---
title: "Phi-index and W-index of isolated storm, 500 ha"
subject: "Engineering Hydrology"
chapter: 3
key: "semester_5/engineering_hydrology:phi-w-index-500ha"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-w-index-500ha
site: BCT Engineering
---

# Phi-index and W-index of isolated storm, 500 ha

Engineering Hydrology · Chapter 3: Hydrological Losses

- Asked in: 2075 Chaitra, 5+1 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/phi-w-index-500ha
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-3

## Question

The mass curve of an isolated storm in a 500 ha watershed is as follows:

| Time from start (h) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 |
|---|---|---|---|---|---|---|---|---|---|---|
| Cumulative rainfall (cm) | 0 | 0.8 | 2.6 | 2.8 | 4.1 | 7.3 | 10.8 | 11.8 | 12.4 | 12.6 |

If runoff measured at the outlet is 0.361 Mm$^3$ and baseflow is negligible, estimate the $\phi$-index of the storm and duration of rainfall excess. Also determine the W-index if the other losses in the storm is 0.1 Mm$^3$.

## Short answer

φ-index found by trial with 2-h blocks: 0.365 cm/h, rainfall excess duration 12 h. W-index accounting additional 2 cm loss is 0.282 cm/h.

## Answer

**Given:** $A = 500$ ha $= 5\times10^6$ m²; runoff volume $= 0.361\ \text{Mm}^3 = 3.61\times10^5$ m³; baseflow negligible; 18 h of rain.

#### Step 1: Runoff depth and rainfall
$$
R = \frac{3.61\times10^5}{5\times10^6} = 0.0722\ \text{m} = 7.22\ \text{cm}
$$

Total rainfall $P = 12.6$ cm. Rainfall in each 2-h block (cm): 0.8, 1.8, 0.2, 1.3, 3.2, 3.5, 1.0, 0.6, 0.2.

#### Step 2: $\phi$-index by trial
Find $\phi$ (cm per 2 h) so that the sum of $(P_i - \phi)$ over the blocks with $P_i > \phi$ is 7.22 cm. Try $\phi = 0.73$ cm per 2 h; the blocks 0.8, 1.8, 1.3, 3.2, 3.5, 1.0 exceed it (0.2, 0.2 and 0.6 do not):

$$
(12.6 - 0.2 - 0.2 - 0.6) - 6\times\phi' = 7.22 \Rightarrow 11.6 - 6\phi' = 7.22 \Rightarrow \phi' = 0.73\ \text{cm per 2 h}
$$

$$
\phi = \frac{0.73}{2} = 0.365\ \text{cm/h}
$$

#### Step 3: Duration of rainfall excess
| Period (h) | Rain (cm) | Excess = rain - 0.73 (cm) |
|---|---|---|
| 0-2 | 0.8 | 0.07 |
| 2-4 | 1.8 | 1.07 |
| 4-6 | 0.2 | 0 |
| 6-8 | 1.3 | 0.57 |
| 8-10 | 3.2 | 2.47 |
| 10-12 | 3.5 | 2.77 |
| 12-14 | 1.0 | 0.27 |
| 14-16 | 0.6 | 0 |
| 16-18 | 0.2 | 0 |
| **Total** | 12.6 | **7.22** |

Rainfall excess occurs in 6 blocks of 2 h, so the duration of rainfall excess $t_r = 6\times2 = 12$ h.

#### Step 4: W-index
Other losses (depression storage, interception) $= 0.1\times10^6/5\times10^6 = 0.02$ m $= 2.0$ cm.

$$
W = \frac{P - R - S_d}{t_r} = \frac{12.6 - 7.22 - 2.0}{12} = \frac{3.38}{12} = 0.282\ \text{cm/h}
$$

**Answer: $\phi$-index = 0.365 cm/h; duration of rainfall excess = 12 h; W-index = 0.282 cm/h.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Runoff depth is:
   A. 7.22 cm
   B. 72.2 cm
   C. 0.361 cm
   D. 12.6 cm
   Answer: A. 3.61x10^5/5x10^6 = 0.0722 m. (Key point: Runoff depth)

2. phi-index is:
   A. 0.73 cm/h
   B. 0.365 cm/h
   C. 1.46 cm/h
   D. 0.282 cm/h
   Answer: B. 0.73 per 2 h gives 0.365 per hour. (Key point: phi)

3. Duration of excess is:
   A. 18 h
   B. 9 h
   C. 12 h
   D. 6 h
   Answer: C. Six 2-h blocks have excess. (Key point: Duration)

4. W-index is:
   A. 0.282 cm/h
   B. 0.365 cm/h
   C. 0.45 cm/h
   D. 0.60 cm/h
   Answer: A. (12.6 - 7.22 - 2.0)/12 = 0.282. (Key point: W-index)

## Close variants set in other exams

- [Phi-index and rainfall excess duration, 0.340 Mm3 runoff](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aphi-index-500ha-0-340mm3) — Engineering Hydrology, chapter 3, asked 2×

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