---
title: "Flood hydrograph from 4-h UH, 50 m3/s base"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:flood-hydrograph-4h-uh-3-5"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/flood-hydrograph-4h-uh-3-5
site: BCT Engineering
---

# Flood hydrograph from 4-h UH, 50 m3/s base

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2076 Chaitra, 12 marks
- Video: [TYPE 2 SUM : UNIT HYDROGRAPH TO FLOOD HYDROGRAPH | UH TO FH NUMERICAL | STEPS & SOLVED EXAMPLE OF UH](https://www.youtube.com/watch?v=mXPu7nYGQII) — Professor Nidhi Trivedi
- 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-3-5
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

The ordinate of 4-h UH are given:

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

A catchment has rainfall of 3.5 cm, 2.5 cm and 4.5 cm in three consecutive two hour period. Assuming an average $\phi$-index of 1.25 cm/hour and base flow of river is 50 m$^3$/s, determine the flood hydrograph of the catchment.

## Short answer

Derive 2-h unit hydrograph from the 4-h UH using S-curve method. Convolve with rainfall excesses of 1 cm (0–2 h) and 2 cm (4–6 h) and add baseflow 50 m³/s. Peak flood is approximately 550 m³/s at 10 h, returning to baseflow 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 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 | 3.5 | 2.5 | 1.00 |
| 2 – 4 h | 2.5 | 2.5 | 0.00 |
| 4 – 6 h | 4.5 | 2.5 | 2.00 |

#### 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
The middle period gives no excess, so

$$
Q(t) = 1.00\,U_2(t) + 2.00\,U_2(t-4) + 50
$$

| t (h) | 2-h UH (m³/s) | 1 × U₂(t) | 2 × U₂(t−4) | Base | Q (m³/s) |
|---|---|---|---|---|---|
| 0 | 0 | 0.0 | 0.0 | 50.0 | 50.0 |
| 2 | 80 | 80.0 | 0.0 | 50.0 | 130.0 |
| 4 | 120 | 120.0 | 0.0 | 50.0 | 170.0 |
| 6 | 180 | 180.0 | 160.0 | 50.0 | 390.0 |
| 8 | 180 | 180.0 | 240.0 | 50.0 | 470.0 |
| 10 | 140 | 140.0 | 360.0 | 50.0 | 550.0 |
| 12 | 100 | 100.0 | 360.0 | 50.0 | 510.0 |
| 14 | 60 | 60.0 | 280.0 | 50.0 | 390.0 |
| 16 | 36 | 36.0 | 200.0 | 50.0 | 286.0 |
| 18 | 24 | 24.0 | 120.0 | 50.0 | 194.0 |
| 20 | 14 | 14.0 | 72.0 | 50.0 | 136.0 |
| 22 | 2 | 2.0 | 48.0 | 50.0 | 100.0 |
| 24 | 0 | 0.0 | 28.0 | 50.0 | 78.0 |
| 26 | 0 | 0.0 | 4.0 | 50.0 | 54.0 |
| 28 | 0 | 0.0 | 0.0 | 50.0 | 50.0 |

**Answer: peak flood ≈ 550 m³/s at t = 10 h; the flow returns to the base flow of 50 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, the loss in each 2-h period is:
   A. 2.5 cm
   B. 1.25 cm
   C. 5.0 cm
   D. 0.625 cm
   Answer: A. Loss = phi x time = 1.25 x 2 = 2.5 cm. (Key point: Loss per period)

2. The rainfall excess in the three periods (3.5, 2.5, 4.5 cm) is:
   A. 3.5, 2.5 and 4.5 cm
   B. 1.0, 0.0 and 2.0 cm
   C. 1.0, 0.0 and 2.5 cm
   D. 1.25, 1.25 and 1.25 cm
   Answer: B. Excess = rain - 2.5 cm in each period, with no negative values: 1.0, 0, 2.0. (Key point: Rainfall excess)

3. Why is the 2.0 cm excess UH shifted by 4 h in the convolution?
   A. The UH base is 4 h
   B. Base flow is delayed by 4 h
   C. It occurs in the 4-6 h period, which starts 4 h after the first excess
   D. The UH duration is 4 h and must be doubled
   Answer: C. Each period's response is lagged by the start time of its rainfall excess; the third period begins at 4 h. (Key point: Convolution lag)

4. The flood hydrograph peak is:
   A. 510 m3/s at t = 10 h
   B. 450 m3/s at t = 8 h
   C. 520 m3/s at t = 12 h
   D. 570 m3/s at t = 12 h
   Answer: D. At t = 12 h: 120 + 400 + 50 base = 570 m3/s. (Key point: Final result)

## Close variants set in other exams

- [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×

## 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
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