Numerical Simulation of Pressure Surge with the Method of Characteristics

Transcription

1 Numerical Simulation of Pressure Surge with the Method of Characteristics R. Fiereder Saint Petersburg, Russia

2 Content Motivation Governing Equations Continuity Equation Momentum Equation Method of Characteristics Implementation Boundary Conditions pplication pressure surge oscillating valve surge chamber oscillations power plant simulation 2

3 Motivation Simulation of interaction of plant components Identification of resonance cases 3

4 Motivation pump station power station III surge chamber power station I power station II 4

5 Governing Equations h Loss y h S h O z v h U L 1D unsteady compressible Continuity equation Momentum equation 5

6 Continuity Equation 1 dρ ρ 1 d v 0 Continuity ( ρ) ( ρv) t Product rule 0 ρ ρ v ρ v ρv ρ t t ρ 1 v 1 1 Total differential d d t d 0 y z d(ρ) d ρ 2 v 2 2 6

7 Continuity Equation 1 d 2R E s R dp ( 2 1 μ ) s σ ϕ Circular Pipe d 2πR dr 1 d ε ϕ d 2 σ a E, R μ No aial displacement / Hooke s law σ ϕ μσ a σ a μσ ϕ εϕ ER s << R d σ ϕ R s dp 2R E F 7

8 Continuity Equation Fluid compressibility 1 dρ ρ 1 E F dp Hooke s law applies E F dp dv V Mass in Control Volume is constant dm Vdρ ρdv 0 dv V dρ ρ 8

9 Continuity Equation p t v p ρa 2 v 0 dp 1 2R 2 EF ERs E Sys v ( 1 μ ) 0 2 a E Sys ρ 9

10 Momentum Equation 1 p ρ g z gj e v v v t 0 Momentum conservation i F i di y Pressure force p F P d p 1 1 τ 0 v p 2 2 Gravity force z F g ρg d z d τ 0 Friction force Fτ 0 2πRτ d τ 0 ρλ v v 8 4J e gr λ v 2 10

11 11 Final Set of Equations First order hyperbolic differential equation system Two dependent variables p, v Two independent variables, t a, ρ, J e are parameters of the system Solution by means of Method of Characteristics 0 2 v a p v t p ρ 0 1 t v v v gj z g p e ρ

12 12 Method of Characteristics Hyperbolic differential equation system Interferences propagate at a certain speed Eistence of constant variables along characteristic lines : G v a p v t p ρ G t v v v gj z g p e ρ G 2 G 1 G ξ 0 z J g dp dv e ξ ρξ ξρ 1 2 a v d ρa ξ 1 ± ( ) p v t p v a v t v ρξ ξ ξρ 0 ± ± C R d Riemann Variables

13 Method of Characteristics t dv 1 dp g J ρa d C v a e z d R 0 t 0 Δt C C - dv C 1 ρa d dp g J v a e z d R 0 t

14 Method of Characteristics Calculate values in P from known values in Point and B Integrate R ± along C ± P 1 P P dv dp g J g a ρ e P z 0 t t 0 Δt P C C - P B 1 dv ρa P B dp g P B J g e P B z 0 Two equations for two unknown variables t 0 B

15 Method of Characteristics-Summary t Solution is only valid on Characteristics t 0 Δt C - rea of influence C B rea of influence t 0 P rea of dependence 0 1 Characteristics separate two different solution domains t t 0 Δt P C C - rea of dependence t 0 B

16 Implementation Simplifications P a >> v d C a straight lines, velocity v is not known along C -> Friction term has to be approimated 2 J e J v Jv v Δt Ο( Δt) J e, Δ aδt Δt t Boundary Condition Boundary Condition Required Computational grid Steady-state solution for the system Boundary condition providing a disturbance Δt steady state Δ 16

17 Implementation Get system parameters Start read in parameters Initialize system Calculate Wave velocity a Generate calculation grid Calculate steady state solution Stationary Hydraulics Calculate solution for time t initialize system calculate steady state solution t t 0 calculate solution at time t write results? Yes write results to file No Yes t Δt t < t ma? No Stop 17

18 Implementation Method for calculating new time level C t P 1 ( vp v) ( pp p) gj e, Δt 0 ρa 1 1 vp pp v p gj e, Δt 0 ρa 1444 ρa K Δt C C - B C - analog i-1 i i1 v P K B K 2 p P ρa( K B K 2 ) 18

