It is possible to formulate equations describing three
conservation laws for quantities that are useful in open-channel flow: mass, momentum, and energy. The governing equations result from considering the dynamics of the
flow velocity vector field {\bf v} with components {\bf v} = \begin{pmatrix} u & v & w \end{pmatrix}^{T}. In
Cartesian coordinates, these components correspond to the flow velocity in the x, y, and z axes respectively. To simplify the final form of the equations, it is acceptable to make several assumptions: • The flow is
incompressible (this is not a good assumption for rapidly-varied flow) • The Reynolds number is sufficiently large such that viscous diffusion can be neglected • The flow is one-dimensional across the x-axis
Continuity equation The general
continuity equation, describing the conservation of mass, takes the form:{\partial \rho\over{\partial t}} + \nabla \cdot (\rho {\bf v}) = 0where \rho is the fluid
density and \nabla \cdot() is the
divergence operator. Under the assumption of incompressible flow, with a constant
control volume V, this equation has the simple expression \nabla \cdot {\bf v} = 0. However, it is possible that the
cross-sectional area A can change with both time and space in the channel. If we start from the integral form of the continuity equation:{d\over{dt}}\int_{V}\rho \; dV = -\int_{V} \nabla\cdot(\rho {\bf v}) \; dVit is possible to decompose the volume integral into a cross-section and length, which leads to the form:{d\over{dt}}\int_{x}\left(\int_{A}\rho \; dA \right) dx = -\int_{x}\left[\int_{A}\nabla\cdot(\rho {\bf v}) \; dA \right] dxUnder the assumption of incompressible, 1D flow, this equation becomes:{d\over{dt}}\int_{x}\left(\int_{A}dA \right) dx = -\int_{x}{\partial\over{\partial x}}\left(\int_{A} u \; dA \right) dxBy noting that \int_{A}dA = A and defining the
volumetric flow rate Q = \int_{A}u \; dA, the equation is reduced to:\int_{x}{\partial A\over{\partial t}} \; dx = -\int_{x}{\partial Q\over{\partial x}} dxFinally, this leads to the continuity equation for incompressible, 1D open-channel flow:{{Equation box 1|cellpadding|border|indent=:|equation= {\partial A\over{\partial t}} + {\partial Q\over{\partial x}} = 0 |border colour=#0073CF|background colour=#F5FFFA}}
Momentum equation The momentum equation for open-channel flow may be found by starting from the
incompressible Navier–Stokes equations :\overbrace{\underbrace{{\partial {\bf v}\over{\partial t}}}_{\begin{smallmatrix} \text{Local} \\ \text{Change} \end{smallmatrix}} + \underbrace{{\bf v}\cdot\nabla {\bf v}}_{\text{Advection}}}^{\text{Inertial Acceleration}} = -\underbrace{{1\over{\rho}}\nabla p}_{\begin{smallmatrix} \text{Pressure} \\ \text{Gradient} \end{smallmatrix}} + \underbrace{\nu \Delta {\bf v}}_{\text{Diffusion}} - \underbrace{\nabla \Phi}_{\text{Gravity}} + \underbrace_{\begin{smallmatrix} \text{External} \\ \text{Forces} \end{smallmatrix}}where p is the
pressure, \nu is the
kinematic viscosity, \Delta is the
Laplace operator, and \Phi = gz is the
gravitational potential. By invoking the high Reynolds number and 1D flow assumptions, we have the equations:\begin{aligned} {\partial u\over{\partial t}} + u{\partial u\over{\partial x}} &= -{1\over{\rho}}{\partial p\over{\partial x}} + F_{x} \\ -{1\over{\rho}}{\partial p\over{\partial z}} - g &= 0 \end{aligned}The second equation implies a
hydrostatic pressure p = \rho g \zeta, where the channel depth \eta(t,x) = \zeta(t,x) - z_{b}(x) is the difference between the free surface elevation \zeta and the channel bottom z_{b}. Substitution into the first equation gives:{\partial u\over{\partial t}} + u{\partial u\over{\partial x}} + g{\partial \zeta\over{\partial x}} = F_{x} \implies {\partial u\over{\partial t}} + u{\partial u\over{\partial x}} + g{\partial \eta\over{\partial x}} - gS = F_{x}where the channel bed slope S = -dz_{b}/dx. To account for shear stress along the channel banks, we may define the force term to be:F_{x} = -{1\over{\rho}}{\tau\over{R}}where \tau is the
shear stress and R is the
hydraulic radius. Defining the friction slope S_{f} = \tau/\rho g R, a way of quantifying friction losses, leads to the final form of the momentum equation:{{Equation box 1|cellpadding|border|indent=:|equation= {\partial u\over{\partial t}} + u{\partial u\over{\partial x}} + g{\partial \eta\over{\partial x}} + g(S_{f}- S) = 0 |border colour=#0073CF|background colour=#F5FFFA}}
Energy equation To derive an
energy equation, note that the advective acceleration term {\bf v}\cdot\nabla {\bf v} may be decomposed as:{\bf v}\cdot\nabla {\bf v} = \omega \times {\bf v} + {1\over{2}}\nabla\|{\bf v}\|^{2}where \omega is the
vorticity of the flow and \|\cdot\| is the
Euclidean norm. This leads to a form of the momentum equation, ignoring the external forces term, given by:{\partial {\bf v}\over{\partial t}} + \omega \times {\bf v} = -\nabla\left({1\over{2}}\|{\bf v}\|^{2} + {p\over{\rho}} + \Phi \right )Taking the
dot product of {\bf v} with this equation leads to:{\partial\over{\partial t}}\left({1\over{2}}\|{\bf v}\|^{2} \right ) + {\bf v}\cdot \nabla \left({1\over{2}}\|{\bf v}\|^{2} + {p\over{\rho}} + \Phi \right ) = 0This equation was arrived at using the
scalar triple product {\bf v}\cdot (\omega \times {\bf v}) = 0. Define E to be the
energy density:E = \underbrace{{1\over{2}}\rho\|{\bf v} \|^{2} }_{\begin{smallmatrix} \text{Kinetic} \\ \text{Energy} \end{smallmatrix}} + \underbrace{\rho\Phi}_{\begin{smallmatrix} \text{Potential} \\ \text{Energy} \end{smallmatrix}}Noting that \Phi is time-independent, we arrive at the equation:{\partial E\over{\partial t}} + {\bf v}\cdot\nabla (E+p) = 0Assuming that the energy density is time-independent and the flow is one-dimensional leads to the simplification:E + p = Cwith C being a constant; this is equivalent to
Bernoulli's principle. Of particular interest in open-channel flow is the
specific energy e = E/\rho g, which is used to compute the
hydraulic head h that is defined as:{{Equation box 1|cellpadding|border|indent=:|equation= \begin{aligned} h &= e + {p\over{\rho g}} \\ &= {u^{2}\over{2g}} + z + {p\over{\gamma}} \end{aligned} |border colour=#0073CF|background colour=#F5FFFA}}with \gamma = \rho g being the
specific weight. However, realistic systems require the addition of a
head loss term h_{f} to account for energy
dissipation due to
friction and
turbulence that was ignored by discounting the external forces term in the momentum equation. == See also ==