Question

The mean annual flood of a river is $600m^3/sec$, and the standard deviation of annual flood time series is $150m^3/sec$. The percent probability of a flood of magnitude $1000m^3/s$ occuring in the river atleast once in 5 years, will be
[Use Gumbel's method and assume the sample size to be very large]

• A. 11.25%
• B. 16.67%
• C. 9.99%
• D. 8.78%

Answer 1

The correct option is D 8.78%

Given;
$\overline{x}=600m^3/\sec and{\sigma }_{n-1}=150m^3/\sec$

We know that,

$x_T=\overline{x}+k{\sigma }_{n-1}$

$\Rightarrow 1000=600+k\left(150\right)$

$\Rightarrow k=2.6667=\frac{y_T-y_n}{s_n}$

$\Rightarrow 2.6667=\frac{y_T-0.577}{1.2825}$

$\Rightarrow y_T=3.9970$

Also,
$y_T=3.9970=-lnln\frac{T}{T-1}$

$\Rightarrow \frac{T}{T-1}=1.01854$

$\Rightarrow$ T = 54.9 years, say 55 years
$\text{Let probability of occurrence of a flood of magnitude}1000m^3/\sec =P$

$\therefore P=\frac{1}{55}=0.0182$

Probability of a flood of magnitude 1000 $m^3/sec$ occuring atleast once in 5 years.

$P_1=1-{(1-P)}^5$

$=1-{\left(0.9818\right)}^5$
= 0.08775
= 8.775% $\simeq$ 8.78%