Gumbel design flood, factor of safety 1.3, 45-year life

(i) Peak discharge by Gumbel's method

Given: $N = 25$ years, $\bar{x} = 8520\ \text{m}^3/\text{s}$, $\sigma = 3900\ \text{m}^3/\text{s}$, $y_n = 0.5309$, $S_n = 1.0915$, design life $n = 45$ years, reliability $= 85\%$.

Return period. Reliability $= (1 - 1/T)^n = 0.85$, so risk $\bar{R} = 1 - 0.85 = 0.15$.

$$\begin{aligned} 1 - \frac{1}{T} &= (0.85)^{1/45} = 0.996395\\ T &= 277.4\ \text{years} \end{aligned}$$

Reduced variate.

$$y_T = -\ln\left[\ln\frac{T}{T-1}\right] = -\ln\left[\ln\frac{277.4}{276.4}\right] = 5.6236$$

Frequency factor.

$$K = \frac{y_T - y_n}{S_n} = \frac{5.6236 - 0.5309}{1.0915} = 4.6658$$

Flood magnitude.

$$x_T = \bar{x} + K\sigma = 8520 + (4.6658)(3900) = 26717\ \text{m}^3/\text{s}$$

Answer (i): recommended peak discharge $\approx$ 26717 m³/s (return period about 277 years).

(ii) Design discharge with factor of safety 1.3

$$Q_{design} = 1.3 \times 26717 = 34732\ \text{m}^3/\text{s}$$

Safety margin $= Q_{design} - x_T = 34732 - 26717 = 8015\ \text{m}^3/\text{s}$ (30 % of $x_T$).

Answer (ii): design discharge = 34732 m³/s; safety margin = 8015 m³/s.