---
title: "Flood hydrograph from 6-hr UH, two storms"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:flood-hydrograph-3-3-5-5cm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-hydrograph-3-3-5-5cm
site: BCT Engineering
---

# Flood hydrograph from 6-hr UH, two storms

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2075 Chaitra, 10 marks
- Video: [Flood hydrograph resulting from successive storms](https://www.youtube.com/watch?v=RUYrdXQKQJA) — The Study Sphere
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-hydrograph-3-3-5-5cm
- 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 12-hr.

| t (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 baseflow40 m³/s, derive6-h UH from DRH (effective rainfall5.02 cm). For12-h storm with3.3 cm then5.5 cm effective rain, superposition gives flood hydrograph peaking at638 m³/s at30 h.

## Answer

**Assumptions:** the 6-h hydrograph is the response to one 6-h effective-rainfall pulse. A 12-h effective storm is two successive 6-h pulses, the first giving 3.3 cm and the second 5.5 cm. The base flow of 40 m$^3$/s (the initial and final flow) is constant.

#### Step 1: Direct runoff and 6-h UH
DRH = Q - 40. Sum of DRH ordinates $= 1790$ m$^3$/s.
$$
V=1790\times6\times3600 = 38.66\times10^6\ \text{m}^3,\qquad d=\frac{V}{A}=\frac{38.66\times10^6}{770\times10^6}=5.02\ \text{cm}
$$
So the observed storm gave about 5.02 cm of effective rainfall, and the 6-h UH is $U = DRH/5.02$.

#### Step 2: Superposition
$$
Q(t)=40+3.3\,U(t)+5.5\,U(t-6)
$$

| t (h) | Observed Q | DRH | UH (m$^3$/s per cm) | 3.3 U(t) | 5.5 U(t-6) | Flood Q (m$^3$/s) |
|---|---|---|---|---|---|---|
| 0 | 40 | 0 | 0.00 | 0.0 | 0.0 | 40.0 |
| 6 | 65 | 25 | 4.98 | 16.4 | 0.0 | 56.4 |
| 12 | 215 | 175 | 34.85 | 115.0 | 27.4 | 182.4 |
| 18 | 360 | 320 | 63.73 | 210.3 | 191.7 | 442.0 |
| 24 | 400 | 360 | 71.69 | 236.6 | 350.5 | 627.1 |
| 30 | 350 | 310 | 61.74 | 203.7 | 394.3 | 638.1 |
| 36 | 270 | 230 | 45.80 | 151.2 | 339.6 | 530.7 |
| 42 | 205 | 165 | 32.86 | 108.4 | 251.9 | 400.4 |
| 48 | 145 | 105 | 20.91 | 69.0 | 180.7 | 289.7 |
| 54 | 100 | 60 | 11.95 | 39.4 | 115.0 | 194.4 |
| 60 | 70 | 30 | 5.97 | 19.7 | 65.7 | 125.4 |
| 66 | 50 | 10 | 1.99 | 6.6 | 32.9 | 79.4 |
| 72 | 40 | 0 | 0.00 | 0.0 | 11.0 | 51.0 |
| 78 | - | - | 0.00 | 0.0 | 0.0 | 40.0 |

**Answer: The flood hydrograph peaks at about 638 m$^3$/s at 30 h; ordinates are in the last column.**

## Check yourself

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

1. The observed 6-h hydrograph gives effective rainfall of 5.02 cm; this is computed from:
   A. Sum DRH x 6 x 3600 / basin area
   B. Peak / area
   C. Sum Q x 6
   D. Sum of observed Q
   Answer: A. V = 1790 x 21600 = 38.66 x 10^6 m^3; d = V/A = 5.02 cm. (Key point: Effective depth)

2. The 6-h UH ordinate is:
   A. DRH x 5.02
   B. DRH / 5.02
   C. DRH - 5.02
   D. DRH / 770
   Answer: B. Dividing DRH by the effective depth gives ordinates per cm. (Key point: UH)

3. A 12-h effective storm of 3.3 and 5.5 cm is treated as:
   A. Two 6-h pulses, 3.3 cm first then 5.5 cm
   B. One 12-h pulse of 8.8 cm
   C. Two pulses of 4.4 cm
   D. Overlapping storms
   Answer: A. Q(t) = 40 + 3.3 U(t) + 5.5 U(t-6). (Key point: 12-h storm)

4. The flood hydrograph peak is about:
   A. 638 m^3/s at 30 h
   B. 400 m^3/s at 24 h
   C. 442 m^3/s at 18 h
   D. 360 m^3/s at 18 h
   Answer: A. At t = 30 h: 40 + 203.7 + 394.3 = 638. (Key point: Peak)

## Related in Engineering Hydrology

- [Flood hydrograph from 6-hr hydrograph, 3.3 and 5.5 cm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-770km2-6hr) — Engineering Hydrology, chapter 5
- [6-h UH and flood from two storms, 700 km2](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-6h-700km2-two-storms) — Engineering Hydrology, chapter 5
- [2-hr UH from figure: area, flood, Tc, two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-2hr-figure-four-parts) — Engineering Hydrology, chapter 5
- [Flood hydrograph from 6-h UH, three storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-6h-three-storms-30cumec) — 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
- [Flood hydrograph from 4-h UH, three 2-hr rainfalls](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-4h-uh-3-2) — Engineering Hydrology, chapter 5, asked 2×