Advanced hydrology

Advanced Hydrology(Web course)

Subhankar KarmakarAssistant Professor

Centre for Environmental Science and Engineering (CESE)Indian Institute of Technology Bombay

Powai, Mumbai 400 076Email: [email protected]

Ph. # +91 22 2576 7857

mailto:[email protected]�

Hydrologic Cycle

Prof. Subhankar KarmakarIIT Bombay

Module 13 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.

Stratosphere19% 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 inthe 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 toFebruary.

► Vertical sun rays shift towardssouthern hemisphere.

► North India experiencesintense cold

► Light wind blow makes thisseason pleasant in southIndia.

► Occasional tropical cyclonevisit eastern coast in thisseason.

Tropical Cyclone

Cold Weather Season

Seasons

Module 1Lecture 1

250C

250C

200C

200C200C

150C

200C

100C`

Temperature-January

(climateofindia.pbworks.com) Module 1

Seasons

Lecture 1

Pressure-January

(climateofindia.pbworks.com)

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


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


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

(climateofindia.pbworks.com) 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 rain2.5 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 MethodSimplest method for determining areal average

where, Pi : rainfall at the ith raingauge station N : total no: of raingauge stations

∑=

=N

iiP

NP

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

iii PA

AP

1

1

mmP 7.2047

302020151012=

×+×+×=

P1

P2

P3

A1

A2

A3

P1 = 10 mm, A1 = 12 Km2

P2 = 20 mm, A2 = 15 Km2

P3 = 30 mm, A3 = 20 km2

3. Isohyetal method

∑=

=N

iii PA

AP

1

1

mmP .2150

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.

( ) ( )2212

2112 yyxxd −+−=

P1=10

P2= 20

P3=30

d1=25

d2=15

d3=10p

∑

∑

=

=

=N

i i

N

i i

i

d

dP

P

12

12

1ˆ mmP 24.25

101

151

251

1030

1520

2510

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 yearover 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

Acc

umul

ated

Pre

cipi

tatio

n of

St

atio

n X,

ΣPx

Average accumulated precipitation of neighbouring stations ΣPav

9089

88

87

86

85

84

8382

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 XPx – Original recorded precipitation at time period t1 at station XMc – corrected slope of the double mass curveMa – 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(0nKAPP −=

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

aDTki+

=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

Inte

nsi

ty, c

m/h

r

Duration, hr

Typical IDF Curve

T = 25 years

T = 50 years

T = 100 years

k = 6.93x = 0.189a = 0.5n = 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 < fcwhere 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

fcf

infiltration

time t

Infiltration

Graphical representation of Horton formula

Measurement of infiltration1. 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 qualityof water on Earth including the hydrologic cycle, water resources andenvironmental 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 precipitationactivity, whereas climate is the term for the average atmospheric conditionsover longer periods of time.

Formation of Precipitation: frontal, convective, cyclonic and orographic

The four different seasons are: Cold weather, Hot weather, South-Westmonsoon and Retreating monsoon

Module 1



Philosophy of Mathematical Models of Watershed Hydrology

Module 22 Lectures

Philosophy of mathematical models of watershed hydrology

Lecture 1

Objectives of this module is to introduce the terms andconcepts in mathematical modelling which will form as a toolfor effective and efficient watershed management throughwatershed 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

MathematicsComputer 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 knowledgeof 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

TypeEmpirical

Based on data analysisMechanistic

Mathematical descriptions based on theory

Time FactorStatic or steady-state

Time-independentDynamic

Describe or predict system behavior over time

Treatment of Data Uncertainty and VariabilityDeterministic

Do not address data variabilityStochastic

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 modelsModule 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 describedby unique values, but rather by probability distributions.

Static:A static model does not account for the element of time, while a dynamic modeldoes.

Dynamic:Dynamic models typically are represented with difference equations or differentialequations.

Discrete:A discrete model does not take into account the function of time and usually usestime-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 variablesspace-time fields of precipitation, temperature, etc.

Parameters Size Shape Physiography Climate Hydrogeology Socioeconomics

State variablesspace-time fields of soil moisture, etc.

Drainage Land use Vegetation Geology and Soils Hydrology

Module 2

Equations variables

Independent variablesspace x

time t

Dependent variablesdischarge 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 termGoals

Module 2

Philosophy of mathematical models of watershed hydrology (contd.)

Lecture 2

time

Prec

ipita

tion

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 andmovement of water between the biosphere, atmosphere, lithosphere, and thehydrosphere.

Module 2

Note:For 90 yrs of record,(2/3) of 90 = 60 yrs for calibrationRemaining (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 ModelModule 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

InterflowDirect Runoff

Surface Storage

BaseflowPercolation

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 Modelβ1β2β3β4

C1C2C3

j

j

i

i

ijC

C

Sβ

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 jii

n

i ijj

CCC

nCCnHere

βββ

ββ

−=∆−=∆

== ∑∑ ==

;

; 11

Mod

el O

utpu

t

Observed Value

x

xx

x

xx

x

xx

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


Mod

el O

utpu

t

Observed Value

x

xx

x

xx

x

xx

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’

xx

x

xx

x

xx

x xx

x

xx

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


Module 36 Lectures

Module 3

Objective of this module is to introduce the watershedconcepts, rainfall-runoff, hydrograph analysis and unithydrograph 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 ingeneral, by one main channel andby tributaries that drain into mainchannel at one or more confluencepoints

A “divide” or “drainage divide” is theline drawn through the highestelevated points within a watershed

Divide forms the limits of a singlewatershed and the boundarybetween 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 adjacentbasins, and,

2. Subsurface water divide –which refers to faults, folds, tilted geologicstrata (rock layers), etc., that cause sub-surface flow to move in onedirection or the other.

Surface water divideSubsurface 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 therainwater with the improve in urbanization and increased pavements.

Main channel and tributary characteristics: It can effect the streamflow response in various ways such as slope, cross-sectionalarea, Manning’s roughness coefficient, presence of obstructions andchannel condition

Physiography: Lands altitude and physical disposition

Socio-economics: Depends on the standard of living of the people andit is important in managing water

Module 3

Characteristics of watershed Contd…

A watershed management approach is one that considers the watershed as awhole, rather than separate parts of the watershed in isolation

Managing the water and other natural resources is an effective and efficientway to sustain the local economy and environmental health

Watershed management helps reduce flood damage, decrease the loss ofgreen 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 resourcemanagement issues

Module 3

Watershed Management

Use an ecological approach that would recover and maintain the biologicaldiversity, ecological Function, and defining characteristics of naturalecosystems

Recognize that humans are part of ecosystems-they shape and are shapedby the natural systems: the sustainability of ecological and societal systemsare mutually dependent

Adopt a management approach that recognizes ecosystems and institutionsare characteristically heterogeneous in time and space

Integrate sustained economic and community activity into the managementof ecosystems

Module 3

Principles for Watershed Management

Provide for ecosystem governance at appropriate ecological and institutionalscales

Use adaptive management as the mechanism for achieving both desiredoutcomes and new understandings regarding ecosystem conditions

Integrate the best science available into the decision-making process, whilecontinuing scientific research to reduce uncertainties

Implement ecosystem management principles through coordinatedgovernment 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 soilsurface and move laterally through the upper soil layers until it enters a streamchannel.

Groundwater Flow or Base Flow: The portion of precipitation that percolatesdownward until it reaches the water table. This water accretion may eventuallydischarge into the streams if the water table intersects the stream channels ofthe basin. However, its contribution to stream flow cannot fluctuate rapidlybecause 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 basiscan be plotted for a given watershed to relate rainfall to direct runoff for agiven 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 unitsA = 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 asR = 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 byTc = ∑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 fromone location to another in a watershed

• Travel time for sheet flow

where,n = Manning’s roughness coefficientL = Flow length (meters)

P2 = 2-yr, 24-hr rainfall (in.) and S is the hydraulic grade line orland 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, hydrologiclosses such as infiltration,depression storage and detentionstorage must be satisfied prior tothe onset of surface runoff

• As the depth of surface detentionincreases, overland flow may occurin portion if a basin

• Water eventually moves into smallrivulets, small channels and finallythe main stream of a watershed

• Some of the water that infiltratesthe soil may move laterally throughupper soil zones (subsurfacestromflow) until it enters a streamchannel

Uniform rainfall

Infiltration

Depression storage

Detention storage

Time (hr)

Run

off (

cfs)

Rai

nfal

l (in

./hr)

Distribution of uniform rainfall


• If the rainfall continues at a constantintensity for a very long period,storage is filled at some point andthen an equilibrium discharge canbe reached

• In equilibrium discharge the inflow andoutflow are equal

• The point P indicates the time atwhich the entire discharge areacontributes to the flow

• The condition of equilibrium dischargeis seldom observed in nature, exceptfor very small basins, because ofnatural variations in rainfall intensityand duration

Rainfall

Equilibrium discharge

Run

off (

cfs)

Rai

nfal

l (in

./hr)

Time (hr)

P

Equilibrium hydrograph

Module 3



• 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)


Module 3


Recession and Base flow separation• In this the hydrograph is divided into

two parts1. Direct runoff (DRO) and2. Base flow (BF)

• DRO include some interflow whereasBF is considered to be mostly fromcontributing ground water

• Recession curve method is used toseparate DRO from BF and can by anexponential depletion equation

qt = qo ·e-kt whereqt = discharge at a later time tqo = specified initial dischargek = recession constant

C

D

BA

Q

Time

N=bA0.2

Base flow separationModule 3


• There are three types of baseflow separation techniques 1. Straight line method2. Fixed base method3. 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 curve2. 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 beneglected, 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 determinerainfall excess include1. Horton infiltration method2. Ø index method

• Note:- In this, the initial loss isincluded for depression storage

Rai

nfal

l and

infil

tratio

n

Depression storage

Net storm rainfall

Ø index

Horton infiltration

Time

Infiltration loss curves

Module 3

Rainfall excess Contd…

• Horton infiltration methodHorton 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 methodIt 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-hdurations on a catchment area 27km2 produced the following hydrograph offlow 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 66Observed 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

129

75 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

Dis

char

ge(m

3 /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 10 5 06 13 8

12 26 2118 21 1624 16 1130 12 736 9 442 7 248 5 054 5 060 4.5 066 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

97

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

Dis

char

ge(m

3 /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 masscurve of rainfall of the storm is as follows:

If the φ-index of the catchment is 0.4cm/h, determine the effective rainfallhyetograph and the volume of direct runoff from the catchment due to thestorm.