19 Implementation Calculate λ Evaluate boundary condition Calculate solution for inner field call from main calculate λ for friction term boundary condition left end i 0 Calculate K and K - boundary condition right end i i ma back to main i 1 Yes calculate v and p for point i i > i? ma No Yes No i i < i ma? i 0 - calculate K and K for point i 19

20 Boundary Conditions Static passive active Reservoir Valve static passive active f ( p, v) f ( p, v, t) Dynamic passive surge chamber dynamic f ( p, dp dv v, ) f dp dv ( p, v,, t) active Francis Turbine 20

21 Boundary Conditions Reservoir p const. v is calculated compression waves are reflected as epansions waves epansions waves are reflected as compression waves Wall v 0 p is calculated waves are reflected 21

22 Boundary Conditions Valve v f( Δp, τ, ζ ) v 1 ζτ 2 p0 p ρ 1 ζ > loss coefficient τ valve pipe τ ζ/ζ ma 0.6 α

23 Boundary Conditions Surge chamber (h) W p W can not be computed directly Q W (h w, ) h W Q W,L Q W,R predictor corrector scheme guess for p W calculate Q W dh W 1 ( h) ( Q Q ) W, L W, R correct p W correct Q W iterate till convergence Δh W h h W, t t 0 Δt W, 0 23

24 Boundary Conditions Turbine Machine data Mass moment of inertia of machine and rotating fluid Hill chart Generator η q 1 n 1 Governor Power Grid ω ω. τ. τ 24

25 T R ρav 0 pplication PRESSURE SURGE Joukowski - pressure surge ( Жуковский ) p(t) p 0 -ρav 0 sudden closure of a Valve kinetic energy of fluid is converted in pressure t[s] p(t) p 0 t[s] 25

26 pplication OSCILLTING VLVE T τ T R Results pressure and velocity in time at the valve /L 0 τ pressure and velocity in time in middle of the pipe T R τ valve pipe τ v/v ma p/p ma /L

27 pplication OSCILLTING VLVE Tτ 2TR Pressure and velocity variation along pipe Tτ 2TR 27

28 pplication SURGE CHMBER OSCILLTION surge chamber surge chamber reduces pressure fluctuations reservoir hydraulic uncoupling of pressure tunnel from penstock Improvement of control m pressure tunnel penstock m surge camber oscillation Interaction of high frequent pressure waves and low frequent inertia waves 28

29 pplication SURGE CHMBER OSCILLTION Model Δ Q 0 z W Q W W 270 m 250 m τ Δ [m] Δt [s] a [m/s] 2 W 79.0 [m ] D 2.5 [m] 3 Q [m /s] 1714m 686m D τ 0 m 0 t 0 t 29

30 pplication SURGE CHMBER OSCILLTION Results 7.5 Elevation of surge chamber water surface Zustandsdiagramm z[m] W z W [m] t [s] Q W [m /s] 30

31 pplication SURGE CHMBER OSCILLTION 3.35 Results pressure at the lower end of the penstock pressure at the upper end of the penstock p [MPa] X Detail X p W p Valve t [s] 31

32 pplication POWER PLNT Model Simulation of plant, governor, grid and machine Francis turbine Isolated grid 500m 50m 0m 500 m. ω, ω,... Gov. τ Geno ω Grid D 0.8 [m] 1 a 1230 [m/s] 1 D 0.6 [m] 2 a 1300 [m/s] 2 Δt 8.6 [ms] P 1.0 [MW] T,ma 100 m 32

33 pplication POWER PLNT 1.2 p T ω 1 Geno Grid 0.8 v t ω 0.6 ω, ω,.... Gov. τ 0.4 P T P Geno start up τ open loop control to overcome mass moment of inertia 0.2 synchronization t [s] closed loop control by governor 33

34 pplication POWER PLNT ω Grid 1 p T v t Geno ω ω, ω,.... Gov. τ 0.4 load rejection 0.2 Sudden load rejection caused by failure in grid or generator 0 P Geno τ P T t [s] 34

35 Thank you for your attention! 35

36 Implementation Method for calculating new time level t t 0 Initial solution constant velocity low pressure Δt

37 Implementation Method for calculating new time level t t 1 valve closes suddenly Δt

38 Implementation Method for calculating new time level t t 1.5 disturbance propagates upstream half time step Δt

39 Implementation Method for calculating new time level t t 2.0 Net time level calculate solution on intersection Point Δt

40 Implementation Method for calculating new time level t t 2.0 Δt

41 Implementation Method for calculating new time level t t 3.0 Δt

42 Implementation Method for calculating new time level t t 3.0 Δt