--- title: "Phi-index and cumulative infiltration curve" subject: "Engineering Hydrology" chapter: 3 key: "semester_5/engineering_hydrology:phi-index-cum-infil-0-2km2" source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-cum-infil-0-2km2 site: BCT Engineering --- # Phi-index and cumulative infiltration curve Engineering Hydrology · Chapter 3: Hydrological Losses - Asked in: 2079 Baisakh, 6 marks - Video: [#Numericals on Phi-Index# II Runoff Part 4II Hydrology II Prof. Sagar Kolekar](https://www.youtube.com/watch?v=aA3SzG9tIus) — Education Beyond the Wall - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-index-cum-infil-0-2km2 - Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3 ## Question For the following rainfall-runoff data, determine the $\phi$-index and ordinates of the cumulative infiltration curve based upon the $\phi$-index. The watershed area is 0.2 km$^2$. | Time (h) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | |---|---|---|---|---|---|---|---| | Rainfall rate (cm/hr) | 1.05 | 1.28 | 0.80 | 0.75 | 0.70 | 0.60 | 0 | | Direct runoff (m$^3$/s) | 0 | 30 | 60 | 45 | 30 | 15 | 0 | ## Short answer The φ-index is 1.003 cm/h. Cumulative infiltration ordinates at 1–7 h are 1.003, 2.006, 2.806, 3.556, 4.256, 4.856 and 4.856 cm, using only hours with rainfall rate above φ. ## Answer **Direct runoff depth:** Total direct runoff volume $= \sum Q\,\Delta t = (30+60+45+30+15)\times3600 = 648\,000$ m³. **Note on area:** with $A = 0.2$ km² the runoff depth ($648000/2\times10^5 = 3.24$ m) is far greater than the rainfall (5.18 cm), which is impossible. The area must be 200 km² (0.2 × 10³ km²); this is assumed. $$ R = \frac{648\,000}{200\times10^6} = 0.00324\ \text{m} = 0.324\ \text{cm} $$ **Total rainfall** $= 1.05+1.28+0.80+0.75+0.70+0.60 = 5.18$ cm, so the total loss $= 5.18 - 0.324 = 4.856$ cm. #### Finding $\phi$ Runoff occurs only when rainfall rate exceeds $\phi$. Try the hours with rain greater than $\phi$: 1.05 and 1.28 cm/h (the others are below $\phi$ and infiltrate fully). $$ (1.05 - \phi) + (1.28 - \phi) = 0.324 \Rightarrow \phi = \frac{2.33 - 0.324}{2} = 1.003\ \text{cm/h} $$ Check: $1.05, 1.28 > 1.003$ and $0.80, 0.75, 0.70, 0.60 < 1.003$. Valid. #### Cumulative infiltration (loss) curve Infiltration in each hour $= \min(\text{rain rate}, \phi)$: | Time (h) | Rain (cm/h) | Infiltration in the hour (cm) | Cumulative infiltration (cm) | |---|---|---|---| | 1 | 1.05 | 1.003 | 1.003 | | 2 | 1.28 | 1.003 | 2.006 | | 3 | 0.8 | 0.800 | 2.806 | | 4 | 0.75 | 0.750 | 3.556 | | 5 | 0.7 | 0.700 | 4.256 | | 6 | 0.6 | 0.600 | 4.856 | | 7 | 0 | 0.000 | 4.856 | The last ordinate (4.856 cm) equals the total loss, confirming the result. **Answer: $\phi$-index = 1.003 cm/h; cumulative infiltration ordinates = 1.003, 2.006, 2.806, 3.556, 4.256, 4.856 and 4.856 cm at 1 to 7 h.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the area taken as 200 km2 instead of 0.2 km2? A. 0.2 km2 gives runoff depth 3.24 m, far above the rainfall B. The question says so C. Rainfall is in metres D. To make phi positive Answer: A. With 0.2 km2 the runoff exceeds rainfall, impossible, so 200 km2 is assumed. (Key point: Area assumption) 2. The direct runoff depth is: A. 3.24 cm B. 0.324 cm C. 5.18 cm D. 4.856 cm Answer: B. 648000 m3 / 200 x 10^6 m2 = 0.00324 m = 0.324 cm. (Key point: Runoff depth) 3. phi is found from: A. (1.05 - phi) + (1.28 - phi) = 0.324 B. 1.05 + 1.28 = phi C. 5.18 - 0.324 = phi D. Average of rainfall rates Answer: A. Only the 1st and 2nd hours exceed phi, so excess equals runoff: phi = 1.003 cm/h. (Key point: Finding phi) 4. Infiltration in each hour is: A. The rainfall rate always B. min(rainfall rate, phi) C. phi always D. Rain minus runoff Answer: B. When rain is below phi, all infiltrates; the last ordinate 4.856 cm equals total loss. 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