---
title: "Flood hydrograph from 6-hr hydrograph, 3.3 and 5.5 cm"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:flood-hydrograph-770km2-6hr"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-hydrograph-770km2-6hr
site: BCT Engineering
---

# Flood hydrograph from 6-hr hydrograph, 3.3 and 5.5 cm

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2080 Bhadra, 8 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/flood-hydrograph-770km2-6hr
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

Following are the ordinates of hydrograph from a catchment area of 770 km$^2$ due to 6-hr rainfall. Derive the ordinates of flood hydrograph due to 3.3 cm and 5.5 cm effective rainfall of duration 6-hr. Make suitable assumptions regarding the base flow.

| Time (hr) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Discharge (m$^3$/s) | 40 | 65 | 215 | 360 | 400 | 350 | 270 | 205 | 145 | 100 | 70 | 50 | 40 |

## Short answer

Separate constant base flow of40 m³/s and derive the6-h UH by dividing DRH by the effective depth 5.021 cm. Convolve the UH with3.3 cm and5.5 cm lagged by6 h, then add base flow. The peak flood is 638 m³/s at 30 h.

## Answer

**Assumption for base flow:** the discharge at the start and end of the storm hydrograph is 40 m³/s, so a constant base flow of 40 m³/s is assumed.

#### Step 1: 6-h UH
Direct runoff ordinates = observed − 40. Volume = (sum of ordinates) × 6 × 3600.

$$
\begin{aligned} \sum DRH &= 1790 \text{ m}^3/\text{s} \\ V &= 1790\times 21600 = 38,664,000 \text{ m}^3 \\ \text{depth} &= \frac{V}{A} = \frac{38,664,000}{770\times 10^6}\times 100 = 5.021 \text{ cm} \end{aligned}
$$

UH ordinate = DRH / 5.021:

| 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 | 0 | 25 | 175 | 320 | 360 | 310 | 230 | 165 | 105 | 60 | 30 | 10 | 0 |
| UH (m³/s per cm) | 0.0 | 5.0 | 34.9 | 63.7 | 71.7 | 61.7 | 45.8 | 32.9 | 20.9 | 11.9 | 6.0 | 2.0 | 0.0 |

#### Step 2: Flood hydrograph
Two successive 6-h effective rains of 3.3 cm and 5.5 cm:

$$
Q(t) = 3.3\,U(t) + 5.5\,U(t-6) + 40
$$

| t (h) | UH (m³/s) | 3.3 cm × UH | 5.5 cm × UH | Base | Q (m³/s) |
|---|---|---|---|---|---|
| 0 | 0.00 | 0.0 | 0.0 | 40.0 | 40.0 |
| 6 | 4.98 | 16.4 | 0.0 | 40.0 | 56.4 |
| 12 | 34.85 | 115.0 | 27.4 | 40.0 | 182.4 |
| 18 | 63.73 | 210.3 | 191.7 | 40.0 | 442.0 |
| 24 | 71.69 | 236.6 | 350.5 | 40.0 | 627.1 |
| 30 | 61.74 | 203.7 | 394.3 | 40.0 | 638.1 |
| 36 | 45.80 | 151.2 | 339.6 | 40.0 | 530.7 |
| 42 | 32.86 | 108.4 | 251.9 | 40.0 | 400.4 |
| 48 | 20.91 | 69.0 | 180.7 | 40.0 | 289.7 |
| 54 | 11.95 | 39.4 | 115.0 | 40.0 | 194.4 |
| 60 | 5.97 | 19.7 | 65.7 | 40.0 | 125.4 |
| 66 | 1.99 | 6.6 | 32.9 | 40.0 | 79.4 |
| 72 | 0.00 | 0.0 | 11.0 | 40.0 | 51.0 |
| 78 | 0.00 | 0.0 | 0.0 | 40.0 | 40.0 |

**Answer: peak flood ≈ 638 m³/s at t = 30 h.**

## Check yourself

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

1. What base flow is assumed?
   A. Zero
   B. 400 m3/s
   C. Constant 40 m3/s, the flow at start and end
   D. Varying linearly from 0 to 400
   Answer: C. The hydrograph begins and ends at 40. (Key point: Base flow assumption)

2. How is the runoff depth of the 6-h UH found?
   A. Sum of DRH / area
   B. Peak DRH x 6 / area
   C. Sum of observed Q x 3600 / area
   D. Sum of DRH x 6 x 3600 / area = 5.02 cm
   Answer: D. V = 1790 x 21600 m3; divide by 770 km2. (Key point: Runoff depth)

3. Which equation is used for 3.3 cm and 5.5 cm successive 6-h rains?
   A. Q = 3.3U(t) + 5.5U(t-6) + 40
   B. Q = 8.8U(t) + 40
   C. Q = 3.3U(t) + 5.5U(t-12) + 40
   D. Q = 5.5U(t) + 3.3U(t-6) + 40
   Answer: A. First rain unlagged; second lagged by 6 h. (Key point: Flood equation)

## Related in Engineering Hydrology

- [Flood hydrograph from 6-hr UH, two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-3-3-5-5cm) — Engineering Hydrology, chapter 5
- [Flood hydrograph from 2-h UH, 5.2 cm rainfall](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-from-2h-uh-5-2cm) — Engineering Hydrology, chapter 5
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