Open-channel flow: Difference between revisions

Revision as of 23:35, 4 April 2022

Revision as of 21:14, 24 February 2023

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=== Continuity equation ===

The general [[continuity equation]], describing the conservation of mass, takes the form:<math display="block">{\partial \rho\over{\partial t}} + \nabla \cdot (\rho {\bf v}) = 0</math>where <math>\rho</math> is the fluid [[density]] and <math>\nabla \cdot()</math> is the [[divergence]] operator. Under the assumption of incompressible flow, with a constant [[control volume]] <math>V</math>, this equation has the simple expression <math>\nabla \cdot {\bf v} = 0</math>. However, it is possible that the [[Cross section (geometry)|cross-sectional area]] <math>A</math> can change with both time and space in the channel. If we start from the integral form of the continuity equation:<math display="block">{d\over{dt}}\int_{V}\rho \; dV = -\int_{V} \nabla\cdot(\rho {\bf v}) \; dV</math>it is possible to decompose the volume integral into a cross-section and length, which leads to the form:<math display="block">{d\over{dt}}\int_{x}\left(\int_{A}\rho \; dA \right) dx = -\int_{x}\left[\int_{A}\nabla\cdot(\rho {\bf v}) \; dA \right] dx</math>Under the assumption of incompressible, 1D flow, this equation becomes:<math display="block">{d\over{dt}}\int_{x}\left(\int_{A}dA \right) dx = -\int_{x}{\partial\over{\partial x}}\left(\int_{A} u \; dA \right) dx</math>By noting that <math>\int_{A}dA = A</math> and defining the [[volumetric flow rate]] <math>Q = \int_{A}u \; dA</math>, the equation is reduced to:<math display="block">\int_{x}{\partial A\over{\partial t}} \; dx = -\int_{x}{\partial Q\over{\partial x}} dx</math>Finally, this leads to the continuity equation for incompressible, 1D open-channel flow:<math{{Equation displaybox 1|cellpadding|border|indent="block":|equation=<math> {\partial A\over{\partial t}} + {\partial Q\over{\partial x}} = 0 </math>|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|incompressible Navier-Stokes equations]] :<math display="block">\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{{\bf F}}_{\begin{smallmatrix} \text{External} \\ \text{Forces} \end{smallmatrix}}</math>where <math>p</math> is the [[pressure]], <math>\nu</math> is the [[kinematic viscosity]], <math>\Delta</math> is the [[Laplace operator]], and <math>\Phi = gz</math> is the [[gravitational potential]]. By invoking the high Reynolds number and 1D flow assumptions, we have the equations:<math display="block">\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}</math>The second equation implies a [[hydrostatic pressure]] <math>p = \rho g \zeta</math>, where the channel depth <math>\eta(t,x) = \zeta(t,x) - z_{b}(x)</math> is the difference between the free surface elevation <math>\zeta</math> and the channel bottom <math>z_{b}</math>. Substitution into the first equation gives:<math display="block">{\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}</math>where the channel bed slope <math>S = -dz_{b}/dx</math>. To account for shear stress along the channel banks, we may define the force term to be:<math display="block">F_{x} = -{1\over{\rho}}{\tau\over{R}}</math>where <math>\tau</math> is the [[shear stress]] and <math>R</math> is the [[hydraulic radius]]. Defining the friction slope <math>S_{f} = \tau/\rho g R</math>, a way of quantifying friction losses, leads to the final form of the momentum equation:<math{{Equation displaybox 1|cellpadding|border|indent="block":|equation=<math> {\partial u\over{\partial t}} + u{\partial u\over{\partial x}} + g{\partial \eta\over{\partial x}} + g(S_{f}- S) = 0 </math>|border colour=#0073CF|background colour=#F5FFFA}}

=== Energy equation ===

To derive an [[energy]] equation, note that the advective acceleration term <math>{\bf v}\cdot\nabla {\bf v}</math> may be decomposed as:<math display="block">{\bf v}\cdot\nabla {\bf v} = \omega \times {\bf v} + {1\over{2}}\nabla\|{\bf v}\|^{2}</math>where <math>\omega</math> is the [[vorticity]] of the flow and <math>\|\cdot\|</math> is the [[Euclidean norm]]. This leads to a form of the momentum equation, ignoring the external forces term, given by:<math display="block">{\partial {\bf v}\over{\partial t}} + \omega \times {\bf v} = -\nabla\left({1\over{2}}\|{\bf v}\|^{2} + {p\over{\rho}} + \Phi \right )</math>Taking the [[dot product]] of <math>{\bf v}</math> with this equation leads to:<math display="block">{\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 ) = 0</math>This equation was arrived at using the [[scalar triple product]] <math>{\bf v}\cdot (\omega \times {\bf v}) = 0</math>. Define <math>E</math> to be the [[energy density]]:<math display="block">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}}</math>Noting that <math>\Phi</math> is time-independent, we arrive at the equation:<math display="block">{\partial E\over{\partial t}} + {\bf v}\cdot\nabla (E+p) = 0</math>Assuming that the energy density is time-independent and the flow is one-dimensional leads to the simplification:<math display="block">E + p = C</math>with <math>C</math> being a constant; this is equivalent to [[Bernoulli's principle]]. Of particular interest in open-channel flow is the [[specific energy]] <math>e = E/\rho g</math>, which is used to compute the [[hydraulic head]] <math>h</math> that is defined as:<math{{Equation displaybox 1|cellpadding|border|indent="block":|equation=<math> \begin{aligned}

h &= e + {p\over{\rho g}} \\

&= {u^{2}\over{2g}} + z + {p\over{\gamma}}

\end{aligned} </math>|border colour=#0073CF|background colour=#F5FFFA}}with <math>\gamma = \rho g</math> being the [[specific weight]]. However, realistic systems require the addition of a [[head loss]] term <math>h_{f}</math> 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==

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