--- title: "Hyetograph and intensity-duration curve, 6:00-12:00 storm" subject: "Engineering Hydrology" chapter: 2 key: "semester_5/engineering_hydrology:hyetograph-intensity-duration-6to12" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/hyetograph-intensity-duration-6to12 site: BCT Engineering --- # Hyetograph and intensity-duration curve, 6:00-12:00 storm Engineering Hydrology · Chapter 2: Precipitation - Asked in: 2070 Chaitra, 2+6 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/hyetograph-intensity-duration-6to12 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-2 ## Question The rainfall depth with time during a storm at a station is as given: | Time | 6:00 | 6:30 | 7:00 | 7:30 | 8:00 | 8:30 | 9:00 | 9:30 | 10:00 | 10:30 | 11:00 | 11:30 | 12:00 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---| | Rainfall (cm) | 0 | 7 | 5 | 8 | 9 | 13 | 10 | 8 | 6 | 5 | 3 | 1 | 0 | i) Construct the hyetograph of this storm for 30 min and 2 hours interval. ii) Compute maximum average intensity of rainfall for 30 min, 1 hour, 2 hour in this storm and plot the resulting intensity duration curve. ## Short answer 30-min hyetograph uses given depths; 2-h hyetograph groups four intervals. Maximum intensities: 30-min = 13 cm (26 cm/h, 8:00-8:30); 1-h = 13+10=23 cm (23 cm/h, 8:00-9:00); 2-h = 8+9+13+10=40 cm (20 cm/h, 7:00-9:00). Plot intensity vs duration. ## Answer The values are the rainfall (cm) recorded in each 30-minute interval ending at the stated time. Total rainfall $= 7+5+8+9+13+10+8+6+5+3+1 = 75$ cm. #### (i) Hyetographs **30-min interval:** the bar chart of the 30-min depths (intensity = depth × 2 cm/h). | Interval | Depth (cm) | Intensity (cm/h) | |---|---|---| | 6:00-6:30 | 7 | 14 | | 6:30-7:00 | 5 | 10 | | 7:00-7:30 | 8 | 16 | | 7:30-8:00 | 9 | 18 | | 8:00-8:30 | 13 | 26 | | 8:30-9:00 | 10 | 20 | | 9:00-9:30 | 8 | 16 | | 9:30-10:00 | 6 | 12 | | 10:00-10:30 | 5 | 10 | | 10:30-11:00 | 3 | 6 | | 11:00-11:30 | 1 | 2 | **2-hour interval** (sum of four 30-min depths): | Interval | Depth (cm) | Intensity (cm/h) | |---|---|---| | 6:00-8:00 | 29 | 14.5 | | 8:00-10:00 | 37 | 18.5 | | 10:00-12:00 | 9 | 4.5 | ```text cm/h (30-min hyetograph; one bar per 30-min interval from 6:00 to 11:30) 26 | ## 22 | ## 18 | ## ## ## 14 |## ## ## ## ## ## 10 |## ## ## ## ## ## ## ## ## 6 |## ## ## ## ## ## ## ## ## ## 2 |## ## ## ## ## ## ## ## ## ## ## +--------------------------------- 6:00 11:00 ``` #### (ii) Maximum average intensities | Duration | Max depth (cm) | Critical period | Max intensity (cm/h) | |---|---|---|---| | 30 min | 13 | 8:00-8:30 | 26.0 | | 1 h | 23 | 8:00-9:00 (13+10) | 23.0 | | 2 h | 40 | 7:00-9:00 (8+9+13+10) | 20.0 | (For 2 h the window 7:00-9:00 holds 8 + 9 + 13 + 10 = 40 cm.) **Intensity-duration curve:** plot duration (x) against maximum intensity (y): points (0.5 h, 26), (1 h, 23), (2 h, 20) joined by a smooth falling curve. **Answer: Maximum intensities = 26, 23 and 20 cm/h for 30 min, 1 h and 2 h.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is intensity found from a 30-min depth? A. Depth / 2 B. Depth x 30 C. Depth / 30 D. Depth x 2 cm/h Answer: D. 30 min is 0.5 h, so intensity = depth/0.5 = depth x 2. (Key point: Intensity from 30-min depth) 2. How is a 2-hour hyetograph bar obtained? A. Sum of four 30-min depths B. Average of four depths C. Largest of four depths D. Last of four depths Answer: A. 6:00-8:00 = 7+5+8+9 = 29 cm, intensity 14.5 cm/h. (Key point: 2-hour interval depth) 3. Which 2-h window is critical? A. 6:00-8:00 with 29 cm B. 7:00-9:00 with 40 cm C. 8:00-10:00 with 37 cm D. 9:00-11:00 Answer: B. 8+9+13+10 = 40 cm gives 20 cm/h. (Key point: Maximum 2-h depth) 4. How does maximum intensity vary: 26, 23, 20 cm/h? A. Rises with duration B. Constant C. Falls with increasing duration D. Random Answer: C. Longer windows include lighter rain, so average falls. 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