---
title: "Flood hydrograph from 2-h UH, 5.2 cm rainfall"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:flood-from-2h-uh-5-2cm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-from-2h-uh-5-2cm
site: BCT Engineering
---

# Flood hydrograph from 2-h UH, 5.2 cm rainfall

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2082 Bhadra, 12 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-from-2h-uh-5-2cm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

Calculate the flood discharge from the outlet of the catchment if the total rainfall of 5.2 cm occurred at an interval of 6 hours using the following 2 UH. Assume base flow = 15 cumes and infiltration index of 0.2 cm/hr.

| Time (hr) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 2hr-UH (cumes) | 0 | 30 | 55 | 90 | 130 | 170 | 180 | 110 | 60 | 35 | 20 | 0 |

## Short answer

Total rainfall 5.2 cm over 6 h with φ-index 0.2 cm/h yields net excess 4.0 cm, divided equally into three 2-h blocks of 1.333 cm. Superpose given 2-h UH ordinates with baseflow 15 m³/s; peak flood discharge 655.0 m³/s at 12 h.

## Answer

**Reading of the problem.** A total rainfall of 5.2 cm falls in 6 hours. The UH is 2-h, so the storm is taken as three consecutive 2-h periods with equal rain.

#### Rainfall excess
- Total loss = $\phi t$ = 0.2 × 6 = 1.2 cm
- Total excess = 5.2 − 1.2 = 4.0 cm
- Per 2-h period: 4.0 / 3 = 1.333 cm (rain per period = 5.2/3 = 1.733 cm; loss per period = 0.4 cm)

#### Superposition
$$
Q(t) = 1.333\,[U(t) + U(t-2) + U(t-4)] + 15
$$

| t (h) | UH (m³/s) | 1.33333 cm × UH | 1.33333 cm × UH | 1.33333 cm × UH | Base | Q (m³/s) |
|---|---|---|---|---|---|---|
| 0 | 0.00 | 0.0 | 0.0 | 0.0 | 15.0 | 15.0 |
| 2 | 30.00 | 40.0 | 0.0 | 0.0 | 15.0 | 55.0 |
| 4 | 55.00 | 73.3 | 40.0 | 0.0 | 15.0 | 128.3 |
| 6 | 90.00 | 120.0 | 73.3 | 40.0 | 15.0 | 248.3 |
| 8 | 130.00 | 173.3 | 120.0 | 73.3 | 15.0 | 381.7 |
| 10 | 170.00 | 226.7 | 173.3 | 120.0 | 15.0 | 535.0 |
| 12 | 180.00 | 240.0 | 226.7 | 173.3 | 15.0 | 655.0 |
| 14 | 110.00 | 146.7 | 240.0 | 226.7 | 15.0 | 628.3 |
| 16 | 60.00 | 80.0 | 146.7 | 240.0 | 15.0 | 481.7 |
| 18 | 35.00 | 46.7 | 80.0 | 146.7 | 15.0 | 288.3 |
| 20 | 20.00 | 26.7 | 46.7 | 80.0 | 15.0 | 168.3 |
| 22 | 0.00 | 0.0 | 26.7 | 46.7 | 15.0 | 88.3 |
| 24 | 0.00 | 0.0 | 0.0 | 26.7 | 15.0 | 41.7 |
| 26 | 0.00 | 0.0 | 0.0 | 0.0 | 15.0 | 15.0 |

**Answer: peak flood ≈ 655.0 m³/s at t = 12 h** (the last column gives the flood hydrograph; it returns to the 15 m³/s base flow after t = 26 h).

## Check yourself

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

1. The 5.2 cm falls in 6 h and the UH is 2 h. How is the storm treated?
   A. As one 6-h block
   B. As two 3-h blocks
   C. As a single 2-h burst of 5.2 cm
   D. As three consecutive 2-h periods of equal rain
   Answer: D. 6 h / 2 h = 3 periods matching the UH duration. (Key point: Storm split)

2. What is the total excess and excess per period?
   A. 4.0 cm total, 1.333 cm per period
   B. 5.2 cm total, 1.733 cm each
   C. 4.0 cm total, 4.0 cm each
   D. 1.2 cm total, 0.4 cm each
   Answer: A. 5.2 - 0.2 x 6 = 4.0 cm; divided by 3. (Key point: Excess)

3. Which equation gives the flood?
   A. Q = 4.0 U(t) + 15
   B. Q = 1.333[U(t) + U(t-2) + U(t-4)] + 15
   C. Q = 1.333 U(t) + 15
   D. Q = 1.333[U(t) x U(t-2) x U(t-4)] + 15
   Answer: B. Equal excess blocks lagged by 2 h each, plus base flow. (Key point: Superposition)

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