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ENGINEERING HYDROLOGY

1 PRECIPITATION

1. INTRODUCTION

1.1. What is hydrology?

Hydrology is the study of water in all its phase, and includes hydraulics, the physics, and
chemistry of water, meteorology, geology and biology.
In a more limited sense hydrology is the study of the hydrologic cycle in which water rises
by process of evaporation and transpiration, is carried over the land where it falls as
precipitation.
1.2. Hydrologic Cycle
The three main phases of the hydrologic cycle are
(i) Evaporation and transpiration.
(ii) Precipitation
(iii) Runoff
The water particles have different residence time in each phase.
1.3. Residence time
• This is the average time taken by a water molecule in crossing one particular phase of
hydrologic cycle.
• The ‘phase’ does not essentially mean solid, liquid and gas.
• Mathematically, it can be expressed as
𝑉𝑜𝑙𝑢𝑚𝑒
𝑅𝑒 𝑠 𝑖𝑑𝑒𝑛𝑐𝑒 𝑡𝑖𝑚𝑒  =
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑑𝑖𝑠𝑐ℎ 𝑎𝑟𝑔 𝑒
1.4. Catchment Area
Area draining into a river or stream is called the catchment area of that stream at a
particular location.
It is also called watershed, drainage area and drainage basin.

Ridge: It is a line which divides one catchment are from its neighbouring catchment area.
It is also called water divide or divide.

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1.5. Hydrologic Equation (Water budget equation)

The hydrologic equation is merely a statement of the law of conservation of matter and
mathematically it can be expressed as
| Mass Inflow – Mass Outflow | = Change in Storage
• Inflow includes following quantities
(i) Precipitation
(ii) Surface inflow
(iii) Ground water inflow
(iv) Import.
• Outflow includes following quantities.
(i) Surface outflow
(ii) Ground water outflow
(iii) Export
(iv) Evaporation
(v) Transpiration
(vi) Interception
• All values should have same units only then we can use this equation.
Example: A small catchment of area 150 ha received a rainfall of 10.5 cm in 90 minutes
due to a storm. At the outlet of the catchment, the stream draining the catchment was
dry before the storm and experienced a runoff lasting for 10 hours with an average
discharge of 1.5 m3/s. The stream was again dry after the runoff event. (a) What is the
amount of water which was not available to runoff due to combined effect of infiltration,
evaporation and transpiration? What is the ratio of runoff to precipitation?
Solution:
The water budget equation for the catchment in a time ∆t is
R=P–L
where L = Losses = water not available to runoff due to infiltration (causing addition to
soil moisture in the groundwater storage), evaporation, transpiration and surface storage.
In the present case ∆t = duration of the runoff = 10 hours.
Note that the rainfall occurred in the first 90 minutes and the rest 8.5 hours the
precipitation was zero.
(a) P = Input due to precipitation in 10 hours
= 150 × 100 × 100 × (10.5/100) = 157,500 m3
R = runoff volume = outflow volume at the catchment outlet in 10 hours
= 1.5 × 10 × 60 × 60 = 54,000 m3
Hence losses L = 157,500 – 54,000 = 103,500 m3
(b) Runoff/rainfall = 54,000/157,500 = 0.343
1.6. World water balance
Total volume of water on Earth = 1386 Mkm3 ≈ 1400 Mkm3

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96.5% → 𝑂𝑐𝑒𝑎𝑛 → 𝑆𝑎𝑙𝑖𝑛𝑒

} 97.5% → 𝑆𝑎𝑙𝑖𝑛𝑒
1% → ℎ𝑎𝑛𝑑 → 𝑆𝑎𝑙𝑖𝑛𝑒
(All values are approximate)

2. PRECIPITATION

Precipitation denotes all form of water that reach the earth from the atmosphere. Precipitation
is one of the three major phases in hydrologic cycle and one of the major items of inflow in the
hydrologic equation. It includes rain, snow, hail, dew and frost.
2.1 Types and Causes
The following are the different types of precipitation.
(i) Rain: It is the most dominant from of precipitation in Indian. It denotes water
droplets of size varying 0.5 to 6 mm. On the basis of which intensity of rainfall is
classified as follows

