
To find the shape of the free surface, take the x-derivative:
Solve for dz / dx:
Module 6

Since subcritical: Fr < 1
supercritical: Fr > 1
Subcritical flow with dh / dx > 1 dz / dx < 1
Supercritical flow with dh / dx > 1 dz / dx > 1
if flow is subcritical if flow is supercritical
Module 6

There is a lot of viscous dissipation ( = head loss ) within the hydraulic jump.
Module 6
Hydraulic Jump Contd…

Apply the momentum equation:
Momentum equation is used here as there is an unknown loss of energy (where
mechanical energy is converted to heat).
But as long as there is no friction along the base of the flow, there is no loss of
momentum involved.
Module 6

Momentum balance:
The forces are hydrostatic forces on each end:
(where and are the pressures at centroids of A1 and A2 )
Module 6

If y1 and Q are given, then for rectangular channel
is the pressure at mid-depth.
Here, entire left-hand side is known, and we also know the first term on the
right-hand side. So we can find V2.
Module 6

1) A rectangular channel 4m wide has a flow discharge of 10.0 m3/s and depth of
flow as 2.5 m. Draw specific energy diagram and find critical and alternate
depth.
2) A triangular channel with side slopes having ratio of 1:1.5 has a discharge
capacity of 0.02 m3/s. Calculate:
a. critical depth
b. Emin
c. Plot specific energy curve
d. Determine energy for 0.25 ft and alternate depth
e. Velocity of flow and Froude number
f. Calculate required slopes if depths from d are to be normal
depths for given flow.
Exercises
Module 6

 Reynolds' transport theorem (Leibniz-Reynolds' transport theorem) is a 3-
D generalization of the Leibniz integral rule.
 Control volume is a definite volume specified in space. Matter in a control
volume can change with time as matter enters and leaves its control surface.
 Reynolds Transport Theorem states that the total rate of change of any
extensive property B of a system occupying a control volume C.V. at time ‘t’ is
equal to the sum of:
a) the temporal rate of change of B within the C.V.
b) the net flux of B through the control surface C.S. that surrounds
the C.V.
Module 6

 St.Venant equations are derived from Navier-Stokes Equations for shallow
water flow conditions.
 The solution of the St. Venant equations is known as dynamic routing,
which is generally the standard to which other methods are measured or
compared.
 Forces acting on the C.V. in an open channel flow are gravity force,
friction force , contraction/expansion force, wind shear force and unbalanced
pressure forces.
 Solutions to St. Venant equations :
 Method of characteristics
 Finite Difference methods : Explicit, Implicit
 Finite Element Methods
Module 6

Hydrological Statistics
IIT Bombay
Module 7
7 Lectures

Objective of this module is to learn the fundamentals
of stochastic hydrological phenomena.
Module 7

Module 7
 Statistical parameter estimation
 Probability distribution
 Goodness of fit
 Concepts of probability weighted moments & l-moments

Lecture 1: Introduction to probability
distribution
Module 7

Introduction
 Deterministic model
– Variables involved are deterministic in nature
– No uncertainty for a given set of inputs
E.g. Newtonian model
(2nd law of motion)
 Probabilistic/Stochastic model
– Variables involved are random in nature.
– There is always uncertainty for a given set of inputs
E.g. Rainfall-runoff model
21
2
s ut ft= +
In a deterministic model, a given input yields the same output always, while a
probabilistic model yields different outputs for multiple trials with the same input.
Module 7

Random Variables
 Random Variable (R.V.) is a variable whose value can not be predicted
with certainty before it actually takes over the value. R.V may be discrete or
continuous.
E.g. rainfall, stream flow, soil hydraulic properties
(permeability, porosity, etc.), evaporation, diffusion, temperature, ground
water level, etc.
Notations:
– Capital letters X, Y, Z indicates R.V.
– Small letter x, y, z indicates value of the R.V.
X x≤
Rainfall 30 mm (value)
If X is a R.V. & Z(X) is function of X ==> Z is a R.V.
E.g. Rainfall is R.V, so Runoff is R.V.
Module 7

 Discrete random variables
 If the set of values for a R.V. can be assumed to be finite (or
countable infinite), then RV is said to be a discrete RV.
E.g. No. of raining days in a month (0,1,2,…,31)
No. of particles emitted by a radio-active material.
 Continuous random variables
 If the set of values for a R.V. can be assumed to be infinite,
then RV is said to be a continuous R.V.
E.g. Depth of rainfall in a period of given month
Random Variables Contd..…..
 By using ‘class interval’ we can convert continuous random
variables to discrete random variables.
Module 7

Probability Distribution
 Discrete Variable
X1 X2 X3 Xn
Splices
P [X=xi] Probability mass function (pmf)
Important conditions:
(i) 0 ≤ P [X=xi] ≤ 1
(ii) P [X=xi] = 1
[Here 1 is likelihood of occurrence]
Cumulative distribution function (CDF)
 CDF is a step function
i
∑
x1 x2 x3 xn
1.0
x
p(x1)
p(x1)
+ p(x2)
for xi ≤ x2
p(x1)
+ p(x2)
+ p(x3)
Notation
P [ X = x] = p(x)
P
[ ] [ ]
i
i i
x x
P X x P X x
≤
≤= =∑
{ }1 2( ) 0 , ,......, nP X x X x x x= = ∀ =
Cumulative distribution function (cdf)
Module 7

Probability Distribution Contd..…..
 Continuous Variable
 Probability density function
(PDF), f(x) does not indicate
probability directly but it indicates
probability density
 CDF is given by
f(x)
a
x
+α
a x
F(x)
+α-α
1.0
( ) [ ] ( )
a
F a P X a f x dx
α−
= ≤ = ∫
Module 7
-α

Piecewise Continuous Distribution
d x
f(x)
0
f1(x)
f2(x)
P[X=d]
spike
1 2( ) + P[X=d]+ ( ) = 1
Then, P[X<d] P[X d];
Simillarly, P[X>d] P[X d];
d
d
f x dx f x dx
α
α−
≠ ≤
≠ ≥
∫ ∫
d x0
f(x)
P[X = 0]
X = 0
Module 7
d x0
Jump ΔF
P[X=d]
X=d
CDF for this case

CDF within an interval
Δx
x
2 2 2
i i i
x x x
x x
∆ ∆ ∆   
− +   
   
xi
f(xi)
i
ii
ii
ii
xxf
x
x
x
x
ervaltheincurvetheunderArea
x
xF
x
xF
x
xX
x
xP
∆=





 ∆
+
∆
−
=
∆
−−
∆
+=
∆
+≤≤
∆
−
*)(
2
,
2
int)
2
()
2
(
]
22
[
Here, interval is given by (x-∆x/2, x+∆x/2)
Module 7

Example Problem
Estimate the expected relative frequencies within each class of interval 0.25.
2
3
( ) ; 0 x 5
125
x
f x= ≤ ≤
x1 x2 x3
0 0.25 0.75 1.25 1.75 5.0
xi f(xi)=3x2
i/125 f(xi) x Δx = expected rel. freq.
0.25 0.0015 0.00075
0.75 0.0135 0.00675
1.25 0.0375 0.01875
1.75 0.0735 0.03675
2.25 0.1215 --
2.75 0.066 --
3.25 -- --
3.75 0.3375 0.16875
4.25 0.4335 0.21675
4.75 0.5415 0.27075
Sum 0.9975 ≈1.0
Module 7

Bivariate Random Variables
 (X,Y) – two dimensional random vector or random variable
e.g. Temperature  Evaporation
Temperature  Radiation Coeff.
Rainfall  Recharge
Rainfall  Runoff
 Variables may or may not be dependent
X = rainfall
y = runoff
X
Y
(X+Y)
Case 1: (X,Y) is a 2-D discrete random vector
The possible values of may be represented as (Xi, Yj), i=1,2,….n & J=1,2,….n
Case 2: (X,Y) is a 2-D continuous random variable
Here, (X,Y) can assume all possible values in some non-countable set.
Module 7

1. Discrete 2-D Random Vector
( ) [ ] ∑∑
≤
∞−
≤
∞−
=≤≤=
xX yY
YXpyYxXPyxF ),(,,
i=1 2 3 4 5 6 P[Y]
X
Y
0 1 2 3 4 5
j=1 0 0 0.01 0.03 0.05 0.07 0.09 0.25 P[Y=0]
2 1 0.01 0.02 0.04 0.05 0.06 0.08 0.26
3 2 0.01 0.03 0.05 0.05 0.05 0.06 0.25
4 3 0.01 0.02 0.04 0.06 0.06 0.05 0.24 P[Y=3]
0.03 0.08 0.16 0.21 0.24 0.28
P[x=0] P[x=5]
( ) 1, =∞∞F
[ ]
( )
1)8,7(
1,)4,7(
35.02,3)2,3(
1),(
6
1
4
1
=
=∞∞=
=≤≤=
=∑∑= =
F
FF
YXPF
YXp
i j
ji
E.g. Calculate the following using the values from the given table.
1) F(3,2)
2) F(7,4)
3) F(7,8)
Module 7

( , ) joint pdf of (X,Y)
f(X,Y) 0
( , ) 1
F(x,y) joint cdf of (X,Y)
= P[X , ]
= ( , )
Probablity of hatched region on the plane
α α
α α
α α
− −
− −
→
≥
=
→
≤ ≤
∫ ∫
∫ ∫
y x
If f X Y
f x y dxdy
x Y y
f x y dxdy
= ( , )∫B
f x y dB
B
f(x,y) plane
2. Continuous 2-D Random Vector
Module 7
F( , )=1
F(- ,y) = F(x, - ) = 0
α α
α α

Example Problem 1
Consider the flows in two adjacent streams.
Denote it as a 2-D RV (X,Y), with a joint pdf,
( , ) if 5,000 x 10,000 and
4,000 y 9,000
Solution: T
= ≤ ≤
≤ ≤
f X Y C
2
o get C,
( , ) 1 1
1
C =
(5000)
,
P(B) = ( , )
α α α α
α α α α− − − −
=⇒ =
∴
∫ ∫ ∫ ∫
∫ ∫B
f x y dxdy C dxdy
For a watershed region B
f x y dxdy
Module 7

9000
5000 5000
9000 9000
5000 5000 5000
90002
5000
etermine P[X Y]
= 1- P[X Y] = 1 - ( , )
= 1 - = 1 - [ 5000]
= 1- C 5000
2
=
≥
≤
−
 
− 
 
∫ ∫
∫ ∫ ∫
y
y
D
f x y dxdy
Cdxdy y dy
y
y
90002 2
5000
(9000) (5000)
1- C 45000000 25000000
2 2
17=
25
 
− − + 
 
Module 7
Example Problem 1 Contd…

2
2 2
2 2 2 2
2
1
F(x,y) = ( , ) =
(5000)
4000
= - dx
(5000) (5000)
4000 (5000) 5000 4000
= -
(5000) (5000) (5000) (5000)
4
= -
(5000) 5 (50
α α α α
α
− − − −
−
 
 
 
×
− +
×
∫ ∫ ∫ ∫
∫
y yx x
x
f x y dydx dydx
y
yx y
x
xy x 4
(Ans.)
500) (5000)
− +
y
2
F(10000,9000)
10000 9000 4(10000) 9000 4
= -
5(5000) 5 (5000) (5000)
= 3.6 - 1.6 -1.8 +0.8
= 1.0
×
− +
×
Module 7
Calculate F(10000,9000):

2
12 1 2 3 2
2
0 0 0 0
f(x,y) x +xy 0 x 1 & 0 y 2
= 0 elsewhere
Find, P[x+y 1], verify f(x,y)dxdy = 1
Solution :
x x yxy(x )dxdy= dy
3 3 6
α α
α α− −
= ≤ ≤ ≤ ≤
≥
 
+ + 
 
∫ ∫
∫∫ ∫
22 2
0 0
1 y 1 y 2 4
= dy= =
3 6 3 12 3 12
= 1 [verified]
    
+ + +    
    
∫
Module 7
Example Problem 2

1 1
2
0 0
11 2
2
0 0
1 2
2
0
1 2 3 2 3
0
[ 1]
1 [ 1]
= 1-
3
=1- dx
6
(1 2 )
= 1- (1 ) dx
6
6 6 2 )
= 1- dx
6
x
x
P X Y
P X Y
xy
x dydx
y x
x y
x x x
x x
x x x x x
−
−
+ ≥
= − + ≤
 
+ 
 
 
+ 
 
 − +
− + 
 
 − + − +
 
 
∫ ∫
∫
∫
∫
1
2 3
0
13 4 2
0
1
=1- (4 5 ) dx
6
1 4 5
=1-
6 3 4 2
1 4 5 1
1
6 3 4 2
1 7
1
6 12
65
72
x x x
x x x
− +
 
− + 
 
 
= − − + 
 
 
= −  
 
=
∫
Module 7

Marginal Distribution Function
1
1
Marginal Distribution of X is [ ] ( , ), i
Marginal Distribution of y is [ ] ( , ),
α
α
=
=
= ∀
= ∀
∑
∑
i i j
j
j i j
i
p x p x y
q y p x y j
α
α
α
α
−
−
∫
∫
Marginal distribution function of x is given by g(x) = ( , )
of y is given by h(y) = ( , )
[g(x) & h(y) are also pdf and should satisfy the conditions]
f x y dy
and f x y dx
Discrete variables
Continuous variables
Module 7

Marginal Distribution Function Contd…
α
α
α α
−
≤ ≤ = ≤ ≤ − ≤ ≤
 
 
 
= ×
∫ ∫
∫
[ ] P[c , ]
= ( , )
= ( )
[ ( , ) ( ) ( ) ......stochastically independent
( ) is original distribution of the 1-D ran
d
c
d
c
P c X d X d y
f x y dy dx
g x dx
f x y g x h y
g x
α−
∫
dom variable X
Remember C.D.F = F(x) = ( )
( ) is same as f(x), i.e. orginal density function of x.
x
g x dx
g x
Module 7

