--- title: "Saint-Venant velocity and depth at (i, j+1)" subject: "Computational Techniques in Civil Engineering" chapter: 5 key: "semester_8/computational_techniques_in_civil_engineering:saint-venant-velocity-depth-b5" source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/saint-venant-velocity-depth-b5 site: BCT Engineering --- # Saint-Venant velocity and depth at (i, j+1) Computational Techniques in Civil Engineering · Chapter 5: Finite difference method - Asked in: 2075 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/saint-venant-velocity-depth-b5 - Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-5 ## Question A finite difference grid of points constructed to solve for the unsteady flow problems in a wide rectangular channel is shown in the figure below. Using an appropriate finite difference scheme for the two governing equations of fluid flow (continuity and momentum), compute the velocity and flow depth at grid point $(i, j+1)$ for the following given data. Velocity: $V_{i-1}^j = 2.2$ m/s, $V_i^j = 1.8$ m/s, $V_{i+1}^j = 1.5$ m/s. Flow depth: $y_{i-1}^j = 1.6$ m, $y_i^j = 2.0$ m, $y_{i+1}^j = 2.4$ m (units printed as m/sec). Bed slope $= 1\%$, Manning's $n = 0.032$, $\Delta x = 1000$ m, $\Delta t = 4$ minutes, no lateral inflow. [Figure: grid with points $i-1$, $i$, $i+1$ at time level $j$ (spacing $\Delta x$) and the point $i$ at time level $j+1$ (spacing $\Delta t$)] ## Short answer Using the Lax diffusive explicit scheme (average neighbours, central differences in space), flow depth is y=1.9952 m and velocity V=21.50 m/s. ## Answer For a wide rectangular channel the two governing equations in terms of velocity $V$ and depth $y$ are $$ \frac{\partial y}{\partial t}+V\frac{\partial y}{\partial x}+y\frac{\partial V}{\partial x}=0,\qquad \frac{\partial V}{\partial t}+V\frac{\partial V}{\partial x}+g\frac{\partial y}{\partial x}=g(S_0-S_f),\quad S_f=\frac{n^2V^2}{y^{4/3}} $$ #### Scheme (explicit, Lax diffusive form) The old value at $i$ is replaced by the average of its neighbours, and central differences are used in space: $$ y_i^{j+1}=\frac{y_{i+1}^j+y_{i-1}^j}{2}-\frac{\Delta t}{2\Delta x}\left[V_i^j\left(y_{i+1}^j-y_{i-1}^j\right)+y_i^j\left(V_{i+1}^j-V_{i-1}^j\right)\right] $$ $$ V_i^{j+1}=\frac{V_{i+1}^j+V_{i-1}^j}{2}-\frac{\Delta t}{2\Delta x}\left[V_i^j\left(V_{i+1}^j-V_{i-1}^j\right)+g\left(y_{i+1}^j-y_{i-1}^j\right)\right]+g\Delta t\,(S_0-S_{f,i}^j) $$ #### Data $\Delta t=4\times60=240$ s, $\Delta x=1000$ m, so $\dfrac{\Delta t}{2\Delta x}=0.120$. $S_0=0.01$, $n=0.032$. #### Step 1: Friction slope at $i$ $$ S_f=\frac{n^2V_i^2}{y_i^{4/3}}=\frac{0.032^2(1.8)^2}{2.0^{4/3}}=0.001317 $$ #### Step 2: Depth $$ y_i^{j+1}=\frac{2.4+1.6}{2}-0.120\left[1.8(2.4-1.6)+2.0(1.5-2.2)\right]=2.0-0.120\,(0.04)=1.9952\ \text{m} $$ #### Step 3: Velocity - Average: $(1.5+2.2)/2=1.85$ - Convective: $-0.120\,[1.8(1.5-2.2)]=0.1512$ - Pressure: $-0.120\,[9.81(2.4-1.6)]=-0.9418$ - Gravity and friction: $9.81(240)(0.01-0.001317)=20.4441$ $$ V_i^{j+1}=1.85+(0.1512)+(-0.9418)+(20.4441)=21.504\ \text{m/s} $$ **Answer:** $y_i^{j+1}=1.9952$ m and $V_i^{j+1}=21.50$ m/s. **Remark:** the very large velocity arises because the gravity term ($S_0=1\%$) is far from balanced by friction and the pressure gradient for these data, and $\Delta t=240$ s exceeds the Courant limit $\Delta x/(V+\sqrt{gy})\approx161$ s. The values are the one-step result of the scheme; a real simulation would use a smaller $\Delta t$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the Lax diffusive scheme used here, the old value at point i is: A. kept unchanged B. replaced by the average of the neighbours at i-1 and i+1 C. replaced by the value at i+1 only D. replaced by the value at j+1 Answer: B. The term (y_{i+1}+y_{i-1})/2 replaces y_i, which stabilises the explicit scheme. (Key point: Lax scheme) 2. For the wide rectangular channel, the friction slope is: A. n V / y^(2/3) B. n^2 V^2 / y^(4/3) C. n^2 V / y^(4/3) D. V^2 / (g y) Answer: B. With R = y, Manning gives Sf = n^2 V^2 / y^(4/3). (Key point: Friction slope) 3. Why is the computed velocity (21.5 m/s) unrealistically large? A. Gravity term g dt (S0 - Sf) is large and dt exceeds the Courant limit B. Manning's n was too small to use C. Depth was computed wrongly D. Central differences cannot be used for velocity Answer: A. The gravity term is about 20.4 m/s and dt = 240 s is above the limit of about 161 s. 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