Intensity (mm/hr)

0 – 2.5 Light Rain

2.7 – 7.5 Medium/Moderate Rain

> 7.5 Heavy Rain

Note:
• In India, rainfall data is collected every day at 8.30 am and if rainfall on a particular
day is more than 2.5 mm then that day is known as rainy day.
• Average rain fall per year in India is 119 cm ≃ 120 cm.
(ii) Drizzle: It denotes fine droplets of water whose size is less than 0.5 mm and
intensity is less than 1 mm/hr
(iii) Snow: These are ice crystals having the density of 0.1 gm/cc
(iv) Glaze: When droplet of water comes in contact with cold ground surface, then
water is converted into ice, which is called as glaze or freezing rain.
(v) Sleet: These are frozen rain drops of transparent nature which forms when rain
falls through subfreezing temperature.
(vi) Hail: These are lumps of ice whose size is greater than 8 mm.
2.2. Weather System for Precipitation
• Convective Precipitation: This type of precipitation occurs due to the temperature
differences. Warm air rises up because of its lesser density and air from cooler
surrounding flows to take up its place, thus setting up a convective cell. The rising

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warm air undergoes cooling and results in precipitation. Areal extent of such rains is
usually small (diameter of around 10km).
• Oro graphic Precipitation: When moist air masses get lifted up due to presence of
mountain barriers, they undergo cooling, condensation and precipitation. Such a
precipitation is known as Orographic precipitation. The windward slopes have heavy
precipitation and the leeward slopes light rainfall.

• Frontal Precipitation: Under certain favourable conditions when a warm air mass
and cold air mass meet, the warmer air mass is lifted over the colder one with the
formation of a front. The ascending warmer air cools adiabatically with the
consequent formation of clouds and precipitation.
• Cyclonic Precipitation: This type of precipitation occurs due to the pressure
differences. A cyclone is a large low-pressure region with circular wind motion. The
rainfall will normally be heavy in the entire region occupied by the cyclone.

3. AVERAGE ANNUAL RAINFALL

The amount of rain collected by a rain gauge in the last 24 hrs is called daily rainfall and the
amount collected in one year is called as annual rainfall. The average value of annual rainfall
for a period of 35 years is called average annual rainfall.

4. INDEX OF WETNESS

The actual precipitation for a given year may differ from the long-time mean. To bring out this
important relationship. We often express the actual precipitation for a given year as a ratio to

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the mean. Such a factor is called the index of wetness. This factor is used to find deficiency in
rainfall and is given as follows:
𝑅𝑎 𝑖𝑛𝑓 𝑎 𝑙𝑙 𝑖𝑛 𝑎 𝑦𝑒𝑎𝑟
𝐼𝑛𝑑𝑒𝑥 𝑜𝑓 𝑤𝑒𝑡𝑛𝑒𝑠𝑠 = × 100
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑛𝑛𝑢𝑎𝑙 𝑟𝑎 𝑖𝑛𝑓 𝑎 𝑙𝑙
• If index of wetness is 100%, it indicates a normal year. If it is greater than 100%, it is
called as good year and if it is less than 100%, it is called bad year.

5. DROUGHT

This is a climatic situation characterised by less availability of moisture. Following are different
types of drought.
(i) Meteorological Drought
The type of drought occurs by deficiency in precipitation. If the deficiency is more than 25% it
is known as drought. If deficiency is in between 25 – 50%, it is known as moderate drought
and it is greater than 50%, it is known as severe drought.
A particular year is drought year if area affected by drought is more than 20% of total area of
the country.
(ii) Hydrological drought
This type of drought denotes below average value of stream flow, water content in lakes,
reservoir, underground water etc.
(iii) Agricultural draught
This type of drought denotes deficiency of water which is required for meeting the
evapotranspiration needs of a crop. It is denoted by a factor known as Aridity Index.
Mathematically it can be expressed as
𝑃𝐸𝑇−𝐴𝐸𝑇
Aridity index = × 100
𝑃𝐸𝑇

Here, PET = potential evapotranspiration.

