Basic equations in open channel hydraulics

Dr. Huidae Cho
Department of Civil Engineering...New Mexico State University

1   Basic equations

  • Continuity equation
  • Momentum equation
  • Energy equation

2   Extensive vs. intensive properties

An extensive property is dependent of the size of the system. For example, the mass of a fluid system.

An intensive property is independent of the size of the system. For example, the density of the same system.

3   Specifications of the flow field

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   Reynolds transport theorem

How can we relate these two specifications of the flow field?

The Reynolds transport theorem!

Change within the system = Change within the control volume + Change across the control surface

4.1   Change of what?

system-vs-control-volume

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{dB_\text{cv}}{dt}+\dot{B}_\text{net out}\\ \end{aligned} \end{equation}

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}

5   Intensive property

Let’s take a closer look at $b$ before we move on. By definition, $b=\frac{B}{m}$, so $B=bm$. Taking the derivative of both sides with respect to $m$, we obtain $\frac{dB}{dm}=b$. Oh! $b=\frac{B}{m}$ and $b=\frac{dB}{dm}$.

6   Continuity equation

  • $B=m$
  • $b=\frac{B}{m}=\frac{m}{m}=1$

Mass doesn’t change over time. What is $\frac{dB}{dt}$? 0

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

Change within the control volume is balanced by the net outflux through the control surface.

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

open-channel-over-dx

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   Momentum equation

  • $\vec{B}=m\vec{V}$
  • $\vec{b}=\frac{\vec{B}}{m}=\frac{m\vec{V}}{m}=\vec{V}$
  • $\frac{d\vec{B}}{dt}=\frac{d(m\vec{V})}{dt}=m\frac{d\vec{V}}{dt}=m\vec{a}=\vec{F}$
  • $\vec{F}$ is the net force of the system acting on the control volume
  • Can be written as $\sum\vec{F}$

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

8   Energy equation

Expand the Bernoulli equation by including the kinetic energy flux correction coefficient $\alpha=\frac{\int_A v_s^3\,dA}{V_s^3A}$ and head loss due to friction $h_f$: \begin{equation} \frac{p_1}{\gamma}+z_1+\alpha_1\frac{V_1^2}{2g} =\frac{p_2}{\gamma}+z_2+\alpha_2\frac{V_2^2}{2g}+h_f \end{equation}

9   Homework: Proof of the equivalence between the Bernoulli equation and work per unit weight

Prove the equivalence between the Bernoulli equation and work per unit weight by deriving equations analytically. Please type your solution using a computer program for clarity and readability, instead of handwriting it. Submit your solution in PDF to the submission folder.