floods - university of asia pacifict 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [by using eq....

Floods To estimate the magnitude of a flood peak the following alternative are available 1. Rational Method 2. Empirical Method 3. Unit-Hydrograph technique, and 4. Flood-Frequency Studies 1. Rational Method

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Page 1: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Floods To estimate the magnitude of a flood peak the following alternative are

available

1. Rational Method

2. Empirical Method

3. Unit-Hydrograph technique, and

4. Flood-Frequency Studies

1. Rational Method

Page 2: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Rainfall Intensity (itc,p)

In which K, a, x and m are constant

Time of Concentration (tc)

Runoff Coefficient (C) The coefficient C represents the integrated effect of the catchment looses

and hence depends upon the nature of the surface. Some typical values of C are indicated in Table 7.1

Page 3: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 4: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

2. Empirical Method

Dickens Formula (1865)

Page 5: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

3. Unit- Hydrograph Technique

4. Flood Frequency Studies

Page 6: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Gumbel’s Method

Page 7: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 8: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Gumbel Probability Paper

Page 9: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 10: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 11: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 12: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 13: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 14: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Confidence Limits For a confidence probability c, the confidence interval of the variate XT is

bounded by values X1 and X2 given by

Page 15: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

Page 16: Floods - University of Asia PacificT 9600 = 10,700m3/s 10088 m3/s.l 9558 m3/s and = [By using Eq. (7.20) to (7.22), = From Tables 7.3 and 7.4, for N = Ž7,yn = 0.5332 andSn = Choosing

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