Basic equations in open channel hydraulics

3.1   Lagrangian specification of the flow field

The observer follows a system of interest (the same fluid particles) in space and time.

3.2   Eulerian specification of the flow field

However, we usually make observations at a fixed region (control volume).

A control volume is a fixed volume in space of interest and its surrounding closed surface is a control surface.

See Figure 1.2 in the text.

4.1   Change of what?

Let $B$ be an extensive property of a fluid.

At time $t$, \begin{equation}B_\text{sys}(t)=B_\text{cv}(t)\end{equation}

At time $t+dt$, \begin{equation}B_\text{sys}(t+dt)=B_\text{cv}(t+dt)+B_\text{out}(t+dt)-B_\text{in}(t+dt)\end{equation}

4.2   Change within the system

\begin{equation} \begin{aligned} \frac{dB_\text{sys}}{dt} &=\frac{B_\text{sys}(t+dt)-B_\text{sys}(t)}{dt}\\ &=\frac{B_\text{cv}(t+dt)+B_\text{out}(t+dt)-B_\text{in}(t+dt)-B_\text{cv}(t)}{dt}\\ &=\frac{B_\text{cv}(t+dt)-B_\text{cv}(t)}{dt}+\frac{B_\text{out}(t+dt)-B_\text{in}(t+dt)}{dt}\\ &=\frac{B_\text{cv}(t+dt)-B_\text{cv}(t)}{dt}+\frac{\left[B_\text{out}(t+dt)-B_\text{in}(t+dt)\right]-\left[B_\text{out}(t)-B_\text{in}(t)\right]}{dt}\\ &=\frac{dB_\text{cv}}{dt}+\dot{B}_\text{net out}\\ \end{aligned} \end{equation} because $B_\text{out}(t)=B_\text{in}(t)$.

Following the convention of Eq. (1.2) in the text, $B\equiv B_\text{sys}$ (system property). \begin{equation} \frac{dB}{dt}=\frac{dB_\text{cv}}{dt}+\dot{B}_\text{net out} \end{equation}

4.3   Change within the control volume

\[ \newcommand{\volsym}{\rlap{\kern.08em–}V} \newcommand{\volsubsym}{\rlap{\scriptsize\kern.08em–}V} \] \begin{equation} \frac{dB_\text{cv}}{dt} =\frac{d}{dt}\int\limits_\text{cv}dB =\frac{d}{dt}\int\limits_\text{cv}d(bm) =\frac{d}{dt}\int\limits_\text{cv}b\,dm =\frac{d}{dt}\int\limits_\text{cv}b\rho\,d\volsym \end{equation} where $m$ is the mass, $\rho$ is the density, $b=\frac{B}{m}$ is the intensive property, and $\volsym$ is the volume.

Remember, $b$ doesn’t change with mass (a fixed quantity regardless of the mass), so we can take $b$ out of $d(bm)$.

4.4   Change across the control surface

For net out, outflux positive and influx negative

\begin{equation} \dot{B}_\text{net out} =b\dot{m}_\text{net out} =b\rho\dot{\volsym}_\text{net out} =\int\limits_\text{cs}b\rho V\,dA =\int\limits_\text{cs}b\rho(\vec{V}\cdot\vec{n})\,dA \end{equation}

4.5   Reynolds transport theorem equation

\begin{equation} \frac{dB}{dt} =\frac{d}{dt}\int\limits_\text{cv}b\rho\,d\volsym +\int\limits_\text{cs}b\rho(\vec{V}\cdot\vec{n})\,dA \end{equation}

6.1   One-dimensional steady flow of an incompressible fluid

In a given time period, the mass of the control volume remains the same ($\frac{d}{dt}\int\limits_\text{cv}\rho\,d\volsym=0$) and what’s going into the control volume should come out of it.

\begin{equation} \int\limits_\text{cs}\rho(\vec{V}\cdot\vec{n})\,dA =\sum Q_\text{out}-\sum Q_\text{in} =0 \end{equation}

At a cross section in one-dimensional flow, $v_s=\vec{V}\cdot\vec{n}$ because flow is assumed to have only one direction along the streamline. \begin{equation} Q =\int\limits_Av_s\,dA =V_sA \end{equation}

I think “cs” in Eq. (1.5) in the text meant to mean “cross section,” not control surface because if you take the integral of $v_s\,dA$ over the control surface in one-dimensional incompressible flow, that’ll be 0 just like in Eq. (1.4). This integration should be over just one cross-sectional area, either $A_\text{in}$ or $A_\text{out}$, not both; hence the difference.

