--- title: "Gumbel flood at given risk, 50-year project life" subject: "Engineering Hydrology" chapter: 6 key: "semester_5/engineering_hydrology:gumbel-risk-63-58-4200" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-risk-63-58-4200 site: BCT Engineering --- # Gumbel flood at given risk, 50-year project life Engineering Hydrology · Chapter 6: Flood Hydrology - Asked in: 2069 Chaitra, 14 marks - Video: [flood discharge for gumbel distribution 2021 spring Pokhara university](https://www.youtube.com/watch?v=2WrHR9_H0Rc) — Engineer with shanta bahadur shahi - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-risk-63-58-4200 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6 ## Question The project life of headworks is 50 years. The flood discharge at risk 63.58303% is 4200 cumes. The average flood is 3500 cumec, which is derived from long term historical data using Gumbel distribution. Calculate the discharge for 500 year return period and risk 39.49939%. Prepare a Gumbel graph paper using normal arithmetic graph paper. Plot these three discharges on Gumbel paper. ## Short answer Risk 39.5% gives T=100 years. From long-term data, mean=3500 m³/s and derived σ=270 m³/s. Discharge at T=100 ≈ 4347 m³/s, at T=500 ≈ 4687 m³/s. ## Answer #### 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 | T (yr) | Risk (50 yr) | $y_T$ | $K=(y_T-0.5772)/1.2825$ | $x_T=3500+K\sigma$ (m³/s) | |---|---|---|---|---| | 50 | 63.58% | 3.902 | 2.592 | 4200 | | 100 | 39.50% | 4.600 | 3.137 | 4347 | | 500 | 9.53% | 6.214 | 4.395 | 4687 | **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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a risk of 63.58% over 50 years correspond to? A. T = 50 years B. T = 100 years C. T = 63 years D. T = 500 years Answer: A. T = 1/(1 - (1 - R)^(1/n)) gives 50 years, so 4200 m^3/s is the 50-year flood. (Key point: Return period from risk) 2. How is sigma obtained from the 50-year flood and the mean? A. sigma = x50/K50 B. sigma = (x50 - mean)/K50 C. sigma = K50 x mean D. sigma = (x50 + mean)/K50 Answer: B. From x_T = mean + K sigma, sigma = (4200 - 3500)/2.592 = 270. (Key point: Finding sigma) 3. For a long record, which reduced mean and standard deviation are used? A. 0 and 1 B. 0.5 and 1.0 C. 1.2825 and 0.5772 D. 0.5772 and 1.2825 Answer: D. For N tending to infinity y_n = 0.5772 (Euler's constant) and S_n = 1.2825. (Key point: Large-sample constants) 4. How is Gumbel paper made from arithmetic paper? A. Log scale on the discharge axis B. Uniform scale for y_T horizontally, labelled with return periods T at their y_T positions C. Both axes labelled with T D. Uniform scale for T horizontally Answer: B. Because discharge is linear in y_T, a uniform y_T axis with T labels makes the T scale non-uniform and data plot straight. 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