AET = actual evapotranspiration.
On the basis of aridity index, regions can be classified as follows.
Aridity index %)
0 – 25 Mild
25 – 50 Moderate
>50 Severe

6. MEASUREMENT OF RAINFALL

Rainfall is expressed in terms of depth to which water would stand on an area if all the rainfall
is collected on it. Rainfall is measured by an instrument called Rain gauge which is also known
as pluviometer, ombrometor, hyetometer.

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Following are the requirements which has to be satisfied prior to installation of rain gauge.
(a) Rain gauge must be surrounded by an open fenced area of at least 5.5 m × 5.5 m
(b) Rain gauge must be installed at a distance of at least 30 m or twice the height of building
or obstruction.
(c) Rain gauge must be installed on a level surface which is free from undulation.
A rain gauge basically consists of cylindrical assembly which is kept in open in order to collect
the rain water. There are basically two types of rain gauges:
(i) Non Recording rain gauge
In India, most commonly used non recording rain gauge is Symon’s rain gauge which has a
collecting diameter of 127 mm. Now a days IMD (Indian meteorological department) is using
fibre glass reinforced polyester type rain gauge. Which comes in two variants having collecting
area of 100 cm2.
(ii) Recording type rain gauge
This type of rain gauge automatically records the rate of precipitation and the time of its
occurrence. It also produces a continuous variation of rainfall with time. Graphically which can
be represented as follows.

In India natural siphon or float type is used as the standard recording type rain gauge.
Latest improvements in rain gauge technology has led to the development of following rain
gauges.
(ii) Telemetering Rain Gauge
It is basically a recording type rain gauge which contains electronic equipment to transfer
rainfall data to the base relation. As such it can be used in far remote and inaccessible locations.
(ii) Radar Based Rain Gauge:
It is used to find aerial extent location and movement of rain storms. Rainfall over a large area
can be measured with good accuracy. Meteorological radar operates at a wave length range
of3 – 10 cm
Note: Typical examples of recording type rain gauges are: Tipping bucket, weighing rain
gauge, Natural siphon etc.

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7. NETWORK DENSITY

World meteorological organisation gave the following recommendation regarding density of

rain gauge:
(i) For flat regions of temperate tropical and Mediterranean zone: 1 station per 600-900 km2
(ii) For mountainous regions of temperature, tropical and Mediterranean zone: 1 station per
100 – 250 km2
(iii) For arid and polar regions: 1 station/1500 – 10000 km2.
According to Indian standard recommendations.
(i) for plain region: 1 station per 520 km2
(ii) for regions having elevation of 1000 m: 1 station per 260-390 km2
(iii) in predominantly hilly areas with heavy rainfall: 1 station per 130 km2
Note: As per recommendation of WMO, at least 10% of the rain gauge should be of recording
type.
7.1. Optimum number of rain gauge
This can be calculated on the basis of variation or deviation in the given rain fall data
and allowable percentage error. Mathematically it is given as

𝐶𝑣 2
𝑁=( )
∈
Here, Cv = coefficient of variation.
𝜎
𝐶𝑣 = × 100
𝑥̄
∑𝑛
𝑖=1 𝑥𝑖 ∑𝑛
𝑖=1(𝑥𝑖 –𝑥̄ )
2
Here, 𝑥̄ =  𝑎𝑛𝑑 𝜎 = √
𝑛 𝑛−1

n = number of rain gauge stations.

ϵ = allowable percentage error.
Also, for given number of rain gauge standard error ϵs is given as
𝐶𝑣
∈𝑠 =
√𝑛
Example: The average Annual rainfall of 5 rain gauge stations are 90, 55, 42, 40 and 50

cm. It the error in the estimation of rainfall should not exceed 10%. How many additional

gauges may be required?

