Open-channel flow

The general continuity equation, describing the conservation of mass, takes the form:

$${\displaystyle {\partial \rho \over {\partial t}}+\nabla \cdot (\rho {\bf {v}})=0}$$

where ${\displaystyle \rho }$ is the fluid density and ${\displaystyle \nabla \cdot ()}$ is the divergence operator. Under the assumption of incompressible flow, with a constant control volume ${\displaystyle V}$, this equation has the simple expression ${\displaystyle \nabla \cdot {\bf {v}}=0}$. However, it is possible that the cross-sectional area ${\displaystyle A}$ can change with both time and space in the channel. If we start from the integral form of the continuity equation:

$${\displaystyle {d \over {dt}}\int _{V}\rho \;dV=-\int _{V}\nabla \cdot (\rho {\bf {v}})\;dV}$$

it is possible to decompose the volume integral into a cross-section and length, which leads to the form:

$${\displaystyle {d \over {dt}}\int _{x}\left(\int _{A}\rho \;dA\right)dx=-\int _{x}\left[\int _{A}\nabla \cdot (\rho {\bf {v}})\;dA\right]dx}$$

Under the assumption of incompressible, 1D flow, this equation becomes:

$${\displaystyle {d \over {dt}}\int _{x}\left(\int _{A}dA\right)dx=-\int _{x}{\partial \over {\partial x}}\left(\int _{A}u\;dA\right)dx}$$

By noting that ${\displaystyle \int _{A}dA=A}$ and defining the volumetric flow rate ${\displaystyle Q=\int _{A}u\;dA}$, the equation is reduced to:

$${\displaystyle \int _{x}{\partial A \over {\partial t}}\;dx=-\int _{x}{\partial Q \over {\partial x}}dx}$$

Finally, this leads to the continuity equation for incompressible, 1D open-channel flow:

The momentum equation for open-channel flow may be found by starting from the incompressible Navier-Stokes equations :

$${\displaystyle \overbrace {\underbrace {\partial {\bf {v}} \over {\partial t}} _{\begin{smallmatrix}{\text{Local}}\\{\text{Change}}\end{smallmatrix}}+\underbrace {{\bf {v}}\cdot \nabla {\bf {v}}} _{\text{Advection}}} ^{\text{Inertial Acceleration}}=-\underbrace {{1 \over {\rho }}\nabla p} _{\begin{smallmatrix}{\text{Pressure}}\\{\text{Gradient}}\end{smallmatrix}}+\underbrace {\nu \Delta {\bf {v}}} _{\text{Diffusion}}-\underbrace {\nabla \Phi } _{\text{Gravity}}+\underbrace {\bf {F}} _{\begin{smallmatrix}{\text{External}}\\{\text{Forces}}\end{smallmatrix}}}$$

where ${\displaystyle p}$ is the pressure, ${\displaystyle \nu }$ is the kinematic viscosity, ${\displaystyle \Delta }$ is the Laplace operator, and ${\displaystyle \Phi =gz}$ is the gravitational potential. By invoking the high Reynolds number and 1D flow assumptions, we have the equations:

$${\displaystyle {\begin{aligned}{\partial u \over {\partial t}}+u{\partial u \over {\partial x}}&=-{1 \over {\rho }}{\partial p \over {\partial x}}+F_{x}\\-{1 \over {\rho }}{\partial p \over {\partial z}}-g&=0\end{aligned}}}$$

The second equation implies a hydrostatic pressure ${\displaystyle p=\rho g\zeta }$, where the channel depth ${\displaystyle \eta (t,x)=\zeta (t,x)-z_{b}(x)}$ is the difference between the free surface elevation ${\displaystyle \zeta }$ and the channel bottom ${\displaystyle z_{b}}$. Substitution into the first equation gives:

$${\displaystyle {\partial u \over {\partial t}}+u{\partial u \over {\partial x}}+g{\partial \zeta \over {\partial x}}=F_{x}\implies {\partial u \over {\partial t}}+u{\partial u \over {\partial x}}+g{\partial \eta \over {\partial x}}-gS=F_{x}}$$

where the channel bed slope ${\displaystyle S=-dz_{b}/dx}$. To account for shear stress along the channel banks, we may define the force term to be:

$${\displaystyle F_{x}=-{1 \over {\rho }}{\tau \over {R}}}$$

where ${\displaystyle \tau }$ is the shear stress and ${\displaystyle R}$ is the hydraulic radius. Defining the friction slope ${\displaystyle S_{f}=\tau /\rho gR}$, a way of quantifying friction losses, leads to the final form of the momentum equation:

To derive an energy equation, note that the advective acceleration term ${\displaystyle {\bf {v}}\cdot \nabla {\bf {v}}}$ may be decomposed as:

$${\displaystyle {\bf {v}}\cdot \nabla {\bf {v}}=\omega \times {\bf {v}}+{1 \over {2}}\nabla \|{\bf {v}}\|^{2}}$$

where ${\displaystyle \omega }$ is the vorticity of the flow and ${\displaystyle \|\cdot \|}$ is the Euclidean norm. This leads to a form of the momentum equation, ignoring the external forces term, given by:

$${\displaystyle {\partial {\bf {v}} \over {\partial t}}+\omega \times {\bf {v}}=-\nabla \left({1 \over {2}}\|{\bf {v}}\|^{2}+{p \over {\rho }}+\Phi \right)}$$

Taking the dot product of ${\displaystyle {\bf {v}}}$ with this equation leads to:

$${\displaystyle {\partial \over {\partial t}}\left({1 \over {2}}\|{\bf {v}}\|^{2}\right)+{\bf {v}}\cdot \nabla \left({1 \over {2}}\|{\bf {v}}\|^{2}+{p \over {\rho }}+\Phi \right)=0}$$

This equation was arrived at using the scalar triple product ${\displaystyle {\bf {v}}\cdot (\omega \times {\bf {v}})=0}$. Define ${\displaystyle E}$ to be the energy density:

$${\displaystyle E=\underbrace {{1 \over {2}}\rho \|{\bf {v}}\|^{2}} _{\begin{smallmatrix}{\text{Kinetic}}\\{\text{Energy}}\end{smallmatrix}}+\underbrace {\rho \Phi } _{\begin{smallmatrix}{\text{Potential}}\\{\text{Energy}}\end{smallmatrix}}}$$

Noting that ${\displaystyle \Phi }$ is time-independent, we arrive at the equation:

$${\displaystyle {\partial E \over {\partial t}}+{\bf {v}}\cdot \nabla (E+p)=0}$$

Assuming that the energy density is time-independent and the flow is one-dimensional leads to the simplification:

$${\displaystyle E+p=C}$$

with ${\displaystyle C}$ being a constant; this is equivalent to Bernoulli's principle. Of particular interest in open-channel flow is the specific energy ${\displaystyle e=E/\rho g}$, which is used to compute the hydraulic head ${\displaystyle h}$ that is defined as:

with ${\displaystyle \gamma =\rho g}$ being the specific weight. However, realistic systems require the addition of a head loss term ${\displaystyle h_{f}}$ to account for energy dissipation due to friction and turbulence that was ignored by discounting the external forces term in the momentum equation.