COAXIAL GRAPHICAL CORRELATION OF RAINFALL RUNOFF

13.4 COAXIAL GRAPHICAL CORRELATION OF RAINFALL RUNOFF

Kohler and Linsley (1951) showed that the rate at which the soil moisture is depleted from a catchment is roughly proportional to the amount of storage and the soil moisture decreases logarithmically with time during periods (days) of no precipitation and

I t =I 0 K t

where I 0 = initial value of the antecedent precipitation index (API, rainfall depth)

I t = reduced value of API t -days later K = recession factor ranging normally between 0.85 and 0.98

putting t = 1 day in Eq. (13.20), gives

I t =I 0 K

i.e., the index for any day is equal to that of the previous day multiplied by the factor K. If rain occurs on any day, the amount of rain (rather precipitation minus runoff) is added to the index. Since K is a function of potential evapotranspiration, it should be related to seasons or calender months. API is a satisfactory concept in estimating of runoff but systematic records of soil moisture are difficult to obtain for large areas.

LINEAR REGRESSION

(i) on 15th July. ANNACIVIL (ii) on 15th July, assuming no rainfall during 1-15 July.

Example 13.3 The API for a station was 50 mm on 1st July 1995; 40 mm rain fell on 6th July,

25 mm on 8th July and 30 mm on 9th July. Assuming a recession constant of 0.9, compute the API

Solution (i) I

t =I 0 K ,I 0 = 50 mm, K = 0.9

12.5 12.5 d ve

year year

2.5 m Stor

0 0 Storm runoff (cm) computed 0 2.5 (3.6) (3.6) 5 7.5 10 12.5

Stor Stor

2 (hr) (hr)

2.6 2.6 pptn. pptn. 3

72 72 6 6 7.5 10 10 (cm) dur (cm) dur

ation ation 0 0 24 24 7.5

Fig. 13.2 Coaxial correlation for Monocacy river, USA BLOG

Stor Stor

(US National Weather Service)

87.5 87.5 75 75 62.5 62.5 37.5 37.5 25 25 12.5 12.5 1 10 10 month month

month month

No No

Rainf Rainf

Runoff (cm) computed

Rainfall in the pre-

d vious month (cm)

ve 25 Given:

3 obser

Rainfall in June = 49.5 cm

50 in July = 67.0 cm

(cm)

From graph: Runoff in July = 41.3 cm

75 Runoff

Fig. 13.3 Coaxial correlation for Kallada basin (after NN Pillai, 1964)

HYDROLOGY

ANNACIVIL (ii) I

on 6-July,

I 6–1 = 50 × 0.9 5 + 40 = 69.52 mm

on 8-July,

I 8–6 = 69.52 × 0.9 2 + 25 = 81.31 mm

on 9-July,

I 9–8 = 81.31 × 0.9 + 30 = 103.18 mm

on 15-July,

I = 103.18 × 0.9 15–9 6 = 54.84 mm = API

15–1 = 50 × 0.9 14 = 11.44 mm = API

Depending upon the API, the time of the year, duration and magnitude of the storm and the altitude, the estimation of runoff can be made by following the data as indicated by the dotted line on the graphical plot for the Monacacy river, USA (Fig. 13.2). Thus, a catchment with an API of 2.5 cm, in the 10th week of the year with the occurrence of storm of 24 hr duration and 12 cm depth of precipitation, will yield a runoff of 3.6 cm. This graphical ap- proach is called coaxial correlation and is preferred to the multivariate linear correlation since many complex characteristics of the basin and storm are involved.

Another coaxial graphical correlation for estimating the monthly runoff from a catch- ment of the river Kallada in south Kerala (south India) as given by Pillai N.N. (1964) is shown in Fig. 13.3. Here the API has been taken as the precipitation of the previous month and the runoff for a particular month can be read on the graphical plot if the precipitation in the previous month is known. Thus, if the surface runoff in the month of July (7th month) is required, given the rainfalls for the months of June and July as 49.5 cm and 67.0 cm, respec- tively, then the runoff during the month of July is 41.3 cm, as indicated by the dotted line. If however, the total yield from the catchment is to be found out, the base flow (estimated as 30 cm) is to be added to the cumulative surface runoff of the whole year.

Though the correlation graph was developed for river Kallada (basin area = 874 km 2 ) during 1952-57, it was also applied to compute the yield of river Pamba (basin area = 1700 km 2 ) in 1953 and of river Achenkoil (basin area = 847 km 2 ) in 1955 and was found to be within ±4% of the observed yield.

BLOG

Example 13.4 Rainfall (P) and Runoff (R) data for a small catchment are given below:

P (mm):

22 26 14 4 30 12 R (mm):

6 12 4 0 18 6 Develop a linear regression equation and find the coefficient of correlation; write a com-

puter program in C-language. Solution R = aP + b x = P, y = R, m = no. of data pairs = 6

m . Σ xy − Σ x . Σ y

b = (Σy – a Σx)/m m . Σ x 2

Correlation coefficient, r =

Σx = 108, Σy = 46, (Σx) 2 = 11664, (Σy) 2 = 1116 Σx 2 = 484 + 676 + 196 + 16 + 900 + 144 = 2416 Σy 2 = 36 + 144 + 16 + 0 + 324 + 36 = 556

Σxy = 132 + 312 + 56 + 0 + 540 + 72 = 1112 Substituting these values in (i), (ii) and (iii),

∴ Good fit Regression equation: R = 0.6P = 3.16.

a = 0.6, b = – 3.16, r = 0.917 → → → → → 1,

LINEAR REGRESSION

ANNACIVIL #include <conio.h>

C-LANGUAGE CODE to fit a linear regression equation, Example 13.4, Chapter-13

/ * program for a linear fit eqn. R = aP + b */

#include <stdio.h> #include <math.h>

main ( ) {

Float sumx = 0, sumy = 0, sumxx = 0, sumyy = 0, sumxy = 0; float x[10], y[10], num, den, dx, dy, a, b, r; int i, m; printf (‘‘Program LINREG/n’’); scanf (‘‘%d’’, &m);/* m = No. of data pairs read */ for (i = 1; i < = m; i + +)

{ scanf (‘‘% f% f’’, & x[i], & y[i]); sumx + = x[i]; sumy + = y[i]; sumxx + = x[i]* x[i]; sumyy + = y[i]* y[i]; sumxy + = x[i]* y[i];