Time from start of

storm (h) 0 2 4 6 8 10 12 14Accumulated 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 04 2 2.8 2.2 0.8 1.4 0.76 2 5.2 2.4 0.8 1.6 0.88 2 6.7 1.5 0.8 0.7 0.35

10 2 7.5 0.8 0.8 0 012 2 9.2 1.7 0.8 0.9 0.4514 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 puretranslation of direct runoff to the outlet, at an uniform velocity, ignoring anystorage effect in the watershed

• The relation ship is defined by dividing a watershed into subareas withdistinct runoff translation times to the outlet

• The subareas are delineated with isochrones of equal translation timenumbered upstream from the outlet

• In a uniform rainfall intensity distribution over the watershed, water first flowsfrom areas immediately adjacent to the outlet, and the percentage of totalarea contributing increases progressively in time

• The surface runoff from area A1 reaches the outlet first followed bycontributions from A2, A3 and A4,

Module 3

Time- Area method

2A1A

3A4A

Isochrone of Equal time to outlet

hr5hr10hr15

jiin ARARARQ 1211 ...+++= −

2R

1R3R

Time, t

Rai

nfal

l

2A

1A

3A

4A

0 5 10 15 20Time, t

Are

a

Outlet

Module 3

Run-off Measurement Contd…

Time- Area method

whereQn = 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 isochronallines and the hydrograph must be further adjusted to represent storageeffects 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 DArea (ac) 100 200 300 100Time to gage G (hr) 1 2 3 4

AB

CD

G

Module 3

Time area histogram method uses Qn = RiA1 + Ri-2A2 +…….+ RiAj

For n = 5, i = 5, and j = 5Q5 = 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./hrNote that 1 ac-in./hr ≈ 1 cfs, hence

Q5 = 350 cfs

Example Problem

Example Problem Contd…

Time(hr)

HydrographOrdinate (R1:Rn)

Basin No.

Time to gage

Basin area A1:An (ac)

R1:An R2:An R2:An R2:An R2:An Stormhydrograph

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 (C

FS)

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 hydrologicsystem.

• First proposed by Sherman (1932), the unit hydrograph (originally namedunit-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 rainfallgenerated uniformly over the drainage area at a constant rate for an effectiveduration.

• Sherman originally used the word “unit” to denote a unit of time. But sincethat time it has often been interpreted as a unit depth of excess rainfall.

• Sherman classified runoff into surface runoff and groundwater runoff anddefined 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

Terminologies1. 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 hydrographBase flow

Direct runoff

Inflection point

TRtp

Effective rainfall/excess rainfall

Q (c

fs)

Module 3


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 0.51-2 2.52-3 2.53-4 0.5

Stream flow data Rainfall data

Module 3

Unit hydrograph

Example Problem

• Empirical unit hydrograph derivation separates the base flow from the observedstream flow hydrograph in order to obtain the direct runoff hydrograph (DRH). Forthis example, use the horizontal line method to separate the base flow. Fromobservation of the hydrograph data, the stream flow at the start of the rising limbof 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 theordinates of the DRH by the VDRH in equivalent units of depth


Direct Runoff Hydrograph (DRH) (m3/s)

Unit Hydrograph (m3/s/cm)

0 100 0 01 100 0 02 300 200 503 700 600 1504 1000 900 2255 800 700 1756 600 500 1257 400 300 758 300 200 509 200 100 25

10 100 0 011 100 0 0

Module 3

Module 3

Unit hydrograph


0

200

400

600

800

1000

1200

0 2 4 6 8 10 12

Q (m

3 /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 UnitHydrograph.

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 inSuccession produce Constant outflow Qe cumec

Time t (hr)trI

Lagged s-curveLagged by tr-hr

S-curvehydrograph

To obtain tr-hr UG multiply the S-curve difference by tr/trI

Constant flow Qe (Cumec)

Successive unit storms of Pnet = 1 cm

Dis

char

ge Q

(Cum

ec)

Inte

nsity

(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 01 752 2503 3004 2755 2006 1007 758 509 2510 0

Module 3

Example Problem

Unit hydrograph

SolutionMake 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 01 75 752 250 0 2503 300 75 3754 275 250 0 5255 200 300 75 5756 100 275 250 0 6257 75 200 300 75 6508 50 100 275 250 0 6759 25 75 200 300 75 675

10 0 50 100 275 250 0 67511 25 75 200 300 75 675


0

100

200

300

400

500

600

700

800

0 2 4 6 8 10 12 14

Q (c

fs)

Time (hr)

S-curve

2 hr UHLagged 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-curveordinate

S-curve lagged 3hr

Difference 3-HR UH ordinate

0 0 0 01 75 75 502 250 250 166.73 375 0 375 2504 525 75 450 3005 575 250 352 2166 625 375 250 166.77 650 525 125 83.38 675 575 100 66.79 675 625 50 33.3

10 675 650 25 16.711 675 675 0 0

Unit hydrograph


Find the one hour unit hydrograph using the excess rainfall hyetograph anddirect 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 1m 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

U1U2 U3 U4 U5

Un-m+1

n-m+1

Unit pulse response applied to P1

Unit pulse response applied to P2

n-m+1Un-m+1

Output Qn

Output ∑ − +=

=m*

n m n m 1m 1

Q P U

Combination of 3 rainfall UH

The set of equations for discrete time convolution

∑ − +=

=m*

n m n m 1m 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 tableyields a set of 11 simultaneous equations

Module 3

Unit hydrograph

− −= = =2 2 1

11

Q P U 1,928 1.93x404U 1,079 cfs/ inP 1.06

Similarly calculate for remaining ordinates and the final UH is tabulated below

n 1 2 3 4 5 6 7 8 9Un (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 dueto single and multiple storms, are measured)

• There are many drainage basins (catchments) for which no stream flowrecords are available and unit hydrographs may be required for such basins

• In such cases, hydrographs may be synthesized directly from othercatchments, which are hydrologically and meteorologically homogeneous,or indirectly from other catchments through the application of empiricalrelationship

• Methods for synthesizing hydrographs for ungauged areas have beendeveloped from time to time by Bernard, Clark, McCarthy and Snyder. Thebest 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


Qp = 640 CpA/tpwhereQp = 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/8where Tb is the time base of hydrographNote: 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 rainfallD= 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 (c

fs)

Time (hr)

Qp

W75

W50 Area drawn to represent 1in. of runoff over thewatershed

W75 = 440(QP/A)-1.08

W50 = 770(QP/A)-1.08

(widths are distributed 1/3 before Qpand 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

orwhere B is given by

Therefore runoff eq. becomes, for 1 in. of rainfall excess,

=

BTvolQ

Rp +=

2

RTB 67.1=

Rp T

volQ 75.0=

Rp T

AQ 484=

whereA= area of basin (sq mi)TR = time of rise (hr)

Module 3

Rp T

AQ )008.1()640(75.0=

22BQTQ

Vol pRp +=

SCS Unit Hydrograph Contd…


• Time of rise TR is given by

whereD= rainfall duration (hr)tp= lag time from centroid of rainfall to QP

Lag time is given by

whereL= length to divide (ft)Y= average watershed slope (in present)CN= curve number for various soil/land use

Module 3

pR tDT +=2

0.5

0.70.8

L

19000y

9CN

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


SolutionFind 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

0.70.8

L

19000y

9CN

1000

−

=pt

Find TR using eq.

Given rainfall duration is 2 hr, TR = 4.36 hr, the rise of the hydrographThen find Qp using the eq, given A= 10 mi2

. Hence Qp = 1.110 cfs

Module 3


Rp T

AQ 484=

pR tDT +=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 hr22BQTQ

Vol pRp +=


0

200

400

600

800

1000

1200

0 2 4 6 8 10 12 14

Q (c

fs)

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 6hours duration each on a basin are given below. The area of the basin is 118.8km2 . Assuming a constant base flow of 20 cumec, derive a 6-hour unithydrograph 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 givenbelow. Derive the ordinates of (i) the S-curve hydrograph, and (ii) the 2-hourunit hydrograph, and plot them, area of the basin is 630 km2

Time (hr) Discharge (cumec)0 02 254 1006 1608 190

10 17012 110

Time (hr) Discharge (cumec)14 7016 3018 2020 622 1.524 0

Module 3

Exercise problems Contd…

3. The following are the ordinates of the 9-hour unit hydrograph for the entirecatchment of the river Damodar up to Tenughat dam site: and the catchmentcharacteristics are , A = 4480 km2, L = 318 km, Lca = 198 km. Derive a 3-hourunit hydrograph for the catchment area of river Damodar up to the head ofTenughat reservoir, given the catchment characteristics as, A = 3780km2, L =284 km, Lca = 184km. Use Snyder’s approach with necessary modifications forthe 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.)


