1. 1 Numerical Hydraulics W. Kinzelbach with Marc Wolf and Cornel Beffa Lecture 4: Computation of pressure surges continued

2. 2 Additions Formation of vapour bubble Branching pipes Different closing functions Pumps and pressure reduction valves …. Consistent initial conditions through steady state computation of flow/pressure distribution

3. 3 Closing function Expressed as Q=Q(t) or by degree of closure , depending from position of valve,  = f(t) Valve closed:  = 0 Valve completely open:  = 1 In between: function corresponding to ratio of loss coefficients  or Q (%) time t 0 100 t close 0 open closed Index 0 refers to open valve

4. 4 Valve as boundary condition Valve directly in front of a downstream reservoir with pressure p B2 Valve at Node N+1 Linear closing function  = 1-t/t close Determine new pressure and velocity at valve From boundary condition From forward characteristic (1) (2) Inserting (1) into (2) yields quadratic equation for For t < t close :

5. 5 Valve as boundary condition Only one of the two solutions is physically meaningful For t > t close :

6. 6 Pump Given characteristic function of pump: Pump at node i: Simply insert into characteristic equation. e.g. forward characteristic:

7. 7 Formation of vapour bubble If the pressure falls below the vapour pressure of the fluid (at temperature T) a vapour bubble forms, which fixes the pressure at the vapour pressure of the fluid. The bubble grows as long as the pressure in the fluid does not rise. It collapses again when the pressure increases above the vapour pressure. Additional equation: Forward characteristic in N: Volume balance of vapour bubble: volume Vol at valve As long as the vapour bubble exists, the boundary condition v = 0 at the valve must be replaced by the pressure boundary condition p = p vapour. If Vol becomes 0 the vapour bubble has collapsed. The velocity in the volume equation is negative, as bubble grows as long as wave moves away from valve.

8. 8 Branching of pipe i-1i i+1 k k+1 Note that continuity requires that A i-1 v i-1 =A k v k +A i v i Characteristics along i -1 … k+1 and along i -1 … i+1 With different lengths of pipes the reflected waves return at different times. At the branching, partial reflection takes place. The pressure surge signal in a pipe grid therefore becomes much more complicated, but at the same time less extreme, as the interferences weaken the maximum.

9. 9 Consistent initial conditions by steady state computation of flow/pressure In the example: Tank 1 Tank 2 connecting pipe In a grid with branchings a steady state computation of the whole grid is required

10. 10 Measures against pressure surges Slowing down of closing process Surge vessel (Windkessel) Surge shaft Special valves air

11. 11 Surge shaft oscillations Task: Write a program in Matlab for the calculation of the surge shaft oscillations Simplified theory: see next page

12. 12 Surge shaft oscillations The following formulae can be used (approximation of rigid water column) : Solve for Z(t), Estimate the frequency under neglection of friction. Data: l = 200 m, d 1 = 1.25 m, d 2 = 4 m, Q = 2 m 3 /s at time t = 0 local losses negligible, = 0.04, computation time from t = 0 to t = 120 s, instantaneous closing of valve at time t = 0.

13. 13 Surge shaft oscillations The surge in the following surge shaft is to be calculated using the program Hydraulic System. Vary parameters and compare! 250 m ü. M. Further data: Closing time 1 s, area surge shaft 95 m 2, roughness pressure duct k=0.00161 m, modulus of elasticity pressure duct = 30 GN/m 2, modulus of elasticity pressure duct = 30 GN/m 2, Loss coefficient valve 2.1 (am ->av) and 2.0 (am<- av) resp., linear closing law, cross-section valve 1.5 m w stands for wall thickness, in the pressure duct in the rock it is assumed as 2 m effectively.