Solution: -
The mean rainfall is obtained as
90 + 55 + 42 + 40 + 50
X or Pm =
5
X = 55 .4 cm

Now,

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(90 − 55.4)2 + (55 − 55.4)2 +

(42 − 55.4)2 + (40 − 55.4)2 + (50 − 55.4)2
2 =
5 −1
 P − Pm 
 2 = i 
 m−1 

σ2 = 410.8
 σ = 20.27
The coefficient of variation is calculated as:

CV =
X
20.27
 CV = = 0.366
55.4
Now, optimum no. of rain gauges, N
2
C 
N= V
  
2
 0.366 
N=  (  = 10%)
 0.10 
N = 13.396 ~ 14
Thus, additional rain gauges required = 14 – 5
= ‘9’
Example: A catchment area has seven rain gauge stations. In a year the annual rainfall
recorded by the gauges are as fallows.
Station P Q R S T U V
Rainfall (cm) 120 14 110 130 150 116 45

(a) Determine the standards error in the estimation of mean rainfall in the existing set
of rain gauges.
(b) for a 5% error in the estimation of mean rainfall calculate the minimum number of
additional rain gauge stations to be establish in the catchment.
Solution: -
For the given data,
No. of rain gauges, m = 7
Mean annual rainfall
120 + 140 + 110 + 130 + 150 + 116 + 145
P =
7

P = 130.142
Now, standard deviation,

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 (P
1
i −P )2
m −1 =
m −1

(120 − 130.14)2 + (140 − 130.14)2

+ (130 − 13014)2 + (150 − 130.14)2
+ (116 − 130.14)2 + (145 − 130.14)2
 m−1 =
7 −1

102.82 + 97.22 + 405.62 + 0.0196

+ 394.42 + 199.92 + 220.82
m −1 =
6

σm–1 = 15.388
 coefficient of variation
m−1
CV =  100
P
15.388
CV =  100 = 11.79
130.42
CV = 12
(a) Standard error new the estimation of the mean ' ' .
CV 12
= = = 4.54%
m 7
(b) When the error is limited to 5%,  = 5 and the optimum number of rain gauges in
the catchment is given by
2 2
C   11 
N= v =  = 4.84 ~ 5
    5 

8. CONSISTENCY OF RAINFALL

Whenever there is change in condition at rain gauge prevailing, there will be inconsistency in
rainfall data available for that period. This inconsistency may be due to the following reasons.
(i) Due to observation error
(ii) Neighbourhood of station undergoing a significant change.
(iii) Due to shifting rain gauge station to new location.
This inconsistency can be found out by a technique called as ‘double mass curve technique’.

9. ESTIMATION OF MISSION RAINFALL DATA

If on a particular day of a year rainfall data at some station ‘x’ could not be recorded (due to
defective rain gauge or otherwise) and it is required to find an approximate value of this missing
data, following methods are adopted.

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Few stations close to problematic station ‘x’ are selected and rainfall value at these stations is
noted down as P1, O2, ….., Pn – 1’ Pn. Px is the missing rainfall data whose approximate value
has to be obtained.
Let N1, N2, N3, … and Nx be the normal precipitation values for station ‘1 to m’ and ‘x’ Normal
precipitation is average value of rainfall for a day, month or a year on the basis of record of
last 30 years.
Case 1: when N1, N2, … Nm differs from Nx by less than 10% the value of Px is given as follows.
𝑃1 + 𝑃2 + 𝑃3 + 𝑃3 +. . . +𝑃𝑚
𝑃𝑥 =
𝑚
Case 2: when one or more of N1, N2,…Nm, differs from Nx by 10% or more, the value of Px is
calculated by:
𝑁𝑥 𝑃1 𝑃2 𝑃𝑚
𝑃𝑥 = [ + +. . . . . . . + ]
𝑚 𝑁1 𝑁2 𝑁𝑚

Example: The normal annual rainfall of stations A, B, C and D in a catchment is 85 mm, 90

mm, 80 mm and 87 mm in the year 2020, the station D was inoperative when stations A, B
and C recorded annual rainfall of 92, 72, 81 mm. estimate the missing rainfall at station D in
the year 2020.
Solution: -
The normal precipitation of the station A, B & C are within 10% of that at station ‘D’

Hence single average arithmetic will be used.