Module 43 Lectures

Module 4

Objective of this module is to learn linear-kinematicwave 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 flowingwater 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 hydrographsand eventually, as stream or channel hydrograph

• This method is the combination of continuity equation and asimplified form of St. Venant equations

(Note:- The complete description of St. Venant equations is providedin Module-6)

Module 4

Q

S0

Ѳ

Z

V2/2g

h

dx

Ff

x

dxxQQ∂∂

+

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 = txtA

∆∆∂∂

where,q= rate of lateral inflow per unit length of channelA = cross- sectional area


Continuity Equation Contd…

Module 4

txqtxxQQ ∆∆+∆•

∆

•∂∂

−2

txxQQ ∆•

∆

•∂∂

−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

qxQ

tA

=∂∂

+∂∂

Module 4

bq

ty

xyv

xvy =

∂∂

+∂∂

+∂∂


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

( ) xxyA

∆∂

∂− γ

AS xγ− ∆0

fAS xγ− ∆

• The rate of change of momentum is expressed from Newton’s second lawas

where the total derivative of v W.R.T t can be expressed

( )mvdtdF =

xvv

tv

dtdv

∂∂

+∂∂

=

………..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 berearranged to yield

Sf = So

………..4.3

………..4.4

Saint Venant equation

Module 4

( )Avq

xyA

Ag

xvv

tv

+∂

∂+

∂∂

+∂∂

tgv

xgvv

xy

∂∂

−∂∂

−∂∂

−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 suchas 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 equationKinematic wave ( study uniform)

Sf = So

Diffusion (non inertia) model Sf = So

Steady no-uniform Sf = So

Unsteady non-uniform Sf = So

xy∂∂

−

Dynamic wave

xv

gv

xy

∂∂

−

∂∂

−

tv

g1

xv

gv

xy

∂∂

−

∂∂

−

∂∂

−

Possible types of open channel flow

Module 4


Kinematic wave Dynamic wave

It is defined as the study of motionexclusive of the influences of massand force

In this the influences of mass and forceare included

When the inertial and pressure forcesare not important to the movement ofwave then the kinematic wavesgoverns the flow

When inertial and pressure forces areimportant then dynamic waves governthe moment of long waves in shallowwater (large flood wave in a wide river)

Force of this nature will remainapproximately uniform all along thechannel (Steady and uniform flow)

Flows of this nature will be unsteadyand non-uniform along the length ofthe channel

Froude No. < 2 Froude No. > 2

Difference between kinematic and dynamic wave

Kinematic modelling methods

Fr =

Froude number

gdv

WhereV= velocity of flowg= acceleration due to gravityd= hydraulic depth of water

Wave celerity (C) gdc =1. Flows with Froude numbers greater than one are classified as supercritical

flows2. Froude number greater than two tend to be unstable, that are classified as

dynamic wave3. 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

21

111

+

∂∂

+∂∂

+∂∂

−=gyqv

tgv

xgvv

xy

SQQ

oo

• 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 planeFr= Froude numbery= depth at the end of the planeS0= slopek= dimensionless kinematic flow number

………..4.8


Module 4

102 ≥=yFr

LSk o 102 ≥=

vLgS

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 withshallow flows can be derived from Eq. 4.5 and manning's Eq. for overlandflow

• 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 slopeyo= mean depth of overland flow

………..4.10

Module 4

3/5ySn

kq o

m=

omooo yq α= o

mo S

nk

=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 rateqo= flow per unit width ( cfs/ft)yo= mean depth of overland flow

………..4.11

………..4.12

Module 4

fix

qt

y oo −=∂∂

+∂∂

fixy

ymt

y omooo

o o −=∂∂

+∂∂ −1α



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

occ q

xQ

tA

=∂∂

+∂∂

cmccc 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

2194.0

+ zz

ns

ns72.0

( )3/249.1 −Wn

s

( )6/1804.0cD

ns

• Determine αc and mc for the case of a triangular prismatic channel

11

ZZ

yc

Example Problem

Module 4

Solution

and yc = channel depth

Wetted perimeter =

hydraulic radius =

Substituting these into manning’s Eq. given by

2cc ZyAArea ==

c

c

PA

R =

Module 4

212 zyP cc +=

3/2

3/549.1

c

cc P

As

nQ =


From Eq.14, . Therefore,

and mc= 4/3

Module 4

( )( ) 3/123/2

3/103/5

159.149.1

Zy

yZs

nQ

c

cc

+=

( ) 3/423/1

2194.0

cc ZyZ

Zsn

Q

+

=

( ) 3/43/1

2194.0

cc AZ

Zsn

Q

+

=

cmccc AQ α=

3/1

2194.0

+

=Z

Zsncα


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


Module 54 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) systemsProtective measuresHydro-system operation

Water Dynamics:Ungauged riversPeak flow estimationRiver-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 TransferFunction

Outflow)( =tQDownstream hydrograph

OutflowQ

Module 5

Flow Routing Analysis Contd…

As flood wave travels downstream, it undergoesPeak 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= inflowQ= outflowS =storaget=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



STEPSFor 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 −

IQ

II

IQ −

I Q

AdvancingFloodWaveI > Q

RecedingFloodWaveQ > 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



St. Venant Equations (Refer to Module 6 for more details)

Kinematic wave routing




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.

Qdischarge and y depth of flow



Module 5

,

dydA

dydQ

dAdQc ==

tOIIO 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

QX***

121

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 completeSt.Venant equation represents the dynamic wave routing. (Refer to Module 6 for moredetails)

Valley storage

Prismstorage

Wedgestorage

Non-conservative form of continuity equation

Module 5

∂∂

+

∂∂

+∂∂

=xyV

xVy

ty0

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 SSgxyg

AQ

xAtQ

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

5927.01)15.01(3.2*21)15.01(*3.2*2

)1(2)1(2

3442.01)15.01(3.2*2

15.0*3.2*21)1(2

2

0631.01)15.01(3.2*2

15.0*3.2*21)1(2

2

3

2

1

=+−−−

=∆+−∆−−

=

=+−

+=

∆+−+∆

=

=+−

−=

∆+−−∆

=

tXKtXKC

tXKKXtC

tXKKXtC

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

Dis

char

ge (c

fs)

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. 503:00P.M 75

9:00 P.M. 1503:00A.M. 4509:00A.M. 10003:00P.M. 8409:00P.M. 7503:00A.M. 6009:00A.M. 3003:00P.M. 1009:00P.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 routing\Muskingum method

Series of reservoir models

Distributed / Hydraulic routing

Flow is calculated as a function of space and time throughout the system



Module 5

Distributed / Hydraulic routing methods


Muskingum Cunge method


Complete solution to St.Venant equations

Module 5



Equations Governing Hydrologic and Hydraulic Routing

Module 65 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 dmdm

β β= =

mass (m) momentum

P.E. (mgh) K.E. (1/2 mv2)

1)(==

dmmdβ )( vm

gh=β

Module 6

vdm

vmd==

)(β2

21 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

relvc

dAVdtdt

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…

III 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’

III

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 term1st term is equivalent to the change in extensive property stored in control volume (as ∆t 0)

Module 6

( ) ( )[ ]

( )

( ) ( )[ ] ( ) ( )[ ]{ }tIttIIItIIttIItsys

tIIIttIIIIItsys

BBBBtdt

dB

eqFrom

dincontainedfluidtheofmassdm

volumeelementaldddmwhere

ddmdBHere

BBBBtdt

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

fluidofpropertyextensiveofchangeofrateTotaldtdB

changeShapepropertiesextensiveinChangeBdtd

onlypropertiesextensiveinChangeBt

Note

ldA

surfacecontrolthethroughgpasfluidofvolumeTotal

sys

sc

+=

=⇒

+⇒

⇒∂∂

∆−= ∫∫ θ

:sAssumptiontconsisi tan)( ρ

Module 6

( )

)(inf)()()(,

),()(.

0.

0,

.var

)2.6......(0.

0

,,.

....

....

..

