1. Numerical Hydraulics W. Kinzelbach with Marc Wolf and Cornel Beffa Lecture 3: Computation of pressure surges

2. The phenomenon Till time t=0: Steady state flow Q At time t = 0: Instantaneous closing of valve Observation: Sudden pressure rise at valve → pressure surge (water hammer) Valve Reservoir

3. And some typical damages Sayano-Shushenskaya plant in southern Siberia Water pipe damage due to pressure surge San Bruno: PG&E Power Outage and Pressure Surge Preceded Blast

4. Where pressure surges occur… water distribution systems waste water transfers storm water rising mains power station cooling systems oil pipelines RAS (activated sludge) onsite pipelines hydropower stations and any fluid system in which the inertia (mass and velocity) of the fluid is significant.

6. The phenomenon Pressure wave propagates with wave velocity c If valve closing time is smaller than the run time of the wave to the reflection point and back the surge is called Joukowski surge Pressure vs. time at valve Damping of amplitude through friction

7. Negative pressure wave The negative pressure wave can not become lower than the vapour pressure of the fluid. If the pressure falls below the vapour pressure, a vapour bubble is formed. The water column separates from the valve. When the pressure increases again the bubble collapses. This phenomenon is called cavitation.

8. Pressure surge with vapour bubble formation

9. Computed and observed pressure surge at two places along a pipe Today the reliable computation of pressure surges is possible

10. Measures against pressure surges Surge shaft

11. Measures against pressure surges Surge vessels (Windkessel) Special valves

12. The equations of unsteady pipe flow Continuity Momentum equation (per unit volume) (  inclination angle of pipe)

13. Further transformations (1) Continuity: As density and cross-sectional area depend on x and t only via the pressure p, the chain rule can be applied. Using the moduli of elasticity of water E W und of the pipe E pipe e is the pipe wall thickness, E‘ is the combined modulus of elasticity of the system

14. Further transformations (2) Compressibility: Definition Using pressure tank formula

15. Further transformations(3) Momentum equation

16. The equations of unsteady pipe flow Continuity Momentum equation 2 PDE with 2 unknown functions p(x,t) und v(x,t) plus initial and boundary conditions (1) (2)

17. Boundary conditions Pressure boundary condition: p given –e.g. water level in reservoir, controlled pump Velocity/Flux boundary condition: v given –e.g. flow controlled (v from Q/A) Combination: Relation between pressure and flux given –Z. B. function of pressure reduction valve, characteristic curve of pump Closing of a valve at the end of a pipe –Initially flow Q, then according to closing function reduction to zero withing closing time of the valve.

18. Linearised equations Delete all terms in (1) and (2) which are non-linear (for convenience:  = 0) : General solution by elimination: -Take partial derivative of first equation with respect to t -Take partial derivative of second equation with respect to x Subtraction yields:

19. Linearised equations Wave equation (for p, analogously for v) which has general solution Wave with wave velocity Example: Modulus of elasticity of steel = 200‘000 MN/m 2, Modulus of elasticity of water = 2‘000 MN/m 2, wall thickness e = 0.02 m, D = 1 m,  = 1000 kg/m 3 yields c = 1333 m/s

20. Joukowski surge Estimate of surge pressure after instantaneous closing of valve (neglecting friction, linearized equations): „Worst case“ General solution: Proof by insertion into linearised equations!!

21. Joukowski surge After t = 0 only the backward running wave F(x+ct) is found in the upstream v at the valve is 0 Maximum  p is given by: Solution: Example continued: c=1333 m/s, Q 0 =1 m 3 /s, L=100 m yields:  p=1.7E6 N/m 2

22. Numerical solution of the complete equations Difference method –Discretisation of space and time –  x and  t Difference equations for time step t, t+  t Problem: Discretisation „softens“ pressure front numerically Way out: Method of characteristics –Follows the pressure signal in moving coordinate system

23. Normal difference method –Softening of pressure front Method of characteristics –Grid is adapted to frontal velocity (feasible, as v<<c, c+v ≈ c-v ≈ c) Method of characteristics Front of pressure wave Front of pressure wave c  t =  x c  t <  x xx xx

24. Method of characteristics Replacing equations (1) and (2) by 2 linear combinations yields:

25. Method of characteristics With total derivative along x(t) the equations have the form: Forward characteristic Backward characteristic (c is actually relative wave velocity with respect to average water movement.)

26. Difference scheme Chose time step such that  x = c  t In every time step there are two unknowns at each of the N+1 nodes: Divide pipe of length L in N sections, length of one section  x = L/N Node 1 section 1 Node N+1 section N x To determine these unknowns 2N+2 equations are required. From quantities at time j quantities at time j+1 are computed. The new times j+1 become the old times j of the next time step. Upper index time step, lower index node

27. Difference scheme xx tt  x = c  t ii-1 i+1 j j +1 3 1 2 1 2 3 Total derivative or derivative along characteristic line space time Using c + v ≈ c - v ≈ c node i communicates within time interval  t with node i-1 via the forward characteristic and with node i+1via the backward characteristic

28. Difference form of equations Equations for nodes 2 to N: 2N-2 equations forward characteristic backward characteristic The pressure loss term is linearised by evaluating it at the old time j Equations can be solved for The further equations are determined by the boundary conditions and the one characteristic which can be used at the respective boundary

29. Method of characteristics Example: Reservoir with pipe which is closed instantaneously at t=0 2 further equations from boundary conditions In the example: 2 further equations from characteristic equations In the example: From forward characteristic for i=N+1 From backward characteristic for i=1

30. Simplified case for basic Matlab-Program  =0, friction neglected, equations nodes 2 to N forward characteristic backward characteristic = 0 Solution by subtracting resp. adding the two equations

31. Simplified case for basic Matlab-Program 2 equations from boundary conditions 2 equations from characteristics for i = 1 and i = N+1 From forward characteristic for i=N+1 From backward characteristic for i=1 = 0

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

33. Example (1) Tank 1 Tank 2 connecting pipe valve L=500 m D = 0.2 m, e = 0.01 m, roughness k = 0.0005 m  = 1000 kg/m 3, E w = 2000 MN/m 2, E piper = 210000 MN/m 2 pressure downstream reservoir 80 mWS, pressure upstream reservoir 90 mWS closing time of valve1 s, Q before closing: 0.2 m 3 /s loss coefficient valve 2, time of calculation 60 s, number of pipe sections n = 10 Use Program „Hydraulic System“

34. Example (2) Tank 1 Tank 2 Valve L=500 m D = 0.2 m, e = 0.01 m, roughness k = 0.0005 m  = 1000 kg/m 3, E w = 2000 MN/m 2, E pipe = 210000 MN/m 2 pressure of both downstream reservoirs 80 mWS, pressure upstream reservoir 90 mWS closing time 1 s, Q before closing of valve: 0.2 m 3 /s loss coefficient of valve 2, computation time 60 s, number of pipe sections n = 10 Tank 3 L=500 m