---
title: "Derive 4-hr UH from triangular direct runoff hydrograph"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:triangular-dro-4hr-uh"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/triangular-dro-4hr-uh
site: BCT Engineering
---

# Derive 4-hr UH from triangular direct runoff hydrograph

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2070 Chaitra, 8 marks
- Video: [6-Hour Unit Hydrograph Derivation | Triangular Unit Hydrograph Numerical Problem #hydrologynumerical](https://www.youtube.com/watch?v=jUhPdquRXz8) — Class 1 to B.Tech(Civil) 
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/triangular-dro-4hr-uh
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

The direct runoff hydrograph due to an effective rainfall event is given by a triangle such that its base is 8 hours and its height at the midpoint of the base is 1 cm/h. The duration and intensity of the effective rainfall are 4 hours and 1 cm/h, respectively. Derive and sketch a 4 hour unit hydrograph.

## Short answer

The DRH is a triangle peaked at1 cm/h at4 h with base8 h, produced by4 cm effective rainfall. Divide ordinates by4: the4-h UH is a triangle of base8 h and peak0.25 h⁻¹ at t=4 h.

## Answer

A UH is the DRH divided by the depth of effective rainfall that produced it.

#### Step 1: Depth of effective rainfall
$$
ER = i\,t_r = 1 \text{ cm/h} \times 4 \text{ h} = 4 \text{ cm}
$$

#### Step 2: Check the DRH volume
The DRH is a triangle: base 8 h, height 1 cm/h. Area = ½ × 8 × 1 = 4 cm. This equals the effective rainfall, so the DRH is consistent (no loss of volume).

#### Step 3: Ordinates of the 4-h UH
The 4-h UH ordinate = DRH ordinate / 4 (per cm of rain). The DRH rises linearly from 0 at t = 0 to 1 cm/h at t = 4 h, then falls linearly to 0 at t = 8 h.

| t (h) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
|---|---|---|---|---|---|---|---|---|---|
| DRH (cm/h) | 0 | 0.25 | 0.50 | 0.75 | 1.00 | 0.75 | 0.50 | 0.25 | 0 |
| 4-h UH (cm/h per cm) | 0 | 0.0625 | 0.125 | 0.1875 | 0.25 | 0.1875 | 0.125 | 0.0625 | 0 |

#### Sketch of the 4-h UH
```text
 U (per h)
 0.25 |          /\
      |        /    \
      |      /        \
      |    /            \
      |  /                \
    0 +--+--+--+--+--+--+--+-- t (h)
      0  1  2  3  4  5  6  7  8
```

**Answer: the 4-h UH is a triangle of base 8 h and peak 0.25 per hour (0.25 cm/h per cm of effective rain) at t = 4 h.** Check: area = ½ × 8 × 0.25 = 1 cm ✓.

## Check yourself

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

1. What is the depth of effective rainfall that produced the DRH?
   A. 1 cm
   B. 4 cm (1 cm/h x 4 h)
   C. 8 cm
   D. 2 cm
   Answer: B. ER = i x tr = 1 x 4 = 4 cm; it matches the triangle area 0.5 x 8 x 1 = 4 cm. (Key point: Effective rainfall depth)

2. How are the 4-h UH ordinates obtained from the DRH?
   A. Multiply DRH ordinates by 4
   B. Divide by 8 h
   C. Divide DRH ordinates by 4 cm
   D. Subtract 4 from each ordinate
   Answer: C. A UH is the DRH per unit (1 cm) of effective rain. (Key point: UH from DRH)

3. What is the peak of the 4-h UH and when does it occur?
   A. 1.0 per hour at t = 4 h
   B. 0.25 per hour at t = 8 h
   C. 0.5 per hour at t = 2 h
   D. 0.25 per hour at t = 4 h, with base 8 h
   Answer: D. Peak DRH 1.0 / 4 = 0.25; area 0.5 x 8 x 0.25 = 1 cm. (Key point: Resulting UH)

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