---
title: "Storm hydrograph from 6-h UH, 3-hr storm"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:storm-hydrograph-6h-uh-3hr-storm"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/storm-hydrograph-6h-uh-3hr-storm
site: BCT Engineering
---

# Storm hydrograph from 6-h UH, 3-hr storm

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2081 Bhadra, 9 marks
- 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/storm-hydrograph-6h-uh-3hr-storm
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

The ordinates of 6 h-UH of a basin are given as under. Derive the storm hydrograph due to a 3 hr storm with a total rainfall of 4 cm, $\phi$-index of 0.5 cm/hr and base flow of 2 m$^3$/s.

| Time (hr) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 |
|---|---|---|---|---|---|---|---|---|---|
| Ordinate (m$^3$/s) | 0 | 8 | 14 | 28 | 16 | 10 | 6 | 4 | 0 |

## Short answer

Rainfall excess = 4 - 0.5×3 = 2.5 cm. Derive 3-h UH from given 6-h UH by S-curve: S(t) sum of 6-h UH lagged; 3-h UH = 2[S(t)-S(t-3)]. Superpose 2.5 cm excess with baseflow 2 m³/s yields flood hydrograph peak 73.1 m³/s at 18 h.

## Answer

The storm lasts 3 h but the UH is for 6 h, so first derive the **3-h UH from the 6-h UH by the S-curve method**, then apply the rainfall excess.

#### Step 1: Rainfall excess
$$
ER = P - \phi t = 4 - 0.5\times 3 = 2.5 \text{ cm}
$$

#### Step 2: S-curve and 3-h UH
The S-curve is the sum of the 6-h UH repeated every 6 h. Its values at 6-h points are 0, 8, 22, 50, 66, 76, 82, 86, 86 (running sum of the UH ordinates). A smooth S-curve is drawn through them and read at 3-h intervals, so that the curve can be lagged by 3 h:

$$
S(t) = U_6(t) + S(t-6)
$$

The 3-h UH is the S-curve lagged by 3 h, subtracted, and multiplied by 6/3:

$$
U_3(t) = \frac{6}{3}\,[S(t)-S(t-3)]
$$

#### Step 3: Storm hydrograph
$$
Q(t) = 2.5\,U_3(t) + 2
$$

| t (h) | 6-h UH | S(t) | S(t−3) | 3-h UH = 2[S(t)−S(t−3)] | 2.5 × UH | Q = +2 |
|---|---|---|---|---|---|---|
| 0 | 0 | 0.0 | 0.0 | 0.00 | 0.0 | 2.0 |
| 3 | – | 3.4 | 0.0 | 6.70 | 16.8 | 18.8 |
| 6 | 8 | 8.0 | 3.4 | 9.30 | 23.2 | 25.2 |
| 9 | – | 13.9 | 8.0 | 11.88 | 29.7 | 31.7 |
| 12 | 14 | 22.0 | 13.9 | 16.12 | 40.3 | 42.3 |
| 15 | – | 35.8 | 22.0 | 27.58 | 68.9 | 70.9 |
| 18 | 28 | 50.0 | 35.8 | 28.42 | 71.1 | 73.1 |
| 21 | – | 59.0 | 50.0 | 18.01 | 45.0 | 47.0 |
| 24 | 16 | 66.0 | 59.0 | 13.99 | 35.0 | 37.0 |
| 27 | – | 71.6 | 66.0 | 11.20 | 28.0 | 30.0 |
| 30 | 10 | 76.0 | 71.6 | 8.80 | 22.0 | 24.0 |
| 33 | – | 79.3 | 76.0 | 6.67 | 16.7 | 18.7 |
| 36 | 6 | 82.0 | 79.3 | 5.33 | 13.3 | 15.3 |
| 39 | – | 84.6 | 82.0 | 5.20 | 13.0 | 15.0 |
| 42 | 4 | 86.0 | 84.6 | 2.80 | 7.0 | 9.0 |
| 45 | – | 86.0 | 86.0 | 0.00 | 0.0 | 2.0 |

The S-curve reaches a constant value of 86 m³/s (sum of the 6-h UH ordinates), as it should.

**Answer: peak of the storm hydrograph ≈ 73.1 m³/s at t = 18 h; it returns to the 2 m³/s base flow at t = 45 h.**

## Check yourself

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

1. What is the excess for 4 cm in 3 h with phi 0.5 cm/h?
   A. 4.0 cm
   B. 2.5 cm
   C. 2.0 cm
   D. 3.5 cm
   Answer: B. ER = 4 - 0.5 x 3 = 2.5 cm. (Key point: Rainfall excess)

2. Why is the S-curve method needed here?
   A. To remove base flow
   B. To convert cm to mm
   C. Storm is 3 h but the UH is 6 h, so a 3-h UH must be derived
   D. To find the phi-index
   Answer: C. A UH can be used directly only for rain of its own duration. (Key point: Why S-curve)

3. How is the 3-h UH obtained from the S-curve of the 6-h UH?
   A. U3 = (3/6)[S(t) - S(t-3)]
   B. U3 = S(t) + S(t-3)
   C. U3 = [S(t) - S(t-6)]/3
   D. U3 = (6/3)[S(t) - S(t-3)]
   Answer: D. Lag by 3 h, subtract, multiply by T/D = 6/3. (Key point: 3-h UH formula)

## Close variants set in other exams

- [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×

## 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
- [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
- [Flood hydrograph from 6-hr hydrograph, 3.3 and 5.5 cm](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-770km2-6hr) — Engineering Hydrology, chapter 5
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- [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×
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