Hydrology

Precipitation
[2]

Dr. Mohammad N. Almasri

1 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation

 Objectives of this Presentation

This presentation will come across the following:

① Forms of precipitation
② Measuring rainfall
③ Rain gauge network
④ Estimation of missing rainfall data
⑤ Consistency of rainfall data
⑥ Double mass curves and hyetographs
⑦ Mean rainfall over an area
⑧ Maximum-intensity and maximum-depth duration curves
⑨ Intensity duration frequency curves
⑩ Probable maximum precipitation
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 Important Definitions Related to Rainfall

Intensity: time rate of precipitation or depth of

precipitation per unit time (mm/h)

Duration: period of time (h) during which rainfall occurs

Depth: the total amount of rainfall (mm) for a given

period of time

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 Rainfall Intensity, Duration and Depth

Intensity = 12 mm/hr
Duration = 15 minutes
Depth = 3 mm

After 15 minutes of rainfall

3 mm
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 Measurement of Rainfall

1 mm of rainfall over an area of 1 km2 represents a

volume of water equal to 103 m3

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 Introduction

The magnitude of precipitation varies with time

and space

It is this variation that is responsible for many

hydrological problems such as floods and droughts

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 Floods

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 Droughts

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 Introduction

The term precipitation denotes all forms of water that reach

the earth from the atmosphere

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 Forms of Precipitation
Drizzle
Drizzle is a fine sprinkle of water droplets of size < 0.5
mm and intensity < 1 mm/h

The drops are so small that they appear to float in the air

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 Forms of Precipitation
Rain

Rain is the principal form of precipitation where water

drops > 0.5 mm and < 6 mm

Type Intensity
Light rain < 2.5 mm/h
Moderate rain 2.5 – 7.5 mm/h
Heavy rain > 7.5 mm/h

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 Forms of Precipitation
Sleet
It is frozen raindrops (5 mm) of transparent grains which
form when rain falls through air at subfreezing
temperature

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 Forms of Precipitation
Snow
Snow consists of ice crystals which usually combine to
form flakes

When fresh, snow has an initial density varying from 0.06

to 0.15 g/cm3 (average density of 0.1 g/ cm3)

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 Forms of Precipitation
Snow

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 Forms of Precipitation
Hail
Called hailstones are irregular balls of ice of size more
than 8 mm
Hails occur in violent storms in which vertical currents
are very strong

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 Measurement of Rainfall
Rain gauge

The rainfall is collected and measured in a rain gauge

Rain gauges can be broadly classified into two

categories as:

① Non-recording rain gauges

② Recording gauges which produce continuous
rainfall intensity with time

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 Rainfall
Recording Tipping Bucket Rain Gauge

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 Rainfall
Recording Tipping Bucket Rain Gauge

The tipping bucket rain gauge consists of a large cylinder

set into the ground

At the top of the cylinder is a funnel that collects and

channels the rain

The rain falls onto one of two small buckets which are
balanced as a scale

After an amount of precipitation falls, the lever tips, other

bucket quickly moves into place to catch the next unit of
rainfall and an electrical signal is sent to the recorder
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 Sample of Rainfall Measurements
Tipping Bucket Rain Gauge
Rainfall depth

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 Measurement of Rainfall
Rain gauge considerations
① The ground must be leveled
② The gauge must be set as
near the ground as possible
to reduce wind effects
③ Must be placed in open
space. No object should be
nearer to the gauge than 30
m or twice the height of the
obstruction

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 Rain Gauge Network

To get a representative
picture of rainfall over a
catchment the number
of rain gauges should be
as large as possible (the
catchment area per
gauge should be small)

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 Density and Distribution of Rain Gauges
1 2 3

A=18 km2

4 5 6

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 Density and Distribution of Rain Gauges

7 8

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 Rain Gauge Network

However, the economic considerations, topography,

and accessibility restrict the number of gauges

Hence one aims at an optimum density of gauges from

which reasonably accurate information about the
storms can be obtained

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 Adequacy of Rain Gauge Stations

