Rainfall Frequency Analysis

1. Frequency Analysis of Rainfall
• Relation between the magnitude of the event and its probability of
exceedance.
• If there are rainfall records for several years, The probability analysis
may be made either by empirical or by analytical methods.

2. Probability P of an event equals to or exceeds,
and recurrence interval T
The probability P of an event equals to or exceeds, and recurrence interval T (also
known as the return period) of the storm magnitude is given by one of the following
equations:
Method Probability P Recurrence interval T
Kimball’s method 𝑃 =
𝑚
𝑛+1
𝑇 =
𝑛+1
𝑚
California method
(1923),
𝑇 =
𝑚
𝑛
𝑇 =
𝑛
𝑚
Hazen’s method (1930)
𝑇 =
𝑚−
1
2
𝑛
𝑇 =
𝑛
𝑚−
1
2

3. Example 6:
The annual rainfall at a place for a period of 21 years is given below.
Year Rainfall (cm) Year Rainfall (cm)
1950 50 1960 40
1951 60 1961 56
1952 40 1962 52
1953 27 1963 42
1954 30 1964 38
1955 38 1965 27
1956 70 1966 40
1957 60 1967 100
1958 35 1968 90
1959 55 1969 43
1970 33

4. Example Cont’d
Draw the rainfall frequency curve and determine:
a) the rainfall of 5-year and 20-year recurrence, interval
b)the rainfall which occurs 50% of the times
c) the rainfall of probability of 0.75
d)the recurrence interval of rainfall of 75 cm and its probability of
occurrence.

5. Solution
n = 21
Year Rainfall (cm) m Probability, P = m/n+1 Return Period (years), T=1/P
1967 100 1 0.045 22.000
1968 90 2 0.091 11.000
1956 70 3 0.136 7.333
1951, 1957 60 5 0.227 4.400
1961 56 6 0.273 3.667
1959 55 7 0.318 3.143
1962 52 8 0.364 2.750
1950 50 9 0.409 2.444
1963 42 10 0.455 2.200
1969 43 11 0.500 2.000
1952, 1960,
1966
40 14 0.636 1.571
1955, 1964 38 16 0.727 1.375
1958 35 17 0.773 1.294
1970 33 18 0.818 1.222
1954 30 19 0.864 1.158
1953, 1965 27 21 0.955 1.048

6. Solution

7. Solution
1. From the frequency-curve, the required values can be obtained as
• T = 5 years, Rainfall = 65 cm.
• T = 20 years, Rainfall = 97.5 cm
1. The rainfall which occurs 50% of the times,
• For F = 50%, P = 0.5 and T = 1/P = 2
• From the graph Rainfall = 43 cm

8. Solution
3. The rainfall of probability of 0.75
• for probability of 0.75, Return period T = 1/P =1/0.75 =1.33, for which
rainfall = 34 cm
4. The recurrence interval of rainfall of 75 cm and its probability of occurrence
• from the graph, the recurrence interval = 8.2 years and P = 1/T, P = 1/8.2 =
0.122