--- title: "6-h UH and flood from two storms, 700 km2" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:uh-6h-700km2-two-storms" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-6h-700km2-two-storms site: BCT Engineering --- # 6-h UH and flood from two storms, 700 km2 Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2067 Mangsir (old course), 16 marks - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-6h-700km2-two-storms - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question The following are the ordinates of the hydrograph of flow from a catchment area of 700 km$^2$ due to a 6-h rainfall. | Time (Hour) | 0 | 6 | 9 | 12 | 18 | 24 | 30 | 33 | 36 | 42 | 48 | 54 | 60 | 66 | 72 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| | Discharge (m$^3$/s) | 40 | 65 | 140 | 215 | 360 | 400 | 350 | 330 | 270 | 205 | 145 | 100 | 70 | 50 | 40 | a) Derive the ordinates of 6-h unit hydrograph. b) Calculate the flood hydrograph for two successive storms of 9.5 and 12.5 cm of 6 hours duration rainfall and an average storm loss of 0.25 cm/hr. ## Short answer Baseflow 40 m³/s subtracted; direct runoff volume gives effective depth 5.52 cm. Scaling DRH yields 6-h UH (peak 65.2 m³/s per cm). For two successive storms 9.5 and 12.5 cm with 1.5 cm loss each, effective depths 8.0 and 11.0 cm; superposition with baseflow gives flood peak 1205.9 m³/s at 30 h. ## Answer #### (a) Ordinates of the 6-h unit hydrograph Base flow: the discharge is 40 m³/s at the start and at the end (t = 0 and 72 h), so a constant base flow of 40 m³/s is assumed and subtracted. Ordinates at 6-h intervals are used (the 9 h and 33 h readings are not on the 6-h grid). Volume of direct runoff = (sum of DRH ordinates) × 6 h × 3600: $$ \begin{aligned} \sum DRH &= 1790 \text{ m}^3/\text{s} \\ V &= 1790 \times 6\times 3600 = 38,664,000 \text{ m}^3 \\ \text{depth} &= \frac{V}{A} = \frac{38,664,000}{700\times 10^6} = 0.0552 \text{ m} = 5.52 \text{ cm} \end{aligned} $$ UH ordinate = DRH ordinate / 5.52 (m³/s per cm): | t (h) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---| | Observed Q | 40 | 65 | 215 | 360 | 400 | 350 | 270 | 205 | 145 | 100 | 70 | 50 | 40 | | DRH = Q − 40 | 0 | 25 | 175 | 320 | 360 | 310 | 230 | 165 | 105 | 60 | 30 | 10 | 0 | | UH (m³/s per cm) | 0.0 | 4.5 | 31.7 | 57.9 | 65.2 | 56.1 | 41.6 | 29.9 | 19.0 | 10.9 | 5.4 | 1.8 | 0.0 | **6-h UH ordinates (m³/s per cm), t = 0 to 72 h: 0.0, 4.5, 31.7, 57.9, 65.2, 56.1, 41.6, 29.9, 19.0, 10.9, 5.4, 1.8, 0.0.** Check: sum × 6 × 3600 / 700 km² = 1.00 cm ✓. #### (b) Flood hydrograph for two successive storms Loss = 0.25 cm/h × 6 h = 1.5 cm in each period. - Storm 1: 9.5 − 1.5 = 8.0 cm - Storm 2: 12.5 − 1.5 = 11.0 cm $$ Q(t) = 8.0\,U(t) + 11.0\,U(t-6) + 40 $$ | t (h) | UH (m³/s) | 8 cm × UH | 11 cm × UH | Base | Q (m³/s) | |---|---|---|---|---|---| | 0 | 0.00 | 0.0 | 0.0 | 40.0 | 40.0 | | 6 | 4.53 | 36.2 | 0.0 | 40.0 | 76.2 | | 12 | 31.68 | 253.5 | 49.8 | 40.0 | 343.3 | | 18 | 57.94 | 463.5 | 348.5 | 40.0 | 852.0 | | 24 | 65.18 | 521.4 | 637.3 | 40.0 | 1198.7 | | 30 | 56.12 | 449.0 | 716.9 | 40.0 | 1205.9 | | 36 | 41.64 | 333.1 | 617.4 | 40.0 | 990.5 | | 42 | 29.87 | 239.0 | 458.0 | 40.0 | 737.0 | | 48 | 19.01 | 152.1 | 328.6 | 40.0 | 520.7 | | 54 | 10.86 | 86.9 | 209.1 | 40.0 | 336.0 | | 60 | 5.43 | 43.5 | 119.5 | 40.0 | 202.9 | | 66 | 1.81 | 14.5 | 59.7 | 40.0 | 114.2 | | 72 | 0.00 | 0.0 | 19.9 | 40.0 | 59.9 | | 78 | 0.00 | 0.0 | 0.0 | 40.0 | 40.0 | **Answer: peak flood ≈ 1206 m³/s at t = 30 h** (ordinates of the flood hydrograph are in the last column). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is base flow chosen from the hydrograph? A. Zero B. The peak 400 m3/s C. The average of all ordinates D. Constant 40 m3/s, the discharge at start and end, subtracted from all ordinates Answer: D. The flow returns to 40 at both ends, so constant base flow is assumed. (Key point: Base flow) 2. How are UH ordinates derived from the DRH? A. Divide DRH by the runoff depth V/A = 5.52 cm B. Divide DRH by the area 700 C. Multiply DRH by 5.52 D. Divide DRH by 6 hr Answer: A. Depth = 1790 x 6 x 3600 / 700e6 = 0.0552 m = 5.52 cm. (Key point: UH from runoff depth) 3. With loss 0.25 cm/h, what are the excess depths for 9.5 and 12.5 cm in 6 h? A. 9.25 and 12.25 cm B. 8.0 cm and 11.0 cm C. 7.0 and 10.0 cm D. 8.0 and 12.5 cm Answer: B. Loss per period = 0.25 x 6 = 1.5 cm. (Key point: Storm losses) 4. What is the combined equation and result? A. Q = 9.5U(t) + 12.5U(t-6) + 40, peak at 24 h B. Q = 19U(t) + 40 at t = 6 h C. Q = 8U(t) + 11U(t-6) + 40, peak about 1206 m3/s at 30 h D. Q = 8U(t) + 11U(t) + 40 with no lag Answer: C. Second storm lagged 6 h; peak occurs at 30 h. 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