If there are already some rain gauge stations in a

catchment, the optimal number of stations that should
exist in order to have an assigned percentage of error
in the estimation of the mean rainfall is obtained by:

where N = optimal number of stations, ε = allowable

degree of error in the estimate of the mean rainfall (a
usual value is 10%) and Cv = coefficient of variation
(standard deviation/mean) of the rainfall values at the
existing m stations (in percent)
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 Adequacy of Rain Gauge Stations
For m stations in the catchment of rainfall values P1, P2, ……, Pi, ... Pm
in a known time, the coefficient of variation Cv is calculated: as:

The higher the coefficient of variation, the

greater the level of dispersion around the mean

The standard deviation is a statistic that measures

the dispersion of a dataset relative to its mean
A high standard
deviation indicates
that the data points
are spread out over a
wider range of values
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 Adequacy of Rain Gauge Stations
Example

A catchment has 6 rain gauge stations. In a year, the

annual rainfall recorded by the gauges are as follows:

Station A B C D E F
Rainfall
82.6 102.9 180.3 110.3 98.8 136.7
(cm)

For a 10% error in the estimation of the mean rainfall,

calculate the optimum number of stations in the
catchment

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 Adequacy of Rain Gauge Stations
Example

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 Adequacy of Rain Gauge Stations
Example

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 Mean Rainfall Over an Area

Rain gauges represent only point sampling of the areal

distribution of a storm

Hydrological analysis requires a knowledge of the

rainfall over an area such as over a catchment

To obtain the average value over a catchment from the

point rainfall values use:
Arithmetical-mean method
Thiessen-polygon method
Isohyetal method
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 Mean Rainfall Over an Area
Arithmetical-Mean Method
When the rainfall measured at various stations in a
catchment show little variation, the average
precipitation over the catchment area is taken as the
arithmetic mean of the station values

Thus if P1, P2, .... Pi, ... Pn are the rainfall values in a
given period in N stations within a catchment, then the
value of the mean precipitation P over the catchment
by the arithmetic-mean method is:

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 Mean Rainfall Over an Area
Arithmetical-Mean Method

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 Mean Rainfall Over an Area
Thiessen Method
In this method the rainfall recorded at each station is
given a weightage on the basis of an area closest to the
station

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 Mean Rainfall Over an Area
Thiessen Method

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 Mean Rainfall Over an Area
Thiessen Method
Station Bounded by Area Weightage
1 abcd A1 A1/A
2 kade A2 A2/A
3 edcgf A3 A3/A
4 fgh A4 A4/A
5 hgcbj A5 A5/A
6 jbak A6 A6/A

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 Mean Rainfall Over an Area
Thiessen Method

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 Mean Rainfall Over an Area
Thiessen Method

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 Mean Rainfall Over an Area
Thiessen Method – Example
In a catchment area, approximated by a circle of
diameter 100 km, four rainfall stations are situated
inside the catchment and one station is outside in its
neighborhood

The coordinates of the center of the catchment and of

the five stations are given below. Also given are the
annual precipitation recorded by the five stations in
1980

Determine the average annual precipitation by the

Thiessen method
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 Mean Rainfall Over an Area
Thiessen Method – Example

Station 1 2 3 4 5
Coordinates 30, 80 70, 100 100, 140 130, 100 100, 70
Rainfall (cm) 85.0 135.2 95.3 146.4 102.2

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 Mean Rainfall Over an Area
Thiessen Method – Example

Boundary of Area Fraction of Rainfall Weighted P

area (km2) total area (cm) (cm)
- - - 85.0 -
abcd 2141 0.2726 135.2 36.86
dce 1609 0.2049 95.3 19.53
ecbf 2141 0.2726 146.4 39.91
fba 1963 0.2499 102.2 25.54
Mean 121.83

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 Mean Rainfall Over an Area
Isohyetal Method

An isohyet is a line joining points of equal rainfall

magnitude

The area between two adjacent isohyets are then

determined

If the isohyets go out of catchment, the catchment

boundary is used as the bounding line

The average value of the rainfall indicated by two isohyets

is assumed to be acting over the inter-isohyet area
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 Mean Rainfall Over an Area
Isohyetal Method

The mean precipitation over the catchment of area A is

given by:

