SUPRA RAIN TECHNIQUE
3.11 SUPRA RAIN TECHNIQUE
Due to complex conditions antecedent and during the rain, and complex catchment character- istics, the use of infiltration method is usually limited to small areas with well-established values of infiltration.
The rainfall in excess of a particular value of φ-index for the entire pattern of storm rainfall is called supra rain. Allowance for areal variation of rainfall and f-capacity is made by dividing into sub areas in the case of large areas. The mean hourly net rains over the whole catchment can be obtained as
Σ AP 1 net-1
P net-mean =
HYDROLOGY
where A 1 ,A 2 , ... are the sub-areas. P net–1 ,P net–2 , .... are the net rains in the sub areas
ΣA 1 = A = total area of the catchment
When a large number of sub-areas are involved the hourly net rains over the whole catchment can be derived by constructing a supra-rain-curve, in which the supra-rain is plot- ted against hypothetical values of the φ-index, Fig. 3.15. The supra-rain-curve thus obtained is valid only for that particular storm from which it is derived. For other storms, new supra rain curves must be prepared. The supra-rain technique is illustrated in the following two exam- ples.
Example 3.7 Hourly rainfalls of 2.5, 6, and 3 cm occur over a 20-ha area consisting 4 ha of φ = 5 cm/hr, 10 ha of φ = 3 cm/hr, and 6 ha of φ = 1 cm/hr. Derive hourly values of net rain.
Solution
= 0.45 cm (P =
1st hour:
P net–mean =
2.5 cm)
2nd hour:
(P = 6 cm)
P net–mean =
= 3.20 cm
3rd hour:
(P = 3 cm)
P net–mean =
= 0.60 cm
Total net rain for the 3-hour storm = 4.25 cm Example 3.8 The successive hourly rains of a 10-hour storm are: 2.5, 6.3, 10, 12, 8, 5, 3, 1.5, 1
cm. Using the supra-rain-curve technique, determine the total net rain and its time distribu- tion for a 20-hr area consisting of 4 ha of φ = 5 cm/hr, 10 ha of φ = 3 cm/hr and 6 ha of φ = 1 cm/hr.
Solution For φ = 1 cm/hr, P net (supra-rain) from the hyetograph—Fig. 3.14 is 41 cm. Similarly, for φ = 0.5, 2, 3, 4, 5, 6, 7, 8, 9 and 10 cm, P net (supra-rain) values are 47, 33.5, 26, 21, 16, 12, 9,
6, 4 and 2 cm, respectively. With these values, the supra-rain-curve is plotted as shown in Fig. 3.15. The supra-rain for the 20-ha area can be obtained by weighing for the sub-areas as follows:
Sub-area φ–index
Product (3) × (4) A 1 (cm /hr)
Sub-areal supra-rain
A 1 /A
(cm) (ha)
P net–1
Total net rain
= 28.5 cm Corresponding to this supra-rain of 28.5 cm, the mean effective φ-index for the entire 20
over basin
ha, from Fig. 3.15, is 2.6 cm/hr. Application of φ = 2.6 cm/hr to the values of hourly rainfalls of the 10-hr storm. Fig. 3.13 gives the values of hourly net rain as 0, 3.4, 0.4, 7.4, 9.4, 5.4, 0.4, 0 and 0 cm, respectively, giving a total of 28.8 cm.
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Now, by working as in example 3.7, the hourly net rains are obtained in Table 3.1, which also gives a total net rain of 28.80 cm, though the hourly net rains are slightly different from those obtained from the supra-rain-curve technique. Thus, the supra-rain-curve technique yields somewhat eroneous values for hourly net rains as compared with those derived by ap- plying Eq. (3.17), though the total net rain for a given storm is the same by both techniques. However, this loss in accuracy may be justified by the time saved, especially when a large number of sub-areas are involved.
12 12 cm/hr 10 10
(cm/hr) i
1 f = 2.6 cm/hr f = 2.6 cm/hr 0 f = 1 cm/hr
0 1 2 3 4 5 6 7 8 9 10 11 Time t (hr)
Fig. 3.14 Hyetograph (Example 3.8)
Supra rain curve (cm/hr) 30
P net rain
a 20 Supr
0 1 2 3 4 5 6 7 8 9 10 11 12 f (cm/hr)
Fig. 3.15 Supra Rain Curve (Example 3.8)
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Table 3.1 Net rain from sub-areas with φ-values, Example 3.8
Hour
Rainfall
Rainfall excess, P net from sub-areas
Weighted P net from sub-areas
P net over a basin
Total net rain over basin = 28.80 cm
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WATER LOSSES