# One-dimensional Saint-Venant equation

In fluid dynamics, the one-dimensional (1-D) Saint-Venant equation was derived by Adhémar Jean Claude Barré de Saint-Venant, and is commonly used to model transient open-channel flow and surface runoff. It is a simplification of the two-dimensional (2-D) shallow water equations, which are also known as the two-dimensional Saint-Venant equations. The 1-D simplification is used exclusively in models including HEC-RAS, SWMM5, InfoWorks, ISIS, Flood Modeller, MIKE 11, and MIKE SHE because it is significantly easier to solve than the full shallow water equations. Common applications of the 1-D Saint-Venant Equation include dam break analyses, storm pulses in an open channel, as well as storm runoff in overland flow.

## Derivation from Navier–Stokes equations

The 1-D Saint-Venant equation can be derived from the Navier–Stokes equations that describe fluid motion. The x-component of the Navier–Stokes equations – when expressed in Cartesian coordinates in the x-direction – can be written as: $\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}+ w \frac{\partial u}{\partial z}= -\frac{\partial p}{\partial x} \frac{1}{\rho} + \nu \left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right)+ f_x,$

where u is the velocity in the x-direction, v is the velocity in the y-direction, w is the velocity in the z-direction, t is time, p is the pressure, ρ is the density of water, ν is the kinematic viscosity, and fx is the body force in the x-direction.

 1 If it is assumed that friction is taken into account as a body force, then ν can be assumed as zero so: $\nu \left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\right)= 0.$ 2 Assuming one-dimensional flow in the x-direction it follows that: $v\frac{\partial u}{\partial y}+ w \frac{\partial u}{\partial z} = 0$ 3 Assuming also that the pressure distribution is approximately hydrostatic it follows that: $p = \rho g h$ or in differential form: $\partial p = \rho g \left(\partial h \right).$ And when these assumptions are applied to the x-component of the Navier–Stokes equations: $-\frac{\partial p}{\partial x} \frac{1}{\rho} = -\frac{1}{\rho}\frac{\rho g \left(\partial h \right)}{\partial x} = -g \frac{\partial h}{\partial x}.$ 4 There are 2 body forces acting on the channel fluid, gravity, and friction: $f_x =f_{x,g} + f_{x,f}$ where fx,g is the body force due to gravity and fx,f is the body force due to friction. 5 fx,g can be calculated using basic physics and trigonometry: $F_{g} = (\sin\theta) gM$ where Fg is the force of gravity in the x-direction, θ is the angle, and M is the mass. File:Block on Inclined Plane with Direction of Gravity, Angle, and Location of Opposite, Adjacent, and Hypotenuse.png Figure 1: Diagram of block moving down an inclined plane. The expression for sin θ can be simplified using trigonometry as: $\sin\theta = \frac{opp}{hyp}.$ For small θ (reasonable for almost all streams) it can be assumed that: $\sin\theta = \tan\theta = \frac{opp}{adj} = S$ and given that fx represents a force per unit mass, the expression becomes: $f_{x,g} = gS.$ 6 Assuming the energy grade line is not the same as the channel slope, and for a reach of consistent slope there is a consistent friction loss, it follows that: $f_{x,f} = S_f g.$ 7 All of these assumptions combined arrives at the 1-dimensional Saint-Venant equation in the x-direction: $\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + g \frac{\partial h}{\partial x} + g (S - S_f) = 0$ $(a)\quad \ \ (b)\quad \ \ \ (c) \qquad (d) \ \ \ (e)$ where (a) is the local acceleration term, (b) is the convective acceleration term, (c) is the pressure gradient term, (d) is the gravity term, and (e) is the friction term.

## Terms

The local acceleration (a) can also be thought of as the “unsteady term” as this describes some change in velocity over time. The convective acceleration (b) is an acceleration caused by some change in velocity over position, for example the speeding up or slowing down of a fluid entering a constriction or an opening, respectively. Both these terms make up the inertia terms of the 1-dimensional Saint-Venant equation.

The pressure gradient term (c) describes how pressure changes with position, and since the pressure is assumed hydrostatic, this is the change in head over position. The gravity term (d) is the acceleration due to slope, while the friction term (e) accounts for losses in energy due to friction

## Common simplifications

### Dynamic wave

The dynamic wave is the term used to describe the full 1-dimensional Saint-Venant equation. It is numerically challenging to solve, but is valid for all channel flow scenarios. The dynamic wave is used for modeling transient storms in modeling programs including HEC-RAS, InfoWorks, MIKE 11, Wash 123d  and SWMM.

### Kinematic wave

For the kinematic wave it is assumed that the flow is uniform, and that the friction slope is approximately equal slope of the channel. This simplifies the full Saint-Venant equation to the kinematic wave: $(S - S_f) = 0.$

The kinematic wave is valid when the change in wave height over distance and velocity over distance and time is negligible relative to the bed slope, e.g. for shallow flows over steep slopes. The kinematic wave is used in HEC-HMS.

