Hydrologic cycle,weather and temperature, Rain gauge, precipitation, discharge measurement,stream flow, hydrographs, flood and flood routing. unit hydrographs,

1. 1
Lecture -3
Recording rain gauge
The recording gauges produce a continuous plot of rainfall
against time and provide valuable data of intensity and
duration of rainfall for hydrological analysis of storms.
The commonly used rainfall recording gauges are as follows:
1. Typing - bucket type
2. Weighing – bucket type
3. Natural – Siphon type

2. 2
Lecture -3
Types of Recording Rain Gauge
Twotypes
1.Telemetering Rain gauges
This is electronic units to
transmit the data on rainfall to a
base station both at regular
intervals and on interrogation.
The typing- bucket type rain
gauge is ideally suited for this
purpose. Telemetering gauges
are most suitable to record
rainfall data from mountainous
and generally inaccessible
places.
2.Radar Rain gauges
The metrological radar is a
powerful instrument for
measuring the real extent,
location and movement of rain
storms. The amounts of
rainfall over large areas can be
determined through the radar
with a good degree of
accuracy. The radar is a
remote-sensing super gauge
covering an aerial extent of as
much as 100,000 km2 .

3. 3
Lecture -3
Rain gauge networks
According to the World Metrological Organization (WMO) the
criteria for establishment of rain gauge stations :
1. In flat regions of temperate, Mediterranean and tropical
zones :
ideal – 1 station for 600- 900 km2
acceptable – 1 station for 900 – 3000 km2
2. in mountainous regions of temperate, Mediterranean and
tropical zones:
ideal – 1 station for 100 – 200 km2
acceptable - 1 station for 250 – 1000 km2 and
3. In arid and polar zones : 1 station for 1500 – 10,000 km2
depending on the feasibility.

4. 4
Lecture -3
Adequacy of Rain gauge Stations
The percentage of error in the estimation of mean rainfall is
obtained by statistical analysis. Optimal number of stations
may be determined as
N = (Cv / Є )2 ......................... Eq – 1
Where,
N = Optimal number of stations
Є = allowable degree of error in the
estimate of the mean rainfall and
Cv = coefficient of variation of the
rainfall values at the existing
m stations (in percent).

5. 5
If there are m stations in the catchment each recording rainfall
values P1, P2, P3, ..., upto Pm in a known time, the coefficient of
variation Cv is calculated as :
Cv = (100 * σm-1 )/ Pavg
Where,
m
σm-1 = √ [{∑ ( Pi– Pavg ) 2 }/{(m-1)}] =Standard deviation ...Eq-2
i=1
Pi = precipitation magnitude in the ith station
m = Number of stations
Lecture -3(contd.)

6. 6
Lecture -3(contd.)
m
Mean precipitation, Pavg = (1/m) ∑ Pi ....... Eq-3
i = 1
It is usual to take Є = 10 % in calculating N from the Eq- 2.
If the value of Є is small, the number of rain gauge stations
will be more .

7. 7
Lecture -3(contd.)
Example 2- 1
A catchment has a six rain gauge stations. In a year the
annual rainfall recorded by the gauges are shown in Data
Table-1 . For a 10% error in the estimation of the mean
rainfall, calculate the optimum number of stations in the
catchment.
Station A B C D E F
Rainfall
(cm)
82.60 102.90 180.30 110.30 98.80 136.70
Data Table-1

8. 8
Solution :
Given
m (number of stations) = 6
error, Є = 10%
Annual rainfall data in Table-1
To be calculated the optimum number of stations in the catchment.
We know optimal number of stations, N as
N = (Cv / Є )2 and...................... Eq- 4
and Cv = (100 * σm-1 )/ Pavg
Lecture -3(contd.)

9. 9
As we know from Eq-3 mean precipitation
m
Pavg = (1/m) ∑ Pi
i = 1
m
∑ Pi = 82.6 + 102.90 + 180.30 + 110.30 + 98.80 + 136.70 = 711.60cm
i = 1
Putting this value in above equation ,
Pavg = 711.60/ 6 = 118.60 cm
Lecture -3(contd.)

