--- title: "24-hr maximum rainfall for 50-year return period" subject: "Engineering Hydrology" chapter: 6 key: "semester_5/engineering_hydrology:gumbel-24hr-max-50yr-semilog" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-24hr-max-50yr-semilog site: BCT Engineering --- # 24-hr maximum rainfall for 50-year return period Engineering Hydrology · Chapter 6: Flood Hydrology - Asked in: 2076 Asoj, 8 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-24hr-max-50yr-semilog - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6 ## Question For a station A, the recorded annual 24 hr maximum rainfalls are given below. Estimate the 24 hr maximum rainfall with return period of 50 years by using provided semi log graph. | Year | 1950 | 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60 | 61 | 62 | 63 | 64 | 65 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| | Ppt (cm) | 13.0 | 12.0 | 7.6 | 14.3 | 16.0 | 9.6 | 8.0 | 12.5 | 11.2 | 8.9 | 8.9 | 7.8 | 9.0 | 10.2 | 8.5 | 7.5 | ## Short answer Rank annual maxima, assign return period by Weibull T = (N+1)/m, plot rainfall vs log T on semi-log paper, fit a straight line, and read at T = 50 yr. The 24-hr maximum rainfall is approximately 20 cm. ## Answer **Method.** Rank the annual maxima, find the return period of each by a plotting-position formula, plot rainfall against T on semi-log paper (T on the log axis), fit a straight line and extend it to T = 50 years. Weibull formula (N = 16 years of data, m = rank in descending order): $$ T=\frac{N+1}{m}=\frac{17}{m}, \qquad P=\frac1T $$ | Rank m | P24 (cm) | T = 17/m (yr) | P = 1/T | |---|---|---|---| | 1 | 16.0 | 17.00 | 0.059 | | 2 | 14.3 | 8.50 | 0.118 | | 3 | 13.0 | 5.67 | 0.176 | | 4 | 12.5 | 4.25 | 0.235 | | 5 | 12.0 | 3.40 | 0.294 | | 6 | 11.2 | 2.83 | 0.353 | | 7 | 10.2 | 2.43 | 0.412 | | 8 | 9.6 | 2.12 | 0.471 | | 9 | 9.0 | 1.89 | 0.529 | | 10 | 8.9 | 1.70 | 0.588 | | 11 | 8.9 | 1.55 | 0.647 | | 12 | 8.5 | 1.42 | 0.706 | | 13 | 8.0 | 1.31 | 0.765 | | 14 | 7.8 | 1.21 | 0.824 | | 15 | 7.6 | 1.13 | 0.882 | | 16 | 7.5 | 1.06 | 0.941 | **Plotting.** Plot the 16 points (T on the logarithmic horizontal axis, 24-h rainfall on the arithmetic vertical axis). Gumbel-distributed data fall close to a straight line on this paper, with equation: $$ P_{24}=a+b\log_{10}T $$ A line fitted through the points gives $b=7.49$ cm per log-cycle and $a=7.33$ cm. ```text P24 (cm) 20 | o <- T=50 (extrapolated) 16 | o 12 | o o 8 | o o o 4 +--+------+-------+--------- T (yr, log) 1.1 2 10 50 ``` **Reading at T = 50 years:** $$ P_{24,50}= 7.33 + 7.49\times\log_{10}(50)=20.05 \text{ cm} $$ **Answer: 24-h maximum rainfall for the 50-year return period ≈ 20 cm.** As a check, Gumbel's analytical method with $\bar y_n=0.5157$, $S_n=1.0316$ (N = 16), mean = 10.31 cm and $\sigma$ = 2.59 cm gives 18.8 cm, which is of the same order. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Return period for the Weibull formula is: A. T = N/m B. T = m/N C. T = (N+1)/m D. T = N/(m+1) Answer: C. With N = 16, T = 17/m. (Key point: Plotting position) 2. Gumbel data plot as a straight line when: A. Both axes are arithmetic B. Both axes are log C. T on arithmetic and rainfall on log D. T is on the log axis and rainfall on the arithmetic axis Answer: D. P = a + b log10 T. (Key point: Semi-log plot) 3. P24 for T = 50 is: A. 7.33 + 7.49 log10(50) = about 20 cm B. 7.49 + 7.33 x 50 C. About 16 cm D. About 10.3 cm Answer: A. log10(50)=1.699; 7.33 + 12.72 = 20.05. 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