--- title: "Full Saint-Venant explicit scheme numerical (50 m)" subject: "Computational Techniques in Civil Engineering" chapter: 5 key: "semester_8/computational_techniques_in_civil_engineering:saint-venant-explicit-k8" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/saint-venant-explicit-k8 site: BCT Engineering --- # Full Saint-Venant explicit scheme numerical (50 m) Computational Techniques in Civil Engineering · Chapter 5: Finite difference method - Asked in: 2078 Kartik, 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/saint-venant-explicit-k8 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5 ## Question Using any explicit finite difference scheme for the full Saint-Venant equations, compute discharge and flow depth at grid $(i, n+1)$ for the following data. Rectangular channel width $= 50$ m, bed slope $= 0.0002$, Manning's $n = 0.035$, no lateral inflow, $\Delta x = 1.0$ km and $\Delta t = 10$ min. Discharge: $Q_{i-1}^n = 38$ m$^3$/s, $Q_i^n = 36$ m$^3$/s, $Q_{i+1}^n = 35$ m$^3$/s. Flow depth: $y_{i-1}^n = 1.87$ m, $y_i^n = 1.82$ m, $y_{i+1}^n = 1.97$ m. ## Short answer Using Lax diffusive explicit scheme: y_i^{n+1}= (1.97+1.87)/2 - Δt/(2BΔx)(35-38)=1.9380 m. Discharge Q_i^{n+1} from average (35+38)/2=36.5, plus convective term 0.902, minus pressure term 26.78, plus gravity-friction 56.37, total 66.99 m³/s. ## Answer A simple explicit scheme is the **Lax diffusive scheme**: forward in time, central in space, with the old value at $i$ replaced by the average of its two neighbours. The equations for a rectangular channel are $$ B\frac{\partial y}{\partial t}+\frac{\partial Q}{\partial x}=0,\qquad \frac{\partial Q}{\partial t}+\frac{\partial}{\partial x}\left(\frac{Q^2}{A}\right)+gA\frac{\partial y}{\partial x}-gA(S_0-S_f)=0 $$ Finite difference form (grid $(i,n+1)$ from $(i-1,n),(i,n),(i+1,n)$): $$ y_i^{n+1}=\frac{y_{i+1}^n+y_{i-1}^n}{2}-\frac{\Delta t}{2B\Delta x}\left(Q_{i+1}^n-Q_{i-1}^n\right) $$ $$ Q_i^{n+1}=\frac{Q_{i+1}^n+Q_{i-1}^n}{2}-\frac{\Delta t}{2\Delta x}\left[\frac{(Q^2)_{i+1}}{A_{i+1}}-\frac{(Q^2)_{i-1}}{A_{i-1}}\right]-gA_i\frac{\Delta t}{2\Delta x}\left(y_{i+1}-y_{i-1}\right)+gA_i\Delta t\,(S_0-S_{f,i}) $$ #### Data $B=50$ m, $S_0=0.0002$, $n=0.035$, $\Delta x=1000$ m, $\Delta t=600$ s, so $\dfrac{\Delta t}{2\Delta x}=0.300$. #### Step 1: Areas and friction slope $A_{i-1}=93.50$, $A_i=91.00$, $A_{i+1}=98.50$ m$^2$. $$ P_i=B+2y_i=53.64\ \text{m},\quad R_i=\frac{A_i}{P_i}=1.6965\ \text{m},\quad S_{f,i}=\frac{n^2Q_i^2}{A_i^2R_i^{4/3}}=9.475\times10^{-5} $$ #### Step 2: Flow depth $$ y_i^{n+1}=\frac{1.97+1.87}{2}-0.00600\,(35-38)=1.9200-(-0.0180)=1.9380\ \text{m} $$ #### Step 3: Discharge - Average term: $\dfrac{35+38}{2}=36.500$ - Convective term: $\dfrac{Q^2}{A}$ at $i+1$ is 12.437, at $i-1$ is 15.444; difference $=-3.0073$; term $=-0.300(-3.0073)=0.902$ - Pressure term: $-gA_i\dfrac{\Delta t}{2\Delta x}(y_{i+1}-y_{i-1})=-9.81(91.00)(0.300)(0.10)=-26.781$ - Gravity and friction: $gA_i\Delta t(S_0-S_f)=9.81(91.00)(600)(0.0002-9.475\times10^{-5})=56.374$ $$ Q_i^{n+1}=36.500+(0.902)+(-26.781)+(56.374)=66.99\ \text{m}^3/\text{s} $$ **Answer:** $y_i^{n+1}=1.9380$ m and $Q_i^{n+1}=66.99$ m$^3$/s. **Remark:** the large jump in $Q$ comes mainly from the gravity-friction term ($56.4$): the given flow is far from uniform ($S_f$ is less than half of $S_0$), and this imbalance is multiplied by a long time step. Also, $\Delta t=600$ s gives a Courant number $(V+\sqrt{gy})\Delta t/\Delta x\approx2.8>1$, so the explicit scheme is outside its stability limit; a time step below about 216 s would be needed to continue the simulation. One step is computed here, as asked. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What characterises the Lax diffusive explicit scheme? A. Backward in time and upwind in space B. Central in time, forward in space C. Implicit in time D. Forward in time, central in space, with the old value at i replaced by the average of its neighbours Answer: D. The Lax scheme uses (f_{i+1} + f_{i-1})/2 instead of f_i^n, which stabilises forward-time central-space. (Key point: Lax scheme) 2. What is y_i^{n+1} from the continuity equation? A. 1.9380 m B. 1.9200 m C. 1.8200 m D. 1.8620 m Answer: A. y = (1.97 + 1.87)/2 - 0.006(35 - 38) = 1.92 + 0.018 = 1.938 m. (Key point: Flow depth) 3. Which terms are added to the neighbour average to get Q_i^{n+1}? A. Only the friction term B. Convective, pressure and gravity-minus-friction terms C. Only the pressure term D. Only the local acceleration Answer: B. The convective, pressure (-gA dt/2dx (y_{i+1} - y_{i-1})) and gA dt (S_0 - S_f) terms are added to the average. (Key point: Momentum terms) 4. Why is the large jump in Q (to 66.99 m^3/s) suspicious? A. Because the channel is rectangular B. Because Manning n is too small C. Because dx is too large D. Because the Courant number (V + sqrt(gy)) dt/dx is about 2.8 > 1, outside the stability limit Answer: D. dt = 600 s is too large for dx = 1 km, so the explicit scheme is unstable and a smaller dt is needed. 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