....

velowtIandveoutflowtQstorageS

wheretItQdAVandSd

dAVddtd

ddtdflowstatesteadyFor

flowunsteadydensityiableofequationtheisThis

dAVddtd

dtmassd

dtdB

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

jtimeatstorageSwhereQISS

dttQdttIdS

dttQdttIdS

tQtIthendtdS

provedhencetQtIdtdSor

tItQdtdSei

j

j

−

−

−

−

∆

∆−

∆

∆−

=

=

−=

=

−=−

−=∴

−=∴

==∴

=−=

=−+

∫∫∫−

flowsteadyandtconsisii tan)( ρ

∑∫ ⋅+∀=CSCV

sys ddtd

dtdB

AV

βρβρ

∑ ⋅=CS

sys

dtdB

AV

βρ

Module 6

,

)()(

),4.6(

)5.6......(),3.(

)4.6......()(,2

)3.6......()(

,1

221102

221101

1

1101

2212

1101

Thus

QIQISSorQIIQSS

inSofvaluengSubstituti

IQSSEqFrom

QISSj

QISS

jIf

−+−+=−=−+−

+−=

−=−=

−=−

=

∑=

−+=j

tttj QISS

10 )(

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 flowSection 2 – gradually varying flowSection 3 – hydraulic jump

Section 4 – weir and waterfallSection 5 – gradually varyingSection 6 – hydraulic drop due to change in channel slope

Module 6

Open Channel Flow

Unsteady Steady

Varied Uniform Varied

Gradually

Rapidly

Gradually

Rapidly

(NPTEL, Computational Hydraulics)

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 watermovement

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 Venantequations

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, zui: u, v, wuj: u, v, w

Module 6

( )

( )

( )8.6

7.6

6.6

zww

ywv

xwu

twa

zvw

yvv

xvu

tva

zuw

yuv

xuu

tua

z

y

x

∂∂

+∂∂

+∂∂

+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=

∂∂

+∂∂

+∂∂

+∂∂

=

j

ij

ii x

uutua

∂∂

+∂∂

=

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

∂∂

+∂∂

+∂∂

+∂∂

−+=

∂∂

+∂∂

+∂∂

+∂∂

−+=

∂∂

+∂∂

+∂∂

+∂∂

−+=

zyxzpgf

zyxypgf

zyxxpgf

zzyzxzzz

zyyyxyyy

zxyxxxxx

τττρ

τττρ

τττρ


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

∂∂

+∂∂

==

∂∂

+∂∂

==

∂∂

+∂∂

==

∂∂

=∂∂

=∂∂

=

yw

zv

zu

xw

xv

yu

xw

xv

xu

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

notationEinsteinxx

uxpgf

zw

yw

xw

zpgf

zv

yv

xv

ypgf

zu

yu

xu

xpgf

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

νρ

νρ

νρ

νρ


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

ij

i gxx

uxp

xuu

tu

+∂∂

∂+

∂∂−

=∂∂

+∂∂ ν

ρ

)20.6(0

0

=∂∂

=∂∂

+∂∂

+∂∂

i

i

xu

zw

yv

xu


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 channelA-Area of cross-section of flowy- Depth of flowS0-Channel bottom slopeSf- Friction slope

0=∂∂

+∂∂

tA

xQ

0)(11 2

=−−∂∂

+

∂∂

+∂∂

fo SSgxyg

AQ

xAtQ

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

dxxQQ∂∂

+

tAdx∂

∂ )(ρ

xQ∂∂

Plan View

Elevation View

Module 6

St. Venant equations

Conservation of MassIn 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=∂∂

+∂∂

tA

xQ 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.

=

∂∂

+∂∂

⇒

−

==

∂∂

+∂∂

∆⇒

=∆∂

∂+∆

∂∂

∆∂

∂−=−=∆

∂∂

∆+

xQ

tA

tioncrossthethrough

edischQAVHerex

AVtAx

xx

AVxtAei

xx

AVAVAVxtA

xxx

ρ

ρρ

ρρρρ

In 1-D open channel flow continuity equation becomes,

0)(=

∂∂

+∂

∂ty

xVy

Non-conservation form (velocity is dependent variable)

0=−∂∂

+∂∂ q

tA

xQ

Conservation form

Module 6

0=∂∂

+∂∂

+∂∂

ty

xVy

xyV

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

AVAVdtdhA

AVAVhAdtd

ddtd

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 forceson the C.V.

Momentum storedwithin the C.V

Momentum flowacross the C. S.

Module 6

∫∫∫∫∫∑ +∀=....

.scvc

dAVVdVdtdF ρρ

∫∫∫∫∫∑ +∀=....

.scvc

dAVVdVdtdF 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 volumeSince 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 SSgxyg

AQ

xAtQ

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:KinematicDiffusionQuasi-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=∂∂

+∂∂

xQc

tQ

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

xySS f ∂∂

−= 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 xgVV

xySS f ∂

∂−

∂∂

−=

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

10080

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 flowHydrostatic pressure distribution in a curved surface is,

yan

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 timeA = 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)(==

dmmdβ

From Reynold’s transport theorem, we have

Module 6

Gradually Varied Flow (Contd..)

Module 6

( )

,var

0

,,

.

flowunsteadydensityiableofequationtheisThis

dtmassd

dtdB

massofonconservatioflawperasNow

massbetoBConsider

sys ==

tconsisand tanρ

txoffunctionareiablestheallthatindicatest

dAVdt scvc

&var

0.....

∂∂

=+∀∂∂

⇒ ∫∫∫∫∫ ρρ

0.....

=+∀∴ ∫∫∫∫∫scvc

dAVddtd ρρ

Control volume (A.dx)

-Longitudinal Section of the Channel

Module 6

( ) )21.6(....0 ∫∫∫∫ ++∂

∂=

outletinlet

dAVdAVt

dxAor ρρρ

yηdη

1 2dx

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

xQ

tAq

∂∂

+∂∂

=

Basic equation or Conservative form of

continuity equation (Applicable for kinematic &

non-prismatic channels)

Module 6

( ) ( )

( )

( )

orxQq

tA

dxbysidesbothDividing

dxxQqdx

tdxA

dxxQqdx

tdxA

dxxQQqdxQ

tdxA

aswrittenrebecanEquation

∂∂

−−∂∂

=

∂∂

−−∂

∂=

∂∂

+−∂

∂=

∂∂

+++−∂

∂=

−

0

,''

,..

..

..0

,)21.6(

ρ

ρρ

ρρρ

ρρρ

dη

η

dy

y

y- η

B’

Module 6

( )

( ) ( )

0,

,;,

.,

=∂∂

∂∂

∂∂

+

∂∂

+

∂∂

∂∂

=

==∂∂

+∂∂

+∂∂

=

∂∂

+∂∂

=

xAchannelprismaticFor

xy

yAV

xVA

ty

yAqor

txyytyfAHerexAV

xVA

tAqor

xVA

tAqNow

Module 6

( ) ( )

∂∂

+

∂∂

+∂∂

=

==

∂∂

+

∂∂

+

∂∂

=

==

=⇒=

∂∂

++

∂∂

+

∂∂

=

≅∂∂=

xyV

xVD

tyor

channeltheofwidthunitBqgAssuxyBV

xVDB

tyBqor

txyytyfAHereDBABADdepthhydraulicNow

channelprismaticforxyBV

xVA

tyBqor

ByASo

dyBdA

0

)(1'&0,min

'.'.'

,;,'.'/)(,

,'0'

',

'

Valley storage

Prismstorage

Wedgestorage

Non-conservative form of continuity equation

Now, from Reynolds transport theorem:

From conservation law of momentum(Newton’s 2nd law of motion) ∑= F

dtdB

momentum )( vm

Module 6

or

yByBBADei

lengthchannelthethroughoutuniformsameisdepthIf

=== '/'.'.,.

,)(

∂∂

+

∂∂

+∂∂

=xyV

xVy

ty0

∫∫∫∫∫ +∀∂∂

=....

.scvc

dAVdtdt

dB βρβρ

vdm

vmd==

)(β

Unsteady Non-uniform flow

Steady Non-uniform flow

0∑ =FFor steady uniform flow,

Module 6

∫∫∫∫∫∑ +∀∂∂

=∴....

.scvc

dAVVdVt

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

yy- 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 dη

η γ η η=

= −∫10

Module 6

( )

( ) ( )

p

y

p

F&F F dxx

&F F F

F FF (F dx ) dxx x

F b.d y dxx

and y y( x,t ) (u sin g Leibnitz rule)

η

η γ η=

∂ = + ∂ = −

∂ ∂ = − + =− ∂ ∂ ∂ ⇒ = − − ∂

=

∫

12 1

1 2

1 11 1

0

1 2

1F dxxFF

∂∂

+ 11

a) Hydrostatic Force Contd…

( ) ( )

{ } ( )

b( t )

a( t )

b( t )

a( t )

[General rule :

F t x,t dx

d dF(t ) x,t dxdt 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 yx dx dx

(b)b y y d dxx x

yb

η

η

η

η

φφ φ

ηγ η η γ γ

γ η η η

γ

=

=

=

=

= + −

∂ = − − − − + − ∂ ∂ ∂ = − − + − ∂ ∂

∂= −

∂

∫

∫

∫

0

0

00

( )

( )

y

y y

by d dxx x x

y bb d y d dxx x

η γ η η

γ η γ η η

∂ ∂ − + − ∂ ∂ ∂ ∂ = + − ∂ ∂

∫

∫ ∫

0

0 0

( )

( ) dxdxbydxA

xy

dxdxbydxdb

xy

y

yy

∂∂

−−

∂∂

−=

∂∂

−−

∂∂

−=

∫

∫∫

ηηγγ

ηηγηγ

0

00

b) Boundary Reaction

Module 6

( )

.

..

,

)2()2(sec

)1()1(sec.

alsoondistributipressurechydrostatitheinChange

dxdxbdx

xdbdx

xA

areaelementalinChange

tionatareaElementalAdxx

A

tionatAdbareaElementary

element

elementelement

element

∴∂∂

=

∂∂

=

∂∂

=

∴

−=+

∂∂

−==

ηη

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 forcefor 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)(

WFdxAWWeight

tconsbeAareaLet

g =∴

=∴

Module 6

( )

depthmeanhydraulicorradiushydraulicRwhere

SRSPAor

balancedentirelyareforcestwoThesesAssumption

stressshearperimeterwettedPwhereSdxAdxPissidesandbottomchannelforforceShear

SdxAForSandofvaluessmallveryFor

g

=

=

=

===

=

=≅

,....

:

,.....

