Stability of Direct Spring Operated Pressure Relief Valves from CFD to spreadsheet

Transcription

1 Stability of Direct Spring Operated Pressure Relief Valves from CFD to spreadsheet 12th September 217 HDR

2 ... with major contributions by Prof. Alan Champneys (University of Bristol, Dept. of Eng. Mathematics) Dr. Csaba Bazsó (BME HDS) István Erdei (BME HDS) Paul Kenneth, Mike McNelly (Pentair, Houston, TX)

3 Table of Contents 1 Introduction, motivation 2 CFD 1D model 3 4 5

4 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence.

5 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set

6 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set Valve reseat pressure p rs p set

7 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set Valve reseat pressure p rs p set Capacity: mass flow p = 1.1p set and x = x max (full lift)

8 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set Valve reseat pressure p rs p set Capacity: mass flow p = 1.1p set and x = x max (full lift) Challenges

9 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set Valve reseat pressure p rs p set Capacity: mass flow p = 1.1p set and x = x max (full lift) Challenges Valve chatter

10 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set Valve reseat pressure p rs p set Capacity: mass flow p = 1.1p set and x = x max (full lift) Challenges Valve chatter API code: 3% rule based on upstream pipe pressure loss is that sufficient?

11 DSOPRV Direct Spring Operated Pressure Relief Valves: safety device to limit system pressure - last line of defence. Valve set pressure p set Valve reseat pressure p rs p set Capacity: mass flow p = 1.1p set and x = x max (full lift) Challenges Valve chatter API code: 3% rule based on upstream pipe pressure loss is that sufficient? Lack of measurements (in the open domain).

12 Example of stable opening data file: 2J3 ON LIQUID tank Test 47.csv f valve: 46.8 Hz (black) lift, % PH, psig TK, psig ampl. TK ampl. PH ampl. lift f pipe: Hz (red) f QW: Hz (blue) frequency, Hz

13 Example of unstable opening data file: 2J3 ON LIQUID tank Test 22.csv f valve: 47 Hz (black) lift, % ampl. lift f pipe: 243 Hz (red) PH, psig 4 2 ampl. PH f QW: 61 Hz (blue) TK, psig 4 2 ampl. TK frequency, Hz

14 Measurement video 1

15 CFD 1D model Table of Contents 1 Introduction, motivation 2 CFD 1D model 3 4 5

16 CFD 1D model Computational Fluid Dynamics ANSYS CFX + Icem Deforming mesh + automatic remeshing Upstream pipe + simple valve model Axisymmetric Valve disc as rigid body High-resolution, lots of information but slow and no qualitative understanding Stable and unstable behaviour reproduced.

17 CFD 1D model CFD

18 CFD 1D model CFD video 2 video 3

19 CFD 1D model 1D model for liquid service x x v A ft (x v ) s m p v k. m v D, A, L v p (ξ,t), p p (ξ,t) ξ. m r,in. V, p r m r,out

20 CFD 1D model 1D model for liquid service valve: 1DoF oscillator mẍ v + kẋ v + s(x + x v ) = F lift, x F lift = A eff (x v )(p v p ) s k x v A ft (x v ) m p v. m v D, A, L v p (ξ,t), p p (ξ,t) ξ. m r,in. V, p r m r,out

21 CFD 1D model 1D model for liquid service valve: 1DoF oscillator mẍ v + kẋ v + s(x + x v ) = F lift, F lift = A eff (x v )(p v p ) reservoir pressure dynamics: ṁ r,in ṁ r,out = V a 2 ṗ r x x v A ft (x v ) s m p v k. m v D, A, L v p (ξ,t), p p (ξ,t) ξ. m r,in. V, p r m r,out

22 CFD 1D model 1D model for liquid service valve: 1DoF oscillator mẍ v + kẋ v + s(x + x v ) = F lift, F lift = A eff (x v )(p v p ) reservoir pressure dynamics: ṁ r,in ṁ r,out = V a 2 ṗ r 1D unsteady pipeline dynamics: p v p t +ρa2 ξ +v ξ = v v t +v ξ = 1 p ρ ξ + λ 2D v v pipe. m r,in x x v A ft (x v ) ξ s m p v. V, p r k m r,out. m v D, A, L v p (ξ,t), p p (ξ,t)