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 Mean Rainfall Over an Area
Isohyetal Method - Example
The isohyets due to a storm in a catchment were
drawn and the area of the catchment bounded by the
isohyets were tabulated as below

Estimate the mean precipitation due to the storm

Isohyets (cm) Area (km2)

Station - 12 30
12 - 10 140
10 - 8 80
8-6 180
6-4 20
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 Mean Rainfall Over an Area
Isohyetal Method - Example

Isohyets Average P Area Area Weighted P

(cm) (cm) (km2) fraction (cm)
Station 12 30 0.0667 0.80
12 - 10 11 140 0.3111 3.42
10 - 8 9 80 0.1778 1.60
8-6 7 180 0.4000 2.80
6-4 5 20 0.0444 0.22
Total 450 Mean 8.8

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 Preparation of Data

Before using the rainfall records of a station, it is

necessary to first check the data for:
① continuity
② consistency

The continuity of a record may be broken with missing

data

The missing data can be estimated by using the data of

the neighboring stations. In these calculations the
normal rainfall is used as a standard of comparison
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 Preparation of Data

Before using the rainfall records of a station, it is

necessary to first check the data for:
① continuity
② consistency

The continuity of a record may be broken with missing

data

The missing data can be estimated by using the data of

the neighboring stations. In these calculations the
normal rainfall is used as a standard of comparison
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 Preparation of Data

The normal rainfall is the average value of rainfall at a

particular date, month or year over a specified 30-year
period

Thus the term normal annual precipitation at station A

means the average annual precipitation at A based on
a specified period of 30 years of records

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 Estimation of Missing Data
Given the annual precipitation values: P1, ,P2, P3,.. Pm at
neighboring M stations 1,2, 3, ..., M respectively, it is
required to find the missing annual precipitation Px at a
station X not included in the above M stations. Further, the
normal annual precipitations Nx N2, ..., Ni ... at each of the
above (M + 1) stations including station X are known

If the normal annual precipitations at various stations are

within about 10% of the normal annual precipitation at
station X, then a simple arithmetic average procedure is
followed to estimate Px

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 Estimation of Missing Data

If the normal precipitations vary considerably (>10%),

then Px is estimated by weighing the precipitation at the
various stations by the ratios of normal annual
precipitations

This method, known as the normal ratio method, gives

Px as:

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 Estimation of Missing Data
Example

The normal annual rainfall at stations A, B, C, and D in a

basin are 80.97, 67.59, 76.28 and 92.01 cm, 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 79.89 cm, respectively

Estimate the rainfall at station D in that year

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 Estimation of Missing Data
Example

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 Checking Consistency of Point Measurements

Changes in type, location, and/or environment of the gauge are

common such as:
① shifting of a rain gauge station to a new location
② the neighborhood of the station undergoing a marked change
(trees, buildings, etc.)
③ change in the ecosystem due to calamities, such as forest
fires, land slides
④ occurrence of observational error from a certain date

So it is important for a hydrologist to determine if the

precipitation record is affected by such artificial alterations of
measurement conditions and to correct them if they are present
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 Checking Consistency of Point Measurements

The checking for inconsistency of a record is done by

the double-mass curve technique

The double-mass curve is a plot of the:

successive cumulative annual precipitation collected at
a gage where measurement conditions MAY have
changed significantly
versus
the successive cumulative of average annual
precipitation for the same period of years collected at
several gages in the same region
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 Checking Consistency of Point Measurements

A change in the proportionality between the

measurements at the suspect station and those of the
region is reflected in a change in the slope of the trend
of the plotted points

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 Checking Consistency of Point Measurements

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 Checking Consistency of Point Measurements

The problem is in
station E

Find the average

for stations A to D
and then the
cumulative

Find the
cumulative for
station E
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 Checking Consistency of Point Measurements

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 Checking Consistency of Point Measurements
Apparently, there is a slope difference

The slopes:
before the change is 0.77
after the change it is 1.05

To reflect the conditions that exist before

the break then multiply by 0.77/1.05 all
the records after change

To reflect the conditions that exist after

the break then multiply by 1.05/0.77 all
the records before the change
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 Presentation of Rainfall Data
Point Rainfall