,

...

tantansin,

000

000

0

0

γγτ

τγτ

γ

θθθθ

c) Body Force Contd…

.,, 0

hereflowuniformunsteadyconsidertoneedweHoweverSSflowuniformsteadyFor f=

d) Frictional Force

e) Wind Force

Module 6

( )

ff

f

f

SdxAFor

SRdxPdxPforceShear

SR

...

......

..

0

0

γ

γτ

γτ

−=

−=

−=∴

=

frrf

w

ww

w

WVVC

channeltheoflengththethroughoutuniformisitandBiswidthtopIfdxBwindforceF

bestressshearwindLet

ρρ

τ

ττ

−=−=

==

2...

)(..

.,

.,

veissoveisVveisVIf

veissoveisVveisVIf

wrr

wrr

+−⇒−

−+⇒+

τ

τ

ffw

f

f

WdxBWdxBFfactorstressshearwindW

tcoefficienstressshearwindC

...)(. ρρ −=−=∴

=

=

f) Eddy Losses:

( )2

2

2

2

22

2

2

)(

∂∂

=∂∂

=

∂∂

=

∂∂

∝

⇒

AQ

xgKv

xgKS

gv

xKSor

gv

xS

SlinegradientenergytheofSlopelossEddy

eee

ee

e

e

ωVω

Vω.cos ω

V(=Q/A)

Module 6

,,

.)(

exp...:sin

12

equationmomentumNowcasestheallinvealwaysisKthatEnsure

VVObviouslytcoefficienansionorncontractioKHere

SdxAFlosseddythegcauforceDrag

e

e

ee

+

>>

=

−= γ

AdVVdVdtdF

sccv

......

ρρ ∫∫∫∫∫∑ +∀=

as vdm

vmd==

)(β

( ) AdVVAdVVdxAVdtdF

outletinlet

......... ρρρ ∫∫∫∫∑ ++=

Module 6

Module 6

( )

( )dxqVQVdxqQand

directionvealongvelocityofcomponentbetoVConsider

factorcorrectionmomentumHere

AdVVAdVVdxAVdtdF

x

x

outletinlet

.......,.

)(

)1(.........

11

1

ρβρβ

β

ρρρ

==

+

=

++= ∫∫∫∫∑

)..(1 QVρβ ( ) )....(1 dxQVx

QV ρρβ∂∂

+

Forces acting on the C.V. (Contd..)

Body force

Friction force

Contraction/Expansion

forceWind force

Hydrostatic force

Boundary Reaction

Module 6

( ) ( )( ) ( )

( ) ( )

:exp,

........

..............

),1(,

1

111

forcetotaltheoutfindtoansionstheallcombiningNow

dxQVx

dxqVdxAVt

dxQVx

QVdxqVQVdxAVdtd

equationNow

x

x

∂∂

+−∂∂

=

∂∂

+++−

ρρβρ

ρρβρβρβρ

( )

( ) dxdxby

dxdxbyAdx

xyBdxWgAdxSgAdxSgAdxS

y

y

fef

ηηγ

ηηγγρρρρ

∂∂

−+

∂∂

−−∂∂

−−−−

∫

∫

0

00

Module 6

( ) ( )

( )

( )

( )

∂∂

+−

∂∂

=∂∂

−−

∂∂

+−∂∂

=∂∂

−−

==

∂∂

+−∂∂

=

∂∂

+−∂∂

=

∂∂

+−∂∂

=

∂∂

−−−−

AQ

xgAgAqV

tQ

gAxySSor

gAbySHRSHLdividingNowA

Qx

qVtQ

xygASSgA

SWgAssuA

Qx

qVtQ

QVx

qVtQ

dxQVx

qVAVt

AxygBWgASgASgASor

dxbySHRSHLdividingNow

xf

xf

ef

x

x

x

fef

211

0

2

110

2

11

11

1

0

.1

:""..&..,

.

,0&0min

.

..

...

:""..&..,

ββ

ββ

ββ

ββ

β

ρ

Module 6

( )

,""..&..

011,10

:)(

0

2

1

gbySHRSHLgMultiplyin

SSxy

AQ

xgAtQ

gAandqIf

channelsprismaticforequationmomentumofformonConservati

f =−−∂∂

−

∂∂

+

∂∂

== β

Local acceleration

term

Convective acceleration

term

Pressure force term

Gravity force term

Friction force term

0)(11 2

=−−∂∂

+

∂∂

+∂∂

fo SSgxyg

AQ

xAtQ

A

Module 6

( )

0)(.2.

0,,

0)(.2.

0)(

.11..21..1

0)(.1.1

.,

0)(11

:

2

tan

tan

2

2

tan

2

22

2

=−−∂∂

+

∂∂

∂∂

+

∂∂

+

∂∂

∂∂

+∂∂

=∂∂

=−−∂∂

+

∂∂

∂∂

+∂∂

+

∂∂

+

∂∂

∂∂

+∂∂

=−−∂∂

+

∂∂

∂∂

+

∂∂

+

∂∂

+

∂∂

+∂∂

=−−∂∂

+

∂∂

+∂

∂

=

=−−∂∂

+

∂∂

+∂∂

=

=

=

fo

tconsy

fotconsy

fo

tconsy

fo

fo

SSgxyg

xy

yA

AV

xVV

ty

yA

AV

tVor

xAchannelprismaticforNow

SSgxyg

xy

yA

xA

AV

xVV

ty

yA

AV

tVor

SSgxyg

xy

yAV

AxAV

AxVVA

AtAV

tVA

Aor

SSgxyg

AVA

xAtVA

Aor

VAQTaking

SSgxyg

AQ

xAtQ

A

EquationMomentumofformonConservati

Kinematic Wave

Diffusion Wave

Dynamic WaveModule 6

0)(

0)(..

2

2

=−−∂∂

+

∂∂

+

∂∂

+

∂∂

+

∂∂

+∂∂

=−−∂∂

+

∂∂

∂∂

+

∂∂

+

∂∂

+

∂∂

+∂∂

fo

fo

SSgxyg

xy

yV

xVV

xVV

ty

yV

tVor

SSgxyg

xy

yA

AV

xVV

xVV

tyB

AV

tVor

0)(.. =−−∂∂

+

∂∂

+

∂∂

+∂∂

+∂∂

+∂∂

fo SSgxyg

xVV

xyV

xVy

ty

yV

tV

Continuity Equation forNon-Conservation form(=0)

0)( =−−∂∂

+∂∂

+∂∂

fo SSgxyg

xVV

tV

Module 6

gAqVSS

xy

xV

gV

tV

g

gbySHRSHLDividingA

qVSSgxyg

xVV

tV

flowernalisthereifBut

xfo

xfo

1

1

)(1,""..&..

)(

,int

β

β

=−−∂∂

+∂∂

+∂∂

=−−∂∂

+∂∂

+∂∂

fo SSxy

xV

gV

tV

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

gAqVSS

xy

xV

gV

tV

g

gAqVSS

xy

xV

gV

gAqVSSor

gAqVSS

flowuniformfortermonacceleraticonvectivexV

flowsteadyfortermonacceleratilocaltVHere

xfo

xfo

xfo

xfo

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)()(

31

222

2

31

222

2

eqmomentumyh

vuvgnyzgh

yhgh

yhv

xhuv

thv

eqmomentumxh

vuugnxzgh

xhgh

yhuv

xhu

thu

eqcontinuityyhv

xhu

th

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

http://www.efm.leeds.ac.uk/CIVE/CIVE3400/stvenant.pdf�

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)

Ups

trea

m b

ound

ary

Dow

nstr

eam

bou

ndar

y

Tim

e, 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 viewin x-t plane

Module 6

Finite Difference Scheme (FDS) Contd…

Module 6

Explicit Implicit

Temporal derivative

Spatial derivative

Temporal derivative

Spatial derivative

j ji iu uu

t t

+ −∂≈

∂ ∆

1

j ji iu uu

x x+ −−∂

≈∂ ∆

1 1

2

tuuuu

tu j

ij

ij

ij

i

∆−−+

≈∂∂ +

++

+

21

11

1

xuu

xuu

xu j

ij

ij

ij

i

∆−

−+∆−

≈∂∂ +

+++ 1

111 )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+1 i-1

Space

h1h3

h5

h7

Q

Q

Q

Centerpoint

Spatial and temporal derivatives use unknown time lines for computation ‘j+1’

Implicit FDS

Module 6

24

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:

2yP

2γ=

)(sin22 12

22

21 vvqRWyy

f −=−+− ρθγγ

2

22

1

21 qv

2yqv

2y

ρ+γ

=ρ+γ

Momentum Analysis in an Open Channel Contd…

Module 6

Specific force plus momentum curve

• i.e, Mg

qv2y

gqv

2y 2

221

21 =+=+

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)

v2

v1

y1

y2

12

Mg

qv2y

gqv

2y 2

221

21 =+=+


Module 6

Lecture 5: Energy equation and numerical problems

Module 6

Energy in Gradually VariedOpen 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 respectivelyv1 and v2 : velocity of flow at sections 1 & 2 respectively

and : energy coefficientsz1 and z2: elevations of channel bottom at sections 1 & 2 respectively w.r.t any datumhL: energy head loss of flow through the channel

1α 2α

Lhzg

vpzg

vp+++=++ 2

22

22

1

21

11

22α

γα

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.

gvyg

vy 2222

2

21

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

gAQyE

org

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 sectionModule 6

gvzyE 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

2gAQyE +=

2

2

2*81.9*231+=E

Specific Energy Diagram

Module 6

The specific energy can be plotted graphically as a function of depth of flow : E = Es + Ekwhere