92 + 72 + 81
 Px =
3
Px = 81.67 mm
Example: Find the missing rainfall at station ‘x’.
Rain gauge Normal Actual
A 1120 875
B 900 1020
C 760 910
x 825 ?

Solution: -
As the normal precipitation of other stations, A, B and C are not within 10% of normal
precipitation at station x
As,

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Hence, we will use normal ratio method.

Px 1 P P P 
i.e. =  A + B + C
Nx n  NA NB NC 

PX 1  875 1020 910 

=  + +
825 3 1120 900 760 
Px = 855.786
Px = 856 mm Ans.

10. AVERAGE PRECIPITATION/RAINFALL

When a catchment or any other area contains more than one rain gauge station, then average
rainfall for the area can be obtained using the following method.
10.1. Arithmetical Mean/Average Method
This method is suitable if rainfall is uniformly distributed and area is not very large.
𝑃1 + 𝑃2 + 𝑃3 +. . . . . +𝑃𝑛
𝑃𝑎𝑣𝑔 =
𝑛
This method does not give very good result and hence is not used very frequently. Any
station outside the area of consideration is not taken into account in this method.
10.2. Thiessen polygon/mean Method
In this method the rainfall recorded at each station is given a weightage on the basis of
an area closest to the station, that is why this method is also known as weightage area
method.
The procedure of determining the weighing are is as follows:
Step 1: The catchment area is drawn to scale and the position of station is also drawn.
Step 2: All the stations are joined together to form a network of triangles.
Step 3: Perpendicular bisectors for each of the sides of the triangle are drawn.
These bisectors form a polygon around each station. Stations are then assigned the
area by which they are surrounded.
𝑃1 𝐴1 + 𝑃2 𝐴2 +. . . +𝑃𝑛 𝐴𝑛
𝑃𝑎𝑣𝑔 =
𝐴1 + 𝐴3 +. . . +𝐴𝑛
∑𝑛
𝑖=1 𝑃𝑖 𝐴𝑖
⇒𝑃𝑎𝑣𝑔 =
𝐴

The ratio Ai/A is called the weightage factor for each station.
This method of finding average rain fall is suitable when area is large and rainfall is non-
uniformly distributed. This method is superior to arithmetical mean method.
10.3. Isohyetal Method
An isohyet is a line joining all the points having same value of rainfall and isohyetal maps
are the one which shows contours of equal rainfall magnitude.

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In isohyetal method, it is assumed that the precipitation in areas between isohyetal lines
is equal to the mean of the precipitation of at isohyetal lines.
Mathematically, various following cases are possible.
Case 1:

𝑃1 + 𝑃2 𝑃 +𝑃 𝑃 +𝑃
( 2 ) 𝐴1 + ( 2 2 2 ) 𝐴2 +. . . + ( 𝑛−12 𝑛 ) 𝐴𝑛−1
𝑃𝑎𝑣𝑔 =
𝐴1 + 𝐴2 + 𝐴3 +. . . +𝐴𝑛−1
Case 2: If isohyets outside the considered area is given.

𝑃0 + 𝑃1 𝑃 +𝑃 𝑃 +𝑃 𝑃 +𝑃 𝑃 +𝑃
( 2 ) 𝐴0 + ( 1 2 2 ) 𝐴1 + ( 2 2 3 ) 𝐴3 + ( 𝑛−12 𝑛 ) 𝐴𝑛−1 + ( 𝑛 2 𝑛+1 ) 𝐴
𝑃𝑎𝑣𝑔 =
𝐴0 + 𝐴1 + 𝐴2 . . . . . +𝐴𝑛−1 + 𝐴𝑛
Case 3:

𝑃1 + 𝑃2 𝑃 +𝑃 𝑃 +𝑃
( 2 ) 𝐴1 + ( 2 2 3 ) 𝐴2 +. . . + ( 𝑛−12 𝑛 ) 𝐴𝑛−1 + 𝑃𝑛+1 𝐴𝑛
𝑃𝑎𝑣𝑔 =
𝐴1 + 𝐴2 + 𝐴3 +. . . +𝐴𝑛−1 + 𝐴𝑛
Limitations of Isohyetal method
(i) This method gives better result only if there are larger number of stations.
(ii) Inter isohyetal area is a function of rainfall value which will change with any change
of rainfall pattern
(iii) This method requires use of technical and skilled man power.
(iv) It is a time-consuming method.
Note: Among all the methods discussed above orographically derived isohyetal method
is the best method of finding average rainfall.