### Diffusive wave

For the diffusive wave it is assumed that the inertial terms are less than the gravity, friction, and pressure terms. The diffusive wave can therefore be more accurately described as a non-inertia wave, and is written as: $g \frac{\partial h}{\partial x} + g (S - S_f) = 0.$

The diffusive wave is valid when the inertial acceleration is much smaller than all other forms of acceleration, or in other words when there is primarily subcritical flow. Models that use the diffusive wave assumption include MIKE SHE and LISFLOOD-FP.

## Example

Following is an example solution of the kinematic wave using an explicit finite difference algorithm.

### Setup

Multiplying the kinematic wave equation by the area of the channel, it follows that: $\frac{\partial Q}{\partial t} + V_w \frac{\partial Q}{\partial x} = 0$

where Q is the channel flow, and Vw is the velocity of wave propagation.

Vw is the celerity, and is equal to the derivative of the discharge with respect to the depth, normalized by the channel topwidth (T): $V_w = \frac{\frac{dQ}{dh}}{T} = \frac{1}{T} \frac{d}{dh} \left(\frac{1.0}{n}AR^{2/3}S^{1/2} \right)$

where n is Manning’s roughness coefficient, and R is the hydraulic radius.

Assuming a wide channel, the hydraulic radius is approximately the same as the channel depth, simplifying the equation above to: $V_w = \frac{5}{3}V.$

To describe the stability of the finite difference scheme the Courant number (C) is defined as: $C = V_w \frac{\Delta t}{\Delta x}$

where Δt is the time discretization, and Δx is the space discretization.

### Solution

File:Illustration of explicit iterative scheme with distance (x) on the x axis and time (t) on the y axis. The known points (dark grey) are used to calculate the unknown point (red).png
Figure 2: Illustration of explicit iterative scheme with distance (x) on the horizontal axis and time (t) on the vertical axis. The known points (dark grey) are used to calculate the unknown point (red).
File:Table for 1-D Saint Venant equation.png
Table 1: Flow prediction (Q, m³/s) using an explicit finite difference formulation of the kinematic wave with a Courant number of 0.9.

Suppose Vw is 1 m/s, the flow of interest is at 200 m intervals over a 1000 m reach, the initial flow for the entire reach is 3 m3/s, and an estimate is used for the hydrograph at location "0". Then the kinematic wave can be used to estimate the flow over time. If C is less than 1 there is some numerical dispersion but the solution is stable, but if C is greater than 1 the solution is numerically unstable. The case of C = 0.9 yields a Δt of 180 seconds, or 0.05 hours.

The notation used for solving the kinematic wave is: $Q^{j+1}_i$

where $Q^j_i$ is the flow at location i at time j. Note that the j-term is not an exponent.

An explicit solution algorithm is used, with all of the flow values known for the initial conditions (t = 0), and at the upstream boundary (x = 0) for all values of t. The solution is explicit because known flows upstream at the previous time step (i-1, j) and at the same location at the previous timestep (i,j) are used to calculate the flow for the next timestep (j+1).

If the forward difference approximation is used then: $Q^{j+1}_i = \left(1-C\right) Q^j_i + C\left(Q^j_{i-1}\right).$

Using this methodology a storm can be routed through the system, as displayed in Table 1.

The hydrographs at the upstream boundary and the downstream boundary of the routing channel are compared in figure 3.

To illustrate how the forward difference algorithm works, the flow at the location 400 m downstream of the upstream boundary after 0.5 hours will be illustrated (see highlighted entries in Table 1).

From table 1 it follows that: $Q^j_i = 9.6 \text{ m}^3/\text{s}$ $Q^j_{i-1} = 10.3 \text{ m}^3/\text{s}$

Therefore: $Q^{j+1}_i = \left(1-0.9\right) 9.6 + 0.9\left(10.3\right) = 10.2 \text{ m}^3/\text{s}.$

### Courant number analysis

To demonstrate the effect of the Courant number (C) on the predicted hydrograph, the routing is repeated for various C values and the hydrograph at the downstream boundary ( x = 1000 m) is displayed. The results for a C of 0.7, 0.8, 0.9 and 1.0 are displayed in figure 4, and the results for a C of 1.0, 1.1, 1.2, and 1.3 are displayed in figure 5. When using an explicit algorithm, if C is less than 1 numerical dispersion occurs; however, the results are stable. This is illustrated by the hydrographs varying more as C decreases. When using an explicit algorithm, if C is greater than 1 numerical instabilities occur and the results become unstable. This is illustrated by the increasingly oscillatory behavior in the downstream hydrograph as C increases.

File:Flow (Q) at the downstream (x = 1000 m) end of the example routing channel for Courant Numbers of 1 or less (a), and a zoom in at the hydrograph peaks (b). A forward difference explicit algorithm is used to solve the kinematic wave.png.svg
Figure 4: Flow at the downstream (x = 1000 m) end of the example routing channel for Courant Numbers of 1 or less (a), and a zoom in at the hydrograph peaks (b).
File:Flow (Q) at the downstream (x = 1000 m) end of the example routing channel for Courant Numbers of 1 or greater (a), and a zoom in at the hydrograph peaks (b). This is for a forward difference explicit approximation of the kinematic wave.svg
Figure 5: Flow at the downstream (x = 1000 m) end of the example routing channel for Courant numbers of 1 or greater (a), and a zoom in at the hydrograph peaks (b).