Open-channel flow: Difference between revisions

Revision as of 07:18, 9 April 2020

Revision as of 07:19, 9 April 2020

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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 display="block">{\partial A\over{\partial t}} + {\partial Q\over{\partial x}} = 0</math><br />

=== 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}