34
PRECIPITATION
INTRODUCTION
The term precipitation as used in hydrology is meant for all forms of moisture
emanating from the clouds and all forms of water like rain, snow, hail and sleet
derived from atmospheric vapors, falling to the ground.
Precipitation is one of the most important events of hydrology. Floods and
droughts are directly related to the occurrence of precipitation. Water resources
management, water supply schemes, irrigation, hydrologic data for design of
hydraulic structures and environmental effects of water resources development
projects are related to precipitation in one way or the other. So it is important to
study various aspects of precipitation.
FORMS OF PRECIPITATION
In the middle latitudes precipitation occurs in many forms, depending on the
existing meteorological conditions. These are following.
i. Drizzle
These are the minute particles of water at start of rain. These consist of water
drops under 0.5 mm diameter and its intensity is usually less than 1.0 mm/hr.
Their speed is very slow and we cannot even feel them. Therefore they cannot
flow over the surface but usually evaporate.
ii. Rain
It is form of precipitation in which the size of drops in this case is more than 0.5
mm and less than 6.25 mm in diameter. It can produce flow over the ground and
can infiltrate and percolate. Both the duration as well as rate of rainfall are
important. If the rainfall per unit time is greater than the rate of infiltration, the
rain water can flow over the surface of earth.
35
iii. Glaze
It is the ice coating formed on drizzle or rain drops as it comes in contact with the
cold surfaces on the ground.
iv. Sleet
Sleet is frozen rain drops cooled to the ice stage while falling through air at
subfreezing temperatures.
v. Snow
Snow is precipitation in the form of ice crystals resulting from sublimation i.e.
change of water vapor directly to ice.
vi. Snowflake
A snowflake is made up of a number of ice crystals fused together.
vii. Hail
Hail is the type of precipitation in the form of balls or lumps of ice over 5 mm
diameter formed by alternate freezing and melting as they are carried up and down
by highly turbulent air currents. The impact of these is also more. A single
hailstone weighing over a pound has been observed.
FACTORS INFLUENCING PRECIPITATION FORMATION
Following four conditions are necessary for the production of precipitation.
a.
A lifting mechanism to produce cooling of the air.
b.
A mechanism to produce condensation of water vapors and formation of
cloud droplets.
c.
A mechanism to produce growth of cloud droplets to size capable of
falling to the ground against the lifting force of air.
d.
A mechanism to produce sufficient accumulation of moisture to account
for observed heavy rainfall rates.
36
a.
Mechanism of Cooling
The pressure reduction when air ascends from near the surface to upper levels in
the atmosphere is the only mechanism capable of producing the degree and rate of
cooling needed to account for heavy rainfall. Cooling lowers the capacity of a
given volume of air to hold a certain amount of water vapor. As large degrees of
super saturation are not known to occur in the atmosphere, excess moisture over
saturation condenses through the cooling process.
b.
Condensation of Water Vapor
Condensation of water into cloud droplets takes place on hygroscopic nuclei
which are small particles having an affinity for water. The source of these
condensation nuclei is the particles of sea salt or such products of combustion of
certain sulfurous and nitrous acids. There appears to be always sufficient nuclei
present in the atmosphere. Condensation will always occur in air, the lower
atmosphere is cooled to saturation, often before the saturation point is reached.
c.
Growth of Cloud Droplets
Growth of droplets is required if the liquid water present in the cloud is to reach
the ground. The two processes regarded as most effective for droplet growth are:
i. Coalescence of droplets through collision due to difference in speed of motion
between larger and smaller droplets.
ii. Co-existence of ice crystals and water droplets.
Co-existence effect generally happens in the temperature range from 100 to 20o F.
If in a layer of clouds there is a mixture of water droplets and ice crystals, the
saturation vapor pressure over ice is lower than that over water. This leads to the
evaporation of water drops and condensation of much of this water on ice crystals
causing their growth and ultimate fall through the clouds. This effect is known as
Bergeron’s effect. The ice crystals will further grow as they fall and collide with
water droplets.
37
d.
Accumulation of Moisture
Heavy rainfall amount over a river basin exceeds by far the amount of water vapor
at the atmospheric volume vertically above the basin at the beginning of the
rainfall. Clearly there must be a large net horizontal inflow of water vapor into the
atmosphere above the basin area. This process is called convergence, which is
defined as the net horizontal influx of air per unit area. The moisture added to the
atmosphere over a basin may be transported very large distance in the lowest layer
of the atmosphere. When this moist current reaches a region of active vertical
motion it rises thousands of feets and loses much of its contained water vapor in
just a few hours.
Classification of Precipitation Based on the Lifting Mechanism
The precipitation is often classified according to the factor responsible for lifting
of air to higher altitudes. Following are the various types of precipitation based on
this classification.
i. Convectional Precipitation
In the case of convectional precipitation the main causative element is thermal
convection of the moisture laden air. For this to occur, solar radiation is the only
source of heat. A major portion of the solar radiation is utilized in heating the
earth. As the earth conducts heat slowly, the heat accumulates at the surface of the
earth and air which comes in its contact is heated up and the lapse rate near the
surface of the earth increases rapidly. With the passage of time as the sun gets
higher and higher the lapse rate increases above that of dry adiabatic and air
becomes unstable. Vertical currents are then set up which carry heat and the
moisture laden air is picked up from the surface to higher levels. Due to
convection, the moist air in the lower levels of the atmosphere rises up to the
condensation level where clouds develop and with further convection these clouds
finally grow into cumunimbus resulting in a thunderstorm. Lightning and thunder,
gusty surface winds, showers and sometimes hails accompany a thunderstorm.
Each thunderstorm is formed of a cell which updraft or down draft (downward
current) turbulence etc. These cells are called under-storm cells. A cell hardly
covers an area more than 1-2 square miles.
38
As the warm moist air from the ground is lifted up, more and more water will
condense and the water drops and ice crystals will increase in size till such time
that these drops are no longer supported by the existing updrafts. The drops will
then begin to fall. Such thunderstorms are called “air mass thunderstorms”. They
usually develop by mid-day and reach maximum intensity by afternoon. By late
evening such storms start dissipating. Thunderstorms may extend to a height of
20,000 to 25,000 feet in temperate latitude and 30,000 to 40,000 feet in tropics.
ii. Orographic Precipitation
In the orographic precipitation, expansion and condensation occurs because
moisture laden air masses are lifted by contact with orographic barriers. This type
of precipitation is most pronounced on the windward side of mountain range,
generally heaviest precipitation occurs where favorable orographic effects are
present. For instance, heaviest precipitation due to south-easterlies in the
subcontinent occurs along the Southern slopes of Himalayas and its other ranges.
It has been found that the monsoon rainfall (June to October) decreases gradually
as the distance from the line of heaviest rainfall increases.
Orographic precipitation also occurs in the inland areas where mountain ranges
rise above the surrounding areas in the path of the moisture laden air masses. On
these areas, however, orographic precipitation is intermingled with the other types
prevalent in the area and, therefore, it becomes difficult to identify quantitatively
the amount of precipitation that has occurred due to orographic effects only. On
inland areas the orographic precipitation is irregular in occurrence and quantity
due to interference by the atmospheric disturbance resulting from cyclonic storm.
Generally, it has been observed that heavy orographic precipitation occurs on the
windward side of the orographic barrier, leaving a relatively dry area leeward side.
This occurs because the moist air has been forced up the windward side and
precipitated its moisture, and upon passing the peak of the barrier, no further
orographic lifting occurs so that the rain fall is the residual of previous
condensation. If the orographic barrier is sufficiently massive and the low regions
on the leeward side are very extensive, the wind may descend on the leeward side,
thereby undergoing compression and heating and becoming still more unfavorable
for precipitation.
