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Complete governing equations of fluid movement

Short answer

The 1D unsteady Saint-Venant equations: continuity ∂A/∂t+∂Q/∂x=q, momentum ∂Q/∂t+∂(Q²/A)/∂x+gA∂y/∂x−gAS₀+gAS_f=0, with S_f=n²Q|Q|/(A²R^{4/3}). In wide rectangular form: ∂y/∂t+V∂y/∂x+y∂V/∂x=0, ∂V/∂t+V∂V/∂x+g∂y/∂x=g(S₀−S_f).

Write down the complete governing equations describing the movement of fluid.

The movement of water in an open channel (1D unsteady flow) is described by the Saint-Venant equations, a pair of equations from conservation of mass and momentum.

Continuity (conservation of mass):

∂A∂t+∂Q∂x=q\frac{\partial A}{\partial t}+\frac{\partial Q}{\partial x}=q

Momentum (conservation of momentum):

∂Q∂t⏟local acc.+∂∂x ⁣(Q2A)⏟convective acc.+gA∂y∂x⏟pressure force−gAS0⏟gravity+gASf⏟friction=0\underbrace{\frac{\partial Q}{\partial t}}_{\text{local acc.}}+\underbrace{\frac{\partial}{\partial x}\!\left(\frac{Q^2}{A}\right)}_{\text{convective acc.}}+\underbrace{gA\frac{\partial y}{\partial x}}_{\text{pressure force}}-\underbrace{gAS_0}_{\text{gravity}}+\underbrace{gAS_f}_{\text{friction}}=0

where AA = flow area, QQ = discharge, yy = flow depth, qq = lateral inflow per unit length, S0S_0 = bed slope, Sf=n2Q∣Q∣A2R4/3S_f=\dfrac{n^2Q|Q|}{A^2R^{4/3}} = friction slope (Manning), gg = gravity. These are the Saint-Venant equations (1D unsteady open-channel flow).

In terms of velocity VV and depth yy for a wide rectangular channel:

∂y∂t+V∂y∂x+y∂V∂x=0,∂V∂t+V∂V∂x+g∂y∂x=g(S0−Sf)\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)

Assumptions: one-dimensional flow, hydrostatic pressure, small bed slope, uniform velocity over the section, and friction given by the steady-flow (Manning) formula. The two equations have two unknowns, QQ (or VV) and yy (or AA), as functions of xx and tt.

✓ Answer checked by Claude Opus 5.5 (high effort)

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