Engineering Hydrology
Md. Moniruzzaman Saukhin
Lecturer
Dept. of Civil Engineering
Dhaka International University
Precipitation
Precipitation:
Precipitation is any form of moisture which falls to the earth. This includes rain,
snow, hail, sleet, etc.
Forms of precipitation:
Rain: The condensed water vapor of the atmosphere falling in drops from clouds.
Snow: Ice crystals resulting from sublimation (i.e. water vapor condensed to ice).
Hail: Small lumps of ice (>5 mm dia.) formed by alternate freezing and melting.
Dew: Moisture condensed from the atmosphere in small drops upon cool surfaces.
Sleet: Rain or melted snow that freezes into ice pellets before hitting the ground.
Rain Gauge:
Rainfall may be measured by a network of rain gauges which may either be of non-
recording or recording type.
Non-recording type:
✓ Usually used Symon’s rain gauge.
✓ Consists of funnel with a circular rim of 12.5 cm diameter and a glass bottle as
receiver.
✓ The cylindrical metal casing is fixed vertically to the masonry foundation with
the level rim 30 cm above the ground surface.
✓ The rain falling into the funnel is collected in the receiver and it can measure
1.25 cm of rain when full. The rainfall is measured in a special measuring glass
graduated in mm.
Fig. Symon’s Rain Gauge
Recording Rain Gauge:
✓ It is also called self-recording, automatic or integrated rain gauge.
✓ This type of rain gauge has an automatic mechanical arrangement
consisting of a clockwork, a drum with a graph paper fixed around it and a
pencil point which draws the mass curve of rainfall.
✓ From this mass curve, depth of rainfall in a given time, the rate of intensity
of rainfall, can be determined.
✓ The gauge is installed on an area of 45 cm square concrete or masonry
platform.
✓ Three types: a) Tipping Bucket gauge, b) Weighing type rain gauge, c)
Float type rain gauge.
Fig. Tipping Bucket Gauge Fig. Weighing Type Bucket Gauge
Fig. Float Type Rain Gauge
Fig. Typical Mass Curve
a.a.r.:
The mean of yearly rainfall observed for a period of 35 consecutive years is
called the average annual rainfall (a.a.r.).
a.a.r. < 40 cm --- Arid climate
40 cm < a.a.r. < 75 cm --- Semi-arid climate
a.a.r. > 75 cm --- Humid climate
Isohyet:
A line joining the place having the same a.a.r. is called an isohyet.
Index of wetness:
The ratio of rainfall in a particular year to the a.a.r. is called the index of
wetness.
Estimates of missing data and adjustment of records:
❖ Station-year Method:
=> For two stations:
PA PB
=
a.a.r.of A a.a.r.of B
Here, PA = Certain year a.a.r. rainfall in station, A
PB = Certain year a.a.r. rainfall in station, B
=> For more than two stations:
1 PA PB
PN = [ x a.a.r. of N + x a.a.r. of N + …….
N−1 a.a.r.of A a.a.r.of B
P(N−1)
+ x a.a.r. of (N-1)]
a.a.r.of (N−1)
Problem-1: The rainfall of station A, B and C is 8.5, 6.7 and 9.0 cm
respectively. If the a.a.r. for the stations are 75, 84, 70 and 90 cm respectively.
Estimate the storm rainfall at station D.
Solution:
1 8.5 6.7 9.0
The average value of PD = [ x 90 + x 90 + x 90] = 9.65 cm.
3 75 84 70
Problem-2: The rainfall of station A, B, C and D is 8.5, 6.7, 9.0 and 9.65 cm
respectively. If the a.a.r. for the stations A, B and C are 75, 84 and 70 cm
respectively. Estimate a.a.r of station D.
Solution:
1 8.5 6.7
The average value of PD = [ x a.a.r of D + x a.a.r of D +
3 75 84
9.0
x a.a.r of D ]
70
1 8.5 6.7
or, 9.65= [ x a.a.r of D + x a.a.r of D +
3 75 84
9.0
x a.a.r of D ]
70
1 8.5 6.7 9.0
or, a.a.r of D x [ + + ] = 9.65
3 75 84 70
Mean Areal Depth of Precipitation:
❖ Arithmetic Average Method:
∑P
Pavg =
n
❖ Thiessen Polygon Method:
∑n
i=1 Ai Pi
Pavg =
∑A
❖ Isohyetal Method:
∑n
i=1 Ai−(i+1) Pi−(i+1)
Pavg =
∑Ai−(i+1)
Problem-3: Calculate the mean areal depth of precipitation from the
following figure by arithmetic mean and Thiessen polygon method.
A (46 cm) B (65 cm)
10 Km
10 Km
F (60 cm) E (70 cm)
D (80 cm) C (76 cm)
Solution: 52 + 5 2 = 5 2
Area of inner square = (5 2 )2 = 50 km2
Area of square = 10 x 10 = 100 km2 A (46 cm) B (65 cm)
1
Area of each corner triangle = (100-50)
4
= 12.5 km2
F (60 cm) E (70 cm)
D (80 cm) C (76 cm)
Area of equilateral triangle = 0.5 x 10 x 10 sin 600 = 25 3 km2
1 1
Area of equilateral triangle = x 25 3 = 14.4 km2
3 3
A (46 cm) B (65 cm)
By Thiessen Polygon Method: F (60 cm) E (70 cm)
Station Area Precipitation AxP Pavg
A P D (80 cm) C (76 cm)
km2 cm 2
km . cm cm
A 12.5+14.4 46 1238
B 12.5 65 813
Pavg = By Arithmetic mean,
C 12.5 76 950 ∑P 397
∑n
i=1 Ai Pi Pavg = =
D 12.5+14.4 80 2152 ∑A n 6
9517 = 66.17 cm
E 50 70 3500 =
143.2
F 14.4 60 864 = 66.3
n=6 143.2 ∑P = 397 ∑AP =
9517
Problem-4: Determine the optimum number of rain-gauge considering the
error in the mean value of rainfall to 10% from the following basin.
Solution:
Station Normal annual Difference Difference2 Satisfaction
rainfall, x (x-𝑥) (x − 𝑥)2 parameters
cm 𝑥, σ, Cv
A 88 -4.8 23 ∑x
𝑥=
n
B 104 11.2 125.4 = 92.8
∑(x−𝑥)2
C 138 45.2 2040 σ=
n−1
D 78 -14.8 219 3767
=
5−1
E 56 -36.8 1360 = 30.7
σ 30.7
2 Cv = = x 100
n=5 ∑ x = 464 ∑(x − 𝑥) = 𝑥 92.8
3767.4 = 33.1 %
𝐶𝑣 2 33.1 2
N=( ) = ( ) = 11.09 ≈ 11 [P = error in the mean value of rainfall = 10%]
𝑃 10
Additional rain gauge required = 11- 5 = 6.
