--- title: "Gumbel 500-year flood from two risk-based floods" subject: "Engineering Hydrology" chapter: 6 key: "semester_5/engineering_hydrology:gumbel-risk-63-6-4200-5800" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-risk-63-6-4200-5800 site: BCT Engineering --- # Gumbel 500-year flood from two risk-based floods Engineering Hydrology · Chapter 6: Flood Hydrology - Asked in: 2066 Magh (old course), 16 marks - Video: [Flood discharge at various return periods using Gumbel's extreme value distribution | Hydrology](https://www.youtube.com/watch?v=RE7Q4NDVPQQ) — The Study Sphere - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-risk-63-6-4200-5800 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6 ## Question The project life of a headworks is 50 years. The flood discharges at the risks 63.6% and 39.5% are 4200 cumecs and 5800 cumecs respectively, which is derived from long term historical data using Gumbel distribution. Calculate the discharge from 500 years return period. Plot these three discharges on Gumbel paper. ## Short answer Risk 63.6% → T=50 yr (4200 m³/s); risk 39.5% → T=100 yr (5800 m³/s). Using Gumbel line slope 2291.6, the 500-year discharge is 9497 m³/s. ## Answer #### Step 1: Return periods from the risk Life n = 50 years; $T=\dfrac{1}{1-(1-R)^{1/n}}$. - R = 63.6%: $T=\dfrac{1}{1-(0.364)^{1/50}}=49.98\approx50$ years, so **4200 m³/s is the 50-year flood.** - R = 39.5%: $T=\dfrac{1}{1-(0.605)^{1/50}}=100.00\approx100$ years, so **5800 m³/s is the 100-year flood.** #### Step 2: Gumbel line $x_T=\bar x+K\sigma=a+b\,y_T$ is a straight line in the reduced variate $y_T=-\ln[\ln(T/(T-1))]$. | T (yr) | $y_T$ | $x_T$ (m³/s) | |---|---|---| | 50 | 3.902 | 4200 | | 100 | 4.600 | 5800 | | 500 | 6.214 | ? | $$ b=\frac{5800-4200}{4.600-3.902}=2291.6 \text{ m}^3/\text{s} $$ $$ x_{500}=5800+2291.6\,(6.214-4.600)=9497 \text{ m}^3/\text{s} $$ **Answer: the 500-year discharge ≈ 9497 m³/s.** #### Step 3: Plotting on Gumbel paper Use arithmetic graph paper with a uniform $y_T$ axis (labelled with the return period T at the positions y = 3.902, 4.600 and 6.214) and an arithmetic discharge axis. Plot the three points (T = 50, 4200), (T = 100, 5800) and (T = 500, 9497); they lie on one straight line. ```text Q (m3/s) 9497| o T=500 5800 | o T=100 4200 | o T=50 +----+----+----+----+----+----+--- y_T 3 4 5 6 7 ``` ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A risk of 39.5% over a 50-year life corresponds to which return period? A. 50 years B. 200 years C. 100 years D. 500 years Answer: C. T = 1/(1 - 0.605^(1/50)) = 100 years, so 5800 is the 100-year flood. (Key point: Risk to return period) 2. How is the slope of the Gumbel line found? A. (4.600 - 3.902)/1600 B. 5800/4.600 C. (5800 - 4200)/(4.600 - 3.902) = 2291.6 D. (5800 - 4200)/(100 - 50) Answer: C. Discharge is linear in y_T, so slope is change in Q over change in y_T. (Key point: Slope of the line) 3. Which gives x500 with y500 = 6.214? A. 5800 + 2291.6 x (500 - 100) B. 5800 x 5 C. 5800 + 2291.6 x (6.214 - 4.600) = 9497 m^3/s D. 4200 + 2291.6 x 6.214 Answer: C. Extend the straight line from the 100-year point by the increase in y_T. (Key point: 500-year flood) 4. What do the three points T = 50, 100, 500 do on Gumbel paper? A. Lie on a vertical line B. Cannot be plotted C. Lie on one straight line D. Lie on a parabola Answer: C. All three come from the same linear x_T versus y_T relation. (Key point: Plotting) ## Related in Engineering Hydrology - [Gumbel flood at given risk, 50-year project life](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-risk-63-58-4200) — Engineering Hydrology, chapter 6 - [Gumbel design flood, 21 years data, 40-year life](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-21yr-life40-rel85-b) — Engineering Hydrology, chapter 6, asked 3× - [Gumbel 500-year flood, 95% limits, 30 years](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-30yr-100-50-1200-1060-2) — Engineering Hydrology, chapter 6 - [Gumbel 200-year flood from 50 and 100-year floods](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-35yr-200yr-flood) — Engineering Hydrology, chapter 6 - [Gumbel 500-year flood and 80% limits, Kamalamai River](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-kamalamai-80cl) — Engineering Hydrology, chapter 6 - [Gumbel analysis from 100 and 50-year floods](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-30yr-100-50-1200-1060) — Engineering Hydrology, chapter 6