)(2 2

2

energyKineticgAQEk =

)( energyStaticyEs =

2

2

2gAQyE +=

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 gravityPerhaps 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 = subcriticalRight = 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 < 1supercritical: Fr > 1

Subcritical flow with dh / dx > 1 dz / dx < 1Supercritical flow with dh / dx > 1 dz / dx > 1

if flow is subcritical if flow is supercritical

Module 6


Hydraulic Jump

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


Module 77 Lectures

Objective of this module is to learn the fundamentalsof 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

212s 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)

NotationP [ X = x] = p(x)P

[ ] [ ]i

i ix 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)

ax

+α

a x

F(x)

+α-α

1.0

( ) [ ] ( ) a

F a P X a f x dxα−

= ≤ = ∫

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 dxα

α−

≠ ≤

≠ ≥

∫ ∫

d x0

f(x)

P[X = 0]

X = 0Module 7

d x0

Jump ΔFP[X=d]X=d

CDF for this case

CDF within an interval

Δx

x 2 2 2

i i i

x x xx x∆ ∆ ∆ − +

xi

f(xi)

i

ii

ii

ii

xxf

xxxx

ervaltheincurvetheunderAreaxxFxxF

xxXxxP

∆=

∆

+∆

−

=∆

−−∆

+=

∆+≤≤

∆−

*)(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. 23( ) ; 0 x 5

125

xf x = ≤ ≤x1 x2 x3

0 0.25 0.75 1.25 1.75 5.0

xi f(xi)=3x2i/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 EvaporationTemperature Radiation Coeff.Rainfall RechargeRainfall 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

==∞∞=

=≤≤=

=∑∑= =

FFF

YXPF

YXpi 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 50002

=

≥

≤

−

−

∫ ∫

∫ ∫ ∫

y

y

D

f x y dxdy

Cdxdy y dy

y y

90002 2

5000

(9000) (5000) 1- C 45000000 250000002 2

17 = 25

− − +

Module 7

Example Problem 1 Contd…

2

2 2

2 2 2 2

2

1F(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 yx

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 22

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 12

0 0

11 22

0 0

1 22

0

1 2 3 2 3

0

[ 1]

1 [ 1]

= 1- 3

=1- dx6

(1 2 )= 1- (1 ) dx6

6 6 2 )= 1- dx6

x

x

P X Y

P X Y

xyx dydx

y xx y

x x xx x

x x x x x

−

−

+ ≥

= − + ≤

+

+

− +− +

− + − +

∫ ∫

∫

∫

∫

12 3

0

13 4 2

0

1=1- (4 5 ) dx6

1 4 5=1-6 3 4 2

1 4 5 116 3 4 2

1 716 12

6572

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 jj

j i ji

p x p x y

q y p x y j

α

αα

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 02

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

54

x yyx y C x for

SC t

sC s [ ] − = ⇒ − − =

∫ ∫

22 2

0 00

1 10 1 2 1ts dt C t dt

[ ]− = ⇒ − =

− = ⇒ =

∫2

2 20

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 ] = 514 4 14 4 2

x y

yx

x

x y prob X x Y y

S t

y xy x yy ty dt xy − −

1 1 1 = 5 = 0.30414 4 2

≤ ≤ii) probability of x 1 & y 1 Find

iii) "marginal desities" P (x) & P (Y) x yFind

=

− −

− −

− − −

∫

∫

2

,02

022

0

P (y) ( , )

1 = (5 ) 2141 = 5

14 2 21 1 = [10 2]= (8 )

14 14

y X YP t y dt

y t dt

yt tt

y y

= − −

− −

− −

−

∫ ∫2 2

,0 0

22

0

P (x) ( , ) = (5 ) /14 2

1 = 514 41 = [10 1 2 ]

141 = (9 2 )

14

x X YsP x s ds x ds

ss xs

x

x

Module 7

α α

α −∴ = ≤

− −

∫ ∫

, ,

, ,0 0

2 2

0

iv) cumulative marginal distributions P (x, ) & Px ( ,Y)

9 2P (x, ) ( )= ( ) = ( ) 14

9 9 = = 14 14

x y x y

x x

x y X

x

Find

xprob X x P t dt dx

x x x x

α − −

= ≤

−=

−=

∫ ∫2

,0 0

2

2

8 8 / 2P ( , ) ( )= ( ) = ( ) = 14 14

9 Check putting x=2 , 114

16& y=1, 128

y y

x y yy y yy prob Y y P s ds ds

x x

y y

− −

−− −

−

/ ,

/ ,

v) "conditional densities" (5 )2 P (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

Findy x

yy x

andx

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 yg x yh y

=

( , )( | ) but g(x)>0( )

f x yh y xg x

=

α α α

α α αα

α

− − −

−

≥

= =

∴ =

∫ ∫ ∫

∫

( | ) 0 as f(x,y) and h(y) are (+ve)( , ) 1 1( | ) 1 = ( , ) = ( ) = 1( ) ( ) ( )

( , ) ( )

g x yf x yg x y dx dx f x y dx h yh 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 xh 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 yx y

x y

f x y eee 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

Examplecx discrete xx

7 takes 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 5y 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+11P[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

xx

e dx

e e

e

Alternatively

ee dx eαα

−

− −

−

−− −

−− −

=

= = −

∫

∫Module 7

2

i) g(x|y) = f(x,y) / h(y)

1 14 (10 2 ) = (5 ) = , 0 x 2, 0 y 2214 (8 ) 2(8 )

Conditional CDF:

(10 )( | ) [ | ]= ( | ) = 2(8 )α−

− −− − × ≤ ≤ ≤ ≤

− −

− −= ≤ =

−∫x

y xy xy y

x xy xg x y P X x Y y g x y dxy

≤ =

− − − −=

−

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 01 1

ii) Now, g(x|y 1] 0 x 2; 0 y 2

1( , ) (5 )14 2 = =

( ) (8 ) /14o o

yf x y dy x dy

h y dy y dy

Module 7

− − − − − − − − −

−

∫

∫

1 12

0 01 12

0

(5 ) 5 152 4 4 = = = 18(8 ) 8 22

(19 4 ) = 30

o

y yx dy y xy x

yy dy y

x

[ ]2

0

2

1 x= 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 xiii F x y dx

Now

x xP 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 variationModule 7

α

α

α

α

α

α

µ

µ

µ

µ

+

−

+

−

+

−

=

= =

=

=

∫

∫

∫

0

0 00

01

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 nn

th

th

x f x dx

o x f x dx

x f x dx simply

nα

α

µ+

−

= −∫ment about the expected value ( ) ( ) ...for population (not sample)nn x u f x dx

Moments of Distribution

dxx

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

xf(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 isgiven by = 0.003 v . Find the expected value of pressure.

Sol :1 f(v) = ; 0 v 10; = 0.003 v

10

E( ) =

≤ ≤ ωω

≤ ≤ 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 0.0274Therefore, 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

12

1 12 2

0.330.3 21

2

00

1or v = 18.257 or 18.257 ( )0.003 2

1 1or 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 wdw

w

ww dw

32

0 10 0 18.257 10

0 w 0.3

= 0.1 (Ans.)

≤ ≤ ≤ ≤ ≤ ≤

vor wor

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 :nj

n

r j :nj

( n j )(n j )...(n j )a xn (n )(n )...(n r )

( j )( j )...( j r )b xn (n )(n )...(n r )

=

=

− − − − +=

− − −− − −

=− − −

∑

∑

1

1

1 1 11 2

1 1 21 2

p r sM( p,r ,s ) E[ X { F( x )} { F( x )} ]= −1

rr M( , ,r ) E[ X{ F( x )} ]α = = −1 0 1

rr 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.

( )rrk

r r k:rkkr ( ) EXλ

−−−

−=

= −∑11

1

01

Module 7

L-moments Contd…

In particular, the first four population L-moments are

The first two of these L-moments have conventional names:

( )( )( )

: :

: : :

: : : :

EXEX EX /

EX EX EX /

EX EX EX EX /

λλ

λ

λ

=

= −

= − +

= − + −

1

2 2 2 1 2

3 3 3 2 3 1 3

4 4 4 3 4 2 4 1 4

22 33 3 4

mean, L mean or L location,L scale

λλ

= − −= −

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

ii

n i n in

ii

n i i n i n in

ii

n i i n i i n i n in

ii

x

x

x

x

−

=

− − −

=

− − − − −

=

− − − − − − −

=

=

= −

= − +

= − + −

∑

∑

∑

∑

1

1 1 1

1 1

1 2 1 11

1 1 1

3 3 2 1 1 21

1 1 1 1

3 4 3 2 1 1 2 31

121 231 3 34

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:

Mo

de:

with ma

α

α

µ+

−

=

=

∂ ∂= <

∂ ∂

∫

∑n

ii

n

i

x f x dx

x x

f fx 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 medd

n

P x dx

Def

(b)

Median

Median

(c)

Median

(a)

[ ]

[ ] [ ]

[ ] [ ] [ ]

α α

α αα 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 2σ=

Module 7

( )( )

=

−=

−

∑2

2 1

2

2

Sample Estimate, s1

= + positive square root of variance

Co-effcient of Variation:

=

CoV = Cv =

i) Var(C) = 0ii) Var(Cx) = C var(x)iii) Var(a + bx

Standard Dev

) =

iation

:

b

n

ii

x x

n

Sx

σ σ

σµ

2 var( )x

Lecture 4: Introduction to unit hydrograph

Module 3

Unit hydrograph (UH)

• The unit hydrograph is the unit pulse response function of a linear hydrologicsystem.