= − − ≤ ≤ ≤ ≤
∴ ≤ ≤ − −
−
∫∫
,
2 2
0 0
2
Function P ( , ) (5 ( ) ) 0 x 2 and 0 y 2 can serve as a
2
bivariate contnuous probablity density function.
i) Find the value of C
Probability (x 2 & y 2) = (5 ( ) ) dsdt=1
2
5
4
x y
y
x y C x for
S
C t
s
C s [ ]
 
− = ⇒ − − = 
 
∫ ∫
22 2
0 00
1 10 1 2 1ts dt C t dt
[ ]− =⇒ − =
− = ⇒ =
∫
2
2 2
0
0
9 2 1 [9 ] 1
1[18 4] 1
14
C t dt C t t
C C
Module 7
Example Problem

= ≤ ≤
− −
 
− − − − 
 
∫∫
∫
,
0 0
2 2 2
0
P ( , ) ( , )
= (5 ( ) )/14 dsdt
2
1 1
= [5 ] = 5
14 4 14 4 2
x y
yx
x
x y prob X x Y y
S
t
y xy x y
y ty dt xy  
− − 
 
1 1 1
= 5 = 0.304
14 4 2
≤ ≤ii) probability of x 1 & y 1Find
iii) "marginal desities" P (x) & P (Y)x yFind
=
− −
 
− − 
 
− − −
∫
∫
2
,
0
2
0
22
0
P (y) ( , )
1
= (5 )
214
1
= 5
14 2 2
1 1
= [10 2]= (8 )
14 14
y X YP t y dt
y t dt
yt t
t
y y
− −
 
− − 
 
− −
−
∫ ∫
2 2
,
0 0
22
0
P (x) ( , ) = (5 )/14
2
1
= 5
14 4
1
= [10 1 2 ]
14
1
= (9 2 )
14
x X Y
sP x s ds x ds
s
s xs
x
x
Module 7

α α
α
−
∴ = ≤
   − −
  
   
∫ ∫
, ,
, ,
0 0
2 2
0
iv) cumulative marginal distributions P (x, ) & Px ( ,Y)
9 2
P (x, ) ( )= ( ) = ( )
14
9 9
= =
14 14
x y x y
x x
x y X
x
Find
x
prob X x P t dt dx
x x x x
α
 − −
= ≤  
 
−
=
 −
= 
 
∫ ∫
2
,
0 0
2
2
8 8 / 2
P ( , ) ( )= ( ) = ( ) =
14 14
9
Check putting x=2 , 1
14
16
& y=1, 1
28
y y
x y y
y y y
y prob Y y P s ds ds
x x
y y
− −
−
− −
−
/ ,
/ ,
v) "conditional densities"
(5 )
2P (x/y) = Px (x,y)/P (y) =
(8 )
(5 )
2P (y/x) = Px (x,y)/P (x) =
(9 2 )
x y x y y
x y x y x
Find
y x
y
y x
and
x
Module 7

Conditional Density Functions
Let (X,Y) is a 2-D random vector, with joint density function of f(x,y). Let g(x)
and h(y) be the marginal density functions of X and Y respectively.
The CDF of X, given Y=y is defined as,
and the CDF of Y, given X=x is defined as,
( , )
( | ) but h(y)>0
( )
f x y
g x y
h y
=
( , )
( | ) but g(x)>0
( )
f x y
h y x
g x
=
α α α
α α α
α
α
− − −
−
≥
= =
 
∴ = 
 
∫ ∫ ∫
∫
( | ) 0 as f(x,y) and h(y) are (+ve)
( , ) 1 1
( | ) 1 = ( , ) = ( ) = 1
( ) ( ) ( )
( , ) ( )
g x y
f x y
g x y dx dx f x y dx h y
h y h y h y
f x y dx h y
Module 7

Independence of random variables
( | ) ( ), when X & Y are statistically independent variables
( , )
g(x|y) ( )
( )
f(x,y) = g(x) x h(y), then x, y are independent
=
= =
g x y g x
f x y
g x
h y
If
( )
( )
( )
0 0
( , ) ; x>0; y>0
g(x) = =
1
= e , x>0 & h(y) = e , y>0
( , ) ( ) h(y) x &y are independent
αα
− +
− +
− +
− −
=
 
∴  
− 
∴ = × ∴
∫
x y
x y
x y
x y
f x y e
e
e dy
f x y g x
For e.g.,
Module 7

Functions of random variables
− +2
1:
: variable, p(x) = f(x) = , x=2,3,4,5, y= x 7 12
Example
c
x discrete x
x
7takes two values 0 & 2, p(y=0) = p(x=3) + p(x=4)=
3 4 12
7p(y=2) = p(x=2) + p(x=5) =
2 5 10
+ =
+ =
c c cy
c c c
x 2 3 4 5
y 2 0 0 2
p(x) c/2 c/3 c/4 c/5
 Simultaneous occurrence  Joint density function
 Distribution of one variable irrespective of the value of the other
variables  Marginal density function
 Distribution of one variable conditioned on the other variable
 Conditioned distribution
Module 7

Functions of Random variables Contd…..
2
0
2 :
functions of RVs:
For continuous curve
f(x) = e ; x>0
y = 2x+1
1
P[y 5] = P [ X ]
2
= P[X 2]
= 1- P[X 2]
= 1-
−
−
−
≥ ≥
≥
≤
∫
x
x
Example
for
y
e dx
2
0
2 0
2
2
2 2
= 1-
= 1- [ ]
1 1
= (Ans.)
(Ans.)
1
x
x
x
e dx
e e
e
Alternatively
e
e dx e
αα
−
− −
−
−
− −
−
− −
=
 
= = 
− 
∫
∫
Module 7

2
i) g(x|y) = f(x,y) / h(y)
1 14 (10 2 )
= (5 ) = , 0 x 2, 0 y 2
214 (8 ) 2(8 )
Conditional CDF:
(10 )
( | ) [ | ]= ( | ) =
2(8 )α−
− −
− − × ≤ ≤ ≤ ≤
− −
− −
= ≤ =
−∫
x
y xy x
y y
x xy x
g x y P X x Y y g x y dx
y
≤ =
− − − −
=
−
2
3, [ 1, ]
2
3 3(10(1) (1)( ) 1 ) 10 1 152 23= a[1| ] = =
2 3 13 262(8 ) 2( )
2 2
Now P X Y
≤ ≤ ≤ ≤ ≤
− −
−
∫ ∫
∫ ∫
1 1
0 0
1 1
ii) Now, g(x|y 1] 0 x 2; 0 y 2
1( , ) (5 )
14 2
= =
( ) (8 )/14
o o
yf x y dy x dy
h y dy y dy
Module 7

 − − − −  − −    
 − − −    
−
∫
∫
1 12
0 0
1 12
0
(5 ) 5 152 4 4= = =
18(8 ) 8 2
2
(19 4 )
=
30
o
y yx dy y xy x
yy dy y
x
[ ]
2
0
2
1x=
2
19 4 19
) | 1 =
30 30 15
,
18 21 1| 1 | 1 =
2 2 15 15
18 2 17
=
2 15 15(4) 30
 − 
≤= −  
   
   ≤ ≤= < −
   
− =
×
∫
x
for
x x x
iii F x y dx
Now
x x
P X Y F Y
x
Module 7

Properties of RV
 Population: A complete assemblage of all possible RV
 Sample: A subset of population
 Realization: A time series of the RVs actually realized. They are continuous.
All samples are not realizations.
 Observation: A particular value of the RV in the realization
Note: ARMA is a synthetically generated realization
Module 7

Lecture 2: Statistical Moments
Module 7

Introduction
 Measures of Central Tendency:
Mean,
 Arithmetic average (for sample),
Mode,
Median
 Measures of Spread or Distribution:
Range [(xmax-xmin)],
Relative Range [=(range/mean)],
Variance,
Standard deviation,
Coefficient of variation
Module 7

α
α
α
α
α
α
µ
µ
µ
µ
+
−
+
−
+
−
=
= =
=
=
∫
∫
∫
0
0 0
0
0
1
n moment of the area ( )
moment of the area ( ) 1
Expected value of x is defined as:E(x)
= first moment about the origin
= ( ) ( )
mo
th n
n
th
th
x f x dx
o x f x dx
x f x dx simply
n
α
α
µ
+
−
= −∫ment about the expected value ( ) ( ) ...for population (not sample)n
n x u f x dx
Moments of Distribution
dx
x
f(x)
+α-α
x
Module 7

Moments of Distribution Contd…
[ ] [ ]
[ ] [ ] [ ]1 2 1 2
Therefore,
i) E(x) = ( )
ii) Expected value of a function of RV
= E[g(x)]
= ( ) ( )
iii) E(C) = C
iv) E C g(x) = C E g(x)
v) E g (x) g (x) E g (x) E g (x)
... "Expe
x f x dx
g x f x dx
α
α
α
α
µ
+
−
+
−
=
± = ±
∫
∫
ctations" is an operator for population [- , + ]α α
dx
x
f(x)
+α-α
x
f(x)
μ = E(x)
Module 7

2
n
2
V is wind velocity with a pdf f(v) = 1/10; 0 v 10. The pressure at point is
given by = 0.003 v . Find the expected value of pressure.
Sol :
1
f(v) = ; 0 v 10; = 0.003 v
10
E( ) =
≤ ≤ ω
ω
≤ ≤ ω
ω
2
2
?
E( ) = g( )d , given = 0.003 v or d = 0.006 vdv
1given, f(v) = & = 0.003 v g( )will be monotonically increasing function.
10
dv 1 0.0274
Therefore, g( ) = f(v)
d 10
=
+α
−α
ω ω ω ω ω ω
ω ∴ ω
ω =×
ω ω
∫
1
22
0.00274( ) as = 0.003 v
−
ω ω
Module 7
Example Problem

1
2
1 1
2 2
0.3
30.3 21
2
0
0
1
or v = 18.257 or 18.257 ( )
0.003 2
1 1
or g(w) = 18.257 ( ) = 0.9129 (w)
10 2
Now, E(w) = 0.9129 (w) = 0.9129
3
2
2
= 0.9129 [0.3]
3
−
− −
−
= = × ×
× × ×
 
   × ×
    
 
× ×
∫
dvw w w
dw
w
w
w dw
3
2
0 10
0 18.257 10
0 w 0.3
= 0.1 (Ans.)
≤ ≤ 
 ≤ ≤
 ≤ ≤
 
v
or w
or
Module 7

Probability weighted moments
Given a random variable X with a cumulative distribution function F, the probability
weighted moments are defined to be:
Two special cases are:
For an ordered sample x1:n <= x2:n <= ... <= xn:n, unbiased estimators of
n
r j:n
j
n
r j:n
j
(n j )(n j )...(n j )
a x
n (n )(n )...(n r )
( j )( j )...( j r )
b x
n (n )(n )...(n r )
=
=
− − − − +
=
− − −
− − −
=
− − −
∑
∑
1
1
1 1 1
1 2
1 1 2
1 2
p r s
M( p,r ,s) E[ X { F( x )} { F( x )} ]= −1
r
r
M( , ,r ) E[ X{ F( x )} ]α= = −1 0 1
r
r
M( ,r, ) E[ X{ F( x)} ]β= =1 0
Module 7

L-moments
 L-moments differ from conventional moments in that they are calculated using
linear combinations of the ordered data.
Population L-moments
For a random variable X, the rth population L-moment is
where Xk:n denotes the kth order statistic (kth smallest value) in
an independent sample of size n from the distribution of X and denotes expected
value.
( )
rr
k
r r k:rk
k
r ( ) EXλ
−−
−
−
=
= −∑
11
1
0
1
Module 7

L-moments Contd…
In particular, the first four population L-moments are
The first two of these L-moments have conventional names:
( )
( )
( )
: :
: : :
: : : :
EX
EX EX /
EX EX EX /
EX EX EX EX /
λ
λ
λ
λ
=
= −
= − +
= − + −
1
2 2 2 1 2
3 3 3 2 3 1 3
4 4 4 3 4 2 4 1 4
2
2 3
3 3 4
mean, L mean or L location,
L scale
λ
λ
= − −
= −
1
2
Module 7

L-moments Contd…
Sample L-moments
Direct estimators for the first four L-moments in a finite sample of n observations are
where xi is the ith order statistic and is a binomial coefficient.
( )
( ) ( ) ( ){ }
( ) ( ) ( )( ) ( ){ }
( ) ( ) ( )( ) ( )( ) ( ){ }
n n
i
i
n i n in
i
i
n i i n i n in
i
i
n i i n i i n i n in
i
i
x
x
x
x
−
=
− − −
=
− − − − −
=
− − − − − − −
=
=
−
= − +
= − + −
∑
∑
∑
∑
1
1 1
1
1 1
1 2 1 1
1
1 1 1
3 3 2 1 1 2
1
1 1 1 1
3 4 3 2 1 1 2 3
1
1
2
1
2
3
1
3 3
4




Module 7

L-moments Contd…
L-moment ratios
A set of L-moment ratios, or scaled L-moments, is defined by
The most useful of these are τ3 ,called the L-skewness, and τ4, the L-kurtosis.
r r
/ ,r , ...τ λ λ= =2
3 4
 There are two common ways that L-moments are used:
 As summary statistics for data.
 To derive estimates for the parameters of probability distributions.
Module 7

Lecture 3: Measures of central tendency
and dispersion
Module 7

Measures of Central Tendency
Module 7
1
1
E(x) = = ( ) , for parameter estimate
Arithmatic Mean ( ) = .......for sample estimate
Most frequently occuring value
1) 0 & 0
2) Value of x associated
Mean:
Mode:
with ma
α
α
µ
+
−
=
=
∂ ∂
= <
∂ ∂
∫
∑
n
i
i
n
i
x f x dx
x x
f f
x x
f ( )
The pdf takes the maximum value at the mode.
x x ix

