Question

(15 marks) 26. For the annual maximum series of 15 min, 30 min, storm depths given in the table below, Determine the average intensities associate with return periods of 5, 10, 25, 50, and 100 years. Assume that extreme value type I Gumbel distribution fits the given rainfall series. Do Not Plot them. Just proved the values. (P = Pmean + Ko) 11 3.321 3.088 Table 2.2. Frequency Factor K for Extreme Value Type Distribution T: (years) 5 10 25 50 0.967 1.703 2.632 0.919 1.625 2.517 3.179 0.888 1.575 2.444 0.866 1.541 2.393 3.026 0.851 1.516 2.354 0.838 1.495 2.326 2.943 0.829 1.478 2.303 0.820 1.466 2.283 0.792 1.423 2.220 0.779 1.401 2.187 2.770 0.719 1.305 2592 3.653 2979 2.913 2.889 2.812 100 2.044 Precipitation (depth) 30 minutes 3150 1600 1 250 1 300 1.100 Precipitation (depth) 15 minutes 20600 13600 1.1000 1.1600 1 0000 1 2700 1 3100 1.0800 1.0800 0.9700 1.0100 1.0600 1:1600 10200 1.410 1.410 1 220 1230 1.020 1.070 1.210 1.270 1.160

Answer 1

Here we need to solve the problem by using type 1 gumbels distribution method.

so P= P_mean+ K$\sigma$

$\sigma$ = Standard deviation of the precipitation sample data

P_mean = Mean precipitation of the sample

K value for each return period can be find from the given table.

Here sample size n=20

15 minutes = 0.25 hour

so intensity = [depth of precipitation]/time

time = 0.25 hour

for 30 minutes , time = 0.5 hour

so we will get the K value for different return period.

Return period	K- value	for 15 min precipitation			Intensity of 15 min precipitation (mm/hour)	for 30 min precipitation			Intensity of 30 min precipitation (mm/hour)
		P mean	s	P		P me an	s	P
5	0.919	1.188	0.2652	1.188+(0.2652*0.919) =2.279772	2.27977/0.25 =9.119088	1.407	0.5308	1.407+(0.5308*0.919) =1.894805	3.7896104
10	1.625	1.188	0.2652	1.188+(0.2652*1.625) =3.1185	(3.1185/0.25) =12.474	1.407	0.5308	1.407+(0.5308*1.625) =2.26955	(2.26955/0.5) =4.5391
25	2.517	1.188	0.2652	1.188+(0.2652*2.517) =4.178196	(4.178196/0.25) =16.712784	1.407	0.5308	1.407+(0.5308*2.517) =2.7430240	(2.7430240/0.5) =5.4860472
50	3.179	1.188	0.2652	1.188+(0.2652*3.179) =4.964652	(4.964652/0.25) =19.858608	1.407	0.5308	1.407+(0.5308*3.179) =3.094413	(3.094413/0.5) =6.1888264
100	3.836	1.188	0.2652	1.188+(0.2652*3.836) =5.745168	22.980672	1.407	0.5308	1.407+(0.5308*3.836) =3.443149	(3.443149/0.5) =6.8862976