Point rainfall refers to the rainfall data of a station

Data can be listed as daily, weekly, monthly, seasonal,

or annual values for various periods

Graphically, these data are represented as plots of

magnitude versus time in the form of a bar diagram

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 Presentation of Rainfall Data
Point Rainfall

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 Presentation of Rainfall Data
Point Rainfall – Moving Average

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 Presentation of Rainfall Data
Point Rainfall – Moving Average

Moving average is a technique that enables the trend

in data to be noticed

Select a window of time range m years. Starting from

the first set of m years of data, the average of the data
for m years is calculated and placed in the middle year
of the range m

The window is next moved sequentially one time unit

(year) at a time and the mean of the m terms in the
window is determined at each window location
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 Presentation of Rainfall Data
Point Rainfall – Moving Average – Example

Annual rainfall values recorded at a station for the

period 1950 to 1979 are given

① Represent this data as a bar diagram

② Identify those years in which the annual rainfall is:
(a) less than 20% of the mean
(b) more than the mean
③ Plot the three-year moving average of the annual
rainfall time series

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 Presentation of Rainfall Data
Point Rainfall – Moving Average – Example

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 Presentation of Rainfall Data
Point Rainfall – Example

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 Presentation of Rainfall Data
Point Rainfall – Example

The mean equals 568.7 mm

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 Presentation of Rainfall Data
Point Rainfall – Example

The mean equals 568.7 mm

20% less than the mean equals 454.9 mm
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 Presentation of Rainfall Data
Point Rainfall – Moving Average – Example

No apparent trend is indicated in this plot

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 Presentation of Rainfall Data
Mass Curve of Rainfall

The mass curve of rainfall is a plot of the accumulated

rainfall against time plotted in chronological order

Mass curves of rainfall are very useful in extracting the

information on the duration and magnitude of a storm

Intensities at various time intervals in a storm can be

obtained by the slope of the curve

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 Presentation of Rainfall Data
Mass Curve of Rainfall

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 Presentation of Rainfall Data
Hyetograph

A hyetograph is a plot of the intensity of rainfall against

time

The hyetograph is derived from the mass curve and is

usually represented as a bar chart

It is a very convenient way of representing the

characteristics of a storm

The area under a hyetograph represents the total

precipitation received in the period
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 Presentation of Rainfall Data
Hyetograph

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 Plotting Position
Example

The records of annual rainfall at Station A covering a

period of 22 years are given
① Estimate the annual rainfall with return periods of
10 years and 50 years
② What would be the probability of an annual rainfall
of magnitude equal to or exceeding 100 cm
occurring at Station A?
③ What is the 75% annual rainfall at station A? (the
value of rainfall that can be expected to be equaled
or exceeded 75% times)

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 Plotting Position
Example
Year Annual rainfall (cm) Year Annual rainfall (cm)
1960 130 1971 90
1961 84 1972 102
1962 76 1973 108
1963 89 1974 60
1964 112 1975 75
1965 96 1976 120
1966 80 1977 160
1967 125 1978 85
1968 143 1979 106
1969 89 1980 83
1970 78 1981 95
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 Plotting Position
Example
Annual Annual
Return period Return period
m rainfall P m rainfall P
(years) (years)
(cm) (cm)
1 160 0.043 23.000 12 90 0.522 1.917
2 143 0.087 11.500 13 89 0.565
3 130 0.130 7.667 14 89 0.609 1.643
4 125 0.174 5.750 15 85 0.652 1.533
5 120 0.217 4.600 16 84 0.696 1.438
6 112 0.261 3.833 17 83 0.739 1.353
7 108 0.304 3.286 18 80 0.783 1.278
8 106 0.348 2.875 19 78 0.826 1.211
9 102 0.391 2.556 20 76 0.870 1.150
10 96 0.435 2.300 21 75 0.913 1.095
11 95 0.478 2.091 22 60 0.957 1.045
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 Plotting Position
Example

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 Depth-Area-Duration Relationships
Frequency of Point Rainfall – Example

① For T = 10 years, the corresponding rainfall magnitude is

obtained by interpolation between two appropriate
successive values those having T = 11.5 and 7.667 years
respectively, as 137.9 cm

