---
title: "Flood hydrograph from 4-h UH, three 2-hr rainfalls"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:flood-hydrograph-4h-uh-3-2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-hydrograph-4h-uh-3-2
site: BCT Engineering
---

# Flood hydrograph from 4-h UH, three 2-hr rainfalls

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2081 Baisakh, 14 marks
- Video: [Unit Hydrograph to Flood Hydrograph || Derivation of Flood Hydrograph from Unit Hydrograph](https://www.youtube.com/watch?v=8ZXviXtKVdk) — Engineering School
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-hydrograph-4h-uh-3-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

The ordinate of 4-Hr UH are given:

| Time (hr) | 0 | 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ordinate of UH (m$^3$/s) | 0 | 30 | 100 | 150 | 200 | 160 | 120 | 80 | 40 | 30 | 20 | 6 | 0 |

A catchment has rainfall of 3.2, 3.0 and 5.0 cm in three consecutive two hours period. Assuming an average $\phi$ index of 1.25 cm/hr and base flow of river is 40 m$^3$/s, determine the flood hydrograph of the catchment.

## Short answer

Derive the 2-h UH from the 4-h UH via the S-curve method. Apply rainfall excesses of 0.70, 0.50, and2.50 cm by superposition, then add base flow of40 m³/s. The flood hydrograph peaks at 678 m³/s at 10 h and returns to base flow at 28 h.

## Answer

**Approach.** The rain falls in three 2-h periods, but the given UH is for a 4-h rain. A 4-h UH cannot be used directly for 2-h excess periods, so first derive the 2-h UH by the S-curve method, then apply the three rainfall excesses by superposition.

#### Step 1: Rainfall excess per 2-h period
Loss per period = φ × 2 h = 1.25 × 2 = 2.5 cm.

| Period | Rain (cm) | Loss (cm) | Excess (cm) |
|---|---|---|---|
| 0 – 2 h | 3.2 | 2.5 | 0.70 |
| 2 – 4 h | 3 | 2.5 | 0.50 |
| 4 – 6 h | 5 | 2.5 | 2.50 |

#### Step 2: 2-h UH from the 4-h UH (S-curve)
The S-curve is $S(t)=U_4(t)+S(t-4)$, using the ordinates at 2-h steps. The 2-h UH is

$$
U_2(t)=\frac{4}{2}\,[S(t)-S(t-2)]
$$

The two interleaved series (t = 0, 4, 8, … and t = 2, 6, 10, …) of the plain sum settle at 480 and 456 m³/s, because the given table is not perfectly consistent. The S-curve therefore zig-zags, and subtracting it directly gives erratic (even negative) 2-h ordinates. So a smooth S-curve is drawn through the points. It must level off at $\sum U_4/2 = 936/2 = 468$ m³/s.

| t (h) | 4-h UH | S(t) plain sum | S(t) smooth curve | 2-h UH = 2[S(t) − S(t−2)] |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 2 | 30 | 30 | 40 | 80 |
| 4 | 100 | 100 | 100 | 120 |
| 6 | 150 | 180 | 190 | 180 |
| 8 | 200 | 300 | 280 | 180 |
| 10 | 160 | 340 | 350 | 140 |
| 12 | 120 | 420 | 400 | 100 |
| 14 | 80 | 420 | 430 | 60 |
| 16 | 40 | 460 | 448 | 36 |
| 18 | 30 | 450 | 460 | 24 |
| 20 | 20 | 480 | 467 | 14 |
| 22 | 6 | 456 | 468 | 2 |
| 24 | 0 | 480 | 468 | 0 |

Check: Σ(2-h UH) = 936 m³/s = Σ(4-h UH), so both UHs carry the same runoff volume of 1 cm.

#### Step 3: Convolution with the 2-h UH
$$
Q(t) = 0.70\,U_2(t) + 0.50\,U_2(t-2) + 2.50\,U_2(t-4) + 40
$$

| t (h) | 2-h UH (m³/s) | 0.7 × U₂(t) | 0.5 × U₂(t−2) | 2.5 × U₂(t−4) | Base | Q (m³/s) |
|---|---|---|---|---|---|---|
| 0 | 0 | 0.0 | 0.0 | 0.0 | 40.0 | 40.0 |
| 2 | 80 | 56.0 | 0.0 | 0.0 | 40.0 | 96.0 |
| 4 | 120 | 84.0 | 40.0 | 0.0 | 40.0 | 164.0 |
| 6 | 180 | 126.0 | 60.0 | 200.0 | 40.0 | 426.0 |
| 8 | 180 | 126.0 | 90.0 | 300.0 | 40.0 | 556.0 |
| 10 | 140 | 98.0 | 90.0 | 450.0 | 40.0 | 678.0 |
| 12 | 100 | 70.0 | 70.0 | 450.0 | 40.0 | 630.0 |
| 14 | 60 | 42.0 | 50.0 | 350.0 | 40.0 | 482.0 |
| 16 | 36 | 25.2 | 30.0 | 250.0 | 40.0 | 345.2 |
| 18 | 24 | 16.8 | 18.0 | 150.0 | 40.0 | 224.8 |
| 20 | 14 | 9.8 | 12.0 | 90.0 | 40.0 | 151.8 |
| 22 | 2 | 1.4 | 7.0 | 60.0 | 40.0 | 108.4 |
| 24 | 0 | 0.0 | 1.0 | 35.0 | 40.0 | 76.0 |
| 26 | 0 | 0.0 | 0.0 | 5.0 | 40.0 | 45.0 |
| 28 | 0 | 0.0 | 0.0 | 0.0 | 40.0 | 40.0 |

Check: Σ(Q − 40) = 3463 m³/s = 3.7 cm (total excess) × 936 m³/s, so the volume is conserved.

**Answer: peak flood ≈ 678 m³/s at t = 10 h; the flow returns to the base flow of 40 m³/s at t = 28 h.** The last column of the table is the flood hydrograph. The ordinates depend slightly on how the smooth S-curve is drawn, so they are approximate.

## Check yourself

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

1. With phi 1.25 cm/h, what loss applies to each 2-h period?
   A. 2.5 cm
   B. 1.25 cm
   C. 5.0 cm
   D. 0.625 cm
   Answer: A. Loss = 1.25 x 2 = 2.5 cm. (Key point: Loss per period)

2. What are the excess depths for 3.2, 3.0 and 5.0 cm?
   A. 3.2, 3.0, 5.0 cm
   B. 0.70, 0.50 and 2.50 cm
   C. 1.95, 1.75, 3.75 cm
   D. 0.70, 0.50, 5.0 cm
   Answer: B. Subtract 2.5 cm from each. (Key point: Excess depths)

## Close variants set in other exams

- [Flood hydrograph from 4-h UH, 50 m3/s base](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-4h-uh-3-5) — 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 UH, two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-hydrograph-3-3-5-5cm) — 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×
- [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
- [Flood hydrograph from 2-h UH, 5.2 cm rainfall](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Aflood-from-2h-uh-5-2cm) — Engineering Hydrology, chapter 5

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