Measures of Central Tendency Contd…
Module 7
It divides the area under the pdf curve into two halves.
i.e Area is 50%
= ( ) = P[X ] = 0.5
:
[ It is the observation such that
half the values lie on either side of it ]
Median:
µ
µ µ
−
≤∫
med
med x med
d
n
P x dx
Def
(b)
Median
Median
(c)
Median
(a)

[ ]
[ ] [ ]
[ ] [ ] [ ]
α α
α α
α α α α
α α α α
− −
− − − −
+ = +
+ +
∴ + = +
∫ ∫
∫ ∫ ∫ ∫
L , E ( ) ( , )
= ( , ) y ( , ) =E E
E E E
et X Y x y f x y dxdy
x f x y dxdy f x y dxdy X Y
X Y X Y
[ ]
α α
α α
α
α
− −
−
= ∫ ∫
∫
(X, Y) an independent RVs with a joint pdf of f(x,y)
E , ( , )
For independent variable f(x,y) = g(x) ( ) ( x and y are indpendent)
= (
X Y xy f x y dxdy
x h y because
xy g x [ ] [ ]
[ ] [ ] [ ]
α α α
α α α− − −
• +
∴ + = ×
∫ ∫ ∫) ( ) = ( ) y h( ) = E E
E E E (If x and Y are independent)
h y dxdy x g x dx y dy X Y
X Y X Y
Module 7
Measures of Central Tendency Contd…

Measure of “Spread” or “Dispersion”
Range : (Max - min) value
α
α
µ
µ
σ
−
−∫
2
2
2
Most important measure of dispersion
= Second moment about
Variance:
the mean
= ( ) ( )
=
x f x dx
= E(X)
= Expected value of x
Module 7

Measure of “Spread” or “Dispersion” Contd…
( )
[ ]
[ ]
[ ]
2 2
2
2 2
2 2
2 2
2 2 2
2 2
2
= ( ) ( )
= E
= E 2
= E 2E E
= E 2 E
= E 2
= E
= E E( )
x f x dx
X
X X
X X
X X
X
X
X X
α
α
σ µ
µ
µ µ
µ µ
µ µ
µ µ
µ
−
−
 −
 
 − + 
   − +   
  − + 
  − + 
  − 
  − 
∫
2 2
σ=
Module 7
( )
( )
=
−
=
−
∑
2
2 1
2
2
Sample Estimate, s
1
= + positive square root of variance
Co-effcient of Variation:
=
CoV = Cv =
i) Var(C) = 0
ii) Var(Cx) = C var(x)
iii) Var(a + bx
Standard Dev
) =
iation:
b
n
i
i
x x
n
S
x
σ σ
σ
µ
2
var( )x

Lecture 5: Commonly used distributions in
hydrology
Module 7

Normal Distribution
µ
σσ
α α
α µ α
σ
 − −=   
∏    
≤ ≤
≤ ≤
>
∗
∗
2
1 1( ) exp
22
0
x
f x
Also, called Saminion Distributon; Bell shaped Distribution
Most popular dis
- x +
- +
Normal Distribution:
µ σ µ σ µ σ∗ 2 2 2
( , ), ( , )N N
tribution in any field
has two-parameters & denoted as X
x= μ
x
f(x)
+α-α
µ
σ
−
= tan var
x
z where z is s dard normal iate
Module 7

Normal Distribution Contd….
µ
α
κ
µ σ
→ ± →
=
→
2
,
0
)
s
1) Symmetirc about x =
2) As x f(x) 0 asymptotically
3) C , as symmetric distrbution
4) = 3
5) If x Normal N( , and y = ax + b linear function
Properties of Normal Distribution:
µ σ+
2 2
)bthen y follows a normal dist. with N(a , a
Module 7

Central Limit Theorem
µ
σ
= + + + +
=
=
1 1 3 n
i i i
2
i i
Here, Sn X X X X
n may not be very large, may be 6 or 7
For most hydrological applications, under some general conditions,
it is shown that if X is an independent with E(X )
and var(X ) ,
µ σ
= =
= + + + +
∑ ∑
1 1 3 n
n n
2
i i
i 1 i 1
then,
Sn X X X X
approches a normal distribution with E(Sn) = and var(Sn) =
i
2
i
2
If Sn is the sum of n i.i.d random variables Xi each having a mean µ and
variance then,
Sn approaches a normal distribution with mean, n and variance n as n
approaches to infinity.
σ
µ σ
Module 7

Uniform Distribution
( )
α α α
β
α
α β
β α
β α β α
α α
α β
β α β α β α
≤ ≤
−
   
= = =   
− −   
 −
−= ≤ ≤ 
− − − 
∫ ∫
∫
1
( )
1
( ) ( )
( )
xx x
f x
x
f x f X dx dx
x x
xf X dx
x
Probability is uniformly distributed
= for x
Expected value of x = E(x) =
β
α
β α
β α β α β α
β α β α β α
β α
 
= − 
− − − 
+ − +
−
2 2 2
2( ) 2( ) 2( )
( )( ) ( )
2( ) 2
x
=
= =
Module 7

Uniform Distribution Contd….
[ ]
( )
( )
( )
( )
( )
( )
( )
( )
22 2
3 32
2
3 3 2
2
3 3 2 2
3 3 2 2 2 3 3 2 2
3 3 2 2
( ) ( )
( )
3
,
( )
3 2
4 3( 2 )( )
12
4 4 3 6 6 3 3 6 3
12
3 3
12
E x E x
x
E x dx
Now
β
α
σ
β α
β α β α
β α β α
σ
β α
β α α αβ β β α
β α
β α α β αβ αβ β α α β β α
β α
α β α β αβ
β α
∴ = −
 −
 ∴ = =
− − 
 
− + 
= −  −  
− − + + −
=
−
− − − − − + + +
=
−
− + + −
=
−
∫
Module 7

Uniform Distribution Contd….
( )
( )
( )( )
( )
( )
( )
( ) ( )
3 3
2 2
2 2
2 2
2 2
2
3 ( )
12
3 ( )
12
3
12
2
12
12 12
s
α β αβ α β
β α
α β α αβ β αβ α β
β α
α β α αβ β αβ
β α
α αβ β
α β α β
− + + −
=
−
− − + + + −
=
−
 − − − − + =
−
 − − + =
− − −
= = =
−
Module 7

Exponential Distribution
λ
λ
λ
α α
λ
α
λ
α
λ λ
λ λ
λ
λ
λ
λ
λ
λ λ
−
−
−
−
−
− −
=
= =
= −
= =
=
 
= − 
− − 
∫ ∫
∫ ∫
∫
∫
0 0
0 0
0
0
( ) x>0; >0
If you set ( ) ( )
1 x>0 & >0
( ) ( )
1.
x
x x
x
x
x
x
x x
f x e
F x f x dx e dx
e
E x xf x dx x e dx
xe dx
xe e
dx
αλ
λ
α
λ λ
λ
λ
λ
λ
λ
λ
−
−
+
 
=− − 
 
 
=− + 
 
 
=− + − − 
 
= =
∴ =
0
0
2
1
1
0 0 0
1 1finally & ;
1variance
x
x
x x
e
xe
x
e e
x
Module 7

Gamma Distribution
( )
η η λ
α
λ
λ η η η
η
η η η
η η η η
η η
− −
− −
Γ=
Γ
Γ = −
Γ + = Γ
Γ =
Γ =Γ = Γ =Π
∫
1
1
0
( ) ; x, , >0 ( ) Gamma function of
( )
1) ( ) 1 ! f
Propertie
or =1,2, ....
2) ( 1) ( ) for >0
3) ( ) for >0
14) (1) (2) 1 )
2
:
; (
s
x
n t
x e
f x
t e dt
Module 7

Gamma Distribution Contd….
 When data are positively skewed we have to use Gamma distribution.
η =0.5
λ = 1
η = 1
λ = 1
η = 3
λ = 4
η = Scale parameter
λ = Shape parameter
η γ
λ η
ησ
λ
= = =
=2
2
2( ) ;
x
E x Coefficient of Skewness
Module 7

Log-normal Distribution
x
 
=  + 
=+ =
2
2
2 2
We say x is log-normally distributed, if y = ln X is normally
1
( 1)
ln(1 )
distributed.
v
x
y v
xy Cv is Coefficient of Variation
Cz
S
S C Cv
x
µ
α
σσ
µ
σσ
=
= + + +
  −
− ∴ = = = ≤ ≤   ∏   
  −
− = ∴ = =    ∏   


1 2 n
1 2 n
2
y y
yy
2
y
yy
If x x x x or
ln(x) ln(x ) ln(x ) ln(x )
y1 1f(y) exp y ln(x) or x e [0 y ]
22
ln(x)dy 1 dy 1 1Now, f(x) f
Here
(y) exp
, post
2dx x
ive
dx 2 .x.
skewn = + 2
s v v sess r 3C C , as Cv increses, r also increses.
Module 7

Extreme Value Distribution
Distribution depends on
 Number of random variables considering the point set of distribution
 Parent distribution of the RVs (X1, X2, ……Xn)
[ ]
[ ]
[ ]
= ≤
= ≤
= ≤ = ≤
≤
If y is the max. value (if it also a RV)
( )
We know ( ) , f(y) corresponding pdf
( ) [ ]
(becomes max. value y)
i
y
x i
y i
F y P Y y
F x P X x
F y P Y y P All X y
Module 7

Extreme Value Distribution Contd…
[ ] [ ] [ ] [ ]
[ ]
→
∴ = ≤ × ≤ × ≤ × ≤
= × × × =
= =
1 2
1 2
1 2
1 2 3
(i.i.d independent and identically distributed)
Let X ,X ,..................X be i.i.d
( ) .........
( ) ( ) ........ ( ) ( )
( ) ( ) ..
n
n
y n
n
X X X X
X X
F y P X y P X y P X y P X y
F y F y F y F y
F y F y = ..... ( )nXF y
[ ] [ ]
[ ]
− −
∂ ∂
 =  ∂ ∂
∂
 = ∂
∴ =
 

1 1
( )
( ) is pdf of y = ( )
n ( ) ( ) n ( ) ( )
( ) ( ) ( )
ny
y y
n n
X y X x
n
y x x
F y
f y F y
y y
F y F y F y f y
y
f y n F y f y
Module 7

Lecture 6: Goodness of fit tests
Module 7

 The goodness of fit tests measure the compatibility of a random sample
with a theoretical probability distribution function. In other words, these tests
show how well the distribution one selected fits to the data.
 Kolmogorov-Smirnov Test
This test is used to decide if a sample comes from a hypothesized
continuous distribution. It is based on the empirical cumulative distribution
function (ECDF). Assume that we have a random sample x1, ... , xn from
some distribution with CDF F(x). The empirical CDF is denoted by
( )n
F ( x ) Number of observations x
n
≤
1
Goodness of Fit
Module 7

The Kolmogorov-Smirnov statistic (K) is based on the largest vertical difference
between the theoretical and the empirical cumulative distribution function:
Hypothesis Testing
The null and the alternative hypotheses are:
H0: the data follow the specified distribution;
HA: the data do not follow the specified distribution.
The hypothesis regarding the distributional form is rejected at the chosen
significance level (α) if the test statistic, K, is greater than the critical value. The
fixed values of (0.01, 0.05 etc.) are generally used to evaluate the null hypothesis
(H0) at various significance levels. A value of 0.05 is typically used for most
applications.
( ) ( )i i
i n
i i
K F x , F x
n n
max≤ ≤
− 
= − − 
 1
1
Module 7

 Anderson-Darling Test
The Anderson-Darling procedure is a general test to compare the fit of
an observed cumulative distribution function to an expected cumulative
distribution function. This test gives more weight to the tails than the
Kolmogorov-Smirnov test.
The Anderson-Darling statistic (A2) is defined as:
( ) ( )( )( )
n
i n i
i
A n i . ln F( x ) ln F x
n
− +
=
 =− − − + − ∑2
1
1
1
2 1 1
Goodness of Fit Contd…
Module 7

 Chi-Squared Test
The Chi-Squared test is used to determine if a sample comes from a
population with a specific distribution. This test is applied to binned data, so
the value of the test statistic depends on how the data is binned. Please note
that this test is available for continuous sample data only.
The Chi-Squared statistic is defined as
where Oi is the observed frequency for bin i, and Ei is the expected frequency for
bin i calculated by, Ei =F(x2)-F(x1), where F is the CDF of the probability
distribution being tested and x1 and x2 are the limits for bin i.
( )n
i i
i
i
O E
,
E
χ
=
−
= ∑
2
2
1
Module 7

 The relative goodness of a model fit may be checked using
 RMSE (Root Mean Square Error),
 AIC (Akaike Information Criterion) and
 BIC (Bayesian Information Criterion).
RMSE
The root mean square error is defined as the square root of mean sum of
square of difference between empirical distribution and theoretical
distribution, i.e. MSERMSE =
Module 7

AIC
The AIC is used to identify the most appropriate probability distribution.
It includes:
(1) the lack of fit of the model and
(2) the unreliability of the model due to the number of model parameters
It can be expressed as:
AIC = -2(log(maximum likelihood for model)) + 2(no of fitted parameters)
or
AIC = N log(MSE) + 2(no of fitted parameters)
Module 7

BIC
The BIC (Schwarz, 1978) is another measurement of model selection which
can be expressed as :
BIC = N log(MSE) + [(no of fitted parameters)* log(N)
The best model is the one which has the minimum RMSE, AIC and BIC values
Module 7

Lecture 7: Statistical parameter estimation
Module 7

Parameter Estimation
Methods of Parameter Estimation
1) Method of Matching Points
2) Method of Moments
3) Maximum Likelihood method


θ
θ
θ
θ θ
→
→
=
Population Parameter
Sample Parameter
Unbiased estimation of parameter:An estimate of a parameter
is said to be unbiased estimate, if E( )
i
i
i
i i
Module 7

