diff --git a/OpenHPL/Waterway/Gate.mo b/OpenHPL/Waterway/Gate.mo index 273642b..3b51736 100644 --- a/OpenHPL/Waterway/Gate.mo +++ b/OpenHPL/Waterway/Gate.mo @@ -74,7 +74,8 @@ Equation numbers and figure numbers given below are in sync with the numbers of
The free flow can be calculated with: -$$Q_A = \\mu_A \\cdot A \\cdot \\sqrt{2g\\cdot h_0} \\tag{8.24} $$ +$$ Q_A = \\mu_A \\cdot A \\cdot \\sqrt{2g\\cdot h_0} \\tag{8.24} $$ + (valid for gate opening higher than the downstream water level)
@@ -86,17 +87,17 @@ With
With:@@ -148,11 +149,12 @@ With-
- \\(a \\ldots\\) Vertical gate opening
-- \\(h_h \\ldots\\) Height of the hinge above gate bottom\"
-- \\(r \\ldots\\) Radius of the gate arm
+- \(a \\ldots\) Vertical gate opening
+- \(h_h \\ldots\) Height of the hinge above gate bottom
+- \(r \\ldots\) Radius of the gate arm
-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: +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. +So when \(\\frac{h_2}{a} \\geq \\frac{h_2^*}{a}\) then we have back-up flow.
-