39
iii. Cyclonic Precipitation
Precipitation in plain regions is generally cyclonic in character. Cyclones are of
two general classes, tropical and extra-tropical, so called depending upon whether
they occur within or beyond the tropics. In as much as all cyclones occurring in
the Indo-Pak Subcontinent are of tropical variety. This kind alone will be
discussed here. Moreover, cause of these storms is not of primary concern except
if it affects the precipitation. Tropical cyclones are violent storms which are
generally formed in the ward maritime air-mass of low latitude where the
temperature is high. These are known as typhoons or cyclones when formed in the
Indian Ocean.
In the center of the cyclonic storm there is small low pressure air. The isobars
around such a low pressure are very nearly circular in shape. Their intensity is
generally greater than the extra-tropical cyclones. On an average the tropical
cyclones have a diameter of over 300 to 400 miles. The wind speed around
tropical cyclones may be as high as 60 to 90 miles per hour.
The winds around a tropical cyclone are practically parallel to the isobars. As the
center or eye of the storm approaches, the high winds suddenly die down as
practically dead calm prevails at the center. The diameter of the eye of the storm
is about 50 miles. The winds around the center of the storm blow almost in a
circular path (especially over the sea or ocean where the frictional force is
minimum) and at the same time air is rising. The vertical motion compels the air
to be cooled adiabatically. This air being humid, condensation takes place rapidly,
resulting copious rainfall and widespread thunderstorm. Over the center of the
storm the weather is usually fair.
In the Indo-Pak Subcontinent, the cyclonic storms form in the Bay of Bengal in
different months. During April, May and June most of these storms do not reach
Pakistan. But some of them affect Bangladesh and give very heavy rainfall there.
During the summer monsoon season, the cyclonic storms reach Pakistan and are
fed with moisture from the Arabian sea resulting in heavy rainfall over the
Northern areas of Pakistan. In September, October and November these storms are
very destructive in Bangladesh. Such storms cause considerable loss of life and
property over the coastal districts. Cyclonic storms also form in Arabian sea but
their number is far less.
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MEASUREMENT OF PRECIPITATION
Amount of Precipitation
The amount of precipitation means the vertical depth of water that would
accumulate on a level surface, if the precipitation remains where it falls. The
amount of precipitation is measured in length units (inches, ft., cm, etc.).
Intensity or Rate of Precipitation
Amount of precipitation per unit time is called the intensity of precipitation or rate
of precipitation.
Both the amount and rate of precipitation are important in hydrologic studies.
The precipitation is measured by rain gauges. There are two types of rain gauges:
a. Non-recording rain gauge. (Standard rain gauge)
b. Recording rain gauge
he main difference between these rain gauges is that with the help of recording
rain gauges we get the rain recorded automatically with respect to time, so
intensity of rain fall is also known whereas an observer has to take readings from
non recording rain gauge for rain and he has to record the time also, for
calculation of intensity of rain fall.
a. Non-Recording Rain Gauges
The standard gauge of U.S. Weather Bureau has a collector of 200 mm diameter
and 600 mm height. Rain passes from a collector into a cylindrical measuring tube
inside the overflow can. The measuring tube has a cross sectional area 1/10th of
the collector, so that 2.5 mm rain fall will fill the tube to 25 mm depth. A
measuring stick is marked in such a way that 1/10 th of a cm depth can be
measured. In this way net rainfall can be measured to the nearest 1 mm. The
collector and tube are removed when snow is expected. The snow collected in the
outer container or overflow can is melted, poured into the measuring tube and
then measured. This type of rain gauge is one of the most commonly used rain
gauge.
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Fig. 3.1 Non-Recording Rain Gauge
Sources of Error
Rainfall measured by the rain gauge might have some errors. For example some
water is used to wet the surface of instrument; the rain recorded may be less than
the actual rainfall due to the direction of the rainfall as affected by wind. Dents in
the collector and tube may also cause error. Some water is absorbed by the
measuring stick. Losses due to evaporation can also take place. The volume of
stick replaces some water which causes some error.
b. Recording Rain Gauges
Recording rain gauges can be divided into the following types:
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i.
Float type
ii.
Weighing type
iii.
Tipping bucket type
i. Float Type Rain Gauge
This type of rain gauge also has a receiver and a float chamber along with some
recording mechanism or arrangement. In this type the rain is led into a float
chamber containing a light, hollow float. The vertical movement of the float as the
level of water rises is recorded on a chart with the help of a pen connected to float.
The chart is wrapped around a rotating clock driven drum. To provide a
continuous record for 24 hours the float chamber has either to be very large, or
some automatic means are provided for emptying the float chamber quickly when
it becomes full, the pen then returning to the bottom of the chart. This is usually
done with some sort of siphoning arrangement. This arrangement activates when
the gauge records a certain fixed amount of rain (mostly 10 mm of rainfall.).
Snow can not be measured by this type of rain gauge. Fig. 3.2 shows line diagram
of float type rain gauge.
Fig. 3.2 FLOAT TYPE RAIN GAUGE
ii. Weighing Type Rain Gauge
The weighing type rain gauge consists of a receiver, a bucket, a spring balance and
some recording arrangement. The weighing type gauge weighs the rain or snow
which falls into a bucket which is set on a lever balance. The weight of the bucket
and content is recorded on a chart by a clock driven drum. The record is in the
form of a graph, one axis of which is in depth units and the other has time. The
records show the accumulation of precipitation. Weighing type gauges operate
from 1 to 2 months without stopping. But normally one chart is enough only for
24 hours. This type of rain gauge has advantage of measuring snow also. The
receiver is removed when snow is expected. Figure 3.3 shows schematic sketch of
a weighing type rain gauge.
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Fig 3.3 WEIGHING TYPE RAIN GAUGE
iii. Tipping Bucket Type Rain Gauge
This type of gauge used at some Weather Bureau First Order Stations, is equipped
with a remote recorder located inside the office which is away from the actual site.
The gauge has two compartments pivoted in such a way that one compartment
receives rain at one time. A certain amount of rain (usually 0.25 mm fills one
compartment and over balances it so that it tips, emptying into a reservoir and
bringing the second compartment of the bucket into place beneath the funnel of
receiver. As the bucket is tipped by each 0.25 mm of rain it actuates an electrical
circuit, causing a pen to mark on a revolving drum. This type of gauge is not
suitable for measuring snow without heating the collector. Plotting is similar to
that of other recording rain gauges. A Tipping Bucket Type Rain Gauge is shown
in Figure 3.4.
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Fig. 3.4 tipping bucket type rain gauge
SOURCES OF ERROR
All sources of error are similar to those of non-recording rain gauges except the
error due to stick. These are mentioned again as below:
a. Dents in the collector.
b. Moistening of inside-surface of the funnel and the tube.
c. Rain drops splashing from the collector.
d. For very intense rain some water is still pouring into the already filled
bucket.
e. Inclination of the gauge may result in catching less or more rain than the
actual amount.
f. Error in measurement due to wind.
Remedial Measures for Errors in Precipitation Measurement
Removal of error due to dents obviously needs repair of the instrument. For rain
recorded with dents a correction should be applied. Errors such as moistening of
the inside surfaces of the gauge, splashing of rainwater from the collector and
pouring of water into the already filled bucket during an intense rain can only be
corrected by some correction factor. Inclined instrument needs to be reinstalled.
The correction factor however can be calculated from the angle of inclination. For
45
wind protection certain wind shields are designed and used which are called
Splash Guards. Proper setting of gauge above ground level is necessary.
Example 3.1
A rain gauge recorded 125 mm of precipitation. It was found later that the gauge
was inclined at an angle of 20 degree with the vertical. Find the actual
precipitation.