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Example: Five rain gauges stations namely A, B, C, D and E are located on a circular
shape basin of diameter 20 km as shown in the fig. given below compute the mean annual
rainfall at stations A, B, C, D and E are 100 cm, 90 cm, 110 cm, 120 cm and 80 cm
respectively.

Solution: -
Diameter of basin = 20 km. we know that in Thiessen polygon method. The polygons are
obtained by joining the perpendicular bisectors of triangles formed, when various rain
gauge stations are joined.
The Thiessen polygon can be drawn as :-

Area under influence of station A (area of region I)

= 10 × 10 = 100 km2
Area under influence of station B, C, D and B.

 (20)
2
− 100
= 4 = 53.54 km2
4
Mean areal rainfall:-
PA A A + PB AB + PC A C + PD AD + PE AE
=
A A + AB + A C + AD + AE

(100  100) + (90  53.54) + (120  53.54)

+ (80  53.54)
=
 4 (20)
2

= 100 cm

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 mean area rainfall

= 100 cm
Example: The Isohyets for annual rainfall over a catchment basin were drawn. The areas
of strips between the Isohyets are given determine the Average precipitation over the
basin
Isohyets (cm) Area in sq. km
75-85 1500
85-95 2500
95-105 2800
105-115 1200
115-135 800
135-155 600

Solution: -

Average precipitation in given as

 P + P2   P + P3 
A1  1  + A2  2  + ... + An−1
 2   2 
 Pn −1 + Pn 
 
 2 
Pavg =
A1 + A 2 + A 3 + ... An

 75 + 85   85 + 95 
1500   + 2500 
 2   2 
 95 + 105   105 + 115 
+ 2800  + 1200 
 2   2 
 115 + 135   135 + 155 
+ 800  + 600 
 2   2 
Pavg =
1500 + 2500 + 2800 + 1200 + 800 + 600
(1500  80) + (2500  90) + (2800  100)
+ (1200  110) + (800  125) + (600  145)
 PAvg =
9400
PAvg = 100.425 cm Ans.
Example: Calculate the average depth of rainfall for the catchment with an area of 200
km2 using (i) Arithmetic method (ii) Thiessen polygon method.

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Station (i) Rainfall (mm) Pi Thiessen polygon area (km2) Description inside (I) outside (0)
1 45 7.5 0
2 20 10.0 0
3 30 45.0 I
4 45 30.0 I
5 50 16.0 0
6 35 45.0 I
7 45 40.0 I
8 40 6.5 0

Solution: -

The average depth of rainfall is determined by: -

(1) Arithmetic mean method

In this method we will consider rain gauges which are inside the catchment only.

P1 + P2 + P3 + P4 . . . Pn
 P =
n

30 + 45 + 35 + 45
P =
4

145
P = = 36.25 mm
4

 volume of rainfall = A  P

= 200 × 36.25 km2 mm

36.25 3
= 200 × 1000 × 1000 × m
100

= 72.5 × 105 m3

(ii) Thiessen polygon method.

We will consider rain guage which are present both inside and outside the catchment
area.
P1 A1 + P2 A2 + ...Pn An
 Pavg =
A1 + A2 + ...

(45  7.5) + (20  10) + (30  45) + (45  30)

+ (50  16) + (35  45) + (45  40) + (40  6.5)
=
200
Pavg = 38.36 mm
Volume of rainfall = Pavg × 200 km2
= 7672.5 km2 mm
= 76.72 × 105 m3
****

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