ANALYSIS OF RAINFALL DATA

2.10 ANALYSIS OF RAINFALL DATA

Rainfall during a year or season (or a number of years) consists of several storms. The charac- teristics of a rainstorm are (i) intensity (cm/hr), (ii) duration (min, hr, or days), (iii) frequency (once in 5 years or once in 10, 20, 40, 60 or 100 years), and (iv) areal extent (i.e., area over which it is distributed).

Correlation of rainfall records —Suppose a number of years of rainfall records observed on recording and non-recording rain-gauges for a river basin are available; then it is possible to correlate (i) the intensity and duration of storms, and (ii) the intensity, duration and fre- quency of storms.

If there are storms of different intensities and of various durations, then a relation may

be obtained by plotting the intensities (i, cm/hr) against durations (t, min, or hr) of the respec- tive storms either on the natural graph paper, or on a double log (log-log) paper, Fig. 2.18(a)

and relations of the form given below may be obtained

a (a) i =

...(2.5) t + b

A.N. Talbot’s formula

(for t = 5-120 min)

k (b) i =

...(2.6) (c) i = kt x

...(2.7) where t = duration of rainfall or its part, a, b, k, n and x are constants for a given region. Since

x is usually negative Eqs. (2.6) and (2.7) are same and are applicable for durations t > 2 hr. By taking logarithms on both sides of Eq. (2.7),

log i = log k + x log t

which is in the form of a straight line, i.e., if i and t are plotted on a log-log paper, the slope, of the straight line plot gives the constant x and the constant k can be determined as i = k when t = 1. Hence, the fitting equation for the rainfall data of the form of Eq. (2.7) can be determined and similarly of the form of Eqs. (2.5) and (2.6).

On the other hand, if there are rainfall records for 30 to 40 years, the various storms during the period of record may be arranged in the descending order of their magnitude (of maximum depth or intensity). When arranged like this in the descending order, if there are a total number of n items and the order number or rank of any particular storm (maximum depth or intensity) is m, then the recurrence interval T (also known as the return period) of the storm magnitude is given by one of the following equations:

(a) California method (1923), T = ....(2.8)

mn

(b) Hazen’s method (1930), T = ...(2.9)

m − 1 2 n +1

(c) Kimball’s method, (Weibull, 1939) T = ...(2.10)

and the frequency F (expressed as per cent of time) of that storm magnitude (having recur- rence interval T) is given by

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Natural paper

i=k

Log-log paper

i (cm/hr)

i (cm/hr)

i = kt

–x = dy dx Intensity

Time t (min or hr) Time t (min or hr)

(a) Correlation of intensity and duration of storms

x = log i 2

i 1 Log-log paper Log-log paper

Decreasing frequency x i= kT e t

T= 2 50-y ear

Natural paper

One log One log

i= kT

(cm/hr)

cycle of T cycle of T

T = 15-y T = 15-y i (cm/hr)

5-year T=1 B T = 5-y T = 5-y

0-year

Intensity i 1 ear ear

T = 5-y

T = 1-year T = 1-year

A : High intensity for short duration Intensity

T = 1-y ear

ear

B : Low intensity for long duration Time t (min or hr)

1 Time t (min or hr)

(b) Correlation of intensity, duration and frequency of storms Fig. 2.18 Correlation of storm characteristics

Values of precipitation plotted against the percentages of time give the ‘frquency curve’. All the three methods given above give very close results especially in the central part of the curve and particularly if the number of items is large.

Recurrence interval is the average number of years during which a storm of given mag- nitude (maximum depth or intensity) may be expected to occur once, i.e., may be equalled or exceeded. Frequency F is the percentage of years during which a storm of given magnitude may be equalled or exceeded. For example if a storm of a given magnitude is expected to occur once in 20 years, then its recurrence interval T = 20 yr, and its frequency (probability of exceedence) F = (1/20) 100 = 5%, i.e., frequency is the reciprocal (percent) of the recurrence interval.

F The probability that a T-year strom 1 and frequency F = × 100% I may not occur in

HG T

KJ

any series of N years is

(N, 0) = (1 – F)

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and that it may occur is

...(2.12a) where P Ex = probability of occurrence of a T-year storm in N-years. The probability of a 20-year storm (i.e., T = 20, F = 5%) will not occur in the next 10 years

P = 1 – (1 – F) N Ex

is (1 – 0.05) 10 = 0.6 or 60% and the probability that the storm will occur (i.e., will be equalled or exceeded) in the next 10 years is 1 – 0.6 = 0.4 or 40% (percent chance).

See art. 8.5 (Encounter Probability), and Ex. 8.6 (a) and (b) (put storm depth instead of flood). If the intensity-duration curves are plotted for various storms, for different recurrence intervals, then a relation may be obtained of the form

kT x

...(2.13) where k, x and e are constants.