Solution
P(measured)
= 125 mm
Angle of inclination (θ) = 20o with the vertical
P(actual) = P(measured)/cos(θ) = 125/cos20o = 133 mm
Measurement of Precipitation by Radar
This is a modern technique for measurement of rainfall rate. It can also detect
local movement of areas of precipitation. The electromagnetic energy released and
received back by radar is a measure of rainfall intensity. The measurement is
appreciably affected by trees and buildings. However extent of rainfall can be
estimated with reasonable accuracy. Use of radar is useful where number of rain
gauges installed in an area is not sufficient.
Rain Gauge Network
The number of rain gauges and their distribution affect the nature of collected
precipitation data. The larger the number of rain gauges the more representative
will be the data colleted. But on the other hand we have to observe other factors
also, like economy of the project, accessibility of certain areas and topography of
the area. So, one has to look for some optimum solution. In this regard the World
Meteorological Organization (WMO) has made following recommendations for
minimum number of rain gauges in a catchment:
a. In comparatively flat regions of temperate, Mediterranean and Tropical
Zones, the ideal is at least one station for 230 – 345 sq. miles. However
one station for 345 – 1155 sq. miles is also acceptable
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b. In mountainous regions of Temperate, Mediterranean and Tropical Zones,
the ideal is at least one station for 35 – 95 sq. miles. However one station
for 95 – 385 sq. miles is also acceptable.
c. In arid and polar zones, one station for 575 – 3860 sq. miles is acceptable
ANALYSIS OF PRECIPITATION DATA
Point Data Analysis
Point precipitation data refers to precipitation of a station. This data could be in
form of hourly record, daily record, monthly precipitation or annual precipitation.
Depending upon the nature of catchment and its area, there could be as many
gauging stations as feasible. Before using records from a rain gauge we should
check its continuity and consistency. Record may not be continuous and consistent
due to many reasons as explained in the following paragraphs. This section deals
with estimation of any missing records at a particular gauging station, checking
consistency of data and its adjustment accordingly.
1. ESTIMATION OF MISSING PRECIPITATION RECORD
Some precipitation stations may have short breaks in the records because of
absence of the observer or because of instrumental failures. It is often necessary to
estimate this missing record. In the procedure used by the U.S. Weather Bureau,
the missing precipitation of a station is estimated from the observations of
precipitation at some other stations as close to and as evenly spaced around the
station with the missing record as possible. There are two methods for estimation
of missing data - Arithmetic mean method and Normal ratio method.
The station whose data is missing is called interpolation station and gauging
stations whose data are used to calculate the missing station data are called index
stations.
If the normal annual precipitation of the index stations lies within 10% of
normal annual precipitation of interpolation station then we apply arithmetic
mean method to determine the missing precipitation record otherwise the normal
ratio method is used for this purpose.
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Consider that record is missing from a station ‘X’.
Now let,
N = Normal annual precipitation. (Mean of 30 years of annual
precipitation data)
P = Storm Precipitation.
Let Px be the missing precipitation for station ‘X’ and Nx , the normal annual
precipitation of this station, Na, Nb and Nc are normal annual precipitations of
nearby three stations, A, B and C respectively while Pa, Pb and Pc are the storm
precipitation of that period for these stations.
Now we have to compare Nx with Na , Nb and Nc separately. If difference of Nx Na, Nx - Nb, Nx - Nc is within 1/10% of Nx then we use simple arithmetic mean
method otherwise the normal ratio method is used.
Simple Arithmetic Mean Method
According to the arithmetic mean method the missing precipitation ‘Px’ is given
as:
Px =
1 in
 Pi
n i 1
where ‘n’ is number of nearby stations, ‘Pi’ is
precipitation at ith station.
In case of three stations 1, 2 and 3,
Px =
(P1 + P2 + P3)/3
and naming stations as A, B and C instead of 1, 2 and 3
Px =
(Pa + Pb + Pc)/3
Where Pa , Pb and Pc are defined above.
Normal Ratio Method
According to the normal ratio method the missing precipitation is given as:
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Px =
1 in N x

Pi
n i 1 N i
Where Px is the missing precipitation for any storm at the interpolation station ‘x’.
Pi is the precipitation for the same period for the same storm at the “ith” station of
a group of index stations and Nx and Ni are the normal annual precipitation values
for the ‘x’ and ‘ith’ stations. For example, for the symbols defined above for three
index stations in a catchment area.
Px=
1 Nx
[ P1  N x P 2  N x P3]
3 N1
N2
N3
Example 3.2
Find out the missing storm precipitation of station ‘C’ given in the following
table:
Station
Storm precipitation (cm)
Normal Annual precipitation (cm)
A
9.7
100.3
B
8.3
109.5
C
---93.5
D
11.7
125.7
E
8.0
117.5
Solution
In this example the storm precipitation and normal annual precipitations at
stations A, B, D and E are given and missing precipitation at station ‘C’ is to be
calculated whose normal annual precipitation is known. We will determine first
that whether arithmetic mean or normal ratio method is to be applied.
10% of Nc = 93.5x10/100 = 9.35
After the addition of 10% of Nc in Nc, we get 93.5 + 9.35=102.85
And by subtracting 10% we get a value of 84.15
So Na, Nb, Nd or Ne values are to be checked for the range 102.85 to 84.15.
If any value of Na, Nb, Nd or Ne lies beyond this range, then normal ratio method
would be used. It is clear from data in table above that Nb, Nd and Ne values are
out of this range so the normal ratio method is applicable here, according to which
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in
Px = 1  N x P i
n i 1 N i
Pc= (1/4 )(93.5 x 9.7/100.3+ 93.5 x 8.3/109.5+ 93.5 x 11.7/125.7+
93.5 x 8.0/117.5) = 7.8 cm
Example 3.3
Precipitation station “X” was inoperative for part of a month during which a storm
occurred. The storm totals at three surrounding stations A, B and C were
respectively 10.7, 8.9 and 12.2 cm. The normal annual precipitation amounts at
stations X, A, B and C are respectively 97.8, 112, 93.5 and 119.9 cm. Estimate
the storm precipitation for station ‘X’.
Solution
Pa = 10.7 cm
Na = 112 cm
Pb = 8.90 cm
Nb = 93.5 cm
Pc = 12.2 cm
Nc = 119.9 cm
Px = ?
Nx = 97.8 cm
10% of Nx = 97.8 x 10/100 = 9.78 cm.

Nx - Na = 97.8 - 112 = -14.2 cm  More than + 10% of Nx (no need of
calculating Nx – Nb and Nx - Nc
So we will use Normal Ratio Method.

Px = (1/3)( 97.8x 10.7/112+ 97.8x 8.90 /93.5 + 97.8x 12.2 /119.9)
Px = 9.5 cm
2. Consistency of Precipitation Data or Double Mass Analysis
In using precipitation in the solution of hydrologic problems, it is necessary to
ascertain that time trends in the data are due to meteorological changes. Quite
frequently these trends are the result of the changes in the gauge location, changes
in the intermediate surroundings such as construction of buildings or growth of
trees, etc. and changes in the observation techniques. Frequently changes in gauge
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location are not disclosed in the published record. Due to such changes the data
might not be consistent. The consistency of the record then is required to be
determined and the necessary adjustments be made. This can be achieved by the
method called the double mass curve technique.
The double mass curve Fig. 3.5 is obtained by plotting the accumulated
precipitation at the station in question along X-axis and the average accumulated
precipitation of a number of other nearby stations which are situated under the
same meteorological conditions along Y-axis. If the curve has a constant slope,
the record of station “X” is consistent. However, if there is any break in the slope
of the curve, the record of the station is inconsistent and has to be adjusted by the
formula.
Where
Pa =
(Sa / So)x Po
Pa =
Adjusted precipitation.
Po =
Observed precipitation.
Sa =
Slope prior to the break in the curve
So =
Slope after the break in the curve.
All values after break are to be adjusted.