23 CFD 1D model 1D model for liquid service valve: 1DoF oscillator mẍ v + kẋ v + s(x + x v ) = F lift, F lift = A eff (x v )(p v p ) reservoir pressure dynamics: ṁ r,in ṁ r,out = V a 2 ṗ r 1D unsteady pipeline dynamics: p v p t +ρa2 ξ +v ξ = v v t +v ξ = 1 p ρ ξ + λ 2D v v pipe res. side: p t = p(, t) + ρ 2 (v(, t))2. m r,in x x v A ft (x v ) ξ s m p v. V, p r k m r,out. m v D, A, L v p (ξ,t), p p (ξ,t)

24 CFD 1D model 1D model for liquid service valve: 1DoF oscillator mẍ v + kẋ v + s(x + x v ) = F lift, F lift = A eff (x v )(p v p ) reservoir pressure dynamics: ṁ r,in ṁ r,out = V a 2 ṗ r 1D unsteady pipeline dynamics: p v p t +ρa2 ξ +v ξ = v v t +v ξ = 1 p ρ ξ + λ 2D v v pipe res. side: p t = p(, t) + ρ 2 (v(, t))2 valve-end: v(l, t)a pipe ρ = C d (x v )A ft (x v ) 2ρ (p(l, t) p ). m r,in x x v A ft (x v ) ξ s m p v. V, p r k m r,out. m v D, A, L v p (ξ,t), p p (ξ,t)

25 CFD 1D model Simulation results pipe length:.5m pipe length: 1.1m pipe length: 1.5m

26 CFD 1D model Simulation results CFD vs. 1D pt,res[bar] t[s] 1 1 xv [%] pe[bar] pv[bar] t[s] t[s]

27 Table of Contents 1 Introduction, motivation 2 CFD 1D model 3 4 5

28 Primary instability types Remember, our model is... Aim: Systematically isolate instability types and give design formulae to avoid them.. m r,in x x v A ft (x v ) ξ s m p v. V, p r k m r,out. m v D, A, L v p (ξ,t), p p (ξ,t)

29 Quarter-wave instability

30 Valve chatter experiments theory valve spring frequency quarter wave frequency self-excited oscillations despite steady-state BCs Quarter-wave frequency of the pipe seems to dominate

31 The Quarter-wave model (QWM) Simplest case (liquid, one mode only): Aim: replace the PDEs describing the pipeline dynamics to ODEs that allow stability analysis.

32 The Quarter-wave model (QWM) Simplest case (liquid, one mode only): Aim: replace the PDEs describing the pipeline dynamics to ODEs that allow stability analysis. Ansatz: p(x, t) = p t (t) ρ ( 2 v(, t)2 + B(t) sin 2π x ) ( 4L v(x, t) = v(l, t) + C(t) cos 2π x ) 4L where v(l, t) = C d A ft (x v ) A pipe 2 ρ (p(l, t) p )

33 The Quarter-wave model (QWM) Simplest case (liquid, one mode only): Aim: replace the PDEs describing the pipeline dynamics to ODEs that allow stability analysis. Ansatz: p(x, t) = p t (t) ρ ( 2 v(, t)2 + B(t) sin 2π x ) ( 4L v(x, t) = v(l, t) + C(t) cos 2π x ) 4L where v(l, t) = C d A ft (x v ) A pipe 2 ρ (p(l, t) p ) Solve the above equations for p(l, t) and v(, t).

34 The Quarter-wave model (QWM) Simplest case (liquid, one mode only): Aim: replace the PDEs describing the pipeline dynamics to ODEs that allow stability analysis. Ansatz: p(x, t) = p t (t) ρ ( 2 v(, t)2 + B(t) sin 2π x ) ( 4L v(x, t) = v(l, t) + C(t) cos 2π x ) 4L where v(l, t) = C d A ft (x v ) A pipe 2 ρ (p(l, t) p ) Solve the above equations for p(l, t) and v(, t). Then use 1-point collocation technique (PDE ODE).