θ
θ
θ
θ
θ
−
−
∴ ≤ ∫ ∫0 0
In a data set, 75 % values are less than, It is assumed to follow the distribution,
f(x) = ; x > 0, estimate the parameter .
P[X 3] = 0.75 = F(x)= ( ) =
x
xx x
e
e
f x dx dx
θ
θ
θ
θ
θ
θ
θ
−
−
−
 
 
− ×
 
−∴
0
= = 1-e
1( )
1 - e = 0.75 or = -1.3863 or = 2.164
x
x
x
x
e
x
Module 7

θ θ θ θi j i j
1 2 n
Given a function f( ,......., ,x) and values ,.......,
we need to find x ,x ,..........x
Generate number of equations by taking moments of the distribution
Take, any distribution, like f(x
θ
θ
θ α α
− −
−
= ∏
⇒
2
1
2
2
(x )
1
22 2
2) (2 ) e - < x < +
Take the 1st moment Mean about the origin
θα
θ
α
θα
θ
α
µ θ
θ
− −
−
−
− −
−
−
= ∏
∏
∫ ∫
∫
2
1
2
2
2
1
2
2
( )
1
22 2
2
( )
1
22 2
2
( ) (2 )
(2 )
x
x
x fx dx x e dx
xe dx
E(X) = =
=
Module 7

2) Method of Moments Contd...
α
α
θ θ θ
θ
−
−
+
∏
∫
2
2
2 1 2
2
1
( )
2
y
y e dxE(X) =
θ
θ
θ θ
θ
−
+
1
2
2 1
2
( )
substituting, y = ( )
x =
dx =
x
y
dy
α α
α α
α
α
θ θ
θ
− −
− −
−
−
 
+ 
∏   
⇓
 
 
∏   
∫ ∫
∫
2 2
2
2 2
2 1
2
1
1
2
1
0
2
y y
y
e ydy e dy
e dy
E(X) =
E(X) = +
θ
θ
θ µ
∏
∏
∴ = =
1
1
1
2
2
( )E x
[As odd multiplier, h(-y) = -h(y)
= 0 + =
Module 7

2) Method of Moments Contd…
Second moment about the mean,
E (X- ) =
=
y = and we will get,
2
1
2
2
2 2 2
( )
1
22 2 2
2
1
2
( ) ( )
( ) (2 )
,
2
x
x f x dx
x e dx
Substituting
x
and
α
α
θα
θ
α
µ σ µ
µ θ
θ µ
µ
θ
−
− −
−
−
  = − 
− ∏
=
−
∫
∫
2
2σ θ=
Module 7

θ
θ θ θ
θ
   
   
   
=   
   
      
 
 
 
1 1
1 2 3
,
We have the following, ( ; ) ( ; ) ( ; )
i i
Sample
x
f x f x f x
x
1 3Product of ( ; ) ( ; ) is "likelihood " L
If L( ; ) ( ; ), then is the estimate preferred,
which maxmizes the likelihood function.
θ θ
θ θ θ
× × ≈
>
∗
∗

i i
f x f x
x f x
θ →( ; ) evaluated at x = xi if x pdf
Module 7

3) Maximum Likelihood method Contd…
{ }
1 2
1 2 ) 1
1 2
1 2 3
(
( ) ; 0 is a parameter
, , , Sample available;
L = ( , ) ( , ) ( , ) ( , )
=
= = (formulation of like
n
n
i
n i
x
n
n
x x x
x
x x xn n
f x e x
x x x
f x f x f x f x
e e e
e e
β
β β β
β
β
β β
β β β β
β β β
β β =
−
− − −
−
− + + +
= >
→
∴ × × × ×
× × ×
∑



lihood function)
Because ln(L) is an increasing function of L, it reaches maximum value
ln( )
at the same pt., as ln(L) does, 0
(When there is no other method feasible, this method is best one)
θ
∂
=
∂
⇒
∗
L
Module 7

Now set L maximum
1
1
1
ln( ) ln
ln( )
0 as we want to set the value of
ln( )
0
1
= Arithmetic average
Max. likelihood estimate
n
i
i
x
n
i
i
n
i
i
L n e
L
L n
x
n
x
x
x
β β
β
β
β β
β
=
=
=
∑
∴ = −
∂
∴ =
∂
∂
∴ =− =
∂
 
∴= =  
 
∑
∑
Module 7

( )
µ
σσ
σ µ
µ
µ σ µ σ µ σ
σσ
µ
σ
σ=
 − −=   
∏    
 
 −   −∴       ∏    
− −=− ∏ −  
 
 
2
2
1 2
2
2
1 1( ) exp
22
[Take, as parameter not S.D. and also]
1 1L = ( , , ) ( , , ) ( , , ) = exp
2
2
1ln( ) ln(2 )
22
i
n n
i
i
x
f x
x
f x f x f x
xn
or L
µ σ
µ σ
µ
µ
µ σ=
∂ ∂
= =
∂ ∂
−∂  
= = ∴ − = ∂  
∑
∑ ∑
1
2
1
ln( ) ln( )
0 set & ,Now
ln( )
0 ( ) 0
n
n
i
i
i
L L
or
xL
X
= 1
n
i
i
x
or x
n
µ =
=
∑
Module 7

( )
2
2
1 2
2
2
1 1( ) exp
22
1 1( , , ) ( , , ) ( , , ) exp
2
2
1ln( ) ln(2 )
22
µ
σσ
σ µ
µ
µ σ µ σ µ σ
σσ
µ
σ
σ
 − −=   
∏    
 
 −   −∴       ∏    
− −=− ∏ −  
 
[Take, as parameter not S.D. and also]
L = =  i
n n
i
x
f x
x
f x f x f x
xn
or L
1
2
1
ln( ) ln( )
0
ln( )
0
( ) 0
µ σ
µ σ
µ
µ σ
µ
=
=
∂ ∂
= =
∂ ∂
−∂  
= =  ∂  
∴ − =
∑
∑
∑
set & ,Now
n
i
n
i
i
i
L L
or
xL
X
Module 7

µ
µσ
µ σ σ
µ
σ σ
µ
σ
=
=
=
=
=
− −∂ ∏ 
==− − ∂ ∏ 
−
+ =
−
=
∑
∑
∑
∑
1
2
2 3
1
2
3
1
2
2 1
( ) ( 2)ln( ) 4 1
0
2 22
( )
0
( )
n
i
i
n
i
i
n
i
i
n
i
i
x
or x
n
xL n
and
xn
or
x
or
n
=
-
[But it is not the best method. It depends upon situation]
Module 7

 In hydrology, most of the phenomena are random in nature.
E.g. rainfall-runoff model
 Random variables involved in a hydrological process may be dependent or
independent.
 The ‘random variables’ X & Y are ‘stochastically independent’ if and only if
their ‘joint density’ is equal to the product of ‘marginal density functions’.
 Joint density function : Simultaneous occurrence
 Marginal density function : Distribution of one variable irrespective of the value
of the other variables
 Conditioned distribution: Distribution of one variable conditioned on the other
variable.
Module 7

 Measures of Central Tendency:
Mean
 Arithmetic average (for sample)
Mode
Median
 Measures of Spread or Dispersion:
Range [(xmax-xmin)]
Relative Range [=(range/mean)]
Variance
Module 7
Standard deviation,
Coefficient of variation

 Measures of Symmetry:
Coefficient of skewness,
Kurtosis
Correlation coefficient shows the degree of linear association between
two random variables
 Commonly used distributions in hydrology :
 Normal distribution
 Uniform distribution
 Exponential distribution
 Gamma distribution
 Log-normal distribution
 Extreme value distribution
Module 7

 Methods of statistical parameter estimation
Module 7

Hydrological Statistics
(contd.)
IIT Bombay
Module 8
5 Lectures

Objective of this module is to learn the concepts in
stochastic hydrology.
Module 8

 Frequency analysis
 Markov process
 Markov chain
 Reliability analysis
Module 8

Lecture 1: Frequency analysis
Module 8

Frequency Analysis
[ ]
[ ]
1 1
1
 
= + = + 
 
= +
v
T T v
We can express X = X+k S, where k can bepositiveor negativeno and S is SD.
k S
X X kC
X
In general, X X k C and T is any
Return Period [T
giv
]:
en time
Expected i
.
.
1
nterval between successive occurences of an event (in a long interval)
p= where p is probability of occurence of the event in any year and
T
T is in years
,
[ ]
1
≥ =P X
Where is magnitude of the T year event (flood)
T T
T
x
T
x
Module 8

Frequency Analysis Contd…
[ ]1= +
=
=
=
Our problem is, given T, how to find X
From , X
K function of distribution and return period
from data
K Frequency factors [Function of distribution and & return
T
T T v
T
v
T
?
X k C
C
[ ]1= +
+
−
= =
period, T]
Normal dist. X
=
(Std. normal variate)
T T v
T
T
T
X k C
X Sk
X X
or k Z
S
Module 8

Treatment of Zeros
[ ] [ ]
1
0 0
=
×
= ≠
∑
(B1, ....., B6) &
Let A be an event within that region.
stat
Mutually exclusive events collectively exhaustive
Total Probability th es:
P[A] =
are
mutually exclusive and co
eorem
lle
n
i i
i
P A | B P B
[ x & x
≠
ctively exhaustive]
Define, x=0 & x 0 as two mutually exclusive
and collectively exhausted events
B2 A B3
B4
B5B6
B1
0
P[x=0]
f(x|x≠0)

Treatment of Zeros Contd…
[ ] [ ] [ ] [ ] [ ]
[ ] [ ]
[ ]
[ ]
[ ]
≥ = ≥ =× =+ ≥ ≠ × ≠
= ≥ ≠ × ≠
=≥ ⇒
= ≤ ≠ ⇒
= ≠
 − =− × 
Using this notation
T
*
*
P X x P X x | X P X P X x | X P X
P X x | X P X
Notations :
i )F( x ) P X x without conditions CDF
ii )F ( x ) P X x | X CDF conditional
iii )K P X
F( x ) F ( x ) k
or F( x
0 0 0 0
0 0
0
0
1 1
= − +
where, F(x) includes all the zeros in the data (non conditional)
and F*(x) does not include zeros (only non-zero data)
*
) k kF ( x )1
Module 8

Example Problem 1
If in a sample there are 95% non-zero values, calculate X .10
[ ]
[ ]
σ
= ≤ ≠ ≠
= ≤
∴
−
= = × + =
*
*
*
F ( x ) . P X x | X
If F ( x )
F ( x ) . P Z z
x
or . or . .
10
2
0 895 0
0 895
1255 1255 15 10 28 33T
t
= given x 0
follows a normal distribution N (10,15 ) given
=
Get z value corresponding to 0.895 z = 1.255
x
x units
[ ] [ ]
Solution:
= = ≠ ≥ = =
= − = ⇒ = − + *
k . P X , P X x .
F( x ) . . . . . F ( x )
10
10 10
1
0 95 0 01
10
1 01 0 9 0 9 1 0 9 5 0 9 5
Module 8

Peak flow data are available for 75 yrs, 20 of the values are zero and the
remaining 55 values have a mean of 100 units and std. deviation of 35.1 units
and are log normally distributed. Estimate the probability of the peak
exceeding 125 units using frequency analysis.
Example Problem 2
[ ]
[ ]
( )
55
0733
75
125 1 1 125 1
125
125 0
1
= = ≠
> = = − = − +
=
≤ ≠
= +

T
T
P[X 0]
log-normally distributed
=
table, frequency table
For normal dist. it is K = S
X For l
*
* *
T
T
T V
k .
P X F( ) F( x ) k kF ( x )
F ( X ) F ( )
P X X
K
X K C og-normal dist.
Module 8

[ ]
125 100 1 0 351
351 100 0 351
0 712
0 21 0 79
1 0 733 0 733 0 79
0 846
1 0 846 0154
VC
But we are interested in F(x)
= +
= = = =
=
∴ ≥ = =
=− + ×
=
≥ =− =
T
T
T
T
or ( . K )
S
CoV . / .
X
or K .
P[ X X ] . .
. . .
.
P X X . . ( Ans )
Module 8

Lecture 2: Markov process and
Thomas Fierring model
Module 8

First Order, Stationary Markov Process
1 1 1( )t x t x iX Xµ ρ µ+ += + − +∈
2
Assuming Dependence to random component with
Stationary previous observation zero mean and variance
= onlyx
σ
µ µ
∈
1
Estimate of X
n
t
i
x
n
µ
=
= =∑
Module 8
Many hydrologic time series exhibit significant serial correlation.
Value of the random variable under consideration, X at one time period (say, t+1)
is correlated with the values of the random variable at earlier time periods.