Example 3.4
Check consistency of the data given in table 3.1 below and adjust it if it is found
to be inconsistent
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Year
Annual
precipitation
at x (mm)
1972
1971
1970
1969
1968
1967
1966
1965
1964
1963
1962
1961
1960
1959
1958
1957
1956
1955
188
185
310
295
208
287
183
304
228
216
224
203
284
295
206
269
241
284
Table 3.1 Precipitation Data
Mean of annual Year
Annual
precipitation of
precipitation
20 surrounding
at x (mm)
stations (mm)
264
228
386
297
284
350
236
371
234
290
282
246
264
332
231
234
231
312
1954
1953
1952
1951
1950
1949
1948
1947
1946
1945
1944
1943
1942
1941
1940
1939
1938
1937
223
173
282
218
246
284
493
320
274
322
437
389
305
320
328
308
302
414
Mean of
annual
precipitation
of 20
surrounding
stations
(mm)
360
234
333
236
251
284
361
282
252
274
302
350
228
312
284
315
280
343
Solution
A double mass curve is plotted by taking cumulative of average precipitation of
surrounding stations along x-axis and accumulative precipitation of station ‘X’
along
y-axis for which consistency of data is being investigated.
The double mass curve is shown in Figure 3.5
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Figure 3.5 Double Mass Curve
The correction for slope is applied to readings beyond break in slope. The
calculations are shown in table 3.2, below.
Slope of Ist line = Sa = 0.854
Slope of deviating line = So = 1.176
Correction to values (multiplying factor) = 0.854/1.176 = 0.70
Now regime changes before 1950. So up to 1950 no correction is required. Before
1950 all readings are multiplied by slopes ratio of 0.7 to get corrected
precipitation. Note that data in latter interval (1973-1950) is considered more
authentic so kept in initial reach of the graph.
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Table 3.2 Adjusted Precipitation
54
Cummulative precipitation
of 20 surrounding stations
(mm)
Corrected Precipitation
1972
1971
1970
1969
1968
1967
1966
1965
1964
1963
1962
1961
1960
1959
1958
1957
1956
1955
1954
1953
1952
1951
1950
1949
1948
1947
1946
1945
1944
1943
1942
1941
1940
1939
1938
1937
188
373
683
978
1186
1473
1656
1960
2188
2404
2628
2831
3115
3410
3616
3885
4126
4410
4633
4806
5088
5306
5552
5836
6329
6649
6923
7245
7682
8071
8376
8696
9024
9332
9634
10048
264
492
878
1175
1459
1809
2045
2416
2650
2940
3222
3468
3732
4064
4295
4529
4760
5072
5432
5666
5999
6235
6486
6770
7131
7413
7665
7939
8241
8591
8819
9131
9415
9730
10010
10353
188
185
310
295
208
287
183
304
228
216
224
203
284
295
206
269
241
284
223
173
282
218
246
190.8
345
224
192
225.4
306
272.3
213.5
224
229.6
215.6
211.4
290
Remark
s
No correction
Cummulative
Annual precipitation
at x (mm)
Precipitation of Station 'X' x 0.7
Year
Areal Precipitation Data Analysis
To find out runoff from a catchment and most of other hydrologic analyses, it is
important to know the average precipitation of a certain part of catchment or for
the whole of the catchment area. To find out average precipitation of watershed
records of precipitation from different rain gauge stations is used. There are many
factors which affect the reliability of average precipitation of watershed
determined by using the data from individual stations in the watershed. For
example : the total number of rain gauges and their distribution in the catchment
(larger the number of rain gauges, the reliable will be the calculated average
precipitation), the size and shape of area of catchment, distribution of rainfall over
the area and topography of the area. Lastly the method used for determining the
average precipitation is also one of these factors. So before using average
precipitation in hydrologic analyses, the user should be aware of these factors.
Estimation of Average Precipitation over a Basin
Conversion of point precipitation of various gauging stations into average
precipitation of that area a great experience and skill is required. There are three
methods to find average precipitation over a basin. Accuracy of estimated average
precipitation will depend upon the choice of an appropriate method. These
methods are described below:
i. Arithmetic Mean Method
In this method the average precipitation over an area is the arithmetic average of
the gauge precipitation values. We take data for only those stations which are
within the boundary. This is the simplest method but can be applicable only for
flat areas and not for the hilly areas i.e. this method is used when:
a. Basin area is flat and
b. All stations are uniformly distributed (within practical limits) over the
area.
c. The rainfall is also nearly uniformly distributed over the area.
According to this method
n
P (average) =  1   P i
 n  i 1
or
Pav = [P1+P2+P3+…………+Pn]/n
55
Where Pi is precipitation at station ‘i’ and there are ‘n’ number of gauges
installed in the catchment area from where the data has been collected.
Example 3.5
Six rain gauges were installed in a relatively flat area and storm precipitation from
these gauges was recorded as 3.7, 4.9, 6.8, 11.4, 7.6 and 12.7 cm respectively
from gauges 1, 2, 3, 4, 5, and 6. Find average precipitation over the catchment.
Solution
As the area is relatively flat so we apply the arithmetic mean method. According
to arithmetic mean method
P(average) = (3.7 + 4.9 + 6.8 + 11.4 + 7.6 + 12.7)/6 = 7.85 cm.
ii. Thiessen Polygon Method:
The fundamental principle followed in this method consists of weighing the
values at each station by a suitable proportion of the basin area. In this method, a
special weighing factor is considered.
The following steps are used to determine average precipitation by Thiessen
Polygon Method.
1. Draw the given area according to a certain scale and locate the stations
where measuring devices are installed.
2. Join all the stations to get a network of non-intersecting system of
triangles.
3. Draw perpendicular bisectors of all the lines joining the stations and get a
suitable network of polygons, each enclosing one station. It is assumed
that precipitation over the area enclosed by the polygon is uniform.
4. Measure area of the each polygon.
5. Calculate the average precipitation. For the whole basin by the formula
56
P (average) =
(P1 A1 + P2 A2 + ...........+ Pn An)/A
Where,
P1 = Precipitation. at station enclosed by polygon of area A1
P2 = Precipitation. at station enclosed by polygon of area A2
and so on
Pn = Precipitation. at station enclosed by polygon of area An
And ‘A’ represents the total area of the catchment.
Example 3.6:
Following is shown map of a catchment having 6 rainfall recording stations.
Fig. 3.6 Average Precipitation by Theissen Polygon Method
Find the Average Precipitation over the whole catchment.
57
The recorded precipitations are shown on the topographic map of the catchment.
The Thiessen’s Polygons are constructed by the method explained above. The
precipitation and polygon area are given below.
Station
Precipitation (mm)
Polygon Area (km²)
Daggar
Besham
Shinkiari
Phulra
Tarbela
Oghi
48
33
25
32
56
30
5,068.76
4,349.17
1,399.25
1,693.80
2,196.33
2,234.29
Solution
The calculations are best done in tabular form as shown in Table 3.3.
Table 3.3 Average Precipitation by Thiessen Polygon Method
Station
Precipitation P
(mm)
Polygon Area A
(km²)
P x A (x106 m³)
Daggar
Besham
Shinkiari
Phulra
Tarbela
Oghi
48
33
25
32
56
30
Total
5,068.76
4,349.17
1,399.25
1,693.80
2,196.33
2,234.29
16,941.60
243.30
143.52
34.98
54.20
122.99
67.03
666.02
in
PiAi
Mean Precipitation = 
i 1
i n
 Ai
i 1
= 666.02x106x10³/16941.60x106
= 39.3 mm
Example 3.7
From the data given in Table 3.4 below, which was obtained from Thiessen
Polygon map of a catchment, find out the average precipitation of the catchment.