10. 10
Putting the value of m, Pi and P in Eq -2 , we find
A, (Pi–Pavg) 2 = (82.6 – 118.6)2 =1296
B, (Pi – Pavg) 2 = (102.9 – 118.6) 2 =246.49
C, (Pi – Pavg) 2 = (180.3– 118.6) 2 = 3806.89
D, (Pi – Pavg) 2 = (110.3 – 118.6) 2 = 68.89
E , (Pi – Pavg ) 2 = (98.8 – 118.6) 2 = 392.00
F, (Pi – Pavg) 2 = (136.7– 118.6) 2 = 327.6
Hence, ∑(Pi–Pavg) 2
=(1296.0 + 246.49 + 3808.89 + 68.89 + 392.00 + 327.6
= 6137.88
Lecture -3(contd.)

11. 11
Lecture -3(contd.)
Now for σm-1
m
σm-1= √ [{∑ ( Pi– Pavg ) 2 }/{(m-1)}] =Standard deviation
i=1
Putting the respective values in the equation above
σm-1 = 6137.88 / (6-1) = 35.04
Again for Cv
Cv = (100 x 35.04 ) / (118.60) = 29.54
Putting all the values of Cv , σm-1 and Pavg in the Eq-1 for
calculation of optimum number of stations,
N = (Cv / Є )2 = (29.54 / 10 )2 = 8.70, Say, 9 nos.

12. 12
Lecture -3(contd.)
Preparation of Data : Estimation of missing Data
Methods of estimation of missing data. Two types
(2) Ratios of normal annual precipitations
(1) Arithmetic mean/Average Method
Assume the annual precipitation values, P1, P2, P3, .......Pm at
neighboring M stations 1,2,3,4.......... M respectively. It is
required to find the missing annual precipitation Px at a station
X not included in the above M stations. The normal annual
precipitations N1, N2, N3 ....... ......... Ni ..... at each of the above
(M+1) stations including station X are known. Then Arithmetic
mean/Average
Px = [1/M ] [ P1+P2+P3+...........+Pm]............... Eq -5
(1) Arithmetic mean

13. 13
Lecture -3(contd.)
If the normal annual precipitations at various stations are
within about 10% of the normal annual precipitation at
station X, than arithmetic average method is followed to
estimate Px .
(2) Ratios of normal annual precipitations
If the normal annual precipitations at various stations are more than
about 10% of the normal annual precipitation at station X then the
Px is estimated by weighing the precipitation at the various stations
by the ratios of normal annual precipitations. This method is known
as the normal ratio method, which will give Px as
Px = (Nx / M )(P1 / N1 + P2 / N2 + P3 / N3 + ........+ Pm / Nm ) .... Eq -6

14. 14
Lecture -3(contd.)
Example 2-2
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 station A, B and C
recorded annual precipitation of 91.11, 72.23 and 79.89 cm
respectively. Estimate the rainfall at station D in that year.
Solution :
Given,
Normal annual rainfall at stations:- A = 80.97 cm
B = 67.59 cm
C = 76.28 cm
D= 92.01 cm

15. 15
Annual precipitations:-
P1 = 91.11 cm
P2 = 72.23 cm
P3 = 79.89 cm
To be estimated annual rainfall at station D
Checking of methods-
The variation among the stations :
76.28-67.59 = 8.69 cm
Which is ,
(8.69 /76.28) x 100 = 11.39 % < 10 %
Lecture -3(contd.)

16. 16
Lecture -3(contd.)
Hence, Ratios of normal annual precipitations
method to be used
Putting respective values in Eq-6
PD = (ND / M )(P1 / N1 + P2 / N2 + P3 / N3 + ........+ Pm / Nm )
PD = (92.01 / 3 )(91.11 / 80.97 + 72.23 / 67.59 + 79.89/ 76.28)
PD = 99.41 cm

17. 17
Lecture -3(contd.)
Presentation of rainfall data
The rainfall data may present in three ways.
1. Mass Curve of Rainfall
The mass curve is a plot of the accumulated precipitation
against time, plotted in chronological order. Records of float
type and weighing bucket type gauges are of these form. For
extracting the information on the duration and magnitude of a
storm, Mass curves rainfall is very much useful. Again the
intensity of the storm can also be calculated by slope of the
curve Figure-2-3.
3.Point Rainfall
2.Hyetograph1. Mass Curve of Rainfall

