Numerical analysis of a free piston problem

Transcription

1 MATHEMATICAL COMMUNICATIONS 573 Mat. Commun., Vol. 15, No. 2, pp (2010) Numerical analysis of a free piston problem Boris Mua 1 and Zvonimir Tutek 1, 1 Department of Matematics, University of Zagreb, Bijenička 30, HR Zagreb, Croatia Received October 28, 2009; accepted June 22, 2010 Abstract. Te problem considered is te Stokes and Navier-Stokes flow troug a system of two pipes in te gravity field; inside a vertical pipe tere is a free eavy piston. Teoretical analysis, te existence and non-uniqueness of solution, as been completed recently by te autors. Here we present numerical analysis, using finite elements metods, of te stationary state wit respect to te angle between te two pipes, diameters of te pipes, searc for solution of a full problem and searc for bifurcation points. Te analysis is carried out for bot, 2D and 3D, and for Stokes and Navier-Stokes case. AMS subject classifications: 65N30, 65N99, 76D05 Key words: free piston problem, Navier-Stokes equations, numerical analysis 1. Introduction 1.1. Notation and geometry of te problem We consider a flow of incompressible Newtonian eavy fluid in a system of two converging pipes. Te system is constituted by a orizontal and a vertical pipe. Te orizontal pipe F 1 is infinite wit a constant cross-section S 1 of diameter d 1. Neverteless, we consider only a control volume Ω 1 of lengt L wic is large enoug to let, in correspondence wit constant pressure gradient, Poiseuille flow develop at te two exits, Σ p and Σ k, respectively. At te center of Ω 1 on a rigid lateral wall te second semi-infinite pipe F 2, wit a constant cross-section of diameter d 2, is connected. Te axis of F 2 is inclined to te vertical by an angle α. Inside te vertical pipe F 2 we ave a eavy piston, wit cross-section of diameter d, wic can translate freely along te vertical pipe F 2 witout rotations. A lower basis of te piston is orizontal (see Figure 1). Terefore, d and d 2 cannot be cosen arbitrary but some compatibility condition must be satisfied. In all numerical examples crosssections S 1 and S 2 will be circular and ten we ave d = d 2 / cos α. Friction between te wall of F 2 and te piston is neglected. Fluid enters te vertical pipe F 2 only up to equilibrium eigt of te piston. Let us call Ω 2 te region of te F 2 filled wit te fluid. Let Σ denote te upper orizontal boundary of Ω 2. Te piston is modeled as a rigid body. It is in equilibrium if and only if te total force iz zero. Since te piston is a rigid body, fluid is at rest on Presented at te Sixt Conference on Applied Matematics and Scientific Computing, Zadar, Croatia, September Corresponding autor. addresses: borism@mat.r (B. Mua), tutek@mat.r (Z. Tutek) ttp:// c 2010 Department of Matematics, University of Osijek

2 574 B. Mua and Z. Tutek Σ. Motion of te fluid is given by Navier-Stokes equations. We will consider bot, a 2D and 3D case. Te goal of tis paper is to numerically determine stationary fluid flow and equilibrium position of te piston and analyse its dependence on geometry, i.e. on an angle α and a ratio d1 d 2. Furtermore, we will analyse non-uniqueness of te stationary flow. Te main realistic example will be blood flow troug arterioles. Figure 1. Ω α Let us now introduce some notations and precise assumption on te geometry. Coordinate x (in 3D case x = (x 1, x 2 ) and in 2D case x = x 1 ) is along te orizontal pipe and y is in te opposite direction of acceleration of gravity. Let be te eigt of te piston in te selected coordinate frame, Ω α Rn, n = 2, 3 te domain occupied by te fluid. More precisely, Ω α = Ω 1 Ω 0 Ω,α 2, were Ω 1 = {(x 1, x 2, y) ; L x 1 L, (x 2, y) S 1 }, S 1 R 2. Let s = cos αe x1 + sin αe y be te direction of te vertical pipe F 2. Ten in te non-ortogonal coordinate frame (e x1, e x2, s), Ω,α 2 as te form: Ω,α 2 = {(z 1, z 2, z 3 ) ; 0 z 3 / cos α, (z 1, z 2 ) Σ}, Σ R 2. Only Ω,α 2 depends on and α. Te lower basis of te piston Σ will be considered as a subset of te y = const plane. Te origin of te coordinate frame is cosen in a suc way tat te lower end of te vertical pipe Σ 0 is a subset of y = 0 plane. We assume tat Σ 0 is a symmetric w.r.t. plane perpendicular to te central axis of te orizontal pipe. Furtermore, we suppose tat Ω 0 Ω 1 (a domain witout te vertical pipe) is also a symmetric w.r.t. plane perpendicular to te central axis of te orizontal pipe; tis is a tecnical assumption and not a restriction. Note tat Ω 0 is an extension of te vertical pipe up to te boundary of Ω 1 ; its sape is complicated in general in a 3D case, in a 2D case it is empty set. Inflow and outflow regions are denoted by Σ p and Σ k, respectively. Γ = Ω α \ (Σ p Σ k Σ ) is a rigid boundary. We suppose tat te domain is locally Lipscitz. Tis is a natural assumption because one cannot expect a smooter domain because we will always ave an angle at te contact of te piston and rigid boundary.