[ ]{ } ( ) [ ]{ } ( )
( ) [ ] ( ) [ ]{ } ( )
( )
2 22 22
1 1 1 1 1
22 22
1 1 1 1 1
2 2 2 2 2 2 2 2
1 1
2 2 2
1
, ( )
2
= 0 ( )
, 1 ...
x t t x t x i t
t x x i t x x i t
x x x x
x
Now E X E X E x E X
E x x E X
So
σ µ ρ µ
ρ µ µ ρ µ µ
ρ σ µ σ µ ρ σ σ
σ σ ρ
+ + + +
+ + +
∈ ∈
∈
   = − = + − + ∈ −   
 = − + + ∈ − + + ∈ −  
+ + + − = +
= −
[ ] ( )1 1 1
1 1
...random component for first order stationary markov process
[Check:
0
so first order stationary]
X N(
t x t x i
x x x
x
E X E x
if
µ ρ µ
µ ρ µ µ ρ
µ
+ +
 = + − + ∈ 
= + − +
=
 2
2
, )
, (0, )
x x
then N
µ σ
σ∈
∈
First Order, Stationary Markov Process Contd…
Module 8

( )
( )
1
N
2
N
2
1
2
1 1 1
1) ,
2) , R (0,1)
0
such that R (0, )
= 1
( ) 1
3) If negative value comes, retain it for gen
t t t
x x
N N x
t x t x N x
Estimate and
Generate random numbers from the distribution N
as N
or R R
X x R
µ σ ρ
σ
σ
σ σ ρ
µ ρ µ σ ρ
∈
∈
∈
+
∈ −
= ∈
∈ = −
∴ = + − + −


erating the next value,
but discard it in application.
4) Generate a large no. of values and discard the first 50-100 values
to ensure that there is no effect of initial value.
Module 8
To generate values for Xt+1

For log-normally distributed data
 Preserves the mean, std. deviation and lag 1 correlation of logarithms of
flows (not the original data)
 This method may be applied on annual runoff, annual rainfall etc. (but not
on 10 day rainfall series)
{ }
( )2
1 1 1
ln
( ) 1
t t
t y t y N y y
Y X
Y y Rµ ρ µ σ ρ+
=
= + − + −
Module 8

First Order Markov Process
(Thomas Fierring Model)
 Consider monthly stream flow data for n years
m: no. of values (months)
Here, i= 1,2…, n and j= 1,2,…,m.
( )
( )
( )1 2
, 1 1 , 1 1
: Serial correlation between jth month and (j+1)th month
when j+1 = (12+1)=13 implies that i=2, j=1
1
j
j
i j j j ij j i j j j
j
X X t
ρ
σ
µ ρ µ σ ρ
σ
+
+ + + +
 
  ∴ = + − + −        
Module 8

Month SD rj
1 15.7 4.14 0.864
2 13.62 4.15 0.302
3 26.21 24.32 0.854
12 18.01 5.08 0.637
X
11 1
One assumption, X X
Xµ= =
correlation
with next
month
This preserves seasonal mean, standard deviation and lag1 correlation.
Non-stationarity in mean and std. deviation
Module 8
Thomas Fierring Model

Lecture 3: Markov chain
Module 8

• A Markov chain is a stochastic process having the property that the value of
the process Xt at time t, depends only on its value at time t-1, Xt-1 and not on
the sequence Xt-2, Xt-3, ……, X0 that the process passed through to arrive at
Xt-1
For 1st order Markov chain or Single step M.C.
Module 8
( )
( )
1 2 3 1 0
1
prob | , , ,...,
prob |
t j t i t k t q
t j t i
X a X a X a X a X a
X a X a
− − −
−
= = = = =
= = =
Markov Chains

Markov Chains Contd….
Xo Xt-2 Xt-1 Xt
Module 8
Diagrammatically, it may be represented as,
We will be able to write this as
Xt-1=ai
t-1
Xt=aj
t time period
State i transited to State j
1
is the probability that it goes in to state j, starting with array i here.
t
ij t j t i
p P X a X a
P
−
 = = = 

Transition probability
It is the probability that state ‘i’ will transit to state ‘j’
[i.e., transition probabilities remain the same across the time]
t-3 t-2 t-1
pt-2
ij
t
pt
ij
then, the series is called homogeneous Markov Chaint,t t
ij ijIf p p τ
τ+
= ∀
Module 8

 Here analysis is done only for:
Single step (1st order) homogeneous M. C.
 If t is month then pij will not be homogeneous (seasonal change)
 pij = transition probability for ‘i’ to ‘j’
i=1,2, ……, m and j=1,2,…….,m
where m is the no. of possible states that the process can occupy.
Module 8
11 12 13 1m
21 22 23 2m
ij
m1 m2 m3 mm
TPM, (Total probability matrix)
p p p p
p p p p
P = p =
p p p p
 
 
     
 
 


    

Probability stating
that i=1 go into
j=2
Sum of each row=1

m
j=1
Each row must add to 1
t,
1,
t t
ij ij
ij
If p p
p i
τ
τ+
= ∀
∴ = ∀∑
Such matrices whose individual rows add up to 1 are called the “stochastic
matrices”
Module 8
2
1
( 1), probability values that need to be estimated
Estimate from historical data
total no. of
ij
ij
mij
ij
j
m m m m
p
n
p
n
∧
∧
=
− = −
=
∑

Historic data
1
3
1 2 1 2
100
Time
period
1 2
States (random)
e.g. No. of times it went into state 1 out of these 50 times = 20,
No. of times it transited to state 2 = 20
No. of times it transited to state 3 = 15
Then
12
15
;
50
p∧
=
Deficit, non-deficit Two states
Drought, non-drought Two states
Module 8
11
20
;
50
p∧
=
13
15
.
50
p∧
=

• p(n)
j : probability that the process will be in state j after n time steps
i j - state
n
Time
interval
t-1
• p(0)
j : Initial probability of being in state j
( ) ( ) ( )
1 2 1
a sum vector, ,............, .....n n n n
m xm
p p p p=   
Probability of being in state 1 in time step n
Module 8

(1)
(1) ( )
11 12 13 1m
21 22 23 2m(0) (0) ( )
1 2
m1 m2 m3 mm
is given
create at t=1
from probability
p p p p
p p p p
= , ,...,
p p p p
→
=
 
 
 ×    
 
 


    

o
o
o
m
P
If p
p
p p
p p pProbability
vector at time 1
+ + +  
(0) (0) (0)
1 11 2 21 1
............ m m
p p p p p p
Probability that event
will start from state ‘2’
Probability of transition
from 2 to 1
Module 8

Probability that the state is 1 in period 1
=   
(1) (1) (1)
1 2
, ,............, m
p p p
Probability that state is
‘2 ‘ in time period ‘1’
Probability that state is ‘m
in time period ‘1’
(2) (1)
(0)
(0) 2
,
.
= . .
=
Simillarly
p p P
p P P
p P
=
( ) ( 0 )
Any time period n,
n n
p p P=
After time n, prob.
to be in particular
state ‘j’
Initial prob. vector
(TPM)
Module 8

(n m) (n)
(n m) (n)
p p ,after a large m then steady state probability
condition is achieved.
Once the steady state is reached,
p p p
So,p p.P
+
+
=
= =
=
Module 8
Example:
• A 2-state Markov Chain; for a sequence of wet and dry spells
i = 1 dry;
i = 2 wet
0.9 0.1
0.5 0.5
P
 
=  
 
d
w
wd
(i) P [day 1 is wet |day 0 is dry]
= P [Xt = 2|Xt-1 = 1]
= p12 = p(1)
2 = 0.1 (Ans)

Example Problem
(ii) P[day 2 is wet |day 0 is dry]
[ ]
[ ]
=
=
(2)
2
(2) (1)
(2)
= 0.9 0.1
0.86 0.14
p
p p P
p
day
wet
Because day 0 is dry
Dry wet
Probability that day 2 will be
wet
= =(2)
2
0.14p
Module 8
0.9 0.1
0.5 0.5
 
 
 

(iii) Prob [day 100 is wet |day 1 is dry]
Example Problem Contd….
(100)
2
. .,
0.9 0.1
0.5 0.5
i e P
P
 
=  
 
Module 8
Here, the fact that day 1 was dry, would not significantly affect the probability of
rain on day 100. So n can be assumed to be large and solve the problem based
on steady-state probabilities

2
4 2 2
8 4 4
16 8 8
To determine steady state,
0.86 0.14
or P . .
0.7 0.3
0.8376 0.1624
P . .
0.8120 0.1880
0.8334 0.1666
P . .
0.8320 0.1672
0.8333 0.1667
P . .
0.8333 0.1667
P P
or P P
or P P
or P P
 
= =  
 
 
= =  
 
 
= =  
 
 
= =  
 
∴
All the rows same
p=(0.8333; 0.1667)
dry wet
Module 8
Example Problem Contd….

Lecture 4: Data generation
Module 8

Data Generation
Module 8
 Data generation from a Markov chain requires only a knowledge of the initial
state and the transitional probability matrix ,
 To determine the state at time 2, a random number is selected between 0
and 1.

P
N
R from N(0,1).
n-1 n-1
N ij ij
j=1 j=1
If this R is between p and p , the state is 'n'∑ ∑

Module 8
Consider a Markov chain model for the amount of water in storage in a reservoir.
Let state 1 represent the nearly full condition, state 2 an intermediate condition and
state 3 the nearly empty condition. Assume that the transition probability matrix
is given by,
Note that it is not possible to pass directly from state 1 to state 3 or from state 3 to
state 1 without going through state 2. Over the long run, what fraction of the
time is the reservoir level in each of the states?
1 2 3
1 0.4 0.6 0
2 0.2 0.6 0.2
3 0 0.7 0.3
P
 
 =  
  
Example Problem 1

Module 8
Solution:
Let p1= 1  p2 = 3 and p3 = 6/7 . After scaling, the solution can be written as:
(p1, p2, p3) = (.2059, .6176, .1765).
Thus, over the long run, the reservoir is nearly full 20.59% of the time, nearly
empty 17.65% of the time and in the intermediate state 61.76% of the time.
( ) ( )1 2 3 1 2 3
1 2 1
1 2 3 2
2 3 3
p.P p
.4 .6 0
p p p .2 .6 .2 p p p
0 .7 .3
.4p .2p 0 p
.6p .6p .7p p
0 .2p .3p p
=
 
  = 
  
+ + =
+ + =
+ + =

Module 8
Example Problem 2
Assume that the reservoir of previous example is nearly full at t=0. Generate a
sequence of 10 possible reservoir levels corresponding to t=1, 2,…, 10.
Solution:
At t=0, state is 1.
Generate cumulative transition probability matrix (CTPM) where
* * *
1
*
,
0.4 1.0 0
0.2 0.8 1.0
0 0.7 1.0
k
ik ik ij
j
P P andP p
Here
P
=
= =  
 
 =  
  
∑
*
P

t
State at t RN State at t+1
Reservoir
level at t
0 1 0.48 2 nearly full
1 2 0.52 2 intermediate
2 2 0.74 2 Intermediate
10 3 nearly empty
Module 8

Lecture 5: Reliability analysis
Module 8

Reliability
 It is defined as the probability of non-failure, ps, at which the resistance of the
system exceeds the load;
where P() denotes the probability.
 The failure probability, pf , is the compliment of the reliability which can be
expressed as
)( RLPps
≤=
sf
pRLPp −=≥= 1)(
The resistance or strength (R) is the ability to accomplish the intended mission
satisfactorily without failure when subjected to loading of demands or external
stresses (L). Failure occurs when the resistance of the system is exceeded by the
load (floods, storms etc.)
Module 8

Measurements of Reliability
 Recurrence interval
T=1/(1-F), F= P(X<xT)
 No consideration for the interaction with the system resistance
 Safety Margin
 It is defined as the difference between the resistance (R) and the
anticipated load (L)
SM=R-L
 Safety Factor
 It is the ratio of resistance to load
SF=R/L
(Tung, 2004)
Module 8

Recurrence Interval
 Assume independence of occurrence of events and the hydraulic
structure design for an event of T-year return period.
 1/T is the probability of exceedance for the hydrologic event in any one
year.
 Failure probability over an n-year service period, pf, is
pf = 1-(1-1/T)n (using Binomial distribution)
or pf = 1-exp(-n/T) (using Poisson distribution)
 Types of problem:
(a) Given T, n, find pf
(b) Specify pf & T, find n
(c) Specify pf & n, find T
Module 8

Probabilistic Approaches to Reliability
 Statistical analysis of data of past failure records for similar systems
 Reliability analysis, which considers and combines the contribution of
each factor potentially influencing the failure with the steps as
(1) to identify and analyze the uncertainties of each contributing factor;
and
(2) to combine the uncertainties of the stochastic factors to determine
the overall reliability of the structure.
Module 8

Uncertainties in Hydraulic Engineering Design
Hydrologic uncertainty
(Inherent, parameter, or model uncertainties)
Hydraulic uncertainty
(Uncertainty in the design and analysis of hydraulic structures)
Structural uncertainty
(Failure from structural weaknesses)
Economic uncertainty
(Uncertainties in various cost items, inflation, project
life, and other intangible factors)

Techniques for Uncertainty Analysis
 Analytical Technique
 Fourier and Exponential Transforms
 Mellin Transform
 Approximate Technique
 First-Order Variance Estimation (FOVE) Method
 Rosenblueth’s Probabilistic Point Estimation (PE) Method
 Harr’s Probabilistic Point Estimation (PE) Method
 Reliability Analysis Methods
Module 8

Reliability Analysis Methods
1. Performance Function and Reliability Index
2. Direct Integration Method
3. Mean-Value First-Order Second-Moment (MFOSM) Method
4. Advanced First-Order Second-Moment (AFOSM) Method
a) First-order approximation of performance function at design point.
b) Algorithms of AFOSM for independent normal parameters.
c) Treatment of correlated normal random variables.
d) Treatment of non-normal random variables.
e) AFOSM reliability analysis for non-normal, correlated random variables.
5. Monte Carlo Simulation Methods

 To enable a quantitative analysis of the reliability of a structure, every failure
mode has to be cast in a mathematical form.
 A limit state function is given by:
W (z, x) = R (z, x) – S (z, x)
where,
z: Vector of design variables;
x: Vector of random input variables;
R: Resistance of the structure;
S: Load on the structure.
The value of the limit state function for given values of x and z denotes the
margin.