For T = 50 years the corresponding rainfall magnitude,

by extrapolation is 190 cm

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 Depth-Area-Duration Relationships
Frequency of Point Rainfall – Example

② Return period of an annual rainfall of magnitude equal

to or exceeding 100 cm, by interpolation, is 2.4 years.
As such the exceedence probability P = 1/2.4 = 0.417

③ 75% annual rainfall at Station A = Annual rainfall with

probability = 0.75, i.e. T = 1/0.75 = 1.33 years. By
interpolation between two successive values having T =
1.28 and 1.35 respectively, the 75% annual rainfall at
Station A is 82.3 cm

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 Important Definitions Related to Rainfall

Return Period: the event with a return period of N

years is the event that is expected to be equaled or
exceeded every N years

Exceedance Probability: the probability that a specific

rainfall intensity is being equaled or exceeded in any
year

Frequency: the average length of time needed for at

least one precipitation event to return with an intensity
equals or exceeds a specific (maximum) value

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 Frequency of Point Rainfall

The probability of occurrence of rainfall in any year whose

magnitude is equal to or in excess of a specified magnitude
X is denoted by P

The return period is defined as:

T = 1/P
Thus if it is stated that the return period of rainfall of 20 cm
in 24 h is 10 years at a certain station A, it implies that on an
average rainfall magnitudes equal to or greater than 20 cm
in 24 h occur once in 10 years, i.e. in a long period of say
100 years, 10 such events can be expected
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 Frequency of Point Rainfall

However, it does not mean that every 10 years one such

event is likely to happen, i.e. periodicity is not implied

The probability of a rainfall of 20 cm in 24 h occurring in

anyone year at station A is 1/T= 1/10 = 0.1

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 Plotting Position

The purpose of the frequency analysis of an annual series is

to obtain a relation between the magnitude of the event
and its probability of exceedence

The probability analysis may be made by:

① arranging the given annual series in descending order of
magnitude
② to assign an order number m

Thus for the first entry m = 1, for the second entry m = 2,

and so on, till the last event for which m = N = number of
years of record
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 Plotting Position

The probability P of an event equaled to or exceeded is

given by the Weibull formula:

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 Rainfall Measurements and Hydrologic Design

Rainfall statistics are most commonly presented in the

form of intensity-duration-frequency (IDF) curves

IDF curves express the relationship between MAXIMUM

rainfall intensity and the time (duration) with a given
probability of occurrence (return period)

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 Intensity – Duration – Frequency Relationships

IDF curves enables the hydrologists to develop

hydrologic systems that consider worst-case scenarios
of rainfall intensity and duration during a given interval
of time

The idea here is that high intensity rainfall in short

periods may cause catastrophic consequences

For instance, in urban watersheds, flooding may occur

such that large volumes of water may not be handled
by the storm water system
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 Rainfall Intensity and Corresponding Depth
In general, we
may have
different rainfall
intensities but
with the same
depth

Apparently,
rainfall duration
plays an
important role in
determining
rainfall depth
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 A Typical IDF Curve

The interpretation
of any point value
obtained from the
IDF curve is that
on the average, for
any given time
duration, storms
having an intensity
(i) for that duration
would have a
recurrence interval
equal to the
corresponding
curve value
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 Interpretation of IDF Curves

For example, in any

time duration of 90
minutes, a location
could experience a
peak 2 in/hr storm
once every 20 years

The 20-yr 90-min

design storm for the
location would have a
depth of P = 3 in

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 Interpretation of IDF Curves
A 20-yr 30-min design
storm would have an
intensity of 4.6 in/hr but
with a depth of only 2.3 in

Although the latter storm

produces less depth, its
high intensity could be the
governing factor in
determining the size of
drainage works. The
probability of occurrence
of both storms would be
the same
89 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation
 Example
The mass curve of rainfall in a storm of a total duration of
270 minutes is given in the table:
① Draw the hyetograph of the storm at 30 minutes time
step
② Plot the maximum intensity-duration curve for this
storm
③ Plot the maximum depth-duration curve for the storm

Time since
start 0 30 60 90 120 150 180 210 240 270
(min)
Cumulative
rainfall 0 6 18 21 36 43 49 52 53 54
(mm)
90 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation
 Example

