---
title: "Peak discharge from 6-h UH with minimum infiltration"
subject: "Engineering Hydrology"
chapter: 5
key: "semester_5/engineering_hydrology:peak-6h-uh-96-minimum-infiltration"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/peak-6h-uh-96-minimum-infiltration
site: BCT Engineering
---

# Peak discharge from 6-h UH with minimum infiltration

Engineering Hydrology · Chapter 5: Hydrograph Analysis

- Asked in: 2073 Shrawan, 4 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/peak-6h-uh-96-minimum-infiltration
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-5

## Question

A 6 h unit hydrograph of a basin has a peak ordinate of 96 m$^3$/s. When the base flow in the stream is 25 m$^3$/s, and when the basin has reached its minimum infiltration capacity of 2.5 mm/h, a 6 h storm with 18.3 cm of total rainfall had occurred on the basin. What is the magnitude of the peak discharge in the flood hydrograph produced by this storm?

## Short answer

Rainfall excess depth = (18.3/6 − 0.25) × 6 = 16.8 cm. Peak discharge = 16.8 × 96 + 25 = 1637.8 m³/s.

## Answer

Direct runoff = rainfall excess; the peak of the flood hydrograph = (ER depth in cm) × (UH peak) + base flow, because a 6 h storm on a 6 h UH needs no lagging.

#### Data
- UH duration = storm duration = 6 h, UH peak = 96 m³/s per cm
- Base flow = 25 m³/s
- Infiltration is at its minimum rate, so it is taken as the φ-index = 2.5 mm/h = 0.25 cm/h

#### Rainfall excess
Rainfall intensity = 18.3 / 6 = 3.05 cm/h

$$
\begin{aligned} ER &= (i-\phi)\,t = (3.05 - 0.25)\times 6 \\ &= 16.8 \text{ cm} \end{aligned}
$$

#### Peak discharge
$$
\begin{aligned} Q_p &= ER \times U_p + Q_b \\ &= 16.8 \times 96 + 25 \\ &= 1637.8 \text{ m}^3/\text{s} \end{aligned}
$$

**Answer: peak discharge ≈ 1637.8 m³/s** (direct runoff 1612.8 m³/s + base flow 25 m³/s).

## Check yourself

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

1. At minimum infiltration capacity of 2.5 mm/h the phi index used is:
   A. 0.25 cm/h
   B. 2.5 cm/h
   C. 0.025 cm/h
   D. 25 cm/h
   Answer: A. 2.5 mm/h = 0.25 cm/h. (Key point: Phi index)

2. ER for 18.3 cm in 6 h is:
   A. 16.8 cm
   B. 18.3 cm
   C. 17.05 cm
   D. 15.3 cm
   Answer: A. (3.05 - 0.25) x 6 = 16.8 cm. (Key point: Effective rainfall)

3. Peak discharge is:
   A. 16.8 x 96 + 25 = 1637.8
   B. 18.3 x 96 + 25
   C. 16.8 x 96
   D. 16.8 x 96 - 25
   Answer: A. Storm duration equals UH duration so ER x Up + base flow. (Key point: Peak)

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