 The reliability index is defined as the ratio of the mean to the standard
deviation of the performance function W(z, x)
where μW and σW are the mean and standard deviation of the
performance function.
 The boundary that separates the safe set and failure set is the failure
surface, defined by the function W(z, x) =0, called the limit state
function.
ww
σµβ /=
1. Performance Function and Reliability Index Contd…
Module 8

System states defined by performance function
1. Performance Function and Reliability Index Contd…
Module 8

 The reliability can be computed in terms of the joint PDF of the load and
resistance as
where fR,L(r, l ) : joint PDF of random load, L, and resistance, R;
r, l : dummy arguments for the resistance and load, respectively;
(r1, r2), (l1, l2) : lower and upper bounds for the resistance and load, respectively.
( ) ( ) dldrlrfdrdllrfp
l
l
r
l
LR
r
r
r
l
LRs ∫ ∫∫ ∫ 





=





=
2
1
22
1 1
,, ,,
Module 8

 Here, the performance function W(z, x), defined on the basis of the load and
resistance functions, S(z, x) and R(z, x), are expanded in a Taylor series at a
selected reference point.
 The second and higher order terms in the series expansion are truncated,
resulting in an approximation that requires the first two statistical moments of
the random variables.
The simplification of Taylor series greatly enhances the practicality of the first
order methods because in many situations, it is difficult to find the PDF of the
variables while it is relatively simple to estimate the first two statistical moments.
Module 8

 Here, it mitigates the deficiencies associated with the MFOSM method, while
keeping the simplicity of the first–order approximation.
 The difference between the AFOSM and MFOSM methods is that the
expansion point in the first–order Taylor series expansion in the AFOSM
method is located on the failure surface defined by the limit state equation,
W(x) =0.
Module 8

5. Monte Carlo Simulation Methods
 It is a general purpose method to estimate the statistical properties of a
random variable that is related to a number of random variables which may or
may not be correlated.
 The values of stochastic parameters are generated according to their
distributional properties and are used to compute the value of performance
function
 Reliability of the structure can be estimated by computing the ratio of the
number of realizations with W ≥0 to the total number of simulated realizations.
 Disadvantage  computational intensiveness.
Module 8

Return period/recurrence interval: The reciprocal of annual exceedance
probability
 First order, Stationary Markov Process states that observed value of
a random event at any time t+1, is a function of a stationary component,
correlated component and a purely random component. Here, the random
component is assumed to be normally distributed with zero mean and
variance.
2
tσ
Module 8

 Reliability is defined as the probability of non-failure, ps, at which the
resistance of the system exceeds the load.
 Failure occurs when the resistance of the system is exceeded by the load
(floods, storms etc.)
 Measurements of reliability are recurrence interval, safety margin
and Safety factor
 Techniques for Uncertainty Analysis:
 Analytical technique
 Approximate technique
 Reliability analysis methods
Module 8

Reliability Analysis Methods:
5. Monte-Carlo Simulation method
Module 8

Module 9
5 Lectures
Hydrologic Simulation Models
IIT Bombay

Objectives of this module is to investigate on various
hydrologic simulation models and the steps in watershed
modeling along with applications and limitations of major
hydrologic models
Module 9

 Hydrologic simulation models
 Steps in watershed modeling
 Major hydrologic models
 HSPF(SWM)
 HEC
 MIKE
Module 9

Lecture 1: Introduction to hydrologic
simulation modelling
Module 9

Watershed Classification
Watershed (ha) Classification
50,000-2,00,000
10,000-50,000
1,000-10,000
100-1,000
10-100
Watershed
Sub-watershed
Milli- watershed
Micro-watershed
Mini-watershed
(Bedient et al., 2008)
Module 9

A hydrologic simulation model is composed of three basic elements, which are:
(1) Equations that govern the hydrologic processes,
(2) Maps that define the study area and
(3) Database tables that numerically describe the study area and model
parameters.
Hydrologic
Simulation Model
Equations Maps Database
Module 9
Hydrologic Simulation Model

 A hydrological simulation model can also be defined here as a
mathematical model aimed at synthesizing a (continuous) record of some
hydrological variable Y, for a period T, from available concurrent records
of other variables X, Z, ... .
 In contrast, a hydrological forecasting model is aimed at
synthesizing a record of a variable Y (or estimating some of its states) in
an interval ∆T, from available records of the same variable Y and/or other
variables X, Z, ... , in an immediately preceding period T.
Module 9
Hydrologic Simulation Model Contd…

 A hydrological simulation model can operate in a "forecasting mode" if
estimates of the records of the independent variables (predictors) X, Z, ..., for the
forecast interval ∆T are available through an independent forecast. Then the
simulation model, by simulating a record of the dependent variable, [Y] ∆T will in
fact produce its forecast.
In short, a hydrological simulation model works in
a forecasting mode whenever it uses forecasted
rather than observed records of the independent
variables.
Module 9
Hydrologic Simulation Model Contd…

http://www.crwr.utexas.edu/gis/gishydro06/WaterQuality/GIStoHSPF/GIStoHSPF.htm
A Typical Watershed Delineation Model
Module 9

Hydrologic
Model
Deterministic
Lumped
Steady
Flow
Unsteady
Flow
Event based Continuous
Distributed
Steady
Flow
Unsteady
Flow
Stochastic
Space
Independent
Time
Independent
Time
Correlated
Space
Dependent
Time
Independent
Time
Correlated
Classification of Hydrologic Models
Module 9

Spatial Scaling of Models
Module 9
Lumped
Parameters assigned to
each sub-basin
Fully-Distributed
Parameters assigned
to each grid cell
Semi-Distributed
Parameters assigned to each grid
cell, but cells with same
parameters are grouped
A1
A2
A3

1. Size
2. Shape
3. Physiography
4. Climate
5. Drainage
6. Land use
Parameters of Watershed
7. Vegetation
8. Geology and Soils
9. Hydrology
10. Hydrogeology
11. Socioeconomics
Module 9

Precipitation
Interception
Storage
Surface
Runoff
Groundwater
Storage
Channel
Processes
Interflow
Direct Runoff
Surface
Storage
Baseflow
Percolation
Infiltration
ET
ET
ET: evapo-transpiration
Module 9
Flowchart of simple watershed model (McCuen, 1989)

Diversity of the current generation of models
There exists a multitude of watershed models, and their diversity is so large
that one can easily find more than one watershed model for addressing any
practical problem.
Comprehensive Nature
Many of the models can be applied to a range of problems.
Reasonable modeling of physical phenomena
In many cases models mimic reasonably well the physics of the underlying
hydrologic processes in space and time.
Strengths of Watershed Models
Module 9

Distributed in Space and Time
Many models are distributed in space and time.
Multi-disciplinary nature
Several of the models attempt to integrate with hydrology :
a) Ecosystems and ecology,
b) Environmental components,
c) Biosystems,
d) Geochemistry,
e) Atmospheric sciences and
f) Coastal processes
 This reflects the increasing role of watershed models in tackling
environmental and ecosystems problems.
Module 9
Strengths of Watershed Models Contd…

The most ubiquitous deficiencies of the models are:
 Lack of user-friendliness,
 Large data requirements,
 Lack of quantitative measures of their reliability,
 Lack of clear statement of their limitations, and
 Lack of clear guidance as to the conditions for their applicability.
Also, some of the models cannot be embedded with social, political, and
environmental systems.
Deficiencies of Watershed Models
Although watershed models have become
increasingly more sophisticated, there is a long way
to go before they become “household” tools.
Module 9

Hydrologic Models
Model Type Example of Model
Lumped parameter Snyder or Clark UH
Distributed Kinematic wave
Event HEC-1, HEC-HMS, SWMM, SCS TR-20
Continuous Stanford Model, SWMM, HSPF, STORM
Physically based HEC-1, HEC-HMS, SWMM, HSPF
Stochastic Synthetic streamflows
Numerical Kinematic or dynamic wave models
Analytical Rational Method, Nash IUH
Module 9

Hydrologic Models Contd…
Models Application Areas
HEC-HMS Design of drainage systems, quantifying the
effect of land use change on flooding
National Weather Service (NWS) Flood forecasting.
Modular Modeling System (MMS) Water resources planning and management
works
University of British Columbia (UBC)
& WATFLOOD
Hydrologic simulation
Runoff-Routing model (RORB)
& WBN
Flood forecasting, drainage design, and
evaluating the effect of land use change
TOPMODEL & SHE Hydrologic analysis
HBV Flow forecasting
Module 9

A List of
Popular Hydrologic Models
Module 9

Model name/acronym Author(s)(year) Remarks
Stanford watershed Model
(SWM)/Hydrologic Simulation
Package-Fortran IV (HSPF)
Crawford and
Linsley (1966),
Bicknell et al.
(1993)
Continuous, dynamic event or
steady-state
simulator of hydrologic and
hydraulic and
water quality processes
Catchment Model (CM) Dawdy and
O’Donnell (1965)
Lumped, event-based runoff model
Tennessee Valley Authority (TVA)
Model
Tenn. Valley
Authority (1972)
Lumped, event-based runoff model
U.S. Department of Agriculture
Hydrograph Laboratory
(USDAHL) Model
Holtan and Lopez
(1971), Holtan et al.
(1974)
Event-based, process-oriented,
lumped hydrograph model
U.S. Geological Survey (USGS)
Model
Dawdy et al. (1970,
1978)
Process-oriented,
continuous/event-based
runoff model
Module 9
Popular Hydrologic Models

Utah State University (USU)
Model
Andrews et al. (1978) Process-oriented, event
/continuous streamflow
model
Purdue Model Huggins and Monke
(1970)
Process-oriented, physically
based, event runoff model
Antecedent Precipitation Index
(API)Model
Sittner et al. (1969) Lumped, river flow forecast
model
Hydrologic Engineering Center—
Hydrologic Modeling
System (HEC-HMS)
Feldman (1981), HEC
(1981, 2000)
Physically-based, semi-
distributed, event-based,
runoff model
Streamflow Synthesis and
Reservoir regulation (SSARR)
Model
Rockwood (1982)
U.S. Army Corps of
Engineers (1987),
Speers (1995)
Lumped, continuous
streamflow simulation model
Module 9
Popular Hydrologic Models Contd…

National Weather service-River
Forecast System (NWS-RFS)
Burnash et al.
(1973a,b),
Burnash (1975)
Lumped, continuous river
forecast system
University of British Columbia
(UBC) Model
Quick and Pipes (1977),
Quick (1995)
Process-oriented, lumped
parameter, continuous
simulation model
Tank Model Sugawara et al. (1974) ,
Sugawara (1995)
Process-oriented, semi-
distributed or lumped
continuous simulation model
Runoff Routing Model (RORB) Laurenson (1964),
Laurenson and Mein
(1993, 1995)
Lumped, event-based runoff
simulation model
Agricultural Runoff Model (ARM) Donigian et al. (1977) Process-oriented, lumped
runoff simulation model
Module 9
Popular Hydrologic Models Contd…

Storm Water Management
Model (SWMM)
Metcalf and Eddy et al.
(1971),
Huber and Dickinson
(1988),
Huber (1995)
Continuous, dynamic event
or steady-state
simulator of hydrologic and
hydraulic and
water quality processes
Areal Non-point Source
Watershed Environment
Response
Simulation (ANSWERS)
Beasley et al. (1977),
Bouraoui et al. (2002)
Event-based or continuous,
lumped parameter runoff
and sediment yield
simulation model
National Hydrology Research
Institute (NHRI)Model
Vandenberg (1989) Physically based, lumped
parameter, continuous
hydrologic simulation
model
Technical Report-20 (TR-20)
Model
Soil Conservation
Service (1965)
Event-based, process-
oriented, lumped
hydrograph model
U.S. Geological Survey (USGS)
Model
Dawdy et al. (1970,
1978)
Lumped parameter, event
based runoff simulation
model
Module 9

Physically Based Runoff
Production Model (TOPMODEL)
Beven and Kirkby
(1976, 1979), Beven
(1995)
Physically based,
distributed, continuous
model
Generalized River Modeling
Package—Systeme
Hydroloque Europeen (MIKE-
SHE)
Refsgaard and Storm
(1995)
Physically based,
hydrologic and hydraulic
simulation model
ARNO(Arno River )Model Todini (1988a,b, 1996) Semidistributed,
continuous rainfall-runoff
simulation model
Waterloo Flood System
(WATFLOOD)
Kouwen et al. (1993),
Kouwen (2000)
Process-oriented,
semidistributed continuous
flow simulation model
Topgraphic Kinematic
Approximation and Integration
(TOPIKAPI)Model
Todini (1995) Distributed, physically
based, continuous
rainfall-runoff simulation
model
Module 9

Soil-Vegetation-Atmosphere
Transfer (SVAT) Model
Ma et al. (1999),
Ma and Cheng (1998)
Macroscale, lumped
parameter, streamflow
simulation system
Systeme Hydrologique Europeen
Transport
(SHETRAN)
Ewen et al. (2000) Physically based,
distributed, water quantity
and quality simulation
model
Daily Conceptual Rainfall-Runoff
Model (HYDROLOG)-Monash
Model
Potter and McMahon
(1976), Chiew and
McMahon (1994)
Lumped, conceptual
rainfall-runoff model
Soil Water Assessment Tool
(SWAT)
Arnold et al. (1998) Distributed, conceptual,
continuous simulation
model
Distributed Hydrological Model
(HYDROTEL)
Fortin et al. (2001a,b) Physically based,
model
Module 9

Lecture 2: Steps in watershed modelling
Module 9

1. Model Selection
Select model based on study objectives and watershed
characteristics, availability of data and project budget
The selection of a model is very difficult and important decision, since the
success of the analysis depends on the accuracy of the results
The selection of a model also depends upon time
scale, hydrologic quantity aiming at and the
processing speed of the computer at hand
Steps in Watershed Modelling
Module 9

2. Input Data
 Obtain all necessary input data-
rainfall, infiltration, physiography, landuse, channel
characteristics, streamflow, design floods and reservoir data.
1. Agricultural Data:
 Vegetative cover,
 Land use,
 Treatment, and
 Fertilizer application
Module 9
Steps in Watershed Modelling Contd…

Source: CSRE, IIT Bombay
Module 9
2. Input Data

2. Hydrometeorologic Data:
Rainfall,
 Snowfall,
 Temperature,
 Radiation,
 Humidity,
 Vapor pressure,
 Sunshine hours,
 Wind velocity, and
 Pan evaporation
Module 9
2. Input Data

3. Pedologic Data:
Soil type, texture, and structure
 Soil condition
 Soil particle size, diameter, porosity
 Soil moisture content and capillary pressure
 Steady-state infiltration,
 Saturated hydraulic conductivity, and
 Antecedent moisture content
Module 9
2. Input Data

