---
title: "First-order kinematic wave, flood peak at community"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:kinematic-wave-peak-flood-community"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-peak-flood-community
site: BCT Engineering
---

# First-order kinematic wave, flood peak at community

Computational Techniques in Civil Engineering · Chapter 5: Finite difference method

- Asked in: 2070 Magh, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-peak-flood-community
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

A flood of 150 m$^3$/s peak discharge passed a gauging station at 12:00 noon on a river. There is a community adjacent to the river 7.2 km downstream. What will be the value of peak discharge at that community at 12:00 noon if the velocity of flow is 1.2 m/s$^2$ [as printed] and the peak discharge at that community at 9:00 A.M. is 100 m$^3$/s? Assume width of river as [?] and use the first order accurate numerical scheme of the kinematic wave equation. Take $\Delta x = 7.2$ km and $\Delta t = 1$ hr.

## Short answer

Apply the first-order implicit scheme Q_{i+1}^{n+1} = (Q_{i+1}^n + C Q_i^{n+1})/(1+C) with c = 1.2 m/s, Δt = 1 h, Δx = 7.2 km, giving C = 0.6. The downstream peak discharge is (100 + 0.6×150)/1.6 = 118.75 m³/s.

## Answer

The first-order accurate (implicit) scheme of the linear kinematic wave equation $\partial Q/\partial t+c\,\partial Q/\partial x=0$ is
$$
\frac{Q_{i+1}^{n+1}-Q_{i+1}^n}{\Delta t}+c\,\frac{Q_{i+1}^{n+1}-Q_i^{n+1}}{\Delta x}=0
\ \Rightarrow\ Q_{i+1}^{n+1}=\frac{Q_{i+1}^n+C\,Q_i^{n+1}}{1+C},\qquad C=\frac{c\Delta t}{\Delta x}
$$

#### Interpretation of the data
- The gauging station is point $i$; the community ($\Delta x=7.2$ km downstream) is $i+1$.
- $Q_i^{n+1}=150$ m$^3$/s (peak at the station at 12:00 noon).
- $Q_{i+1}^n=100$ m$^3$/s is the flow already at the community at the earlier level (taken as the 9:00 A.M. value, with $\Delta t=1$ h per step).
- The velocity given as 1.2 (m/s) is used as the wave celerity $c$; the channel width is then not needed, because the equation is written in terms of $Q$.

#### Calculation
$$
C=\frac{c\,\Delta t}{\Delta x}=\frac{1.2\times3600}{7200}=0.6
$$
$$
Q_{i+1}^{n+1}=\frac{100+0.6\times150}{1+0.6}=\frac{190}{1.6}=118.75\ \text{m}^3/\text{s}
$$

**Answer:** the discharge at the community at 12:00 noon is about **118.75 m$^3$/s** (the peak is delayed and attenuated numerically). The data are incomplete in the question (width missing, unit of velocity printed as m/s$^2$), so this reading is assumed.

## Check yourself

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

1. The first-order scheme gives Q(i+1,n+1) as:
   A. (Q_{i+1}^n + C Q_i^{n+1})/(1+C)
   B. Q_i^n + C(Q_{i+1}^n - Q_i^n)
   C. Q_{i+1}^n - C Q_i^{n+1}
   D. (Q_i^{n+1} + C Q_{i+1}^n)/(1+C)
   Answer: A. It follows from backward differences in time and space at the new level. (Key point: Scheme)

2. With c = 1.2 m/s, dt = 3600 s, dx = 7200 m, C is:
   A. 0.6
   B. 1.67
   C. 0.3
   D. 1.2
   Answer: A. 1.2 x 3600/7200 = 0.6. (Key point: Courant number)

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