--- title: "Hyetograph and intensity-duration curve, Lumle tipping bucket" subject: "Engineering Hydrology" chapter: 2 key: "semester_5/engineering_hydrology:tipping-bucket-lumle" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/tipping-bucket-lumle site: BCT Engineering --- # Hyetograph and intensity-duration curve, Lumle tipping bucket Engineering Hydrology · Chapter 2: Precipitation - Asked in: 2081 Baisakh, 4+6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/tipping-bucket-lumle - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2 ## Question Following are the data of a storm as recorded in a tipping bucket located at Lumle (Lat: 28.3$^\circ$, Long: 83.8$^\circ$) for 1.5 hours: | Time from beginning of storm (minute) | 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 | |---|---|---|---|---|---|---|---|---|---| | Cumulative rainfall (mm) | 1.6 | 6.6 | 13.8 | 20.2 | 24.2 | 38.4 | 46.8 | 49.4 | 50.8 | (i) Plot the hyetograph of the storm. (ii) Plot the maximum intensity duration curve of the storm. ## Short answer Incremental depths from cumulative data give intensities (depth×6). Peak 10-min intensity is 85.2 mm/h in 50–60 min. Maximum intensity-duration values are 85.2, 67.8, 53.2, 49.5, 48.2, 45.2, 41.0, 37.1, 33.9 mm/h for durations 10 to 90 min. ## Answer The tipping bucket gives cumulative rainfall at 10-min intervals. The incremental depth in each interval is the difference of successive values, and intensity = depth / (10/60 h) = depth × 6. #### (i) Hyetograph | Interval (min) | Cum. rain (mm) | Incremental depth (mm) | Intensity (mm/h) | |---|---|---|---| | 0-10 | 1.6 | 1.6 | 9.6 | | 10-20 | 6.6 | 5.0 | 30.0 | | 20-30 | 13.8 | 7.2 | 43.2 | | 30-40 | 20.2 | 6.4 | 38.4 | | 40-50 | 24.2 | 4.0 | 24.0 | | 50-60 | 38.4 | 14.2 | 85.2 | | 60-70 | 46.8 | 8.4 | 50.4 | | 70-80 | 49.4 | 2.6 | 15.6 | | 80-90 | 50.8 | 1.4 | 8.4 | ```text mm/h (one bar per 10-min interval; each row is approximate) 80 | ## 60 | ## 40 | ## ## ## 20 | ## ## ## ## ## ## ## 5 |## ## ## ## ## ## ## ## ## +--------------------------- 10 20 30 40 50 60 70 80 90 min ``` (The peak of 85.2 mm/h occurs in the 50-60 min interval.) #### (ii) Maximum intensity-duration curve For each duration the maximum depth is the greatest difference of cumulative ordinates that far apart. | Duration (min) | Max depth (mm) | Period (min) | Max intensity (mm/h) | |---|---|---|---| | 10 | 14.2 | 50-60 | 85.2 | | 20 | 22.6 | 50-70 | 67.8 | | 30 | 26.6 | 40-70 | 53.2 | | 40 | 33.0 | 30-70 | 49.5 | | 50 | 40.2 | 20-70 | 48.2 | | 60 | 45.2 | 10-70 | 45.2 | | 70 | 47.8 | 10-80 | 41.0 | | 80 | 49.4 | 0-80 | 37.1 | | 90 | 50.8 | 0-90 | 33.9 | Example: 30 min: 46.8 (at 70 min) - 20.2 (at 40 min) = 26.6 mm, intensity 53.2 mm/h. Plot duration (x) against maximum intensity (y). The curve falls from 85.2 mm/h at 10 min to 33.9 mm/h at 90 min. **Answer: Peak 10-min intensity = 85.2 mm/h; maximum intensities are 85.2, 67.8, 53.2, 49.5, 48.2, 45.2, 41.0, 37.1 and 33.9 mm/h for 10 to 90 min.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is incremental depth found? A. Sum B. Product C. Difference of successive cumulative values D. Cumulative divided by 10 Answer: C. E.g. 6.6-1.6 = 5.0 mm. (Key point: Increment) 2. Intensity for 10-min depth in mm/h? A. Depth / 6 B. Depth x 10 C. Depth / 10 D. Depth x 6 Answer: D. 10 min = 1/6 h. (Key point: Intensity factor) 3. What is peak 10-min intensity and when? A. 85.2 mm/h at 50-60 min B. 43.2 at 20-30 C. 50.4 at 60-70 D. 30 at 10-20 Answer: A. 14.2 mm x 6 = 85.2. (Key point: Peak) 4. How is 30-min max found? A. 50.8/3 B. 46.8 - 20.2 = 26.6 mm, 53.2 mm/h C. 85.2/3 D. 13.8 x 2 Answer: B. Greatest difference 30 min apart. 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