--- title: "Full Saint-Venant explicit scheme numerical (10 m)" subject: "Computational Techniques in Civil Engineering" chapter: 5 key: "semester_8/computational_techniques_in_civil_engineering:saint-venant-explicit-m3" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/saint-venant-explicit-m3 site: BCT Engineering --- # Full Saint-Venant explicit scheme numerical (10 m) Computational Techniques in Civil Engineering · Chapter 5: Finite difference method - Asked in: 2073 Magh, 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-m3 - 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 $= 10$ m, bed slope $= 0.0002$, Manning's $n = 0.04$, no lateral flow, $\Delta x = 1$ km and $\Delta t = 5$ min. Discharge: $Q_{i-1}^n = 40$ m$^3$/s, $Q_i^n = 38$ m$^3$/s, $Q_{i+1}^n = 37.5$ m$^3$/s. Flow depth: $y_{i-1}^n = 1.9$ m, $y_i^n = 1.85$ m, $y_{i+1}^n = 2.0$ m. ## Short answer Applying the Lax explicit scheme to Saint-Venant equations gives depth y=1.9875 m and discharge Q=−197.25 m³/s at (i,n+1). The negative Q arises from friction-gravity imbalance and a time step exceeding the Courant limit. ## 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 $B=10$ m, $S_0=0.0002$, $n=0.04$, $\Delta x=1000$ m, $\Delta t=300$ s, so $\dfrac{\Delta t}{2\Delta x}=0.150$. #### Step 1: Areas and friction slope $A_{i-1}=19.00$, $A_i=18.50$, $A_{i+1}=20.00$ m$^2$. $$ P_i=B+2y_i=13.70\ \text{m},\quad R_i=\frac{A_i}{P_i}=1.3504\ \text{m},\quad S_{f,i}=\frac{n^2Q_i^2}{A_i^2R_i^{4/3}}=0.00452 $$ #### Step 2: Flow depth $$ y_i^{n+1}=\frac{2.0+1.9}{2}-0.01500\,(37.5-40)=1.9500-(-0.0375)=1.9875\ \text{m} $$ #### Step 3: Discharge - Average term: $\dfrac{37.5+40}{2}=38.750$ - Convective term: $\dfrac{Q^2}{A}$ at $i+1$ is 70.312, at $i-1$ is 84.211; difference $=-13.8980$; term $=-0.150(-13.8980)=2.085$ - Pressure term: $-gA_i\dfrac{\Delta t}{2\Delta x}(y_{i+1}-y_{i-1})=-9.81(18.50)(0.150)(0.10)=-2.722$ - Gravity and friction: $gA_i\Delta t(S_0-S_f)=9.81(18.50)(300)(0.0002-0.00452)=-235.358$ $$ Q_i^{n+1}=38.750+(2.085)+(-2.722)+(-235.358)=-197.25\ \text{m}^3/\text{s} $$ **Answer:** $y_i^{n+1}=1.9875$ m and $Q_i^{n+1}=-197.25$ m$^3$/s. **Remark:** the huge change in $Q$ (even to a negative value) comes from the friction term: $S_f=0.00452$ is about 23 times $S_0$, so the given flow is far from uniform, and this imbalance is multiplied by a long time step ($-235.4$). Also, $\Delta t=300$ s gives a Courant number $(V+\sqrt{gy})\Delta t/\Delta x\approx1.9>1$, so the explicit scheme is outside its stability limit; a time step below about 158 s would be needed to continue the simulation. A negative discharge has no physical meaning here; it is simply the one-step result of the scheme for these data. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The explicit scheme used is the: A. Lax diffusive scheme: forward in time, central in space with neighbour averaging B. fully implicit backward scheme C. Newton-Raphson scheme D. method of characteristics Answer: A. y and Q at (i, n+1) come from i-1, i and i+1 at level n. (Key point: Scheme) 2. With B = 10 m, dt/(2dx)=0.15, the new depth is: A. 1.9500 m B. 1.9875 m C. 1.9125 m D. 2.0250 m Answer: B. y = 1.95 - (0.15/10)(37.5-40) = 1.95 + 0.0375 = 1.9875 m. (Key point: Depth result) 3. Why is the discharge change so large? A. Courant number is about 1.9 > 1 so the step is outside the stability limit B. Width was ignored C. Manning's n is too large D. Lateral inflow was included Answer: A. dt = 300 s is too large; a smaller dt is needed for continued simulation. 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