58
Table 3.4 Precipitation Data
Sr
No
Gauge
precipitatio
n (cm)
1
2
3
10.2
8.1
12.7
Area of Thiessen
Polygon enclosing
the station
(sq. km)
416
260
650
Sr
No
Gauge
precipitation
(cm)
4
5
6
9.4
15.2
7.6
Area of Thiessen
Polygon enclosing
the station
(sq. km)
520
390
325
Solution
According to Thiessen Polygon Method
P (average) =
(P1 A1 + P2 A2 + ...........+ Pn An)/A
The calculations are shown in tabular form in Table 3.5
Table 3.5 Average Precipitation by Thiessen Method
Gauge precipitation
(cm)
(1) = Pi
Area of Thiessen Polygon enclosing
the station (sq. km)
(2) = Ai
Volume = Pi x Ai (x104 m³)
10.2
8.1
12.7
9.4
15.2
7.6
Total
416
260
650
520
390
325
2561
4243.20
2106.00
8255.00
4888.00
5928.00
2470.00
27890.20
So
(3) = (1)x(2)
P (average) = 27890.20 ÷ 2561=10.9 cm
Example 3.8:
There are 10 observation stations, 7 being inside and 3 in neighborhood of a
catchment. Thiessen Polygons were drawn for a storm data from these observation
stations and the data given in Table 3.6 below was obtained. Find out the average
precipitation of the catchment.
59
Table 3.6 Average Precipitation by Thiessen Polygon Method
Sr No
Gauge precipitation (cm)
1
2
3
4
5
6
7
5
3
4
3.5
4.7
6
4
Area of Thiessen Polygon enclosing
the station (sq. km)
100
160
200
215
250
175
100
Solution
According to Thiessen Polygon Method
P (average) =
(P1 A1 + P2 A2 + ...........+ Pn An)/A
Gauge precipitation
(cm)
Area of Thiessen Polygon enclosing the
station (sq. km)
Volume = PixAi (x104 m³)
(1) = Pi
(2) = Ai
(3) = (1)x(2)
5
3
4
3.5
4.7
6
4
Total
100
160
200
215
250
175
100
1200
500
480
800
752.50
1175
1050
400
5157.50
So
P (average) = 5157.5 ÷ 1200 = 4.3 cm
iii. Isohyetal Method
The most accurate method of averaging precipitation over an area is the isohyetal
method.
For estimation of average precipitation of the catchment by isohyetal method the
following steps are used:
60
1. Draw the map of the area according to a certain scale.
2. Locate the points on map where precipitation measuring gauges are
installed.
3. Write the amount of precipitation for stations.
4. Draw isohyets (Lines joining points of equal precipitation).
5. Measure area enclosed between every two isohyets or the area enclosed by
an isohyet and boundary of the catchment.
6. Find average precipitation by the formula.
P (average) =
(P1 A1 + P2 A2 + ...........+ Pn An)/A
Where,
P1= Mean precipitation of two isohyets 1 and 2
A1= Area between these two isohyets.
P2 = Mean precipitation of two isohyets 2 and 3
A2 = the area b/w these two isohyets.
and, so on
Pn = Mean precipitation of isohyets n-1 and n
An = the area between these two isohyets.
It may be noted that the last and first areas mentioned above may be between an
isohyet and boundary of the catchment. In this case the precipitation at the
boundary line is required which may be extrapolated or interpolated.
Example 3.9
From the data given in table 3.7 below, which was obtained from isohyetal map of
a catchment, find out the average precipitation of the catchment.
61
Table E3.7 Data from Isohyetal Map.
Isohyet
No.
1
Isohyetal precipitation (cm)
2.5
Area enclosed between two
isohyets. (sq km)
390
2
5.0
520
3
7.5
650
4
10.0
390
5
10.0
390
6
7.5
442
7
5.0
546
8
2.5
Note that the isohyet No. 1 and 8 were out of the boundary of the catchment. The
area between isohyet No. 2 and the boundary was estimated to be 312 sq. km and
that of between isohyet No. 8 and boundary was 494 sq. km. Precipitation on
these boundaries was interpolated as 3.0 and 3.1 cm, respectively.
Solution
In isohyetal method we have to calculate the average precipitation of every two
consecutive isohyets. This is given in Table E3.8 below:
62
Isohyet
No.
(1)
Table E3.8 Average Precipitation by Isohyetal Method
Isohyetal
Average of
Area enclosed
Volume
precipitation
precipitation of
between two
(x104 m³)
(cm)
two consecutive
isohyets
isohyets (cm)
(sq km)
(2)
(3)
(4)
(5) = (3) x (4)
1
2.5
2
3
5.0
7.5
4
5
6
7
10.0
10.0
7.5
5.0
8
2.5
3 (for isohyet
and boundary)
6.25
8.75
312 (for isohyet
and boundary)
520
650
936.00
10.0
8.75
6.25
3.1 (for isohyet
and boundary)
390
390
442
546 (for isohyet
and boundary)
3900.00
3412.50
2762.50
1692.60
∑
3250
21641.1
3250.00
5687.50
P (average) =
(P1 A1 + P2 A2 + ...........+ Pn An)/A
= 21641.1/3250=6.66 cm
Example 3.10
In a catchment of area 1,000 sq km, there are 8 rain gauges, 5 inside the area and 3
outside, in its surroundings. Isohyets were drawn from the data of these rain
gauges for a storm. From the isohyetal map the following information was
obtained: areas between 1 and 2 cm isohyets, 2 and 3 cm, 3 and 4 cm and 4 and 5
cm isohyets was 105, 230, 150 and 220 sq. km, respectively. The area between
one end boundary which has 0.75 cm rainfall and 1 cm isohyet was 120 sq. km
and the other end boundary which has precipitation of 5.5 cm and isohyet of 5 cm
was 175 sq. km. Find average precipitation.
Solution
In the isohyetal method we have to calculate the average precipitation of every
two consecutive isohyets. This is given in Table 3.9 below.
63
Table E3.9 Average Precipitation by Isohyetal Method
Isohyet
No
Isohyetal
precipitation
(cm)
Average of
precipitation of two
consecutive isohyets
(cm)
0.875 (for isohyet
and boundary)
Area enclosed
between two
isohyets
(sq km)
120 (for isohyet
and boundary)
Volume
(x104 m³)
Boundary
0.75
1
1
1.5
105
157.50
2
2
2.5
230
575.00
3
3
3.5
150
525.00
4
4
4.5
220
990.00
5
5
5.25 (for isohyet and
boundary)
175
918.75
Boundary
5.5
∑
1000.00
3271.25
P (average) =
(P1 A1 + P2 A2 + ...........+ Pn An)/A
= 3271.25/1000
=3.27 cm
64
105.00
Example 3.11:
From the isohyetal map shown in Fig. 3.3 below find out average precipitation.
Fig. 3.7 Isohyetal Map
Solution:
The isohyets are drawn on the topographic map by interpolating rainfall depths at
given stations. Once isohyets are drawn, the area enclosed between consecutive
isohyets is determined either by planimeter or other suitable more precise method.
The calculations for average precipitation are given in table 3.10 below.
65
Table 3.10 Average Precipitation by Isohyetal Method
Isohyte value
(mm)
Boundary and 25
25 and 30
30 and 35
35 and 40
40 and 45
45 and 50
50 and 55
55 and Boundary
Av. Isohyte
Value (mm)
25.0
27.5
32.5
37.5
42.5
47.5
52.5
55
Total
Volume (x106 m³)
Area Between
Consecutive
Isohytes (km²)
310.53
2220.71
2968.38
2231.86
2303.52
2731.90
2689.70
1484.99
16,941.60
7.76
61.07
96.47
83.69
97.90
129.77
141.21
81.67
699.54
Mean Precipitation Depth = Volume/Area
= 699.54x106x10³/16941.60x106
= 41.29 mm
INTENSITY OF PRECIPITATION
The rate of occurrence of precipitation is called intensity of precipitation or
precipitation occurring in unit time is known as intensity of precipitation.