• First proposed by Sherman (1932), the unit hydrograph (originally namedunit-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 rainfallgenerated uniformly over the drainage area at a constant rate for an effectiveduration.

• Sherman originally used the word “unit” to denote a unit of time. But sincethat time it has often been interpreted as a unit depth of excess rainfall.

• Sherman classified runoff into surface runoff and groundwater runoff anddefined 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


Module 3

Terminologies1. 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 hydrographBase flow

Direct runoff

Inflection point

TRtp

Effective rainfall/excess rainfall

Q (c

fs)

Module 3


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.


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 0.51-2 2.52-3 2.53-4 0.5

Stream flow data Rainfall data

Module 3

Unit hydrograph

Example Problem

• Empirical unit hydrograph derivation separates the base flow from the observedstream flow hydrograph in order to obtain the direct runoff hydrograph (DRH). Forthis example, use the horizontal line method to separate the base flow. Fromobservation of the hydrograph data, the stream flow at the start of the rising limbof 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 theordinates of the DRH by the VDRH in equivalent units of depth


Direct Runoff Hydrograph (DRH) (m3/s)

Unit Hydrograph (m3/s/cm)

0 100 0 01 100 0 02 300 200 503 700 600 1504 1000 900 2255 800 700 1756 600 500 1257 400 300 758 300 200 509 200 100 25

10 100 0 011 100 0 0

Module 3

Module 3

Unit hydrograph


0

200

400

600

800

1000

1200

0 2 4 6 8 10 12

Q (m

3 /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 UnitHydrograph.

Module 3

Unit hydrograph


Lecture 5: Commonly used distributions in hydrology

Module 7

Normal Distribution

µσσ

α α

α µ α

σ

− −= ∏

≤ ≤

≤ ≤

>

∗

∗

21 1( ) exp 22

0

xf x

Also, called Saminion Distributon; Bell shaped Distribution

Most popular dis

- x +

- +

Normal Distribut

ion:

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

µσ−

= tan varxz 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 )b then y follows a normal dist. with N(a , a

Module 7

Central Limit Theorem

µσ

= + + + +

==

1 1 3 n

i i i2

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 n2

i ii 1 i 1

then,

Sn X X X X

approches a normal distribution with E(Sn) = and var(Sn) =

i2i

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 napproaches to infinity.

σµ σ

Module 7

Uniform Distribution

( )

α α α

β

α

α ββ α

β α β α

α α α ββ α β α β α

= ≤ ≤−

= = = − −

−− = ≤ ≤ − − −

∫ ∫

∫

1( )

1( ) ( )

( )

xx x

f x

xf 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 αβ α β α β α

β α β α β αβ α

= − − − −

+ − +−

2 2 2

2( ) 2( ) 2( )

( )( ) ( )2( ) 2

x =

= =

Module 7

Uniform Distribution Contd….

[ ]

( )( )

( )( )

( )( )

( )

( )

22 2

3 322

3 3 22

3 3 2 2

3 3 2 2 2 3 3 2 2

3 3 2 2

( ) ( )

( )3

,( )

3 2

4 3( 2 )( )

124 4 3 6 6 3 3 6 3

123 3

12

E x E x

xE x dx

Now

β

α

σ

β α

β α β α

β α β ασβ α

β α α αβ β β α

β αβ α α β αβ αβ β α α β β α

β αα β α β αβ

β α

∴ = −

− ∴ = =

− −

− + = − −

− − + + −=

−− − − − − + + +

=−

− + + −=

−

∫

Module 7

Uniform Distribution Contd….

( )( )

( )( )( )

( )( )

( ) ( )

3 3

2 2

2 2

2 2

2 22

3 ( )

123 ( )

123

122

12

12 12s

α β αβ α β

β αα β α αβ β αβ α β

β αα β α αβ β αβ

β αα αβ β

α β α β

− + + −=

−− − + + + −

=−

− − − − + =−

− − + =

− − −= = =

−

Module 7

Exponential Distribution

λ

λ

λ

α αλ

αλ

αλ λ

λ λ

λ

λ

λ

λ

λλ λ

−

−

−

−

−

− −

=

= =

= −

= =

=

= − − −

∫ ∫

∫ ∫

∫

∫

0 0

0 0

0

0

( ) x>0; >0

If you set ( ) ( )

1 x>0 & >0

( ) ( )

1.

x

x xx

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

αλλ

α

λ λ

λ

λ

λ

λλ

λ

−−

+

= − −

= − +

= − + − −

= =

∴ =

0

0

2

1

1 0 0 0

1 1 finally & ;

1 variance

xx

x x

exe

xe e

x

Module 7

Gamma Distribution

( )

η η λ

α

λ λ η η ηη

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 ef x

t e dt

Module 7

Gamma Distribution Contd….

When data are positively skewed we have to use Gamma distribution.

η =0.5λ = 1

η = 1λ = 1

η = 3λ = 4

η = Scale parameterλ = Shape parameter

η γλ η

ησλ

= = =

=22

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 )

d

istributed.

v

xy v

xy Cv is Coefficient of VariationCz

SS C Cvx

µα

σσ

µσσ

=

= + + + −− ∴ = = = ≤ ≤ ∏

−− = ∴ = = ∏

1 2 n

1 2 n2

y y

yy2

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 = + 2s 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

nX 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 yF y F y = ..... ( )

nXF y

[ ] [ ]

[ ]

− −

∂ ∂ = ∂ ∂

∂ = = ∂

∴ =

1 1

( )( ) is pdf of y = ( )

n ( ) ( ) n ( ) ( )

( ) ( ) ( )

nyy y

n nX y X x

ny x x

F yf y F y

y y

F y F y F y f yy

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

( )nF ( x ) Number of observations xn

= ≤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 ii n

i iK F x , F xn nmax

≤ ≤

− = − − 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 ii

A n i . ln F( x ) ln F xn − +

=

= − − − + − ∑21

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.

( )ni i

ii

O E,

Eχ

=

−= ∑

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

θθ 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

θθ

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

ef x dx dx

θθ

θ

θ θ

θθ

−−

−

− ×

−∴

0

= = 1-e1( )

1 - e = 0.75 or = -1.3863 or = 2.164

xx

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(xθ

θθ α α− −

−= ∏

⇒

212

2

( x )1 22 2

2) (2 ) e - < x < +

Take the 1st moment Mean about the originθα

θ

α

θαθ

α

µ θ

θ

− −−

−

− −−

−

= ∏

∏

∫ ∫

∫

212

2

212

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 θθ

−

−

+∏ ∫

2

22 1 2

2

1 ( )2

y

y e dx E(X) =

θθ

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 22 1

21

12

1 02

y y

y

e ydy e dy

e dy

E(X) =

E(X) = +

θθ

θ µ

∏∏

∴ = =

11

1

22

( )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,

212

2

2 2 2

( )1 22 2 2

2

1

2

( ) ( )

( ) (2 )

,

2

x

x f x dx

x e dx

Substituting

xand

α

α

θαθ

α

µ σ µ

µ θ

θ µ

µθ

−

− −−

−

= −

− ∏

=

−

∫

∫

22σ θ=

Module 7


θ

θ θ θ

θ

=

1 1

1 2 3

,

We have the following, ( ; ) ( ; ) ( ; )

i i

Samplex

f x f x f x

x

1 3Product of ( ; ) ( ; ) is "likelihood " L

If L( ; ) ( ; ), then is the estimate preferred,

which maxmizes the likelihood func

tion.

θ θ

θ θ θ

× × ≈

>

∗

∗

i i

f x f x

x f x

θ →( ; ) evaluated at x = x i 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

in i

x

n

n

x x x

xx 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

ββ

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 be

st 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

ii

x

n

ii

n

ii

L n e

L

L n x

n xxx

β β

ββ

β β

β

=

=

=

∑∴ = −

∂∴ =

∂

∂∴ = − =

∂

∴ = =

∑

∑

Module 7


( )

µσσ

σ µ

µµ σ µ σ µ σ

σσ

µσ

σ=

− −= ∏

− −∴ ∏

− −= − ∏ −

2

2

1 2

22

1 1 ( ) exp 22 [Take, as parameter not S.D. and also]

1 1L = ( , , ) ( , , ) ( , , ) = exp 22

1 ln( ) ln(2 ) 22

in n

i

i

xf x

xf x f x f x

xnor L

µ σµ σ

µµ

µ σ=

∂ ∂= =

∂ ∂

−∂ = = ∴ − = ∂

∑

∑ ∑

1

21

ln( ) ln( ) 0 set & ,Now

ln( ) 0 ( ) 0

n

ni

ii

L Lor

xL X

= 1

n

ii

xor x

nµ = =

∑

Module 7


( )

2

2

1 2

22

1 1( ) exp 22

1 1( , , ) ( , , ) ( , , ) exp 22

1ln( ) ln(2 ) 22

µσσ

σ µ

µµ σ µ σ µ σ

σσ

µσ

σ

− −= ∏

− −∴ ∏

− −= − ∏ −

[Take, as parameter not S.D. and also]

L = =

in n

i

xf x

xf x f x f x

xnor L1

21

ln( ) ln( )0

ln( ) 0

( ) 0

µ σµ σ

µµ σ

µ

=

=

∂ ∂= =

∂ ∂

−∂ = = ∂

∴ − =

∑

∑

∑

set & ,Now

n

i

ni

i

i

L Lor

xL

X

Module 7


µ

µσµ σ σ

µσ σ

µσ

=

=

=

=

=

− −∂ ∏ = = − − ∂ ∏

−+ =

−=

∑

∑

∑

∑

1

2

2 31

2

31

2

2 1

( ) ( 2)ln( ) 4 102 22

( )0

( )

n

ii

ni

i

ni

i

n

ii

xor x

n

xL nand

xnor

xor

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.)


