SIC^2: Simulation and Integration of Control for Canals

Saint-Venant's equations

David Dorchies — 2009-10-22T15:33:51Z

The canal is divided into homogenous zones, the reaches. We will first examine how to compute the unsteady flow water surface profile in a single reach. We will thereafter examine how to solve the problem for the whole hydraulic network.
The same hypotheses as for Unit 2 are applicable. Furthermore, only smooth transient phenomena are considered. The propagation of a surge cannot be (...)

- Saint-Venant's equations

Continuity equations

David Dorchies — 2009-10-22T15:33:51Z

This equation accounts for the conservation of mass of the water. Consider the variation of the water volume contained between two sections at the abscissa x and x+Dx (see figure 23) during time Dt.

Figure 23

Inflow water mass:

r.Q(x,t).Dt + r.q.Dx.Dt

Outflow water mass:

r.Q(x+Dx,t).Dt

Change in storage:

r Vt+Dt - r Vt = r(A.Dx)t+Dt - r(A.Dx)t

The equation expressing the conservation of mass is written:

r(A.Dx)t+Dt - r(A.Dx)t = r(Q.Dt)x + rq.Dx.Dt - r(Q.Dt)x+Dx

At the limiting conditions, one obtains the equation [28]:

+ = q [28]

Dynamic equation

David Dorchies — 2009-10-22T15:33:51Z

We apply the momentum principle to a volume of water contained between the two sections located at the abscissa x and x+Dx (figure 24).

Consider the projection of the momentum equation on the canal axis:

= S Fext/x [29]

Figure 24

During the time Dt, the water volume V, contained between sections 1 et 2 gets deformed and moves into the volume V' contained between sections 1' et 2'. We have to estimate the variation of momentum projected on the x axis, corresponding to the left hand side of the equation [29].

Estimation of the momentum:

The momentum lost corresponds to the volume V1

i.e. rV(x,t).Dt.A(x,t).V(x,t)

The momentum gained corresponds to the volume V2

i.e. rV(x+Dx,t).Dt.A(x+Dx,t).V(x+Dx,t)

The variation in momentum in the common portion is:

rA(x,t+Dt).Dx.V(x,t+Dt) - rA(x,t).Dx.V(x,t)

The variation of momentum due to lateral inflows or outflows is:

rq(x,t).Dx.Dt.V(x,t)

in the case of an outflow (velocity is equal to the flow velocity), and

+ rq(x,t).Dx.Dt.0

in the case of an inflow (velocity is equal to zero projected on the x axis).

This can be written globally as:

krq(x,t).Dx.Dt.V(x,t)

with:

k = 1 for an outflow
k = 0 for an inflow

Therefore:

D(mVx) = (rV.Dt.A.V) (x+Dx,t) - (rV.Dt.A.V)(x,t) + (rA.Dx.V) (x,t+Dt)

(rA.Dx.V)(x,t) - k(rq.Dx.Dt.V) (x,t)

This can be written as follows:

= rDx [ + - kqV]

It is now necessary to examine the right hand side term of equation [29] which represents the resultant of external forces projected on the x axis. Only the effects of gravity, pressure and friction are considered.

Estimation of external forces:

The resultant of gravity forces is:

Fgx = r.g.A.Dx.sin(S0)

As the flow conditions are quasi-horizontal, sin(S0) » S0 (canal slope).

Therefore, Fgx = rg.A.Dx.S0

The resultant of the pressure forces can be obtained making use of the hydrostatic pressure distribution hypothesis. The resultant of the pressure forces applied on the water mass contained between x and x+Dx (figure 25) is the same as in the static case; that is when the free surface is horizontal. We can thus write:

Figure 25

Fg = - rg.A.Dx = -Fp (resultant of the forces is nil)

Then: |Fp| = rg.A.Dx and Fp is perpendicular to the free surface.

The normal to the free surface has the following components:

since the free surface equation in the system of axes linked to the canal bed is:

h - y (x,t) = 0

One supposes that |n| = 1 because the free surface is almost horizontal. The resultant of the pressure forces is then written as:

Therefore: Fpx = -rg.A.Dx

Friction forces are estimated from the MANNING-STRICKLER formula:

Ffx = - rg.A.Dx.Sf

with: Sf =

The resultant of the external forces projected on the canal axis is then:

S Fext/x = rg.A.Dx.S0 - rg.A.Dx. - rg.A.Dx.Sf

= rg.A.Dx [S0 - Sf - ]

Equation [29] may be written in the following form:

rDx [ + - kqV] = rg.A.Dx [S0 - Sf - ]

Therefore:

+ + g.A = g.A [S0 - Sf] + kqV

This can be written in the form:

+ + g.A = - g.A Sf + kqV [30]

Initial and boundary conditions

David Dorchies — 2009-10-22T15:33:51Z

Partial differential equations must be completed by initial and boundary conditions in order to be solved. The boundary conditions are the hydrographs at the upstream nodes of the reaches and a rating curve at the downstream node of the model (because subcritical flow conditions prevail). The initial condition is the water surface profile resulting from the steady flow computation (Unit (...)

- Saint-Venant's equations