35 The Quarter-wave model (QWM) Simplest case (liquid, one mode only): Aim: replace the PDEs describing the pipeline dynamics to ODEs that allow stability analysis. Ansatz: p(x, t) = p t (t) ρ ( 2 v(, t)2 + B(t) sin 2π x ) ( 4L v(x, t) = v(l, t) + C(t) cos 2π x ) 4L where v(l, t) = C d A ft (x v ) A pipe 2 ρ (p(l, t) p ) Solve the above equations for p(l, t) and v(, t). Then use 1-point collocation technique (PDE ODE). One can perform the same computation for arbitrary wave modes.

36 The simplest QWM x v = v v v v = κv v (x v + δ) + Ãeff (p + B) p = β (q µ(v end + C)) B = π α 2 γ C ( C 2p 2 + φ + 2Cv end + ) 2vend 2 2 C = π 1 2 αγ B 2v end v end = σx p + B

37 The simplest QWM x v = v v v v = κv v (x v + δ) + Ãeff (p + B) p = β (q µ(v end + C)) B = π α 2 γ C ( C 2p 2 + φ + 2Cv end + ) 2vend 2 2 C = π 1 2 αγ B 2v end v end = σx p + B... and one can also add pipe friction, convective terms, more modes (however, you might want to use a computer algebra system).

38 Analytical stability criteria Assume large reservoir (β ) y 3 konstans.

39 Analytical stability criteria Assume large reservoir (β ) y 3 konstans. The pipeline dynamics is ( ) π/2 2 B + B = konst. α γ γ σ d ( Y P + x ) + B +O(β), dτ P

40 Analytical stability criteria Assume large reservoir (β ) y 3 konstans. The pipeline dynamics is ( ) π/2 2 B + B = konst. α γ γ σ d ( Y P + x ) + B +O(β), dτ P and the valve dynamics is Y = Y + B κy.

41 Analytical stability criteria Assume large reservoir (β ) y 3 konstans. The pipeline dynamics is ( ) π/2 2 B + B = konst. α γ γ σ d ( Y P + x ) + B +O(β), dτ P and the valve dynamics is Y = Y + B κy. Close to the stability boundary: B(τ; τ 2 ) = A(τ 2 ) cos(ω p τ),

42 Analytical stability criteria Assume large reservoir (β ) y 3 konstans. The pipeline dynamics is ( ) π/2 2 B + B = konst. α γ γ σ d ( Y P + x ) + B +O(β), dτ P and the valve dynamics is Y = Y + B κy. Close to the stability boundary: B(τ; τ 2 ) = A(τ 2 ) cos(ω p τ), with which the valve displacement becomes: Y = 1 1 ωp 2 B + O(κ)

43 Analytical stability criteria (cont d) The pipe dynamics is: ( π/2 B + γ ) 2 B = konst. α ( γ σ X P ) 2 P ωp 2 1 }{{}!> B

44 Analytical stability criteria (cont d) The pipe dynamics is: ( π/2 B + γ ) 2 B = konst. α ( γ σ X P ) 2 P ωp 2 1 }{{}!> For small q values, the equilibrium (X, P ) can be expanded into Taylor series and given in closed form. B

45 Analytical stability criteria (cont d) The pipe dynamics is: ( π/2 B + γ ) 2 B = konst. α ( γ σ X P ) 2 P ωp 2 1 }{{}!> For small q values, the equilibrium (X, P ) can be expanded into Taylor series and given in closed form. After some algebra, one can arrive ar q 2 δ 3/2 µσ((ω p(l)) 2 1), which is straightforward to implement even in a spreadsheet software. B

46 CFD (red/blue) vs. 1D model (black) vs. QWM analytical (blue)

47 Stability diagram - QWM vs. meas. 2J3 valve 15 2J3, ring: 5 and 25 3% rule pipe length, foot 1 5 QWM prediction flow, percent of capacity

