# Dynamic equation

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:

n

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:

Fp

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]**