... Sherman

‘i vs. t’ plotted on a natural graph paper for storms of different recurrence intervals yields curves of the form shown in Fig. 2.18 (b), while on a log-log paper yields straight line plots. By taking logarithms on both sides of Eq. (2.13),

log i = (log k + x log T) – e log t

which plots a straight line; k = i, when T and t are equal to 1. Writing for two values of T (for the same t) :

log i 1 = (log k + x log T 1 ) – e log t log i 2 = (log k + x log T 2 ) – e log t

Subtracting, log i 1 – log i 2 = x (log T 1 – log T 2 ) ∆ log i

or,

x= ∆ log T

∴ x = charge in log i per log-cycle of T (for the same value of t) Again writing for two values of t (for the same T):

log i 1 = (log k + x log T) – e log t 1 log i 2 = (log k + x log T) – e log t 2

Subtracting

log i 1 – log i 2 = – e(log t 1 – log t 2 )

e = – slope = ∆

log t

∴ e = change in log i per log cycle of t (for the same value of T). The lines obtained for different frequencies (i.e., T values) may be taken as roughly

parallel for a particular basin though there may be variation in the slope ‘e’. Suppose, if a 1- year recurrence interval line is required, draw a line parallel to 10–year line, such that the distance between them is the same as that between 5-year and 50-year line; similarly a 100- year line can be drawn parallel to the 10-year line keeping the same distance (i.e., distance per log cycle of T). The value of i where the 1-year line intersects the unit time ordinate (i.e., t = 1 min, say) gives the value of k. Thus all the constants of Eq. (2.13) can be determined from the log-log plot of ‘i vs. t’ for different values of T, which requires a long record of rainfall data. Such a long record, will not usually be available for the specific design area and hence it

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becomes necessary to apply the intensity duration curves of some nearby rain gauge stations and adjust for the local differences in climate due to difference in elevation, etc. Generally, high intensity precipitations can be expected only for short durations, and higher the intensity of storm, the lesser is its frequency.

The highest recorded intensities are of the order of 3.5 cm in a minute, 20 cm in 20 min and highest observed point annual rainfall of 26 m at Cherrapunji in Assam (India). It has been observed that usually greater the intensity of rainfall, shorter the duration for which the rainfall continues. For example, for upper Jhelum canals (India) maximum intensities are

17.8 and 6.3 cm/hr for storms of 15 and 60 min respectively. Example 2.5 (a) In a Certain water shed, the rainfall mass curves were available for 30 (n)

consecutive years. The most severe storms for each year were picked up and arranged in the descending order (rank m). The mass curve for storms for three years are given below. Establish

kT x

a relation of the form i =

t e , by plotting on log-log graph paper.

Time(min) 5 10 15 30 60 90 120 Accumulated depth (mm) for m = 1

n +1 Time t (min)

5 10 15 30 60 90 120 T-yr = m

Intensity i (mm/hr)

The intensity-duration curves (lines) are plotted on log-log paper (Fig. 2.18 (c)), which yield straight lines nearby parallel. A straight line for T = 1 – yr is drawn parallel to the line T = 10-yr at a distance equal to that between T = 30–yr and T = 3-yr. From the graph at T = 1-

yr and t = 1 min, k = 103. The slope of the lines, say for T = 30-yr is equal to the change in log i per log cycle of t,

i.e., for t = 10 min and 100 min, slope = log 68 – log 17 = 1.8325 – 1.2304 = 0.6021 ~ −

0.6 = e.

200 Log-log paper Log-log paper T&t

Log cycle Log cycle

68 68 Ñ Log cycle 50 55 55 of T, log i = 0.34 = x

i (mm/hr)

1 10 100 t (min)

Fig. 2.18. (c) Intensity-duration relationship, (Ex. 2.5 (a))

At t = 10 min, the change in log i per log cycle of T, i.e., between T = 3–yr and 30–yr lines (on the same vertical), log 68 – log 31 = 1.8325 – 1.4914 = 0.3411 ~ −

0.34 = x. Hence, the intensity-duration relationship for the watershed can be established as

104 T 0.34

i = t 0.6

For illustration, for the most severe storm (m = 1, T = 30–yr), at t = 60 min, i.e., after 1 hr of commencement of storm,

0.6 = 28 mm/hr

which is very near to the observed value of 22 mm/hr.

A more general Intensity-Duration–Frequency (IDF) relationship is of the form

KT x

Sherman

n ( , i in cm/hr, t in min, T yr. t + a )

where K, x, a and n are constants for a given catchment. The rainfall records for about 30 to 50 years of different intensities and durations on a basin can be analysed with their computed recurrence interval (T). They can be plotted giving trial values of ‘a’ for the lines of best fit as

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shown in Fig. 2.18 (b). The values of a and n may be different for different lines of recurrence interval.

The constants can also be obtained by multiple regression model based on the principle of least squares and solutions can be obtained by computer-based numerical analysis; confi- dence intervals for the predictions can be developed.

Extreme point rainfall values of different durations and recurrence interval (return period) have been evaluated by IMD and the ‘isopluvial maps’ (lines connecting equal depths of rainfall) for the country prepared.

Example 2.5 (b) A small water shed consists of 2 km 2 of forest area (c = 0.1), 1.2 km 2 of culti- vated area (c = 0.2) and 1 km 2 under grass cover (c = 0.35). A water course falls by 20 m in a length of 2 km. The IDF relation for the area may be taken as

80 T 0.2

, i in cm/hr, t in min and T yr

( t + 12 ) 0.5

Estimate the peak rate of runoff for a 25 yr frequency. Solution Time of concentration (in hr)

t c = 0.06628 L 0.77 S –0.385 ,

Kirpich’s formula, L in km

F 20 I HG = 0.667 hr × 60 = 40 min.

2 × 1000 KJ

i =i c when t = t c in the given IDF relation

40 12 0.5 ( = 21.1 cm/hr + )

Q peak = 2.78 C i c A ,

rational formula, CA = ΣC i A i

= 2.78 × 21.1 × (0.1 × 2 + 0.2 × 1.2 + 0.35 × 1) = 46.4 cumec