18. 18
Lecture -3(contd.)
0 1 2 3 4
AccumulatedPrecipitation(cm)
02468101214
1st storm = 10 cm
2nd storm
= 4 cm
Time (days)
Figure 2-3 -Mass curve of rainfall

19. 19
0 8 16 24 30 36 42 48 56 62
00.10.20.30.4
Rainfallintensity(cm/hr)
Hyetograph of the first storm in
Total depth = 10 cm
Duration = 56 hrs.
Figure2-4: Hyetograph of a storm
(2) Hyetograph
A plot representated by the intensity of rainfall against the time
interval is called hyetograph. The hyetograph is derived from the
mass curve and is usually represented as a bar chart as shown in
figure 2-4. It is particularly important in the development of design
storms to predict extreme floods. The area under hyetograph
represents the total precipitation received in the period. In larger
catchments the time interval is 6hrs.
Lecture -3(contd.)

20. 20
Lecture -3(contd.)
(3) Point Rainfall
Point rainfall is also known as station rainfall refers to the
rainfall data of a station. Depending on the need, data listed
as
• daily
• weekly
• monthly
• seasonal
• annual values for various periods.
Graphically these data are represented as plots of magnitude
vs chronological time in the form of a bar diagram.

21. 21
Lecture -3(contd.)
Mean Precipitation over an area.
To convert the point rainfall values at various stations into
average value over a catchment- there are three methods :
(1) Arithmetical mean method
(2) Thiessen –polygon method
(3) Isohyetal method

22. 22
Lecture -3(contd.)
(1) Arithmetical mean/average method
Assume P1, P2, P3 ......Pn are the precipitation values of the
stations N for the given period. Then the average value of
the mean precipitation p over the catchment will be
arithmetical mean of the precipitation as follows :
Pavg = (P1 + P2 + P3 + ....... + Pi... + Pn )/N
N
Pavg = 1/N ( ∑ Pi )
i = 1
where. i = 1,2, 3 ....... N

23. 23
Lecture -3(contd.)
2) Thiessen - mean method
The rainfall recorded at each station is given a weightage on
the basis of an area closest to the station.
Procedure for determining the weighting area
Consider a catchment area has three rain gauge stations. And
three stations outside the catchment but the positions in its
neighborhood. The catchment area is drawn to scale and the
positions of the six stations marked on it. Station 1 to 6 are
joined to form a network of triangles. Perpendicular bisectors
for each of the sides of the triangle are drawn. These
bisectors form a polygon around each station. The boundary
of the catchment, if it cuts the bisectors is taken as the outer
limit of the polygon. Thus the station 1, the bounding polygon
is abcd. For station 2, kade is taken as the bounding polygon.
These bounding polygons are called Thiessen

24. 24
Lecture -3(contd.)
The boundary of the catchment, if it cuts the bisectors is taken
as the outer limit of the polygon. Thus the station 1, the
bounding polygon is abcd.
For station 2, kade is taken as the bounding polygon. These
bounding polygons are called Thiessen polygons.
Figure: Thiessen Polygon

25. 25
Lecture -3(contd.)
The areas of those six Thiessen Polygons are determined
either with a planimeter or by using an overlay grid.
Calculations of mean precipitation
If P1, P2, P3 ......P6 are the rainfall magnitudes recorded by the
stations 1,2, 3, ............. 6 respectively and A1, A2, A3 ......A6
are the respective areas of the Thiessen Polygons.
The total area = A1 +A2+ A3 +...... +A6 = A
The weightage area for each stations = A1 /A
A2 /A
... A6 / A

26. 26
Lecture -3(contd.)
then the average rainfall Pavg over the catchment is given
by
Pavg ={ (A1 /A)P1 + (A2/A) P2) + (A3/A)P3 )+..... + (A6/A)P6)}
In general for M stations,
M
Pavg = ∑Pi Ai / A
i where, i = 1,2,3.....M
The ratio Ai / A is called the weightage factor for each station.