--- title: "Phi-index, excess rainfall hyetograph, runoff volume" subject: "Engineering Hydrology" chapter: 3 key: "semester_5/engineering_hydrology:phi-hyetograph-5km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-hyetograph-5km2 site: BCT Engineering --- # Phi-index, excess rainfall hyetograph, runoff volume Engineering Hydrology · Chapter 3: Hydrological Losses - Asked in: 2079 Baisakh, 6 marks - Video: [ENGINEERING HYDROLOGY - BASIN -1- Determine value of the Φ Index and Construct the Hyetograph](https://www.youtube.com/watch?v=N-zGoQkfUiY) — ACIVILENGINEER - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-hyetograph-5km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3 ## Question A catchment area of 5 km$^2$ had the following rainfall pattern. | Time (h) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | |---|---|---|---|---|---|---|---|---| | Cumulative rainfall (cm) | 0 | 0.60 | 2.80 | 5.20 | 6.60 | 7.50 | 9.20 | 9.60 | If $\phi$-index is 0.40 cm/hr, construct the excess rainfall hyetograph and also find the volume of direct runoff. ## Short answer With φ = 0.40 cm/h, excess depths in successive 2-h periods are 0, 1.4, 1.6, 0.6, 0.1, 0.9, 0 cm. Total excess = 4.60 cm, direct runoff volume = 2.3×10⁵ m³. ## Answer Rainfall in each 2-hour period is the difference of cumulative rainfall; the intensity is depth/2 h. Loss rate $\phi = 0.40$ cm/h. | Period (h) | Rain depth (cm) | Intensity (cm/h) | Excess rate = i - 0.40 (cm/h) | Excess depth (cm) | |---|---|---|---|---| | 0-2 | 0.6 | 0.3 | 0 (below $\phi$) | 0 | | 2-4 | 2.2 | 1.1 | 0.7 | 1.4 | | 4-6 | 2.4 | 1.2 | 0.8 | 1.6 | | 6-8 | 1.4 | 0.7 | 0.3 | 0.6 | | 8-10 | 0.9 | 0.45 | 0.05 | 0.1 | | 10-12 | 1.7 | 0.85 | 0.45 | 0.9 | | 12-14 | 0.4 | 0.2 | 0 (below $\phi$) | 0 | | **Total** | 9.6 | | | 4.6 | **Excess rainfall hyetograph:** bars of excess depth for 2-h periods (bar heights are labelled, not drawn to scale): ```text excess (cm) 1.6 | #### 1.4 | #### #### 0.9 | #### #### #### 0.6 | #### #### #### #### 0.1 | #### #### #### #### #### +------------------------------------------------- 0-2 2-4 4-6 6-8 8-10 10-12 12-14 h ``` (Bar heights, 2-4 h: 1.40; 4-6 h: 1.60; 6-8 h: 0.60; 8-10 h: 0.10; 10-12 h: 0.90; there is no excess in 0-2 h and 12-14 h.) **Volume of direct runoff:** $$ V = 4.60\ \text{cm}\times 5\ \text{km}^2 = 0.046\ \text{m}\times 5\times10^6\ \text{m}^2 = 2.30\times10^5\ \text{m}^3 $$ **Answer: Total excess rainfall = 4.60 cm; volume of direct runoff = 2.3 × 10⁵ m³ (0.23 million m³).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is intensity for a 2-h period found? A. Cumulative at end B. Rain depth in period / 2 h C. Rain depth x 2 D. Depth divided by phi Answer: B. Differences of cumulative rain give depth; divide by 2 h. (Key point: Intensity) 2. Why is excess zero for 0-2 h? A. Intensity 0.3 cm/h is below phi 0.40 B. Cumulative rain is zero C. Duration is too short D. Area is small Answer: A. Excess = intensity - phi only when positive. (Key point: Excess) 3. Excess depth in 2-4 h is: A. 1.4 cm B. 0.7 cm C. 1.1 cm D. 2.2 cm Answer: A. (1.1 - 0.4) x 2 = 1.4 cm. (Key point: 2-4 h excess) 4. Direct runoff volume is: A. 4.6 x 10^5 m3 B. 2.3 x 10^5 m3 C. 2.3 x 10^3 m3 D. 9.6 x 10^5 m3 Answer: B. 0.046 m x 5 x 10^6 m2 = 2.3 x 10^5 m3. 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