---
title: "Maximum time step, kinematic wave routing"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:kinematic-wave-max-time-step"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-max-time-step
site: BCT Engineering
---

# Maximum time step, kinematic wave routing

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

- Asked in: 2072 Asoj, 6 marks
- Video: [Which HEC-HMS Routing Method Should You Use? (Channel & Reservoir Routing)](https://www.youtube.com/watch?v=DL-9mHpq2HY) — Water Edu Hub
- 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-max-time-step
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

A channel with a width of 40 m, bed slope 2% and Manning's $n = 0.03$ carries a discharge of 100 m$^3$/s through a section. If $\Delta x$ is taken as 1500 meters, recommend the maximum time step for kinematic wave routing in this condition. Assume hydraulic radius equal to flow depth.

## Short answer

For kinematic wave routing, maximum time step from Courant condition Δt≤Δx/c. Normal depth y=0.684 m, velocity V=3.66 m/s, celerity c=5/3 V=6.096 m/s, so Δt_max=246 s (≈4.1 min).

## Answer

For an explicit kinematic wave scheme the time step is limited by the Courant condition:
$$
C=\frac{c\,\Delta t}{\Delta x}\le1\ \Rightarrow\ \Delta t_{max}=\frac{\Delta x}{c}
$$
where $c=dQ/dA$ is the kinematic wave celerity.

#### Step 1: Normal depth
With $R\approx y$ and $A=By$, Manning's equation gives
$$
Q=\frac1nBy\,y^{2/3}S_0^{1/2}\ \Rightarrow\ y=\left(\frac{nQ}{BS_0^{1/2}}\right)^{3/5}=\left(\frac{0.03\times100}{40\times\sqrt{0.02}}\right)^{0.6}=0.6835\ \text{m}
$$

#### Step 2: Velocity and celerity
$$
V=\frac{Q}{By}=\frac{100}{40\times0.6835}=3.6577\ \text{m/s}
$$
Since $Q=\alpha A^{5/3}$, $c=\dfrac{dQ}{dA}=\dfrac53V=\dfrac53(3.6577)=6.0962\ \text{m/s}$.

#### Step 3: Maximum time step
$$
\Delta t_{max}=\frac{\Delta x}{c}=\frac{1500}{6.0962}=246.1\ \text{s}\approx4.1\ \text{min}
$$

**Answer:** the time step should not exceed about **246 s (about 4.1 minutes)**; a practical choice is slightly less, e.g. 4 minutes (240 s).

## Check yourself

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

1. The maximum explicit time step follows from:
   A. dt = c dx
   B. dt_max = dx/c from C = c dt/dx <= 1
   C. dt = V/dx
   D. dt = dx/(2c)
   Answer: B. Courant condition C <= 1 gives dt_max = dx/c. (Key point: Time step limit)

2. For this wide channel, c is found as:
   A. V
   B. (5/3) V
   C. sqrt(g y)
   D. (3/5) V
   Answer: B. c = dQ/dA = (5/3)V = 6.096 m/s. (Key point: Celerity)

3. The maximum time step is about:
   A. 246 s
   B. 148 s
   C. 410 s
   D. 61 s
   Answer: A. 1500/6.0962 = 246 s, about 4.1 min. (Key point: Result)

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