Module 85 Lectures

Objective of this module is to learn the concepts instochastic 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 kCX

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

xT

xModule 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

TT

X k C

X SkX Xor 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 ii

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 XNotations :

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

xor . or . .

10

2

0 895 0

0 895

1255 1255 15 10 28 33Tt

= 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

10 95 0 0 110

1 0 1 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 073375

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 0 790 846

1 0 846 0154

V

C

But we are interested in F(x)

= +

= = = =

=

∴ ≥ = =

= − + ×=

≥ = − =

T

T

T

T

or ( . K )

SCoV . / .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

σµ µ

∈

1Estimate of X

nt

i

xn 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 22 221 1 1 1 1

22 221 1 1 1 1

2 2 2 2 2 2 2 21 1

2 2 21

, ( )

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 ρµ

+ + = + − +∈ = + − +

=

2

2

, ), (0, )

x x

then Nµ σ

σ∈∈

First Order, Stationary Markov Process Contd…

Module 8

( )( )

1

N

2N

21

21 1 1

1) ,

2) , R (0,1)

0such that R (0, )

= 1

( ) 1

3) If negative value comes, retain it for gent 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

σσ

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)

{ }

( )21 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

ji j j j ij j i j j j

j

X X t

ρ

σµ ρ µ σ ρ

σ+

+ + + +

∴ = + − + −

Module 8

Month SD rj1 15.7 4.14 0.8642 13.62 4.15 0.3023 26.21 24.32 0.854

12 18.01 5.08 0.637

X

11 1One assumption, X X Xµ= =

correlation with next month

This preserves seasonal mean, standard deviation and lag1 correlation.

Non-stationarity in mean and std. deviationModule 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.

tij t j t ip 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-2ij

t

ptij

then, the series is called homogeneous Markov Chain t,t tij 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 2mij

m1 m2 m3 mm

TPM, (Total probability matrix)p p p pp 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 tij 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

ijmij

ijj

m m m m

p

np

n

∧

∧

=

− = −

=∑


Historic data

1 31 2 1 2

100Time 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 statesDrought, 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

nTime interval

t-1

• p(0)j : Initial probability of being in state j

( ) ( ) ( )1 2 1

a sum vector, ,............, .....n n n nm 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 probabilityp p p pp p p p

= , ,...,

p p p p

→

=

×

o

o

om

P

If pp

p p

p p pProbability vector at time 1

+ + + (0) (0) (0)1 11 2 21 1............ m mp 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, ,............, mp 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

, .

= . . =

Simillarlyp p P

p P Pp P

=( ) ( 0 )

Any time period n,

n np 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 probabilitycondition is achieved.Once the steady state is reached,p p pSo,p p.P

+

+

=

= =

=

Module 8

Example:

• A 2-state Markov Chain; for a sequence of wet and dry spellsi = 1 dry;i = 2 wet

0.9 0.10.5 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

pp p P

p

day

wetBecause day 0 is dry

Dry wet

Probability that day 2 will be wet

= =(2)2 0.14p

Module 8

0.9 0.10.5 0.5

(iii) Prob [day 100 is wet |day 1 is dry]

Example Problem Contd….

(100)2. .,0.9 0.10.5 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 0.30.8376 0.1624

P . .0.8120 0.18800.8334 0.1666

P . .0.8320 0.16720.8333 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

NR 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 31 0.4 0.6 02 0.2 0.6 0.23 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 0p p p .2 .6 .2 p p p

0 .7 .3.4p .2p 0 p.6p .6p .7p p0 .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 00.2 0.8 1.00 0.7 1.0

k

ik ik ijj

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 full1 2 0.52 2 intermediate2 2 0.74 2 Intermediate3 2 0.15 1 intermediate4 1 0.27 1 nearly full5 1 0.03 1 nearly full6 1 0.49 2 nearly full7 2 0.02 1 intermediate8 1 0.97 2 nearly full9 2 0.96 3 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.

( ) ( ) dldrlrfdrdllrfpl

l

r

lLR

r

r

r

lLRs ∫ ∫∫ ∫

=

=

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.

2tσ


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 95 Lectures

Hydrologic Simulation Models


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

LumpedParameters assigned to each sub-basin

Fully-DistributedParameters assigned to each grid cell

Semi-DistributedParameters 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

InterflowDirect Runoff

Surface Storage

BaseflowPercolation

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 ModelLumped parameter Snyder or Clark UHDistributed Kinematic waveEvent HEC-1, HEC-HMS, SWMM, SCS TR-20

Continuous Stanford Model, SWMM, HSPF, STORMPhysically based HEC-1, HEC-HMS, SWMM, HSPFStochastic Synthetic streamflowsNumerical Kinematic or dynamic wave modelsAnalytical Rational Method, Nash IUH

Module 9

Hydrologic Models Contd…

Models Application AreasHEC-HMS Design of drainage systems, quantifying the

effect of land use change on floodingNational Weather Service (NWS) Flood forecasting.Modular Modeling System (MMS) Water resources planning and management

worksUniversity 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 analysisHBV Flow forecasting

Module 9

A List of Popular Hydrologic Models

Module 9

Model name/acronym Author(s)(year) RemarksStanford watershed Model (SWM)/Hydrologic Simulation Package-Fortran IV (HSPF)

Crawford and Linsley (1966), Bicknell et al. (1993)

Continuous, dynamic event or steady-statesimulator of hydrologic and hydraulic andwater 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-basedrunoff model

Module 9

Popular Hydrologic Models

Model name/acronym Author(s)(year) RemarksUtah State University (USU) Model

Andrews et al. (1978) Process-oriented, event /continuous streamflowmodel

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 ModelingSystem (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…

Model name/acronym Author(s)(year) Remarks

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-statesimulator of hydrologic and hydraulic andwater quality processes

Areal Non-point Source Watershed Environment ResponseSimulation (ANSWERS)

Beasley et al. (1977),Bouraoui et al. (2002)

Event-based or continuous, lumped parameter runoff and sediment yieldsimulation 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, continuoushydrologic simulation model

Generalized River Modeling Package—SystemeHydroloque Europeen (MIKE-SHE)

Refsgaard and Storm (1995)

Physically based, distributed, continuoushydrologic and hydraulic simulation model

ARNO(Arno River )Model Todini (1988a,b, 1996) Semidistributed, continuous rainfall-runoffsimulation 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, continuousrainfall-runoff simulation model

Module 9


Soil-Vegetation-Atmosphere Transfer (SVAT) Model

Ma et al. (1999),Ma and Cheng (1998)

Macroscale, lumped parameter, streamflowsimulation system

Systeme Hydrologique EuropeenTransport(SHETRAN)

Ewen et al. (2000) Physically based, distributed, water quantityand 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, distributed, continuous hydrologic simulation 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


http://www.cabnr.unr.edu/saito/intmod/docs/tootle-hydrologic-modeling.ppt�

Digital Elevation Model (DEM)

Module 9

0.002.004.006.008.0010.0012.0014.0016.0018.0020.0022.0024.0026.0028.0030.0032.0034.0036.0038.00

(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

OutputObservations


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 conductivityb) 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

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 2.02. HEC-GeoRAS 10 (EAP) 3. HEC-GeoHMS 10 (EAP) 4. HEC-GeoEFM 1.0 5. HEC-SSP 2.0 6. SnoTel 1.2 Plugin7. 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

HEC-GeoRAS

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 BankFlow Path Centerlines

Land Use Themes

Cross SectionCut 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 100IO 4KK SUB1KMPI 0.2 1.5 2 1 0.5BA 2.5LS 66 0UC 2.5 5.5KK SUB2KMBA 2.7LS 58 0UC 2.8 7.5KK AKMHC 2

KMRM 1 3 0.15KK SUB3KMBA 3.3LS 58 0UI 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 Hydrograph2. Distributed Runoff3. Grid-Based Transformation

Methods:a. Clark b. Snyder c. SCSd. Input Ordinates e. ModClarkf. 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

SH

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

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 9IHE, 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 9IHE, 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

(Source: http://www.crwr.utexas.edu/gis/gishyd98/dhi/mike11/M11_main.htm)

Mike 11 main menu

Module 9

Extraction from DEM

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

Import to MIKE 11

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

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+1 i-1

Space

h3

h5

h7

4

6

Q

QQ

Centerpoint

inii

nii

nii QhQq

tA

xQ δγβα =++⇒=

∂∂

+∂∂ +

+++

−1

111

1

inii

nii

nii hQh

ARCQgQ

xhgA

xA

Q

tQ δγβα

α=++=+

∂∂

+∂

∂

+∂∂ +

+++

−1

111

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 SHEModule 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 nameReservoir IDInitial water levelPriority of inflow connectionsPriority of down-stream user(s)Down-streams user(s) reduction factorDown-stream user(s) loss factorRule curvesHeight, Volume, AreaPrecipitationEvaporation

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 RunoffOverflow

Re-inflow after flood

Flow-capacity excess

Sewage/Rainfall waterSewage/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

ValidationPreparationGIS

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 Step1secMax 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 andDatabase 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

http://www.climateofindia.pbworks.com/�

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

Engineering

Transcript of Advanced hydrology