48 Table of Contents 1 Introduction, motivation 2 CFD 1D model 3 4 5

49 Simplified model without pipe Close-coupled valve Small reservoir The resulting model is: y 1 = y 2 y 2 = κy 2 (y 1 + δ) + y 3 y 3 = ˆβ (ˆq y 1 y3 )... with the impact law at y 1 = : (y 1, y 2, y 3 ) T (y 1, ry 2, y 3 ) T. m r,in x x v A ft (x v ) ξ s m p v. V, p r k m r,out. m v D, A, L v p (ξ,t), p (ξ,t)

50 Bifurcation diagram 12 1 ütközéses periodikus pálya, stabil periodikus nincs egyensúlyi pályák ütközés helyzet 8 6 Hopf bifurkáció 4 2 grazing bifurkáció m6=.5 m5=1. 2 m4= m3=6.5 m2=7.1 m1=8 9 1

51 Some orbits: 1 y y y y y y 2 y y y y y y 2

52 Continuation strategy Formulate the problem as a BVP, i.e. y = TF (y) with y 1 () = y 1 (1) = ry 2 () = y 2 (1) y 3 () = y 3 (1)

53 Continuation strategy Formulate the problem as a BVP, i.e. y = TF (y) with y 1 () = y 1 (1) = ry 2 () = y 2 (1) y 3 () = y 3 (1) Use pseudo-arclength cont. to track periodic orbits (+1 BC).

54 Continuation strategy Formulate the problem as a BVP, i.e. y = TF (y) with y 1 () = y 1 (1) = ry 2 () = y 2 (1) y 3 () = y 3 (1) Use pseudo-arclength cont. to track periodic orbits (+1 BC). Stability: solve variational equation to compute monodromy matrix and apply correction at the impact (see Bernardo, M., Budd, C., Champneys, A.R., Kowalczyk, P.: Piecewise-smooth Dynamical Systems: Theory and Applications, Springer, 28, ISBN ).

55 Continuation strategy (cont d) Continuation of grazing points: y 2 () = ( BC)

56 Continuation strategy (cont d) Continuation of grazing points: y 2 () = ( BC) Continuation of period doublings: one of the characteristic multipliers is -1. Problems with accuracy!

57 Continuation strategy (cont d) Continuation of grazing points: y 2 () = ( BC) Continuation of period doublings: one of the characteristic multipliers is -1. Problems with accuracy! Implemented in Matlab, using bvp5c.

58 Qualitative bifurcation diagram Br6 Br5 Br4 Br3 y3 y3 y3 y3 y1 y2 GR3 GR4 Br7 PD4 Br2 Br5 Br6 GR1 y3 PD PD PD2 Br4 GR2 PD1 PD3 Br3 Br2 HB Br1 x

59 T (periódus) Shilnikov-like orbit.5 (8) 15.5 (7) (6) (4) (5) (3) (1) (2).4 y 1.5 y y y 2

60 Shilnikov-like orbit.5 y y 2.1 y y y 2 y t

61 Table of Contents 1 Introduction, motivation 2 CFD 1D model 3 4 5

62 Modeling levels CFD: few hours full 3D, transient 1D unsteady model: few minutes 1D, transient QWM: seconds 1D, only close to equilibrium (no large-amplitude oscillations) Analytical:??? only around the stability boundary, assumptions need to be checked Impacting periodic orbits Relatively new mathematical results. Standard nonlin. dyn. toolkit can be used. Surprisingly rich dynamics.

63 Thank you for you attention!

64 Effective area A simple yet accurate estimate for the fluid force is essential: F fluid = p(a)da + F imp (ṁ, β) (A)

65 Effective area A simple yet accurate estimate for the fluid force is essential: F fluid = p(a)da + F imp (ṁ, β) (A) Define effective area as A ft valve disc A eff A F fluid = A eff (x) p v f,j β v f,v xv % of full lift D

66 Effective area theory vs. CFD Aeff/Apipe [-] x [%]

67 Effective area two more examples Cf [ ] p = 5 bar.6 p = 1 bar measurement X [ ] Cf [ ] p = 5 bar.6 p = 1 bar X [ ]