---
title: "Non-linear kinematic wave, trapezoidal river"
subject: "Computational Techniques in Civil Engineering"
chapter: 5
key: "semester_8/computational_techniques_in_civil_engineering:kinematic-wave-trapezoidal-river"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-trapezoidal-river
site: BCT Engineering
---

# Non-linear kinematic wave, trapezoidal river

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

- Asked in: 2079 Shrawan, 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-trapezoidal-river
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5

## Question

A river which can be generalized as a trapezoidal channel is 350 m wide with side slope 6:1 and a bed slope 1.5% and Manning's $n = 0.0359$. The initial discharge through the river is 420 cumecs. Due to a flood observed at upstream, the value of discharge rises to 580 cumecs. Calculate the discharge that will occur at 4.65 km downstream. Take $\Delta x = 4650$ m and take $\Delta t = 1.5$ hours. Use the non-linear kinematic wave solution.

## Short answer

Using the nonlinear kinematic wave model with second-order four-point scheme: channel geometry yields α=5.0280, β=0.6. With upstream flow rising from 420 to 580 m³/s, Δx=4650 m, Δt=5400 s, Newton iteration solves (Δt/Δx)Q+αQ^0.6=K, giving Q=522.6 m³/s at 4.65 km downstream after 1.5 h.

## Answer

The non-linear kinematic wave model is used with $A=\alpha Q^\beta$ and the second-order four-point scheme (centred in space and time), solved by Newton's method.

#### Step 1: Channel properties at the mean flow
Mean discharge $=(420+580)/2=500$ m$^3$/s. Trapezoid: bottom width $b=350$ m, side slope $z=6$, $S_0=0.015$, $n=0.0359$.
$$
Q=\frac{S_0^{1/2}}{n}\frac{A^{5/3}}{P^{2/3}},\quad A=(b+zy)y,\quad P=b+2y\sqrt{1+z^2}
$$
Solving for $Q=500$: $y=0.592$ m, $A=209.30$ m$^2$, $P=357.20$ m.

#### Step 2: Kinematic coefficients
$$
\alpha=\left(\frac{nP^{2/3}}{S_0^{1/2}}\right)^{0.6}=5.0280,\qquad \beta=0.6
$$
Then $A_i^n=\alpha(420)^{0.6}=188.51$ m$^2$ and $A_i^{n+1}=\alpha(580)^{0.6}=228.80$ m$^2$. (Wave celerity $c\approx3.98$ m/s, so $C=c\Delta t/\Delta x\approx4.6>1$: an explicit scheme would be unstable, the four-point scheme is not.)

#### Step 3: Finite difference equation
Known: $Q_i^n=420$ (initial flow), $Q_{i+1}^n=420$, $Q_i^{n+1}=580$ (upstream flood). $\Delta x=4650$ m, $\Delta t=1.5\times3600=5400$ s, $\Delta t/\Delta x=1.1613$, no lateral inflow.
$$
\frac{\Delta t}{\Delta x}x+\alpha x^{0.6}=K,\qquad
K=\alpha\left[420^{0.6}+420^{0.6}-580^{0.6}\right]-\frac{\Delta t}{\Delta x}\left(420-580-420\right)=821.779
$$
with $x=Q_{i+1}^{n+1}$.

#### Step 4: Newton iteration
$f(x)=1.1613x+5.0280x^{0.6}-821.779$, $f'(x)=1.1613+3.0168x^{-0.4}$.

| Iteration | $x_k$ | $f(x_k)$ | $x_{k+1}$ |
|---|---|---|---|
| 1 | 420.0000 | -145.5231 | 521.7220 |
| 2 | 521.7220 | -1.1957 | 522.5711 |
| 3 | 522.5711 | -0.0001 | 522.5711 |
| 4 | 522.5711 | -0.0000 | 522.5711 |

**Answer:** $Q_{i+1}^{n+1}=522.6$ m$^3$/s at 4.65 km downstream after 1.5 h (it rises from 420 towards the upstream value of 580 m$^3$/s).

## Check yourself

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

1. How are alpha and beta found for the non-linear kinematic model?
   A. alpha = (n P^(2/3) / S_0^(1/2))^0.6, beta = 0.6, using P at the mean discharge
   B. alpha = n/S_0, beta = 5/3
   C. alpha = S_0 / n, beta = 2
   D. alpha = 1, beta = 0.6 for any channel
   Answer: A. From Manning, A = alpha Q^0.6 with alpha as given, evaluated at the mean flow (500 m^3/s). (Key point: Kinematic coefficients)

2. Why is the four-point scheme used rather than an explicit one here?
   A. Because the channel is trapezoidal
   B. Because C = c dt/dx is about 4.6 > 1, so an explicit scheme would be unstable
   C. Because dt is very small
   D. Because C is less than 1
   Answer: B. With c about 3.98 m/s, dt = 5400 s and dx = 4650 m, C is about 4.6, beyond the explicit limit. (Key point: Courant number)

3. What equation is solved by Newton's method and what starting value is used?
   A. f(x) = x - K, x0 = 0
   B. f(x) = (dt/dx) x + alpha x^0.6 - K = 0, x0 = Q_{i+1}^n = 420
   C. f(x) = alpha x^0.6, x0 = 580
   D. f(x) = x^2 - K, x0 = 1
   Answer: B. The unknown x = Q_{i+1}^{n+1} satisfies the non-linear equation, starting from the known value 420 m^3/s. (Key point: Newton iteration)

4. What discharge is obtained at 4.65 km downstream after 1.5 h?
   A. 580 m^3/s
   B. 420 m^3/s
   C. 500 m^3/s
   D. 522.6 m^3/s
   Answer: D. Newton converges to 522.57 m^3/s, between the initial 420 and upstream 580 m^3/s. (Key point: Result)

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