Gumbel flood at given risk, 50-year project life

Step 1: Return periods from the risk

Risk over n = 50 years: $R=1-(1-1/T)^{n}$, so $T=\dfrac{1}{1-(1-R)^{1/n}}$.

• R = 63.58303%: $T=\dfrac{1}{1-(0.3641697)^{1/50}}=50.00$ years, so 4200 m³/s is the 50-year flood.
• R = 39.49939%: $T=\dfrac{1}{1-(0.6050061)^{1/50}}=100.00$ years, so this is the 100-year flood.

Step 2: Gumbel parameters

The average flood is the mean $\bar x=3500$ m³/s. For a long record, $\bar y_n=0.5772$ and $S_n=1.2825$.

$$y_{50}=-\ln[\ln(50/49)]=3.902,\qquad K_{50}=\frac{3.902-0.5772}{1.2825}=2.592$$ $$\sigma=\frac{x_{50}-\bar x}{K_{50}}=\frac{4200-3500}{2.592}=270.0 \text{ m}^3/\text{s}$$

Step 3: Discharges

Answer: discharge at 39.49939% risk (T = 100 years) ≈ 4347 m³/s; discharge for T = 500 years ≈ 4687 m³/s.

Step 4: Gumbel probability paper from ordinary graph paper

1. Draw the horizontal axis with uniform spacing for $y_T$ (e.g. 1 cm = 1 unit of y, range −1 to 7).
2. Mark the return periods at their $y_T$ positions (T = 2, 5, 10, 20, 50, 100, 200, 500, 1000 at $y_T$ = 0.367, 1.500, 2.250, 2.970, 3.902, 4.600, 5.296, 6.214, 6.907) and label them T. This makes the T scale non-uniform.
3. Draw the vertical axis for discharge (arithmetic scale, 3000 to 8000 m³/s).
4. Plot the points; Gumbel data fall on a straight line.

 Q (m3/s)
  4700 |                              o  T=500
  4347 |                 o  T=100
  4200 |         o  T=50
       +----+----+----+----+----+----+--- y_T
            3    4    5    6    7

The three plotted points (T = 50, 100, 500 years) lie on one straight line, which can be extended to any other T.