---
title: "Non-linear kinematic wave discharge at 1 km"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:kinematic-wave-implicit-numerical"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-implicit-numerical
site: BCT Engineering
---

# Non-linear kinematic wave discharge at 1 km

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

- Asked in: 2071 Bhadra, 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-implicit-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

Using the finite difference equation developed in the previous part (first order accurate implicit non-linear kinematic wave), compute the discharge at 1 km downstream of location X at time 14:00 hrs, for the following data: rectangular channel, width = 20 m, bed slope = 0.001, Manning's $n = 0.03$. Discharge at location X at time 14:00 hrs = 14 m$^3$/s; discharge at location X at time 13:45 hrs = 12 m$^3$/s; discharge at 1 km downstream of location X at time 13:45 hrs = 11 m$^3$/s. No lateral flow, wetted perimeter approximately equal to width of channel.

## Short answer

With Δt=900 s, Δx=1000 m, γ=Δt/Δx=0.9, upstream Q=14 m³/s, previous downstream Q=11 m³/s, α=3.2113, the implicit nonlinear scheme solved by Newton-Raphson gives Q at 1 km downstream = 12.67 m³/s.

## Answer

Use the first-order implicit scheme
$$
\frac{\Delta t}{\Delta x}Q_{i+1}^{n+1}+\alpha\left(Q_{i+1}^{n+1}\right)^{\beta}=\frac{\Delta t}{\Delta x}Q_i^{n+1}+\alpha\left(Q_{i+1}^n\right)^\beta
$$
with $q=0$. Here $i$ is location X and $i+1$ is the point 1 km downstream; $n$ is 13:45 and $n+1$ is 14:00.

#### Data
$\Delta x=1000$ m, $\Delta t=15\times60=900$ s, $\dfrac{\Delta t}{\Delta x}=0.90$, $Q_i^{n+1}=14$, $Q_i^n=12$, $Q_{i+1}^n=11$ m$^3$/s.

#### Step 1: Kinematic coefficients ($P\approx B=20$ m)
$$
\alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6}=\left(\frac{0.03\times20^{2/3}}{\sqrt{0.001}}\right)^{0.6}=3.2113,\qquad \beta=0.6
$$

#### Step 2: Equation to solve
Right side: $0.90(14)+3.2113(11)^{0.6}=26.1369$
$$
f(x)=0.90\,x+3.2113\,x^{0.6}-26.1369=0,\qquad f'(x)=0.90+1.9268\,x^{-0.4}
$$

#### Step 3: Newton iteration from $x_0=Q_{i+1}^n=11$

| Iteration | $x_k$ | $f(x_k)$ | $x_{k+1}$ |
|---|---|---|---|
| 1 | 11.0000 | -2.7000 | 12.6480 |
| 2 | 12.6480 | -0.0341 | 12.6693 |
| 3 | 12.6693 | -0.0000 | 12.6693 |
| 4 | 12.6693 | -0.0000 | 12.6693 |

**Answer:** the discharge 1 km downstream of X at 14:00 hrs is $Q_{i+1}^{n+1}\approx12.67$ m$^3$/s.

## Check yourself

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

1. With dt = 15 min and dx = 1 km, dt/dx is:
   A. 0.90
   B. 1.50
   C. 0.15
   D. 9.0
   Answer: A. 900 s / 1000 m = 0.90. (Key point: Time-space ratio)

2. The known right side is:
   A. 0.9(12) + alpha (11)^0.6
   B. 0.9(14) + alpha (11)^0.6
   C. 0.9(14) + alpha (12)^0.6
   D. 0.9(11) + alpha (14)^0.6
   Answer: B. It uses Q_i^{n+1}=14 and Q_{i+1}^n=11, giving 26.1369. (Key point: Right side)

3. Newton iteration gives Q at 1 km downstream at 14:00 as about:
   A. 11.0 m^3/s
   B. 12.67 m^3/s
   C. 14.0 m^3/s
   D. 9.5 m^3/s
   Answer: B. The iteration converges to 12.6693 m^3/s. (Key point: Result)

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