---
title: "Unit hydrograph and UH for multiple storms"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:uh-define-multiple-storm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-define-multiple-storm
site: BCT Engineering
---

# Unit hydrograph and UH for multiple storms

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2082 Bhadra, 2+4 marks; 2081 Bhadra, 2+4 marks (asked 2 times)
- Video: [Unit Hydrograph concept-derivation and Numericals](https://www.youtube.com/watch?v=FbPqxFotoeI) — Chandrakant Jejurkar
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-define-multiple-storm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

What is unit hydrograph (UH)? How do you prepare UH for multiple storm?

## Short answer

Unit hydrograph is the DRH from 1 cm effective rainfall uniformly over a basin in duration D. For multiple storms, deconvolute: solve Q_n = Σ P_m U_{n-m+1} successively, starting U1=Q1/P1, U2=(Q2−P2U1)/P1, etc., and check volume equals 1 cm over area.

## Answer

#### Unit hydrograph
A **unit hydrograph** is the direct runoff hydrograph produced by 1 cm of effective rainfall, falling uniformly over the basin at a constant rate during a specified duration $D$. It assumes linearity and time invariance.

#### Preparing a UH from a multiple (complex) storm
When a storm has several periods of effective rainfall, the observed direct runoff hydrograph (DRH) is the sum of the responses to each period. The UH is obtained by **deconvolution**.

**Steps**
1. Separate the baseflow from the storm hydrograph to get the DRH ordinates $Q_n$.
2. Compute the effective rainfall for each period of duration $D$ by deducting losses ($\phi$-index): $P_1, P_2, \dots, P_M$.
3. The DRH ordinate at time $n$ is
$$
Q_n=\sum_{m=1}^{\min(n,M)} P_m\,U_{n-m+1}
$$
4. Solve for the UH ordinates $U_n$ successively:
$$
U_1=\frac{Q_1}{P_1},\quad U_2=\frac{Q_2-P_2U_1}{P_1},\quad U_3=\frac{Q_3-P_2U_2-P_3U_1}{P_1},\ \dots
$$
5. Check that the area under the UH equals 1 cm over the basin: $\sum U\,\Delta t = 1\ \text{cm}\times A$.
6. For more accuracy, use many storms and the least-squares method, and average the UH ordinates.

#### Example
Two periods of 1 h with effective rainfall $P_1=2$ cm and $P_2=3$ cm. Observed DRH: $Q_1=20$, $Q_2=90$, $Q_3=130$, $Q_4=80$, $Q_5=30$ m$^3$/s.

| n | Working | $U_n$ (m$^3$/s) |
|---|---|---|
| 1 | 20/2 | 10 |
| 2 | (90 - 3x10)/2 | 30 |
| 3 | (130 - 3x30)/2 | 20 |
| 4 | (80 - 3x20)/2 | 10 |
| 5 | (30 - 3x10)/2 | 0 |

The 1-h UH ordinates are 10, 30, 20, 10 m$^3$/s at 1, 2, 3, 4 h.

## Check yourself

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

1. The UH from a storm with several periods of effective rainfall is obtained by:
   A. Dividing the peak by total rainfall
   B. Deconvolution of the DRH using the effective rainfall of each period
   C. Averaging the rainfall ordinates
   D. Integrating the baseflow
   Answer: B. The DRH is the sum of responses to each rain period, so the UH ordinates are solved successively from the convolution equation. (Key point: Deconvolution idea)

2. With effective rain P1, P2 and DRH ordinates Q1, Q2, the first two UH ordinates are:
   A. U1 = Q1/P1, U2 = (Q2 - P2 U1)/P1
   B. U1 = Q1/P2, U2 = (Q2 - P1 U1)/P2
   C. U1 = Q1 P1, U2 = Q2 P2
   D. U1 = Q1/P1, U2 = Q2/P2
   Answer: A. Q2 = P1 U2 + P2 U1, so U2 = (Q2 - P2 U1)/P1. (Key point: First UH ordinates)

3. Before computing UH ordinates from an observed storm hydrograph, you must first:
   A. Separate the baseflow to get the DRH
   B. Compute the S-curve
   C. Add the baseflow to the DRH
   D. Convert to a synthetic UH
   Answer: A. Only direct runoff ordinates Q_n are used in the equations, so baseflow is removed first. (Key point: First step)

4. With P1 = 2 cm, P2 = 3 cm, Q1 = 20 and Q2 = 90 m^3/s, U2 equals:
   A. 45
   B. 35
   C. 30
   D. 25
   Answer: C. U1 = 20/2 = 10; U2 = (90 - 3x10)/2 = 30. (Key point: Worked example)

## Related in Engineering Hydrology

- [Derive UH from storm, peak flow from two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-from-storm-then-peak-two) — 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
- [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
- [2-h UH by S-curve from storm hydrograph](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Ascurve-2h-uh-4successive-hrs) — Engineering Hydrology, chapter 5
- [Storm hydrograph from 6-h UH, 3-hr storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Astorm-hydrograph-6h-uh-3hr-storm) — Engineering Hydrology, chapter 5, asked 2×
- [Storm hydrograph from 6-h UH, 15 cm storm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Astorm-hydrograph-6h-uh-15cm) — Engineering Hydrology, chapter 5, asked 2×