.OpenHPL.Waterway.Gate

Information

Implementation

The calculation of the flow through the gate is approximated for two different regions and is based on [Bollrich2019]. Equation numbers and figure numbers given below are in sync with the numbers of [Bollrich2019].

Free flowing

Fig. 8.13: Free flow through the tainter gate (source: [Bollrich2019], page 376)

The free flow can be calculated with: $$Q_A = \mu_A \cdot A \cdot \sqrt{2g\cdot h_0} \tag{8.24} $$ (valid for gate opening higher than the downstream water level)

With

Opening area

$$ A = a\cdot b $$

Discharge coefficient

$$ \mu_A = \frac{\psi}{\sqrt{1+\frac{\psi\cdot a}{h_0}}} \tag{8.23}$$

Contraction coefficient sluice gate (\(\alpha=90^\circ\))

$$ \psi_{90^\circ}= \frac{1}{1+0.64\cdot \sqrt{1-(a/h_0)^2}} \tag{8.25}$$

Contraction coefficient radial gate (for \(a/h_0 \rightarrow 0\))

$$ \psi_0(\alpha)= 1.3 -0.8\cdot\sqrt{1-\left(\frac{\alpha -205^\circ}{220^\circ}\right)^2} \tag{8.25a}$$

The edge angle \(\alpha\) of the gate

$$ \alpha = \left( \frac{\pi}{2} - \arcsin(\frac{h_h-a}{r})\right) \cdot \frac{180^\circ}{\pi} $$

With:

• \(a \ldots\) Vertical gate opening
• \(h_h \ldots\) Height of the hinge above gate bottom"
• \(r \ldots\) Radius of the gate arm

Backed-up discharge

Fig. 8.16: Backed-up flow through the tainter gate (source: [Bollrich2019], page 379)

$$Q_A = \chi \cdot \mu_A \cdot A \cdot \sqrt{2g\cdot h_0} \tag{8.29} $$ With

Back-up coefficient

$$ \chi = \sqrt{ \left( 1 + \frac{\psi\cdot a}{h_0} \right) \cdot \left\{ \left[ 1 - 2\cdot\frac{\psi\cdot a}{h_0} \cdot \left( 1-\frac{\psi\cdot a}{h_2} \right) \right] - \sqrt{ \left[ 1 - 2 \cdot \frac{\psi\cdot a}{h_0} \cdot \left( 1-\frac{\psi\cdot a}{h_2} \right) \right]^2 + \left( \frac{h_2}{h_0} \right)^2 - 1 } \right\} } \tag{8.28}$$

Boundary between free and backed-up flow

The boundary of the height of the water level \(h_2\) behind the gate from which on the calculation switches to the backed-up flow (8.29) can be derived from: $$ \frac{h_2^*}{a} = \frac{\psi}{2} \cdot \left( \sqrt{ 1 + \frac{16}{\psi\cdot\left(1+\frac{\psi\cdot a}{h_0}\right)}\cdot\frac{h_0}{a}} - 1 \right) \tag{8.26}$$ So when \(\frac{h_2}{a} \geq \frac{h_2^*}{a}\) then we have back-up flow.