To find out intensity of a certain interval the points on graph of accumulative
precipitation vs time are chosen in such a way that we get the maximum
difference for the given interval. It is explained in the following example.
Example 3.12
Find out intensity of precipitation of 5, 10 & 15 minutes for rainfall data given in
Table 3.11.
Table 3.11 Rainfall Data
Time (Minutes)
0
05
10
15
20
66
P (cm)
00
12.5
20
42.5
62.5
Solution
For 5-minutes interval the maximum difference is 22.5
so, intensity for 5-minutes interval = 22.5 / 5 = 4.5 cm/min.
For 10-minutes intensity = 42.5 / 10 = 4.25 cm/min.
For 15-minutes intensity = 50 / 15 = 3.33 cm/min.
Depth - Area Relationships
The distribution of rainfall is usually not uniform over the area. The precipitation
is maximum at the centre of storm and decreases as we move away from the
centre of storm. For rainfall of a given duration, the average depth decreases with
the area in an exponential manner.
Mass Curve
Precipitation recorded by a weighing type and float type recording rain gauge is in
form of a graph. It is a plot of cumulative precipitation against time in
chronological order. This is called mass curve of rainfall data. Intensity of rainfall
for a certain duration is determined from this graph. This graph gives a complete
history of a storm regarding the duration and magnitude of precipitation. In the
case of non-recording rain gauges the mass curve is to be plotted from the data in
which both duration and magnitude of precipitation have been observed for
different time intervals during storm.
Depth-Area-Duration Curves
Analysis of both the time and areal distribution of a storm is required in many
hydrologic studies. Depth area duration curves provide requisite information for
such studies. It is necessary to have information on the maximum amount of
precipitation of various durations occurring over various sizes of areas. The
development of a relationship between maximum depth-area-duration for a region
is called depth area duration analysis (DAD analysis). In this analysis first the
isohyetal maps and mass curves of worst storms which have occurred in past in
the region are prepared. For a storm with a single major centre the isohyets are
taken as the boundaries of individual area. The average storm precipitation within
each isohyet is computed. The storm total is distributed through successive
67
increments of time (say 3 hours) in accordance with the distribution record at
nearby stations. This gives data showing the time distribution of average rainfall
over areas of various sizes. From this data the maximum rainfall for various
durations (3, 6, 9, 12 hours) can be selected for each size of area. The maximum
values for every duration plotted versus area gives what are called depth area
duration curves.
Typical depth-area-duration curves are shown in figure 3.8.
Fig. 3.8 Typical Depth Area Curves
EXAMPLE 3.13:
The isohyetal map shown in Fig 3.9 is for 10 hour storm over a catchment area.
Area enclosed between two consecutive isohyets is shown on the map and
isohyetal interval is 5 cm with storm center having precipitation of 30 cm. Find:
a. The average precipitation of the catchment by isohyetal method
b. The Equivalent Uniform Depth of rain for depth area duration curve.
68
Fig. 3.9 Isohyet Map
Solution
The calculations are performed in Table 3.13. The average precipitation is found
by summing up area enclosed by consecutive isohyets multiplied by average
isohyte value and whole sum divided by total area.
The EUD is found by dividing cumulative volume by cumulative area. The Fig.
3.10 shows variation of EUD with area.
Table 3.13 Finding EUD
Isohyte (cm)
Area
enclosed
(km²)
Cumm.
Area
enclosed
(km²)
Mean
Isohyte
(cm)
Volume
(x106)
Cumm.
Volume
(x106)
EUD
(cm)
1
30
30 & 25
25 & 20
20 & 15
15 & 10
10 & 5
2
60
100
90
130
200
400
3
60
160
250
380
580
980
4
30.00
27.50
22.50
17.50
12.50
7.50
5
18.00
27.50
20.25
22.75
25.00
30.00
6
18.00
45.50
65.75
88.50
113.50
143.50
7 = 6÷3
30.00
28.44
26.30
23.29
19.57
14.64
Mean Precipitation = 14.64 cm
69
Fig. 3.10 Depth-Area Curve
EXAMPLE 3.14:
Plot depth area duration curves from the rainfall data for six stations given in
Table 3.14
70
DATE
16-02-2003
17-02-2003
STATION & PRECIPITATION
TIME
(HOURS
)
BESHAM
PHULRA
DAGGAR
OGHI
SHINKIARI
TARBELA
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
1
1
2
2
3
3
2
2
1
2
1
1
1
2
3
5
4
5
5
8
3
1
3
3
2
2
7
4
6
4
5
5
5
5
4
5
5
1
2
3
3
2
1
2
1
1
2
3
3
3
4
3
3
1
1
1
6
4
4
5
3
4
4
5
6
3
5
3
4
4
2
2
1
1
1
2
1
1
1
1
2
2
2
1
1
3
3
4
5
6
3
2
2
6
11
6
9
3
6
8
9
10
9
9
3
2
2
1
2
3
2
3
3
3
2
3
4
3
3
3
2
1
1
4
3
3
2
9
9
6
5
7
8
6
6
4
5
5
4
5
1
1
3
2
2
3
1
1
3
4
5
3
2
3
2
3
1
3
3
4
3
5
1
3
4
4
4
6
2
3
4
2
1
4
1
1
8
5
5
4
8
4
6
8
9
8
9
10
15
9
71
DATE
18-02-2003
19-02-2003
72
STATION & PRECIPITATION
TIME
(HOURS
)
BESHAM
PHULRA
DAGGAR
OGHI
SHINKIARI
TARBELA
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
6
6
7
5
5
3
2
2
1
3
5
3
1
1
3
4
5
5
10
4
1
1
2
3
2
1
3
2
1
2
1
1
1
2
1
2
3
1
2
-
5
8
9
6
8
5
4
5
5
7
2
5
3
4
2
2
5
4
2
4
6
5
3
1
1
1
2
1
2
1
1
1
2
2
2
1
-
8
7
6
3
3
2
2
1
1
1
1
1
1
3
5
2
1
2
1
4
3
3
2
2
1
1
1
1
1
2
2
2
1
1
-
7
5
4
5
4
6
5
7
10
6
9
10
9
5
4
1
1
5
3
4
2
1
3
2
4
2
2
2
5
1
1
2
1
1
1
3
4
2
1
1
-
6
4
5
4
6
8
9
1
5
6
3
6
7
4
6
3
1
2
4
5
1
5
4
3
4
5
3
1
3
1
1
1
3
1
-
11
6
4
4
1
1
1
6
5
6
9
6
1
3
4
2
3
5
4
3
1
1
4
4
5
3
1
1
2
1
DATE
20-02-2003
TIME
(HOURS
)
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
STATION & PRECIPITATION
BESHAM
PHULRA
DAGGAR
OGHI
SHINKIARI
TARBELA
3
1
2
1
1
3
1
1
-
1
1
1
1
1
1
-
0
0
1
0
0
2
1
1
-
1
1
1
1
2
1
1
1
1
1
2
1
-
5
1
-
1
1
2
-
Solution
Step 1:
First we plot average un-weighted mass curve for total rainfall over the whole
catchment. The calculations are shown in table 3.15 and figure 3.11. Total
precipitation is found by summing up precipitation recorded at individual stations
for each hour of record. Similarly average for precipitation is found by averaging
precipitation recorded at six stations. Now the average un-weighted mass curve is
plotted by taking time along x-axis and cumulative average precipitation along yaxis. The purpose of this plot is to find maximum precipitation for 6 hours, 12
hours and 24 hours rain over the whole catchment.