4. Geologic Data:
 Data on stratigraphy, lithology, and structural controls,
 Depth, and areal extent of aquifers.
 For confined aquifers, hydraulic
conductivity, transmissivity, storativity, compressibility, and porosity are
needed.
 For unconfined aquifers, data on specific yield, specific storage, hydraulic
conductivity, porosity, water table, and recharge are needed.
Module 9
2. Input Data

5. Geomorphologic Data:
 Topographic maps showing:
 elevation contours,
 river networks,
 drainage areas,
 slopes and slope lengths, and
 watershed area
Module 9
2. Input Data

Module 9
2. Input Data

Source: CSRE, IIT Bombay Module 9
2. Input Data

Digital Elevation Model (DEM)
• Digital file that stores the elevation of the land surface in a specified grid cell
size (e.g., 30 meters)
Module 9
http://www.cabnr.unr.edu/saito/intmod/docs/tootle-hydrologic-modeling.ppt.
2. Input Data

Digital Elevation Model (DEM)
Module 9
0.00
2.00
4.00
6.00
8.00
10.00
12.00
14.00
16.00
18.00
20.00
22.00
24.00
26.00
28.00
30.00
32.00
34.00
36.00
38.00
(Source: http://www.crwr.utexas.edu/gis/gishyd98/dhi/mike11/M11_main.htm)

6. Hydraulic Data:
 Roughness,
 Flow stage,
 River cross sections, and
 River morphology
Module 9
2. Input Data

7. Hydrologic Data:
 Flow depth,
 Streamflow discharge,
 Base flow,
 Interflow,
 Stream-aquifer interaction,
 Potential,
 Water table, and
 Drawdown
Design Point
1
5
6
3
2
Streamflow
http://www.cabnr.unr.edu/saito/intmod/docs/tootle-hydrologic-modeling.ppt.
Module 9
2. Input Data

3. Evaluation & Refinement of Objectives
Evaluate and refine study objectives in terms of simulations to be
performed under various watershed conditions.
4. Selection of Methodology
Choose methods for determining sub-basin hydrographs and channel
routing
Module 9

5. Calibration & Verification of Model
Model calibration involves selecting a measured set of input data
(rainfall, channel routing, landuse and so on) and measured output hydrographs
for model application.
Calibrate model using historical rainfall, streamflow and existing watershed
conditions.
Verify model using other events under different conditions while maintaining
same calibration parameters.
Module 9

Watershed Model Calibration
Module 9
Watershed Model
(Representation)
Calibration
Model Results
Parameters
Adjust
Calibrated
Model Results
Accept
Output
Observations

6. Simulations using Model
Perform model simulations using historical or design rainfall, various conditions
of landuse and various control schemes for reservoirs, channels or diversions.
7. Sensitivity Analysis of Model
Perform sensitivity analysis on input rainfall, routing parameters and hydraulic
parameters as required.
1) Precipitation
2) Soil parameters
a) Hydraulic conductivity
b) Soil water holding capacity
3) Evaporation (for continuous simulation)
4) Flow routing parameters (for event-based)
Module 9

8. Model Validation
Evaluate usefulness of the model and comment on needed changes or
modifications.
Uncertainties
 Precipitation
 Extrapolation of point to other areas
 Temporal resolution of data
 Soils information
 Surveys are based on site visits and then extrapolated
 Routing parameters
 Usually assigned based on empirical studies
Module 9

Lecture 3: Major hydrologic models-HSPF,
HEC and MIKE
Module 9

Major Hydrologic Models
 HSPF (SWM)
 HEC
 MIKE
Module 9

HSPF is a deterministic, lumped parameter, physically based, continuous
model for simulating the water quality and quantity processes that occur in
watersheds and in a river network.
Commercial successor of the Stanford Watershed Model (SWM-IV)
(Johanson et al., 1984):
Water-quality considerations
Kinematic Wave routing
Variable Time Steps
Module 9
Hydrological Simulation Program-Fortran
(HSPF)

Data Requirements of HSPF:
Rainfall
Infiltration
Baseflow
Streamflow
Soils
Landuse
HSPF incorporates watershed-scale ARM (Agricultural Run-off
Management) and NPS (Non-Point Source) models into a basin-scale analysis
framework
 fate and transport of pollutants in 1-D stream channels.
Module 9
HSPF Contd…

HSPF is one of the most complex hydrologic models which simulates:
Infiltration: Philip's equation, a physically based method which uses
an hourly time step
Streamflow: Chezy – Manning’s equation
HSPF can simulate temporal scales ranging from minutes to days
Due to its flexible modular design, HSPF can model systems of varying size
and complexity;
Module 9
HSPF Contd…

Stanford Watershed Model (AquaTerra, 2005)
To stream
Actual ET
Output
1
Module 9

CEPSC : interception storage capacity
LSUR : length of the overland flow plane
SLSUR : slope of the overland flow plane
NSUR : Manning's roughness of the land surface
INTFW : interflow inflow
INFILT : index to the infiltration capacity of the soil
UZSN : nominal capacity of the upper-zone storage
IRC : interflow recession constant
LZSN : nominal capacity of the lower-zone storage
LZETP : lower-zone evapotranspiration
AGWRC : basic ground-water recession rate
AGWETP : fraction of remaining potential evapotranspiration that can be satisfied from active ground-water
storage
KVARY : indication of the behavior of ground-water recession flow
DEEPFR : fraction of ground-water inflow that flows to inactive ground water
BASETP : fraction of the remaining potential evapotranspiration that can be satisfied from base flow
(Kate Flynn, U.S. Geological Survey,
written commun., 2004)
Module 9

HEC Models
Modeling of the rainfall-runoff process in a watershed based on watershed
physiographic data
 a variety of modeling options in order to compute UH for basin areas.
 a variety of options for flood routing along streams.
 capable of estimating parameters for calibration of each basin based on
comparison of computed data to observed data
Module 9
1. HEC-GridUtil 2.0
2. HEC-GeoRAS 10 (EAP)
3. HEC-GeoHMS 10 (EAP)
4. HEC-GeoEFM 1.0
5. HEC-SSP 2.0
6. SnoTel 1.2 Plugin
7. HEC-HMS 3.5
8. HEC-FDA 1.2.5a
9. HEC-DSSVue 2.0.1
10. HEC-RAS 4.1
11. HEC-DSS Excel Add-In
12. HEC-GeoDozer 1.0
13. HEC-EFM 2.0
14. HEC-EFM Plotter 1.0
15. HEC-ResSim 3.0
16. HEC-RPT 1.1

HEC-GridUtil is designed to provide viewing, processing, and analysis capabilities for
gridded data sets stored in HEC-DSS format (Hydrologic Engineering Center's Data
Storage System).
http://www.hec.usace.army.mil/software/hec-gridutil/documentation.html
HEC-GridUtil 2.0
Module 9

GIS extension  a set of procedures, tools, and utilities for the preparation of GIS
data for import into HEC-RAS and generation of GIS data from RAS output.
HEC-GeoRAS 10 (EAP)
• ArcGIS w/ extensions
 3D & Spatial Analyst
 HEC-GeoHMS
 HEC-GeoRAS
• HEC-RAS
– Simulates water surface profile of a stream reach
Module 9

Data Requirements
• Triangular Irregular Network
(TIN)
• DEM (high resolution)
– use stds2dem.exe if
downloading from USGS
• Land Use / Land Cover
– Manning’s Coefficient
Module 9
CRWR image, Texas University
(Source: “GIS – Employing HEC-GeoRAS”, Brad Endres
, 2003)

Major Functions of GeoRAS
• Interface between ArcView and HEC-RAS
• Functions:
– PreRAS Menu - prepares Geometry Data necessary for HEC-RAS modeling
– GeoRAS_Util Menu – creates a table of Manning’s n value from land use
shapefile
– PostRAS Menu – reads RAS import file; delineates flood plain; creates
Velocity and Depth TINs
Module 9

Demonstration of Capabilities
• Load TIN
• Create Contour
Lines
Module 9
3-D Scene
3-D Scene

Demonstration of Capabilities Contd…
• Create Stream Centerline
• Create Banks Theme
• Create Flow Path Centerlines
• Create Cross Section Cut Lines
• Add/Create Land Use Theme
• Generate RAS Import File
Module 9

Stream
Centerline
Right Bank
Flow Path
Centerlines
Land Use
Themes
Cross Section
Cut Lines
Module 9

 Generate RAS GIS import file
 Open HEC-RAS and import RAS GIS file
 Complete Geometry, Hydraulic, & Flow Data
 Run Analysis
 Generate RAS Export file
Module 9

RAS GIS import file
Module 9

RAS GIS export file
Module 9

• New GIS data
• PostRAS features
 Water Surface TIN
 Floodplain Delineation – polygon & grid
 Velocity TIN
 Velocity Grid
Module 9

Floodplain Delineation (3-D Scene)
Module 9

Depth Grid (Darker = Deeper) Velocity Grid (Darker = Faster)
Module 9

Employing ArcView, GeoRAS, and RAS for Main Channel
Depth Analysis (1968)
Module 9
PreRAS PostRAS
13.5 ft

Employing ArcView, GeoRAS, and RAS for Main Channel
Depth Analysis (1988)
PreRAS PostRAS
21.0 ft
Module 9

Overall Benefits
 Elevation data is more accurate with TIN files
 Better representation of channel bottom
 Rapid preparation of geometry data (point and click)
 Precision of GIS data increases precision of geometry data
 Efficient data transport via import/export files
 Velocity grid
 Depth grid
Module 9

 Floodplain maps can be made faster
• several flow scenarios
 Both steady & unsteady flow analysis
 GIS tools aid engineering analysis
• Automated calculation of functions (Energy Equation)
• Structural validation of hydraulic control features
• Voluminous data on World Wide Web
 Makes data into visual event – easier for human brain to process!
Module 9
Overall Benefits Contd…

Overall Drawbacks
 Time required to learn several software packages
 Non-availability of TIN or high resolution data
 Estimation of Manning’s Coefficient
• Few LU/LC files have this as attribute data
 Velocity distribution data may not be calculated
• HEC-RAS export file without velocity data means no velocity TIN or
grid
Module 9

HEC-HMS
HEC-HMS simulates rainfall-runoff for the watershed
(Source: ftp://ftp.crwr.utexas.edu) Module 9

HEC-HMS Background
Purpose of HEC-HMS
 Improved User Interface, Graphics, and Reporting
 Improved Hydrologic Computations
 Integration of Related Hydrologic Capabilities
Importance of HEC-HMS
 Foundation for Future Hydrologic Software
 Replacement for HEC-1
Module 9

Ease of Use
 projects divided into three components
 user can run projects with different parameters instead of creating new
projects
 hydrologic data stored as DSS files
 capable of handling NEXRAD-rainfall data and gridded precipitation
Converts HEC-1 files into HMS files
Module 9
Improvements over HEC-1

HEC-1
EXERCISE PROBLEM
A small undeveloped watershed has the parameters listed in the following tables.
A unit hydrograph and Muskingum routing coefficients are known for subbasin 3,
shown in Fig.1(a). TC and R values for subbasins 1 and 2 and associated SCS
curve numbers (CN) are provided as shown. A 5-hr rainfall hyetograph in in./hr is
shown in Fig.1(b) for a storm event that occurred on July 26, 2011. Assume that
the rain fell uniformly over the watershed. Use the information given to develop a
HEC-1 input data set to model this storm. Run the model to determine the
predicted outflow at point B.
SUBBASIN
NUMBER
TC
(hr)
R
(hr)
SCS CURVE
NUMBER
% IMPERVIOUS
(% )
AREA
(mi2
)
1 2.5 5.5 66 0 2.5
2 2.8 7.5 58 0 2.7
3 -- -- 58 0 3.3
UH FOR
SUBBASIN 3:
TIME (hr) 0 1 2 3 4 5 6 7
U (cfs) 0 200 400 600 450 300 150 0
Module 9

Fig.1(a) Fig.1(b)
Muskingum coefficients: x = 0.15, K = 3 hr, Area = 3.3 sq mi
Module 9

ID ****
ID ****
ID ****
ID ****
IT 60 60 25-Jul-07 1200 100
IO 4
KK SUB1
KM
PI 0.2 1.5 2 1 0.5
BA 2.5
LS 66 0
UC 2.5 5.5
KK SUB2
KM
BA 2.7
LS 58 0
UC 2.8 7.5
KK A
KM
HC 2
KM
RM 1 3 0.15
KK SUB3
KM
BA 3.3
LS 58 0
UI 0 200 400 600 450 300 150
MUSKINGUM ROUTING FROM A TO B
RUNOFF FROM SUBBASIN 3
KKA TO B
EXAMPLE PROBLEM
HEC-1 INPUT DATA SET
COMBINE RUNOFF FROM SUB 1 WITH RUNOFF FROM SUB 2 AT A
Solution : The input data set is as follows:
Module 9

Using HEC-HMS Contd…
Three components
 Basin model - contains the elements of the basin, their connectivity, and
runoff parameters ( It will be discussed in detail later)
 Meteorologic Model - contains the rainfall and evapotranspiration data
 Control Specifications - contains the start/stop timing and calculation
intervals for the run
Module 9

Project Definition
 It may contain several basin models, meteorological models, and control
specifications
 It is possible to select a variety of combinations of the three models in order to
see the effects of changing parameters on one sub-basin
Module 9

Basin Model
 GUI supported
 Click on elements from left and drag
into basin area
 Works well with GIS imported files
 Actual locations of elements do not
matter, just connectivity and runoff
parameters
Module 9

1. Basin Model Elements
• subbasins- contains data for subbasins (losses, UH transform, and
baseflow)
• reaches- connects elements together and contains flood routing
data
• junctions- connection point between elements
• reservoirs- stores runoff and releases runoff at a specified rate
(storage-discharge relation)
Module 9

