---
title: "2-hr UH from figure: area, flood, Tc, two storms"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:uh-2hr-figure-four-parts"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-2hr-figure-four-parts
site: BCT Engineering
---

# 2-hr UH from figure: area, flood, Tc, two storms

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2082 Baisakh, 2+3+1+6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-2hr-figure-four-parts
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

A 2-hr unit hydrograph is shown in the figure below.

[Figure: 2-hr unit hydrograph, Q (cumecs, axis 0 to 14) versus time (hr, axis 0 to 30). The curve starts at 0 at t = 0, rises through about 2 cumecs near t = 3 hr, about 4.5 cumecs near t = 5 hr and about 9 cumecs near t = 8 hr, peaks at about 12 cumecs near t = 10 hr, falls to about 9 cumecs near t = 15 hr, about 4.5 cumecs near t = 18 hr and about 2 cumecs near t = 22 hr, and returns to 0 near t = 27 hr. Values read approximately from the plotted points.]

(i) Determine the catchment area of the basin.
(ii) Determine the resulting flood hydrograph for a net rainfall of 1.5 cm/hr for the duration of 2 hr if the constant base flow is 2 cumecs.
(iii) What will be the time of concentration for the basin?
(iv) Two different storms, with rainfall of 0.47 cm/hr and 0.64 cm/hr occur for 1 hour. If the $\phi$ = 0.12 cm/hr for the basin, what will be the peak discharge for the resulting flood hydrograph?

## Answer

#### (i) Catchment area
Ordinates of the 2-h UH read from the curve at 2-h intervals (m³/s per cm):

| t (h) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 | 26 | 28 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| UH | 0.00 | 1.33 | 3.25 | 6.00 | 9.00 | 12.00 | 10.80 | 9.60 | 7.50 | 4.50 | 3.25 | 2.00 | 1.20 | 0.40 | 0.00 |

A UH represents 1 cm of runoff, so

$$
\begin{aligned} \sum Q &= 70.83 \text{ m}^3/\text{s} \\ V &= 70.83 \times 2\times 3600 = 509,976 \text{ m}^3 \\ A &= \frac{V}{0.01\ \text{m}} = 50,997,600 \text{ m}^2 \end{aligned}
$$

**Answer: A ≈ 51 km²** (readings from the figure are approximate).

#### (ii) Flood hydrograph for 1.5 cm/h for 2 h
The duration (2 h) equals the UH duration, so scale the UH by the net rain depth:

$$
ER = 1.5\times 2 = 3.0 \text{ cm}, \qquad Q = 3.0\,U + 2
$$

| t (h) | UH (m³/s) | 3 cm × UH | Base | Q (m³/s) |
|---|---|---|---|---|
| 0 | 0.00 | 0.0 | 2.0 | 2.0 |
| 2 | 1.33 | 4.0 | 2.0 | 6.0 |
| 4 | 3.25 | 9.8 | 2.0 | 11.8 |
| 6 | 6.00 | 18.0 | 2.0 | 20.0 |
| 8 | 9.00 | 27.0 | 2.0 | 29.0 |
| 10 | 12.00 | 36.0 | 2.0 | 38.0 |
| 12 | 10.80 | 32.4 | 2.0 | 34.4 |
| 14 | 9.60 | 28.8 | 2.0 | 30.8 |
| 16 | 7.50 | 22.5 | 2.0 | 24.5 |
| 18 | 4.50 | 13.5 | 2.0 | 15.5 |
| 20 | 3.25 | 9.8 | 2.0 | 11.8 |
| 22 | 2.00 | 6.0 | 2.0 | 8.0 |
| 24 | 1.20 | 3.6 | 2.0 | 5.6 |
| 26 | 0.40 | 1.2 | 2.0 | 3.2 |
| 28 | 0.00 | 0.0 | 2.0 | 2.0 |

**Answer: peak flood = 3.0 × 12 + 2 = 38 cumecs at t = 10 h.**

#### (iii) Time of concentration
Time of concentration is the time from the end of the effective rain to the end of direct runoff, i.e. the time taken by water from the farthest point to reach the outlet: $t_c = t_b - D$.

$$
t_c = 27 - 2 = 25 \text{ h}
$$

**Answer: $t_c \approx$ 25 h.**

#### (iv) Two storms of 1 h with φ = 0.12 cm/h
Excess rain in each hour = intensity − φ:

- Storm 1: 0.47 − 0.12 = 0.35 cm
- Storm 2: 0.64 − 0.12 = 0.52 cm

The two successive 1-h storms together make 2 h (equal to the UH duration), with excess 0.35 + 0.52 = 0.87 cm. The peak occurs at the UH peak (12 cumecs):

$$
Q_p = 0.87\times 12 = 10.44 \text{ cumecs}
$$

**Answer: peak discharge ≈ 10.44 cumecs** (about 12.44 cumecs if the 2 cumecs base flow of part (ii) is added).

## Check yourself

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

1. How is catchment area found from the 2-h UH?
   A. Peak ordinate x base time
   B. Sum of ordinates x 2 h x 3600 divided by 0.01 m of runoff (about 51 km2)
   C. Sum of ordinates x 2 only
   D. Sum of ordinates / 3600
   Answer: B. UH volume = 1 cm over the area; 509,976 m3 / 0.01 m. (Key point: Catchment area)

2. What multiplier is applied to the 2-h UH?
   A. 1.5, giving peak 20 cumecs
   B. 2.0, giving peak 26 cumecs
   C. 3.0 (net rain depth 1.5 x 2 cm), peak 3 x 12 + 2 = 38 cumecs
   D. 3.0 but base flow excluded, peak 36 cumecs
   Answer: C. Duration equals the UH duration so scale by depth and add base flow. (Key point: Net rain 1.5 cm/h for 2 h)

3. How is tc obtained from the UH?
   A. tc = tb = 27 h
   B. tc = time to peak = 10 h
   C. tc = D = 2 h
   D. tc = tb - D = 27 - 2 = 25 h
   Answer: D. tc is the time from end of effective rain to end of direct runoff. (Key point: Time of concentration)

4. Why can the two 1-h storms be combined for the peak?
   A. Together they last 2 h, the UH duration, with excess 0.35 + 0.52 = 0.87 cm, so Qp = 0.87 x 12 = 10.44
   B. Each storm's rain is used without subtracting phi
   C. They are averaged: 0.435 x 12
   D. The peaks occur at different times so they cannot be added
   Answer: A. Excess = intensity - phi per hour; sum gives depth for a 2-h UH. (Key point: Two 1-h storms)

## 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
- [S-curve 3-h UH and flood from two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Ascurve-2h-3h-flood-8cm-7cm) — Engineering Hydrology, chapter 5
- [2-hr UH and flood hydrograph from storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-2h-from-4h-three-storms) — 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×
- [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
- [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