--- title: "Linear kinematic wave, 30 m channel" subject: "Computational Techniques in Civil Engineering" chapter: 5 key: "semester_8/computational_techniques_in_civil_engineering:kinematic-wave-linear-30m" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/kinematic-wave-linear-30m site: BCT Engineering --- # Linear kinematic wave, 30 m channel Computational Techniques in Civil Engineering · Chapter 5: Finite difference method - Asked in: 2077 Chaitra, 6 marks - Video: [Hydraulic Wave Models Explained | Kinematic, Diffusion, and Dynamic Wave](https://www.youtube.com/watch?v=dzXAdsTCjuk) — 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-linear-30m - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5 ## Question For a 30 m wide and 0.015 bed slope rectangular channel the following flow rates are given: $Q_i^n = 20$ m$^3$/s, $Q_i^{n+1} = 28$ m$^3$/s and $Q_{i+1}^n = 18$ m$^3$/s. Taking Manning's $n = 0.025$, $\Delta x = 1200$ m and $\Delta t = 10$ min, determine $Q_{i+1}^{n+1}$ using the finite difference scheme for the linear kinematic wave model. Assume lateral inflow to be zero. Take wetted perimeter approximately equal to the width of the channel. ## Short answer From Manning with P≈B, representative Q=22 m³/s gives y=0.320 m, V=2.29 m/s, celerity c=3.82 m/s, C=1.91. Using linear kinematic wave four-point scheme: Q_{i+1}^{n+1}=20+(1-1.91)/(1+1.91)*(18-28)=23.13 m³/s. ## Answer The linear kinematic wave equation $\dfrac{\partial Q}{\partial t}+c\dfrac{\partial Q}{\partial x}=0$ is solved with the second-order four-point scheme using the three known values $Q_i^n,\ Q_i^{n+1},\ Q_{i+1}^n$. #### Step 1: Celerity (wide channel, $P\approx B$) Manning: $Q=\dfrac{1}{n}S_0^{1/2}B\,y^{5/3}$ (with $A=By$, $R\approx y$). A representative discharge is taken as the mean of the three known values: $Q=(20+28+18)/3=22$ m$^3$/s. $$ y=\left(\frac{nQ}{S_0^{1/2}B}\right)^{3/5}=0.3200\ \text{m},\qquad V=\frac{Q}{By}=2.2918\ \text{m/s} $$ $$ c=\frac{dQ}{dA}=\frac53V=3.8197\ \text{m/s},\qquad C=\frac{c\Delta t}{\Delta x}=\frac{3.8197\times600}{1200}=1.9099 $$ #### Step 2: Scheme $$ \frac{Q_i^{n+1}+Q_{i+1}^{n+1}-Q_i^n-Q_{i+1}^n}{2\Delta t}+c\,\frac{Q_{i+1}^{n+1}+Q_{i+1}^n-Q_i^{n+1}-Q_i^n}{2\Delta x}=0 $$ $$ Q_{i+1}^{n+1}=Q_i^n+\frac{1-C}{1+C}\left(Q_{i+1}^n-Q_i^{n+1}\right) $$ #### Step 3: Substitution $$ Q_{i+1}^{n+1}=20+\frac{1-1.9099}{1+1.9099}(18-28)=20+(-0.3127)(-10)=23.13\ \text{m}^3/\text{s} $$ **Answer:** $Q_{i+1}^{n+1}\approx23.13$ m$^3$/s. (Because $C>1$ the four-point scheme is still stable, as it is unconditionally stable; the value depends on the celerity assumed, here from the mean of the known flows.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a wide rectangular channel with Manning's equation, the kinematic wave celerity c = dQ/dA equals: A. V B. (5/3) V C. (2/3) V D. sqrt(g y) Answer: B. Since Q is proportional to A^(5/3), dQ/dA = (5/3)Q/A = (5/3)V. (Key point: Celerity) 2. With c = 3.8197 m/s, dt = 600 s and dx = 1200 m, the Courant number C is: A. 0.95 B. 1.91 C. 3.82 D. 0.52 Answer: B. C = c dt/dx = 3.8197 x 600/1200 = 1.91. (Key point: Courant number) 3. Which formula gives Q(i+1, n+1) in the four-point scheme with zero lateral inflow? A. Q_i^n + (1-C)/(1+C) (Q_{i+1}^n - Q_i^{n+1}) B. Q_i^n - C (Q_i^n - Q_{i-1}^n) C. (Q_{i+1}^n + C Q_i^{n+1})/(1+C) D. Q_i^{n+1} + C (Q_{i+1}^n - Q_i^n) Answer: A. Centring both derivatives on the cell and solving gives this expression, which gave 23.13 m^3/s. 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