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
Hydrologic flow routing

Advanced hydrology

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