--- title: "Flood hydrograph from 6-h UH, three storms" subject: "Engineering Hydrology" chapter: 5 key: "semester_5/engineering_hydrology:uh-6h-three-storms-30cumec" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-6h-three-storms-30cumec site: BCT Engineering --- # Flood hydrograph from 6-h UH, three storms Engineering Hydrology · Chapter 5: Hydrograph Analysis - Asked in: 2068 Chaitra, 14 marks - Video: [Unit Hydrograph to Flood Hydrograph || Derivation of Flood Hydrograph from Unit Hydrograph](https://www.youtube.com/watch?v=8ZXviXtKVdk) — Engineering School - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-6h-three-storms-30cumec - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5 ## Question The ordinates of a 6-h unit hydrograph are given below. A storm had three successive 6-h intervals of rainfall magnitude of 3.0, 5.0 and 4.0 cm respectively. Assuming an index of 0.20 cm/h and a base flow of 30 m$^3$/s, determine and plot the resulting hydrograph of flow. | Time (hr) | 0 | 3 | 6 | 9 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| | 6-h UH ordinate (m$^3$/s) | 0 | 150 | 250 | 450 | 600 | 800 | 700 | 600 | 450 | 320 | 200 | 100 | 50 | 0 | ## Short answer Subtract φ-index loss1.2 cm per6-h period gives excesses1.8,3.8,2.8 cm. Applying superposition with the6-h UH and adding30 m³/s base flow yields a flood hydrograph peaking at6010 m³/s at24–30 h and returning to30 m³/s at78 h. ## Answer **Step 1: Rainfall excess.** Each 6-h period loses $\phi t = 0.20\times 6 = 1.2$ cm. | Period | Rain (cm) | Loss (cm) | Excess (cm) | |---|---|---|---| | 0 – 6 h | 3.0 | 1.2 | 1.8 | | 6 – 12 h | 5.0 | 1.2 | 3.8 | | 12 – 18 h | 4.0 | 1.2 | 2.8 | (Intensities are 0.5, 0.83 and 0.67 cm/h, all above 0.20 cm/h, so every period produces runoff.) **Step 2: UH ordinates at 6-h interval** (the given 3 h and 9 h values are not needed): t = 0, 6, 12, …, 66 h → 0, 250, 600, 800, 700, 600, 450, 320, 200, 100, 50, 0 m³/s. **Step 3: Superposition.** The three excess depths are applied to the 6-h UH, each lagged by 6 h, and added with the base flow of 30 m³/s: $$ Q(t) = 1.8\,U(t) + 3.8\,U(t-6) + 2.8\,U(t-12) + 30 $$ | t (h) | UH (m³/s) | 1.8 cm × UH | 3.8 cm × UH | 2.8 cm × UH | Base | Q (m³/s) | |---|---|---|---|---|---|---| | 0 | 0.00 | 0.0 | 0.0 | 0.0 | 30.0 | 30.0 | | 6 | 250.00 | 450.0 | 0.0 | 0.0 | 30.0 | 480.0 | | 12 | 600.00 | 1080.0 | 950.0 | 0.0 | 30.0 | 2060.0 | | 18 | 800.00 | 1440.0 | 2280.0 | 700.0 | 30.0 | 4450.0 | | 24 | 700.00 | 1260.0 | 3040.0 | 1680.0 | 30.0 | 6010.0 | | 30 | 600.00 | 1080.0 | 2660.0 | 2240.0 | 30.0 | 6010.0 | | 36 | 450.00 | 810.0 | 2280.0 | 1960.0 | 30.0 | 5080.0 | | 42 | 320.00 | 576.0 | 1710.0 | 1680.0 | 30.0 | 3996.0 | | 48 | 200.00 | 360.0 | 1216.0 | 1260.0 | 30.0 | 2866.0 | | 54 | 100.00 | 180.0 | 760.0 | 896.0 | 30.0 | 1866.0 | | 60 | 50.00 | 90.0 | 380.0 | 560.0 | 30.0 | 1060.0 | | 66 | 0.00 | 0.0 | 190.0 | 280.0 | 30.0 | 500.0 | | 72 | 0.00 | 0.0 | 0.0 | 140.0 | 30.0 | 170.0 | | 78 | 0.00 | 0.0 | 0.0 | 0.0 | 30.0 | 30.0 | **Step 4: Plot.** Plot Q (m³/s) against t (h). The hydrograph rises from 30 m³/s at t = 0 to the peak, and falls back to the base flow of 30 m³/s at t = 78 h. ```text t(h) Q (m3/s) 0 | 30.0 6 |### 480.0 12 |############ 2060.0 18 |######################### 4450.0 24 |################################## 6010.0 30 |################################## 6010.0 36 |############################# 5080.0 42 |####################### 3996.0 48 |################ 2866.0 54 |########### 1866.0 60 |###### 1060.0 66 |### 500.0 72 |# 170.0 78 | 30.0 ``` **Answer: peak flood = 6010 m³/s at t = 24 h to 30 h (equal ordinates at both); base flow 30 m³/s is restored at 78 h.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With phi-index 0.20 cm/h, what excess comes from the 3.0 cm in the first 6 h? A. 2.8 cm B. 3.0 cm C. 0.6 cm D. 1.8 cm (3.0 - 0.20 x 6) Answer: D. Loss = phi x t = 1.2 cm per 6-h period. (Key point: Rainfall excess) 2. What are the three excess depths for 3.0, 5.0, 4.0 cm? A. 1.8, 3.8, 2.8 cm B. 2.8, 4.8, 3.8 cm C. 1.8, 3.8, 3.8 cm D. 3.0, 5.0, 4.0 cm Answer: A. Subtract 1.2 cm from each period. (Key point: Excess depths) 3. Which equation gives the flood hydrograph? A. Q = 1.8U(t) + 3.8U(t) + 2.8U(t) + 30 B. Q = 1.8U(t) + 3.8U(t-6) + 2.8U(t-12) + 30 C. Q = (1.8+3.8+2.8) U(t) + 30 D. Q = 1.8U(t) x 3.8U(t-6) x 2.8U(t-12) Answer: B. Each excess depth multiplies the UH lagged by 6 h, plus base flow. (Key point: Superposition) 4. Why are the 3 h and 9 h UH values not used? A. They are errors in the data B. They are negative C. The UH is a 6-h UH, so superposition uses ordinates at 6-h intervals D. They are used to estimate the base flow Answer: C. Lags are multiples of 6 h, so only the 6-h grid ordinates are needed. 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