Time since Cumulative Incremental Intensity

start (min) rainfall (mm) depth (mm) (mm/h)
0 0
30 6
60 18
90 21
120 36
150 43
180 49
210 52
240 53
270 54
91 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation
 Example

Time since Cumulative Incremental Intensity

start (min) rainfall (mm) depth (mm) (mm/h)
0 0
30 6 6 12
60 18 12 24
90 21 3 6
120 36 15 30
150 43 7 14
180 49 6 12
210 52 3 6
240 53 1 2
270 54 1 2
92 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation
 Example

93 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation

 Example

Time since Cumulative rainfall Incremental depth (mm) for various durations (min)
start (min) (mm) 30 60 90 120 150 180 210 240 270
0 0
30 6 - - - - - - - -
60 18 - - - - - - -
90 21 - - - - - -
120 36 - - - - -
150 43 - - - -
180 49 - - -
210 52 - -
240 53 -
270 54

94 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation

 Example

Time since Cumulative rainfall Incremental depth (mm) for various durations (min)
start (min) (mm) 30 60 90 120 150 180 210 240 270
0 0
30 6 6 - - - - - - - -
60 18 12 18 - - - - - - -
90 21 3 15 21 - - - - - -
120 36 15 18 30 36 - - - - -
150 43 7 22 25 37 43 - - - -
180 49 6 13 28 31 43 49 - - -
210 52 3 9 16 31 34 46 52 - -
240 53 1 4 10 17 32 35 47 53 -
270 54 1 2 5 11 18 33 36 48 54

95 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation

 Example

Duration (min) 30 60 90 120 150 180 210 240 270

Maximum depth (mm) 15 22 30 37 43 49 52 53 54
Maximum intensity
30.0 22.0 20.0 18.5 17.2 16.3 14.9 13.3 12.0
(mm/hr)

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 Example

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 Additional Terms

① The probability that an event F will be equaled or exceeded

in any year:
P = 1/T
② The probability that F will NOT be equaled or exceeded in
any year:
q = 1 – P = 1 – 1/T
③ The probability that F will NOT be equaled or exceeded in
any of n successive years:
qn = (1 – 1/T)n
④ The probability R (risk) that F will be equaled or exceeded
at least once in n successive years:
R = 1 – (1 – 1/T)n
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 Frequency of Point Rainfall
If the probability of an event occurring is P, the
probability of the event not occurring in a given year is
q = (1 — P)

The binomial distribution can be used to find the

probability of occurrence of the event exactly r times
in n successive years. Thus:

where Pr,n = probability of rainfall of a given magnitude

and exceedence probability P occurring r times in n
successive years
99 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation
 Frequency of Point Rainfall
For example:
① The probability of an event of exceedence probability
P occurring exactly 2 times in n successive years is:

② The probability of the event not occurring at all in n

successive years is:

③ The probability of the event occurring at least once in

n successive years:

100 [2] Spring– 2021 – Hydrology Dr. Mohammad N. Almasri Precipitation

 Frequency of Point Rainfall
Example

Analysis of data on maximum one-day rainfall depth

at Madras indicated that a depth of 280 mm had a
return period of 50 years

Determine the probability of a one-day rainfall depth

that equals to or greater than 280 mm at Madras
occurring:
exactly once in 20 successive years
exactly two times in 15 successive years
at least once in 20 successive years
at least twice in 20 successive years
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 Frequency of Point Rainfall
Example

0.0322

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 Frequency of Point Rainfall
Example

Pat least twice = 1 – [P(0) + P(1)]

We know that:
P(0) = 0.9820 = 66.7% (exactly none)
P(1) = 27.2% (exactly once)
Pat least twice = 1 – [66.7% + 27.2%] = 6.1%

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 Probable Maximum Precipitation

The probable maximum precipitation (PMP) is

theoretically the greatest depth of precipitation for a
given duration that is physically possible at a specific
location at a certain time of the year

This rainfall will practically produce a flood flow with

virtually no risk of being exceeded

When constructing a hydrologic (or hydraulic)

structure, the failure probability will be zero

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