---
title: "2-h UH by S-curve from storm hydrograph"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:scurve-2h-uh-4successive-hrs"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/scurve-2h-uh-4successive-hrs
site: BCT Engineering
---

# 2-h UH by S-curve from storm hydrograph

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2074 Asoj, 8 marks
- Video: [Numerical on 'S' Curve Hydrograph (SSCJE/GATE/ESE/PSU's/DFCCIL/Railways)-Civil Engineering](https://www.youtube.com/watch?v=ChzzFVxNlm0) — Engineering Ace- By Varun Sir
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/scurve-2h-uh-4successive-hrs
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

In a storm, the rainfall of depth 0.7 cm, 0.9 cm, 0.2 cm, 1.0 cm occurred in four successive hours. The storm hydrograph due to this storm has the following hourly ordinates: 0.5, 44.5, 110.5, 85.5, 102.8, 94.0, 38.4, 18.6, 10.9, 5.3, 2.9, 0.5 m$^3$/s. If the average losses are 0.2 cm/hr, estimate the hourly ordinates of the unit hydrograph. Assume suitable value of base flow. Calculate the 2-h UH using S-curve method.

## Short answer

Baseflow was taken as 0.5 m³/s; effective rainfall after 0.2 cm/h losses was 0.5, 0.7, 0, 0.8 cm. The 1-h UH was deconvolved, then the 2-h UH by S-curve averaging: ordinates at 1–9 h are 44.0, 92.4, 65.7, 25.0, 13.0, 8.3, 4.5, 1.6, 0.2 m³/s.

## Answer

#### Step 1: Baseflow and effective rainfall
The storm hydrograph begins at 0.5 m$^3$/s and ends at 0.5 m$^3$/s, so a constant base flow of **0.5 m$^3$/s** is assumed.

Losses $=0.2$ cm/h:

| Hour | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| Rain (cm) | 0.7 | 0.9 | 0.2 | 1.0 |
| ER = rain - 0.2 (cm) | 0.5 | 0.7 | 0 | 0.8 |

Total ER = 2.0 cm.

#### Step 2: Hourly (1-h) UH by successive deconvolution
Direct runoff $DRH(t)=Q-0.5$ and
$$
DRH(t)=0.5\,U(t)+0.7\,U(t-1)+0\cdot U(t-2)+0.8\,U(t-3)
$$
$$
U(t)=\frac{DRH(t)-0.7U(t-1)-0.8U(t-3)}{0.5}
$$

Examples: $U(1)=44/0.5=88$; $U(2)=(110-0.7\times88)/0.5=96.8$; $U(3)=(85-0.7\times96.8)/0.5=34.5$; $U(4)=(102.3-0.7\times34.5-0.8\times88)/0.5=15.5$.

#### Step 3: S-curve and 2-h UH
For the 1-h UH, $S(t)=\sum U$ (lag 1 h), and
$$
U_2(t)=\frac{1}{2}\left[S(t)-S(t-2)\right]=\frac{U(t)+U(t-1)}{2}
$$

| t (h) | Q (m$^3$/s) | DRH | 1-h UH, $U$ | 2-h UH, $U_2$ |
|---|---|---|---|---|
| 0 | 0.5 | 0 | 0 | 0.00 |
| 1 | 44.5 | 44 | 88 | 44.00 |
| 2 | 110.5 | 110 | 96.8 | 92.40 |
| 3 | 85.5 | 85 | 34.5 | 65.65 |
| 4 | 102.8 | 102.3 | 15.5 | 25.00 |
| 5 | 94.0 | 93.5 | 10.4 | 12.95 |
| 6 | 38.4 | 37.9 | 6.1 | 8.25 |
| 7 | 18.6 | 18.1 | 2.8 | 4.45 |
| 8 | 10.9 | 10.4 | 0.3 | 1.55 |
| 9 | 5.3 | 4.8 | 0 | 0.15 |
| 10 | 2.9 | 2.4 | 0 | 0.00 |
| 11 | 0.5 | 0 | 0 | 0.00 |

The tiny negative and positive values after 8 h come from rounding of the given ordinates, and they are taken as zero. Check: $\sum DRH = 508.4$ and total ER $=2.0$ cm, so the sum of the 1-h UH ordinates should be $508.4/2.0 = 254.2$. The computed sum is 254.4, so the UH is consistent.

**Answer: 1-h UH ordinates at t = 0 to 8 h: 0, 88, 96.8, 34.5, 15.5, 10.4, 6.1, 2.8, 0.3 m$^3$/s. 2-h UH ordinates at t = 1 to 9 h: 44.0, 92.4, 65.7, 25.0, 13.0, 8.3, 4.5, 1.6, 0.2 m$^3$/s.**

## Check yourself

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

1. With loss 0.2 cm/h, the effective rainfall of the four hours (0.7, 0.9, 0.2, 1.0) is:
   A. 0.5, 0.7, 0, 0.8
   B. 0.5, 0.7, 0.2, 0.8
   C. 0.7, 0.9, 0.2, 1.0
   D. 0.5, 0.7, -0.0, 1.0
   Answer: A. Subtract 0.2 each hour; the third hour (0.2) gives 0, not negative. (Key point: Effective rainfall)

2. The first 1-h UH ordinate is:
   A. U(1) = 44/0.5 = 88
   B. U(1) = 44.5/0.5
   C. U(1) = 44 x 0.5
   D. U(1) = 44/2.0
   Answer: A. DRH = Q - 0.5 baseflow = 44, divided by first ER 0.5 cm. (Key point: Deconvolution)

3. Baseflow assumed is 0.5 m^3/s because:
   A. The hydrograph starts and ends at 0.5
   B. It is the peak
   C. Given in problem
   D. It is zero
   Answer: A. Initial and final flows equal 0.5. (Key point: Baseflow)

4. 2-h UH ordinate from the 1-h UH is:
   A. [U(t) + U(t-1)]/2
   B. U(t) + U(t-1)
   C. 2 U(t)
   D. U(t) - U(t-1)
   Answer: A. U2 = (1/2)[S(t) - S(t-2)] = (U(t) + U(t-1))/2. (Key point: 2-h UH from 1-h UH)

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