The maximum precipitation during selected intervals of 6, 12 and 24 hours can
also be found by calculations. For example for 6-hour maximum rain total
precipitation is added for every 6 consecutive hours. Once it has been done for all
73
data maximum value is noted along with the period in which it occurred. In this
case, this value comes out to be 224 mm. Similarly total precipitation is summed
for consecutive 12 and 24 hours. These values come out to be 418 mm and 697
mm. Note that these are rains for all six given sections for selected time interval.
Table 3.15 Depth Area Duration Analysis
DATE
16-02-2003
17-02-2003
74
6 STATION PRECIP. (mm)
TIME
(HOURS)
TIME
INTERVAL
CUMM. TIME
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
9
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
6-HR
12-HR
24-HR
TOTAL
AV.
CUMM. AV.
PRECIP.
PRECIP.
PRECIP.
1
1
1
6
9
16
16
11
12
6
8
6
5
4
6
14
18
13
17
18
14
20
14
9
10
24
22
0.16
0.16
0.16
1.33
1.5
2.67
2.67
1.5
2.16
1.16
1.33
1
0.83
0.83
1.16
1.83
3.16
2.16
3
3
2.5
3.5
2.33
1.5
1.67
4
3.67
0.16
0.32
0
0.64
1.81
3.31
5.98
8.65
10.15
12.31
13.47
14.8
15.8
16.63
17.96
18.62
20.45
23.61
25.77
28.77
31.77
34.27
37.77
40.14
41.6
43.27
47.27
50.94
20
35
50
59
71
70
69
59
48
44
38
42
53
60
73
86
94
104
99
95
87
93
100
89
94
98
103
109
112
122
119
121
130
132
146
152
155
160
179
194
241
249
258
262
303
DATE
18-02-2003
19-02-2003
6-HR
12-HR
24-HR
TOTAL
AV.
CUMM. AV.
PRECIP.
PRECIP.
PRECIP.
34
23
3.83
54.77
102
206
318
35
29
4.83
59.6
117
216
338
1
36
31
5.16
64.76
139
234
358
1
37
30
5
69.76
159
246
367
14
1
38
34
5.67
75.43
169
262
392
15
1
39
31
5.16
80.59
178
278
410
16
1
40
35
5.83
86.42
190
292
438
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
33
34
35
41
36
43
36
35
27
27
24
23
16
17
16
23
27
25
21
23
24
20
16
26
23
8
14
20
17
14
13
9
13
11
8
7
5.5
5.67
5.83
6.83
6
7.16
6
5.83
4.5
4.5
4
3.8
2.6
2.8
2.7
3.8
4.5
4.2
3.5
3.8
6
3.3
2.8
4.3
3.83
1.33
2.33
3.33
2.83
2.33
2.16
1.5
2.16
1.83
1.33
1.16
91.92
97.59
103.42
110.25
116.25
123.41
129.41
135.24
139.74
144.24
148.24
152.07
154.74
157.57
160.24
164.07
168.57
172.74
176.24
180.07
186.07
189.4
192.23
196.56
200.4
201.73
204.06
207.4
210.23
212.56
214.72
216.22
218.38
220.21
221.54
222.7
194
197
202
209
214
222
225
311
336
361
378
392
412
419
226
218
204
192
172
152
134
123
119
122
124
129
135
143
140
130
131
133
118
108
108
108
96
87
86
86
77
68
61
423
420
413
406
394
377
360
341
323
314
296
281
269
266
259
252
255
262
253
251
248
238
227
219
205
194
185
176
157
463
491
521
557
586
618
635
657
666
675
684
686
688
696
702
701
706
TIME
(HOURS)
TIME
INTERVAL
CUMM.
TIME
10
1
11
1
12
13
6 STATION PRECIP. (mm)
708
700
692
686
674
658
649
639
613
592
571
552
523
500
474
460
444
428
412
75
DATE
20-02-2003
76
TIME
(HOURS)
TIME
INTERVAL
CUMM.
TIME
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
6 STATION PRECIP. (mm)
TOTAL
AV.
CUMM. AV.
5
5
6
10
14
16
11
5
5
3
1
2
2
4
1
2
4
8
4
3
2
7
2
1
1
2
2
-
0.83
0.83
1
1.67
2.33
2.67
1.83
0.83
0.83
0.5
0.16
0.33
0.33
0.67
0.16
0.33
0.67
1.33
0.67
0.5
0.33
1.16
0.33
0.16
0.16
0.33
0.33
-
223.53
224.36
225.36
227.03
229.36
232.03
233.86
234.7
235.52
236.02
236.02
236.02
236.02
236.02
236.02
236.02
236.18
236.51
236.84
237.51
237.67
238
238.67
240
240.67
241.17
241.15
242.66
243
243.16
243.32
243.32
243.65
243.98
243.98
243.98
243.98
243.98
243.98
243.98
243.98
243.98
243.98
243.98
6-HR
PRECIP.
12-HR
PRECIP.
24-HR
PRECIP.
53
49
42
41
47
56
62
63
61
54
40
24
13
8
3
1
3
5
9
10
12
15
21
23
22
23
28
26
19
16
13
13
8
6
5
4
4
2
-
139
136
128
118
115
117
115
111
103
95
87
80
75
70
64
54
41
27
18
17
13
12
16
24
28
31
33
40
41
40
39
35
36
36
32
24
20
17
15
8
6
5
4
4
401
389
379
366
353
344
334
316
297
280
263
237
214
206
192
172
156
144
133
128
116
107
103
104
103
101
97
94
82
67
57
52
49
48
48
48
48
48
48
48
47
45
43
39
Figure 3.11 Mass Curve
Step 2:
Find the contribution of each rain station toward mass of total rainfall during
selected time periods. Mark these precipitation values on topographic map of the
catchment having point rain stations. Draw isohyets by interpolation between rain
recording stations. This is shown in figures 3.12, 3.13 and 3.14.
Step 3:
Find the area between consecutive isohyets and determine Equivalent Uniform
Depth (EUD) for 6, 12 & 24 hours maximum rainfall intervals. These calculations
are shown in table 3.16, 3.17 and 3.18. The EUD is found by dividing cumulative
volume by the cumulative area.
Step 4:
Now we can plot Depth Area Duration Curves. Take cumulative area along x-axis
and EUD along y-axis. The curves are shown in figure 3.15 below:
77
Fig. 3.12, 6-Hour Maximum Rain Isohyet Map
Fig. 3.13, 12-Hour Maximum Rain Isohyet Map
78
Fig. 3.14, 24-Hour Maximum Rain Isohyet Map
79
Table 3.16 : 6 - Hour Rainfall
Isohyet Value (mm)
Av.
Isohyet
Value
Area
Between
Consecutive
Isohyets
(mm)
(km²)
25
27.5
32.5
37.5
42.5
47.5
52.5
55
Total
310.53
2220.71
2968.38
2231.86
2303.52
2731.9
2689.7
1484.99
16,941.60
Boundary and 25
25 and 30
30 and 35
35 and 40
40 and 45
45 and 50
50 and 55
55 and Boundary
Cumulative
Area (km²)
310.53
2531.24
5499.62
7731.48
10035
12766.9
15456.6
16941.6
Volume
Cumulative
Volume
Equivalent
Uniform
Depth
(x106 m³)
(x106 m³)
(mm)
7.76
61.07
96.47
83.69
97.9
129.77
141.21
81.67
699.54
7.76
68.83
165.3
248.99
346.89
476.66
617.87
699.54
24.99
27.2
30.06
32.2
34.57
37.34
39.97
41.3
Average Precipitation over the area = 699.54x1000 ÷ 16941.60 =
41.29 mm
Table 3.17 : 12 - Hour Rainfall
Isohyet Value
(mm)
Av.
