---
title: "Derive 4-hr UH from observed streamflow"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:uh-4hr-600km2-derive"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-4hr-600km2-derive
site: BCT Engineering
---

# Derive 4-hr UH from observed streamflow

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2078 Bhadra, 8 marks
- Video: [Unit  Hydrograph Math Problems|| Derivation of Unit Hydrograph from Flood Hydrograph](https://www.youtube.com/watch?v=mvwdjGLxLtI) — Engineering School
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/uh-4hr-600km2-derive
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

Observed streamflow in a catchment (Area = 600 km$^2$) from a storm event of 4 hours' duration tabulated hereunder. If baseflow is 10 m$^3$/s, derive 4 hr unit hydrograph.

| Time (hr) | 0 | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 | 40 | 44 | 48 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Observed flow (m$^3$/s) | 10 | 100 | 230 | 190 | 130 | 100 | 70 | 60 | 50 | 30 | 20 | 15 | 10 |

## Short answer

Subtract baseflow (10 m³/s) to get DRH; effective rainfall depth is 2.124 cm (volume 12.744×10⁶ m³ over 600 km²). Divide DRH ordinates by 2.124 to obtain 4-h UH; peak 103.6 m³/s at 8 h.

## Answer

**Method:** subtract the baseflow (10 m$^3$/s) to get the direct runoff hydrograph (DRH), find the volume and the effective rainfall depth, and divide the DRH ordinates by this depth to get the 4-h UH (ordinates per 1 cm).

| t (h) | Observed Q (m$^3$/s) | DRH = Q - 10 | UH ordinate = DRH / 2.124 |
|---|---|---|---|
| 0 | 10 | 0 | 0.00 |
| 4 | 100 | 90 | 42.37 |
| 8 | 230 | 220 | 103.58 |
| 12 | 190 | 180 | 84.75 |
| 16 | 130 | 120 | 56.50 |
| 20 | 100 | 90 | 42.37 |
| 24 | 70 | 60 | 28.25 |
| 28 | 60 | 50 | 23.54 |
| 32 | 50 | 40 | 18.83 |
| 36 | 30 | 20 | 9.42 |
| 40 | 20 | 10 | 4.71 |
| 44 | 15 | 5 | 2.35 |
| 48 | 10 | 0 | 0.00 |
| **Sum** | | **885** | |

#### Direct runoff volume
$$
V=\sum DRH\times\Delta t = 885\times4\times3600 = 12,744,000\ \text{m}^3 = 12.744\times10^6\ \text{m}^3
$$

#### Effective rainfall depth over the catchment
$$
d=\frac{V}{A}=\frac{12,744,000}{600\times10^6}=2.124\ \text{cm}
$$

#### Unit hydrograph
$$
U(t)=\frac{DRH(t)}{2.124}
$$
Check: $\sum U\times\Delta t = 416.7\times14400 = 6,000,000$ m$^3$ = 1 cm over 600 km$^2$ (6,000,000 m$^3$).

**Answer: The 4-h UH has a peak of 103.6 m$^3$/s at 8 h and ordinates as shown in the last column (0, 42.4, 103.6, 84.8, 56.5, 42.4, 28.2, 23.5, 18.8, 9.4, 4.7, 2.4, 0 m$^3$/s at 0, 4, ..., 48 h).**

## Check yourself

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

1. The first step in deriving the UH from the observed flow is:
   A. Divide observed flow by area
   B. Subtract the 10 m^3/s baseflow to get the DRH
   C. Add 10 m^3/s to flows
   D. Compute the S-curve
   Answer: B. DRH = Q - baseflow. (Key point: DRH)

2. The effective rainfall depth is found from:
   A. Volume of DRH divided by basin area
   B. Peak divided by area
   C. Sum of observed Q
   D. Baseflow times duration
   Answer: A. d = V/A = 12.744 x 10^6 / 600 x 10^6 m^2 = 2.124 cm. (Key point: Effective rainfall depth)

3. With sum of DRH = 885 m^3/s at 4 h intervals, the volume is:
   A. 885 x 3600
   B. 885 x 4 x 3600 m^3
   C. 885 x 4
   D. 885 / (4 x 3600)
   Answer: B. V = sum DRH x Delta t = 885 x 14400 = 12.744 x 10^6 m^3. (Key point: Volume)

4. The UH ordinates are obtained by:
   A. Multiplying DRH by 2.124
   B. Dividing DRH by 2.124
   C. Subtracting 2.124
   D. Dividing DRH by 600
   Answer: B. UH is the response per 1 cm, so DRH is divided by the effective depth in cm. (Key point: UH ordinates)

## Related in Engineering Hydrology

- [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
- [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×
- [Derive 4-hr UH from triangular direct runoff hydrograph](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Atriangular-dro-4hr-uh) — 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×
- [S-curve: 2-hr hydrograph from 4-hr UH](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Ascurve-4hr-to-2hr-12-5) — Engineering Hydrology, chapter 5
- [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

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