1. Basin Model Elements Contd…
• sinks- has an inflow but no outflow
• sources- has an outflow but no inflow
• diversions- diverts a specified amount of runoff to an element
based on a rating curve - used for detention storage elements or
overflows
Module 9

a) Loss rate
b) Transform
c) Baseflow methods
Module 9
2. Basin Model Parameters

2a) Abstractions (Losses)
1. Interception Storage
2. Depression Storage
3. Surface Storage
4. Evaporation
5. Infiltration
6. Interflow
7. Groundwater and Base Flow
Module 9
2. Basin Model Parameters Contd…
1. Unit Hydrograph
2. Distributed Runoff
3. Grid-Based Transformation
Methods:
a. Clark
b. Snyder
c. SCS
d. Input Ordinates
e. ModClark
f. Kinematic Wave
2b) Transformation

2c) Baseflow Options
a. recession
b. constant monthly
c. linear reservoir
d. no base flow
Module 9
2. Basin Model Parameters Contd…

Stream Flow Routing
 Simulates Movement of Flood Wave Through Stream Reach
 Accounts for Storage and Flow Resistance
 Allows modeling of a watershed with sub-basins
Module 9
a) Simple Lag
b) Modified Puls
c) Muskingum
d) Muskingum Cunge
e) Kinematic Wave
Reach Routing

 Hydraulic Methods - Uses partial form of St Venant Equations
 Kinematic Wave Method
 Muskingum-Cunge Method
 Hydrologic Methods
 Muskingum Method
 Storage Method (Modified Puls)
 Lag Method
Module 9
Methods for Stream Flow Routing

 Developed Outside HEC-HMS
 Storage Specification Alternatives:
 Storage versus Discharge
 Storage versus Elevation
 Surface Area versus Elevation
 Discharge Specification Alternatives:
 Spillways, Low-Level Outlets, Pumps
 Dam Safety: Embankment Overflow, Dam Breach
Module 9
Reservoir Routing

Reservoirs
Q(cfs)
I=Q
time
Q(cfs)
Inflow
Outflow
I - Q = dS
dt
Level Pool Reservoir
Q (weir flow)
Q (orifice flow)
I
S
H
S = f(Q) Q = f(H)
Orifice flow:
Q = C * 2gH
Q
I
I
Weir Flow:
Q = CLH3/2
Q
Pond storage with
outflow pipe
Orifice flow
Weir flows
Inflow and Outflow
Module 9

 Initial Conditions to be considered
 Inflow = Outflow
 Initial Storage Values
 Initial Outflow
 Initial Elevation
 Elevation Data relates to both Storage/Area and Discharge
 HEC-1 Routing routines with initial conditions and elevation data
can be imported as Reservoir Elements
Module 9
Reservoir Data Input

Module 9
Reservoir Data Input Window

User selects:
1. Basin model
2. Meteorologic model
3. Control ID for the
HMS run
Running a project
Module 9

 To view the results:
• right-click on any basin element, results will be for that point
 Display of results:
• hydrograph- graphs outflow vs. time
• summary table- gives the peak flow and time of peak
• time-series table- tabular form of outflow vs. time
 Comparing computed and actual results:
• plot observed data on the same hydrograph to by selecting a discharge
gage for an element
Module 9
Viewing Results

hydrograph
Module 9
Viewing Results Contd…

HEC-HMS Output
1. Tables
 Summary
 Detailed (Time Series)
2. Hyetograph Plots
3. Sub-Basin Hydrograph Plots
4. Routed Hydrograph Plots
5. Combined Hydrograph Plots
6. Recorded Hydrographs - comparison
Module 9

Summary table
Time series table Module 9
Viewing Results

Sub-Basin Plots
 Runoff Hydrograph
 Hyetograph
 Abstractions
 Base Flow
Module 9

Junction Plots
Module 9
a. Tributary Hydrographs
b. Combined Hydrograph
c. Recorded Hydrograph

Lecture 4: MIKE models
Module 9

a) Flood Management
• MIKE 11 For Analyzing Open Channel Flow
• MIKE 21 For Analyzing surface complicated overflow
b) River Basin Management
• MIKE BASIN
c) Hydrological Cycle
• MIKE SHE
d) Urban Drainage
• MOUSE For Analyzing Urban Sewage
MIKE Models
Module 9

 1 Dimensional
• MIKE 11
 2 Dimensional
• MIKE 21
 3 Dimensional
• MIKE SHE
• MIKE 3
Module 9
MIKE Hydrological and Hydrodynamic Models

MIKE Zero-fication!
MIKE ZERO
MIKE
3
MIKE
11
MIKE
SHE
MIKE
21
Module 9

ArcMap with MIKE ZERO
Module 9

MIKE 11 GIS
 Fully integrated GIS based flood modelling
 Developed in ArcView GIS
 Pre-processing:
 Post-processing:
 Analysis with other GIS data
Floodplain schematization
•Flood depth maps
•Comparison maps
•Duration maps
Module 9
 Mike 11 Modules
 HD : hydrodynamic - simulation of unsteady flow in a network of
open channels. Result is time series of discharges and water
levels;
 AD : advection dispersion
 WQ : water quality

 Saint Venant equations (1D)
 continuity equation (mass conservation)
 momentum equation (fluid momentum conservation)
 Assumptions
 water is incompresible and homogeneous
 bottom slope is small
 flow everywhere is paralel to the bottom ( i.e. wave lengths are large
compared with water depths)
Module 9
Open channel flow

Discretization - branches
Module 9
IHE, 2002

Module 9
IHE, 2002
Discretization – branches Contd...

Discretization - cross sections
 Required at representative locations throughout the branches of the river
 Must accurately represent the flow changes, bed slope, shape, flow
resistance characteristics
Module 9
IHE, 2002

 Friction formulas
 Chezy
 Manning
 For each section a curve is made with wetted area, conveyance factor,
hydraulic radius as a function of water level
Module 9
h
R
Discretization – cross sections Contd...
IHE, 2002

Module 9
Typical Model Schematisation
Minor river
Major River
Floodplains
Spill channel
Spill channel

Extraction from DEM
Module 9(Source: http://www.crwr.utexas.edu/gis/gishyd98/dhi/mike11/M11_main.htm)

Import to MIKE 11
Module 9

Menus and input files editors
Module 9

Network editor
Module 9
Tabular view
Graphical view

Module 9
River network -branches
connection
Network editor Contd...

River network -structures
Module 9
Network editor Contd...

Parameter file editor
Module 9

Parameter file - coefficients
Module 9
h1
Time step n+1/2
Time step n
Time
Time step n+1
i i+1i-1
Spac
e
h3
h5
h7
4
6
Q
Q
Q
Center
point
i
n
ii
n
ii
n
ii QhQq
t
A
x
Q
δγβα =++⇒=
∂
∂
+
∂
∂ +
+
++
−
1
1
11
1
i
n
ii
n
ii
n
ii hQh
ARC
QgQ
x
h
gA
x
A
Q
t
Q
δγβα
α
=++=+
∂
∂
+
∂






∂
+
∂
∂ +
+
++
−
1
1
11
12
2
0
2

Because of its numerical limitations, MIKE 11 cannot model the supercritical
flow downstream of the weir.
For the low-flow case, the downstream water level is over-estimated by
a factor of 8 .This high tailwater, impacts on the flow conditions on the
weir, causing a significant error in the upstream water level.
 The incorrect tailwater has less impact for the high-flow case. There is
still significant error in the predictions across the weir, but the upstream
water level is almost correct.
Limitations of MIKE 11
Module 9

Advanced integrated hydrological modeling system
Simulates water flow in the entire land based phase of the hydrological cycle
rainfall to river flow, via various flow processes such as,
overland flow,
infiltration into soils,
evapotranspiration from vegetation, and
groundwater flow
MIKE SHE
Module 9

Integrated:
Fully dynamic exchange of water between all major hydrological components is
included, e.g. surface water, soil water and groundwater
Physically based:
It solves basic equations governing the major flow processes within the study area
Fully distributed:
The spatial and temporal variation of meteorological, hydrological, geological and
hydrogeological data
Modular:
The modular architecture allows user only to focus on the processes, which are
important for the study
MIKE SHE Features
Module 9

Hydrological Processes simulated by MIKE SHE
Module 9

Schematic view of the process in MIKE SHE, including the
available numeric engines for each process.
The arrows show the
available exchange
pathways for water
between the process
models.
(V.P. Singh & D.K. Frevert, 199
Module 9

Flow System Inputs and Outputs
Module 9

Reservoir name
Reservoir ID
Initial water level
Priority of inflow connections
Priority of down-stream user(s)
Down-streams user(s) reduction factor
Down-stream user(s) loss factor
Rule curves
Height, Volume, Area
Precipitation
Evaporation

 Hydrodynamic Integrated model with 1D and 2D
 The result of either 1D or 2D model can be transferred as input
of another model
 This model simulates simultaneously the flow in the sewer, the
drainage system and the surface flow
Module 9
MIKE FLOOD

Conceptual Representations
Rainfall Runoff
Surface Runoff
Overflow
Re-inflow after flood
Flow-capacity excess
Sewage/Rainfall water
Sewage/Rainfall water
Module 9

Lecture 5: Urban Flood Risk Mapping using
MOUSE, MIKE 21 and MIKE FLOOD
Module 9

MIKE FLOOD
Set up
Coupling
Preprocessing for MOUSE
Run Model
Check the result and evaluation
Set up
Validation
MIKE21 Model
MOUSE Model
Set up
Validation
Preparation
GIS
Rainfall
Analysis
Module 9

Urban Drainage Network
 Import of drainage network into MOUSE
 Setting up Urban Drainage model with MOUSE
 Validation
Module 9

 Import Ascii data-set as bathymetry
 Setting up Urban Bathymetry with MIKE21
 Validation
Option Value
Time Step
1sec
Max Cr=0.4
Grid Size 10m
Urban Drainage Modelling
Module 9

 Setting up MIKE Flood model
 Coupling : Link MOUSE Manholes to MIKE21
 Preprocessing for MOUSE Model
 Running Model
 Check the results
(Source: www.hydroasia.org)
MIKE FLOOD Modeling
Module 9

Catchment boundary
Gajwa-WWTP
Cityhall
Ganseok st.
Juan st.
Incheon gyo
Pump Station
Flooded
area
Results Analysis
(Source: www.hydroasia.org) Module 9

Comparing the Result (Modeled VS Reported)
Hwasu Reservoir basin
Around Dohwa and Juan metro station
Seoknam Canal Basin
Gajwa girls juniorhigh
Sipjeong & Ganseok dong
Seoknam Canal
Hwasu Reservoir basin
Around Dohwa and Juan metro station
Seoknam Canal Basin
Gajwa girls juniorhigh
Sipjeong & Ganseok dong
Seoknam Canal
(Source: www.hydroasia.org)
Module 9

 Basic elements of hydrologic simulation model
Equations govern the hydrologic processes,
Maps define the study area and
Database  numerically describe the study area and model parameters.
 Strengths of Watershed Models
 Diversity of the current generation of models
 Comprehensive Nature
 Reasonable modeling of physical phenomena
 Distributed in Space and Time
 Multi-disciplinary nature
Module 9

 Deficiencies of Watershed Models
 Lack of user-friendliness,
 Large data requirements,
 Lack of quantitative measures of their reliability,
 Lack of clear statement of their limitations, and
 Lack of clear guidance as to the conditions for their applicability.
 Steps in Watershed Modelling
 Model Selection
 Input Data
 Agricultural Data, Hydrometeorologic Data, Pedologic Data, Geologic
Data, Geomorphologic Data, Hydraulic Data, Hydrologic Data
Module 9

 Evaluation & Refinement of Objectives
 Selection of Methodology
 Calibration & Verification of Model
 Simulations using Model
 Sensitivity Analysis of Model
 Model Validation
 Major Hydrologic Models
 HSPF (SWM)
 HEC
 MIKE
Module 9

Text/References
1.Abraham, K.R.,Dash, S.K., and Mohanty, U.C.1996. Simulation of monsoon
circulation and cyclones with different types of orography; Mausam, 47, 235-
248.
2.Bedient et al. “Hydrology and Floodplain Analysis”, 2008.
3.Bhalme, H.N. and Jadhav, S.K. 1984. The southern oscillation and its
relation to the monsoon rainfall. J.Climatol., 4, 509-520.
4.Bras, R. L., and Rodriguez-Iturbe. 1994. Random Functions and Hydrology,
Dover Publications, New York.
5.Chow, V. T., D. R. Maidment, and L. W. Mays. “Applied Hydrology”, McGraw
Hill International Editions.
6.Haan, C. T.. 2002. “Statistical Methods in Hydrology”, 2nd ed., Blackwell
Publishing, Ames, IA.

Text/References Contd…
7. Hoskings, J. R. M. and J. R. Wallis. 1997. “Regional Frequency
Analysis, An Approach Based on L-Moments”, Cambridge University
Press, New York.
8. Subramanya K, “Engineering Hydrology”, Tata McGraw-Hill.
9. Viessman Jr., W. and G. L. Lewis. “Introduction to Hydrology”, 4th
ed., Harper-Collins, New York, 1996.
10.Vijay P. Singh, F. and David A. Woolhiser, M. 2002. Mathematical Modeling
of Watershed Hydrology, Journal of Hydrologic Engineering, Vol. 7, No.
4, July 1, 2002.
11.www.climateofindia.pbworks.com/
12.http://www.crwr.utexas.edu/gis/gishyd98/dhi/mikeshe/Mshemain.htm

Related study materials available through NPTEL
http://nptel.iitm.ac.in/courses/105104029/
1. Advanced Hydrology (Video Course) by Prof. Ashu Jain, IIT Kanpur
2. Stochastic Hydrology (Video Course) by Prof. P.P. Mujumdar,
IISc Bangalore
3. Probability Methods in Civil Engineering (Video Course) by Dr. Rajib Maity,
IIT Kharagpur

Advanced hydrology

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