6.2   One-dimensional unsteady flow

Volume change: $\frac{d}{dt}\int\limits_\text{cv}\rho\,d\volsym=\frac{\partial\volsym}{\partial t}=\frac{\partial(A\,dx)}{\partial t}$

In and out: $\int\limits_\text{cs}\rho(\vec{V}\cdot\vec{n})\,dA=\left(Q+\frac{\partial Q}{\partial x}dx\right)-Q$

\begin{equation} 0 =\frac{d}{dt}\int\limits_\text{cv}\rho\,d\volsym +\int\limits_\text{cs}\rho(\vec{V}\cdot\vec{n})\,dA =\frac{\partial A}{\partial t}+\frac{\partial Q}{\partial x} \end{equation}

7.1   One-dimensional steady flow of an incompressible fluid

\begin{equation} \frac{d}{dt}\int\limits_\text{cv}\vec{V}\rho\,d\volsym=0 \end{equation}

\begin{equation} \sum F_s =\int\limits_\text{cs}\rho v_s(\vec{V}\cdot\vec{n})\,dA =\sum(\beta\rho QV_s)_\text{out}-\sum(\beta\rho QV_s)_\text{in} \end{equation} where the momentum flux correction coefficient $\beta=\frac{\int_Av_s^2\,dA}{V_s^2A}$.

• $\beta=1$ in a uniform flow
• $\beta$ is not significantly greater than 1 in turbulent flow in prismatic channels
• $\beta\neq 1$ in a nonuniform flow

7.2   One-dimensional unsteady flow

See Figure 1.2d. In a differential control volume ($dA\,ds$), the accelerating resultant force of the system is balanced by the sum of external forces (pressure force and weight).

Forces positive in the streamline direction

Pressure force: \begin{equation} F_p =p\,dA-\left(p+\frac{\partial p}{\partial s}ds\right)dA =-\frac{\partial p}{\partial s}ds\,dA \end{equation}

Weight: \begin{equation} F_g=-\rho g\,dA\,ds\frac{dz}{ds}=-\rho g\,dA\,dz \end{equation}

Resultant force: \begin{equation} F_a=ma_s=\rho\,dA\,ds\frac{dv_s}{dt} =\rho\,dA\,ds\frac{\frac{\partial v_s}{\partial t}dt+\frac{\partial v_s}{\partial s}ds}{dt} =\rho\,dA\,ds\left(\frac{\partial v_s}{\partial t}+\frac{\partial v_s}{\partial s}\frac{ds}{dt}\right) =\rho\,dA\,ds\left(\frac{\partial v_s}{\partial t}+\frac{\partial v_s}{\partial s}v_s\right) \end{equation}

7.3   One-dimensional unsteady flow momentum equation

\begin{equation} F_p+F_g=F_a \end{equation}

\begin{equation} -\frac{\partial p}{\partial s}ds\,dA -\rho g\,dA\,dz =\rho\,dA\,ds\left(\frac{\partial v_s}{\partial t}+\frac{\partial v_s}{\partial s}v_s\right) \end{equation}

Dividing both sides by the volume $dA\,ds$ yields Euler’s equation: \begin{equation} -\frac{\partial p}{\partial s}-\rho g\frac{\partial z}{\partial s} =\rho\frac{\partial v_s}{\partial t}+\rho v_s\frac{\partial v_s}{\partial s} \end{equation}

7.4   One-dimensional steady flow momentum equation

Steady means $\frac{\partial v_s}{\partial t}=0$.

\begin{equation} -\frac{\partial p}{\partial s}-\rho g\frac{\partial z}{\partial s} =\rho v_s\frac{\partial v_s}{\partial s} \end{equation}

Integrate both sides along $ds$ and rearrange it: \begin{equation} \Delta p+\rho g\Delta z+\rho\bar{v}_s\Delta v_s =(p_2-p_1)+\rho g(z_2-z_1)+\rho\frac{v_1+v_2}{2}(v_2-v_1) =0 \end{equation}

Dividing both sides by $\gamma=\rho g$ yields the Bernoulli equation: \begin{equation} \frac{p_2-p_1}{\gamma}+(z_2-z_1)+\frac{(v_1+v_2)(v_2-v_1)}{2g} =\frac{p_2-p_1}{\gamma}+(z_2-z_1)+\frac{v_2^2-v_1^2}{2g} =0 \end{equation}

\begin{equation} \frac{p_1}{\gamma}+z_1+\frac{v_1^2}{2g} =\frac{p_2}{\gamma}+z_2+\frac{v_2^2}{2g} \end{equation}

What’s the meaning of this equation? Work or energy per unit weight (work-energy equation)

Why?