Isohyet
Value
Area Between
Consecutive
Isohyets
(mm)
(km²)
Cumulative
Area (km²)
Volume
Cumulative
Volume
Equivalent
Uniform
Depth
(x106 m³)
(x106 m³)
(mm)
Boundary and
45
45 and 50
25
675.27
675.27
16.88
16.88
25.00
47.5
698.88
1,374.15
33.20
50.08
36.44
50 and 55
52.5
805.13
2,179.29
42.27
92.35
42.38
55 and 60
57.5
1,360.77
3,540.06
78.24
170.59
48.19
60 and 65
62.5
960.18
4,500.24
60.01
230.60
51.24
65 and 70
67.5
1,396.98
5,897.21
94.30
324.90
55.09
70 and 75
72.5
1,777.90
7,675.11
128.90
453.80
59.13
75 and 80
77.5
1,871.56
9,546.67
145.05
598.84
62.73
80 and 85
82.5
1,863.69
11,410.36
153.75
752.60
65.96
85and 90
87.5
1,677.95
13,088.30
146.82
899.42
68.72
90 and 95
95 and
Boundary
92.5
1,875.49
14,963.79
173.48
1,072.90
71.70
95
1,977.81
16,941.60
187.89
1,260.79
74.42
Total
16,941.60
1,260.79
Average Precipitation over the area = 1260.79x1000 ÷ 16941.60 =
80
74.42 mm
Table 3.18 : 24 - Hour Rainfall
Isohyet
Value (mm)
Area
enclosed
by 145
145 and
140
140
and
Av.
Isohyet
Value
(mm)
Area
Between
Consecutive
Isohyets
(km²)
Cumulative
Area
(km²)
Volume
(x106 m³)
Cumulative
Volume
(x106 m³)
Equivalent
Uniform
Depth
(mm)
145
138.50
138.50
20.08
20.08
145.00
142.5
83.10
221.59
11.84
31.92
144.06
137.5
592.76
814.36
81.50
113.43
139.29
135
135
and
130
130
and
132.5
893.50
1,707.85
118.39
231.82
135.74
127.5
1,164.95
2,872.80
148.53
380.35
132.40
125
125
and
120
120 and
122.5
1,582.02
4,454.82
193.80
574.14
128.88
117.5
3,303.32
7,758.14
388.14
962.28
124.04
115
115
and
110
110 and
112.5
1,205.31
8,963.45
135.60
1,097.88
122.48
107.5
1,179.19
10,142.64
126.76
1,224.65
120.74
102.5
1,620.80
11,763.44
166.13
1,390.78
118.23
100
5,178.16
16,941.60
517.82
1,908.59
112.66
Total
16,941.60
105
105
and
100
100 and
Boundary
1,908.59
Average Precipitation over the area = 1908.59x1000 ÷ 16941.60 =
112.66 mm
Figure 3.15 Depth-Area-Duration Curves for 6, 12 & 24 hours rainfall
81
QUESTIONS
1. What is meant by Precipitation?
2. Define the terms ‘Amount of Rainfall’ and ‘Intensity of Rainfall’?
3. What are the types of precipitation based on the mechanisms that produce
it?
4. What are different forms of precipitation on the basis of lifting
mechanism?
5.
How precipitation is measured?
6. Explain how a network of rain-gauge stations is designed?
7. Categorize different instruments used for measurement of precipitation
with respect to their merits and limitations?
8. Explain various methods used for analysis of precipitation data. What is
meant by reliable data?
9. Describe the ‘Double Mass Curve Technique’ for checking the consistency
of precipitation data?
10. How missing precipitation data is interpolated?
11. How precipitation data is checked for consistency and homogeneity?
12. Explain the utility of “Mass Curve” and “Double Mass Curve”?
13. What are Depth-Area-Duration Curves? What important information is
extracted from these curves?
14. Explain procedure for Depth-Area-Duration Analysis?
15. Define Probable Maximum Precipitation (PMP) of a catchment?
82
EXERCISE
Problem 3.1
The names and locations of the rain gauge stations with their mean annual
precipitation are as given in the table below. Calculate average precipitation of the
catchment area by Isohyetal method.
Name of
Station
Latitude
Degrees
Minutes
Longitude
Degrees
Minutes
Mean Annual
Precipitation
(in)
A
B
C
D
E
F
G
H
I
J
K
L
16
16
14
17
17
17
14
16
15
17
15
16
03
35
58
25
27
10
53
25
15
26
45
26
103
104
102
104
101
104
103
101
104
102
102
102
04
44
07
50
44
09
29
08
53
46
02
51
39.85
56.58
45.73
85.12
47.51
70.59
52.67
50.83
60.57
56.22
41.50
46.81
Latitude and Longitude of the Catchment Area are 140 30’ to 180 30’ and 1010
00’ to 1040 00’, respectively.
Problem 3.2
Assuming the rain falling vertically, express the catch of a gauge inclined 15°
from the vertical as a percentage of the catch of a gauge installed vertically.
Problem 3.3
A 3-hour storm occurred at a place and the precipitation in the neighboring raingauge stations P, Q and R were measured as 3.8, 4.1 and 4.5 cm respectively. The
precipitation in the neighboring station S could not be measured since the rain
gauge was damaged. The normal annual precipitation in the four stations P, Q, R
and S were 45, 48, 53 and 50 cm, respectively. Estimate the storm precipitation at
station ‘S’.
83
Problem 3.4
In a catchment one precipitation station ‘A’, was inoperative for part of a month
during which storm occurred. The respective storm totals at three surrounding
stations 1, 2 and 3 are 35, 40 and 30 mm, respectively. The normal annual
precipitations at A, 1, 2 & 3 are respectively 985, 1125, 940 and 1210 mm. Estimate
the storm precipitation for Station ‘A’.
Problem 3.5
The average annual precipitation for the four sub-basins constituting a large river
basin is 58, 67, 85 and 80 cm. The sub-basin areas are 900, 690, 1050 and 1650
km², respectively. What is the average annual precipitation for the basin as a
whole?
Problem 3.6
The annual precipitation at station ‘A’ and the average precipitation at
15 surrounding stations are given in the Table 3.19 given below:
a. Determine the consistency of the record at station ‘A’.
b. In which year, there is a regime change indicated?
Table 3.19 Data Table
Year
1971
1972
1973
1974
1975
1976
1977
1978
1979
1980
Annual Precipitation
Station A
15 Station
(cm)
Average (cm)
50.5
90.0
16.0
21.5
50.5
62.5
69.5
36.0
42.0
42.0
71.5
57.0
27.5
25.0
60.0
22.0
55.0
57.0
36.5
19.5
Year
1981
1982
1983
1984
1985
1986
1987
1988
1989
1990
Annual Precipitation
Station A
15 Station
(cm)
Average (cm)
36.0
42.0
18.0
30.0
54.0
48.0
12.0
36.0
42.0
36.0
27.5
60.5
55.0
38.5
38.5
47.5
49.5
24.0
44.0
60.5
Problem 3.7
The Table 3.20 given below gives the annual rainfall at a station ‘A’ and the
average annual rainfall of 10 stations in the vicinity for a period of 17 years. It is
suspected that there has been a change in the location of the rain gauge at Station
‘A’ during the period of this record. Determine the year when this change
occurred and the corrected rain gauge readings prior to this year.
84
Table 3.20 Rainfall Data
Rainfall
Year
Station A
(mm)
Average
Annual
Rainfall of 10
Stations (mm)
1931
1932
1933
1934
1935
1936
1937
1938
1939
1940
505
900
60
215
505
625
695
360
420
360
615
570
275
250
605
220
550
570
365
275
Year
Rainfall
Average
Station A
Annual
(mm)
Rainfall of 10
Stations (mm)
1941
1942
1943
1944
1945
1946
1947
420
180
300
540
480
120
360
605
550
385
385
475
495
240
Problem 3.8
Point rainfalls due to a storm at several rain-gauge stations in a basin are shown in
Fig. 3.16. Determine the mean areal depth of rainfall over the basin.
Figure 3.16 Isohyetal Map
85