İ.T.Ü.
CIVIL ENGINEERING FACULTY HYDRAULICS DIVISION
HYDROLOGY
Examples –3 Precipitation
1.
a) On the river basin illustrated the precipitation heights observed during a
strom are given below. Determine the mean precipitation height using the
arithmetic mean, Thiessen and isohyetal metods (isohyets will pass every
5mm).
b) Determine the local distibution of this precipitation and draw
“Precipitation height-area” curve.
Solution 1-a) ARITHMETIC AVERAGE METHOD: The simplest method is to take the average
of the readings of all instruments.
N
Pmean 
P
i
i 1
N

PA  PD  PE  PF  PH  PI 135,4

 22.6mm.
6
6
THIESSEN METHOD: The area is divided such that each point lies in the subregion of the gage to
which it is closest. This is done by joining the neighboring gages by straight lines and drawing
their perpendicular bisectors (normals at their midpoints).
Table 1:
Gage
Precipitation Pi(mm)
Thissen Polygon Area Ai (km2)
PiAi
A
8.7
233.10
2027.97
B
17.8
644.91
11479.40
C
18.3
481.74
8815.84
D
17.7
186.48
3300.70
E
21.7
85.47
1854.70
F
23.7
828.80
19642.56
G
24.9
160.58
3998.44
H
34.3
297.85
10216.26
I
29.3
903.91
26484.56
J
33.2
297.85
9888.62
K
35.5
248.64
8826.72
Total
4369.33
106535.77
N
Pmean 
P A
i 1
N
i
i
A
i 1
i

106535.77
 24.4mm.
4369.33
ISOHYETAL METHOD: Isohyets (curves of equal precipitation depth) are drawn by joining the
points of equal precipitation.
Table 2
isohyetals
Pi (mm)
35-30
32.5
30-25
27.5
25-20
22.5
20-15
17.5
15-10
12.5
<10
7.5
Total
Ai (km2)
1103.34
818.44
1186.22
924.63
300.44
36.26
4369.33
A=Ai (km2)
1103.34
1921.78
3108.00
4032.63
4333.07
4369.33
PiAi
35858.55
22507.10
26689.95
16181.03
3755.50
271.95
105264.08
 PiAi
35858.55
58365.65
85055.60
101236.63
104992.13
105264.08
Pmean (mm)
32.5
30.4
27.4
25.1
24.2
24.1
N
Pmean 
P A
i
i 1
N
i
A
i 1

105264.08
 24.1mm.
4369.33
i
Precipitation – Area Curve
Yağış Yüksekliği-Alan Eğrisi
Port(mm)
Ortalama
Yağış Yüksekliği
Mean
Precipitation
(mm)
34
32
30
28
26
24
22
20
0
1000
2000
3000
4000
5000
22
Area (km) )
Alan,A(km
2. The reservoir surface area of a dam is 32 km2 the heat energy reaching the
reservoir within a day is 500 cal/cm2. The albedo of water surface is %10.
Compute the daily evaporation volume on the reservoir lake (The heat brought by
the incoming river and the heat taken by the out going river will be neglected and
it will be assumed that the temperature degree on the lake will not change during
the day.)
E
Hi  H0  H
L(1  R )
E is the volume of evaporated water. L equals 590 cal/cm3 at normal atmospheric
pressure. Hi is the heat that enters the mass (sum of the heat from the sun and the heat of
inflow), Ho is the sum of the heat of outflow and the reflected heat, R is the Bowen ratio
and H is the heat to change the temperature of the mass of water.
Hi = 500 cal/cm2
Albedo = %10
H0 = 500*0.1=50 cal/cm2
H=0
R = Bowen Ratio = 0.2-0.3
L = 590 cal/cm3 in normal atm. pressure
E
Hi  H0  H 500  50  0

 0.61cm daily evoporation.
L(1  R )
590(1  0.25)
daily evoporation volume = 0.0061m*32*106m2=195200 m3