3 Numerical analysis of a free piston problem Formulation of te problem Since te fluid is modeled by te Navier-Stokes equations, te stress tensor is given by T = pi +2µ sym( u), were u is velocity of te fluid, p pressure and µ viscosity. Total fluid force on te piston in direction s is given by formula: F α () = T n s, (1) Σ were n denotes te unit outer normal; ere n = e y. For simplicity, we will omit index α in te unknowns. Differential formulation of our problem is: find (u, p, ) H 1 (Ω α )3 L 2 (Ω α ) R + suc tat µ u + ϱ( u)u + p = gϱe y in Ω α, div u = 0 in Ω α, u = 0 on Γ, u = 0 on Σ, u n = 0, p = P p/k gϱy on Σ p/k, F α () = P 0. (2) Here ϱ is te density of te fluid, µ dynamic viscosity, g gravity constant and P 0 constant tat takes into account te weigt of te piston and atmosperic pressure. Te angle α is given. Te first two equations are just te Navier-Stokes equations for stationary flow of an incompressible Newtonian fluid. Boundary conditions (2) 3 and (2) 4 are no slip boundary conditions on rigid boundary. Condition (2) 6 is a balance of forces on te piston. F α () is well defined because wit our coice of function spaces we ave T L 2 (Ω α ) 3 3 and div T L 3 2 (Ω α ) 3. P p and P k are constants and since flow is driven by te difference of pressures on inflow and outflow boundary, we can assume tat P p = P k (wit a possible redefinition of constant P 0 ). Fixing te pressure does not affect total force on te piston since relevant quantity is te difference between te atmosperic pressure and fluid pressure. Term gϱy in (2) 5 comes from te ydrostatic pressure. Tese types of non-standard boundary conditions involving pressure were studied in [1] and [3]. Te problem as two non-linearities. One tat comes from te Navier-Stokes equations is classical (see [6]). Te second comes from te fact tat te domain is unknown and terefore F α is a nonlinear function. We will also consider Stokes case were te fluid is modeled wit te Stokes equations. We will often refer to te Stokes or Navier-Stokes problem in a fixed domain, more precisely for R + fixed, find (u, p) H 1 (Ω α )3 L 2 (Ω α ) suc tat (2) 1 5 olds. Tis problem will be denoted by (S ), (NS ) for Stokes and Navier-Stokes case, respectively. 2. Overview of teoretical results In tis capter we will briefly summarize teoretical results; details can be found in [4]. Since domain Ω α is only locally Lipscitz and as concave points, solution

4 576 B. Mua and Z. Tutek (u, p ) of Navier-Stokes system (NS ) as only H 1 L 2 regularity. However, by using tecniques from [3] it can be proven tat formula (1) can be understood in an H 1/2 (Σ ) sense. Lemma 1. Let (u, p ) be te solution of Stokes (S ) or Navier-Stokes (NS ) system in Ω α. Ten for every 0 we ave F α () = cos α p µ sin α Σ y (u ) x1, Σ were te integral is taken in an H 1/2 (Σ ) sense. Furtermore, we ave an existence result. Teorem 1. In te Stokes case, tere exists P R suc tat for every P 0 P problem (2) as at least one solution (u, p, ) H 1 (Ω ) L 2 (Ω ) R +. In te Navier-Stokes case we ave te same conclusion provided tat data P p and P k are small enoug. In general, even in te Stokes case we do not ave a uniqueness result and bifurcation penomena can occur. More precisely, we ave te following result: Teorem 2. Tere exist α and P 0 wit corresponding stationary state (u 0, p 0, 0 ) in wic we ave a turning point. More precisely, (u 0, p 0, 0 ) is a solution of te Stokes case of problem (2) and all solutions of tis problem in some neigborood of (u 0, p 0, 0 ) belong to some curve (X(s), P (s)) wit X(0) = (u 0, p 0, 0 ) and P (0) = P 0. Furtermore, tangent at (X(0), P (0)) is (V, 0) and P does not ave a saddle point at 0. In te next capter we will give a numerical illustration of tis teorem. Numerical results suggest uniqueness of te bifurcation point. One of te goals of tis paper is a numerical searc for te bifurcation point and analysis of its dependence on geometry. In order to do tat, we will need some additional teoretical results about function F. As a corollary in proof of te Teorem 2 we concluded tat function F defined in (1) belongs to C 1 (R + ). Furtermore, we can get an expression for its derivative wit respect to : (F α ) () = cos α P µ sin α ( y U x1 1 ) Σ Σ y(u 0 ) x1. Here P is a solution of te problem:

5 Numerical analysis of a free piston problem 577 find (U, P ) V L 2 (Ω α ) suc tat U v P v = 1 ( u 0 v p 0 v Ω α Ω α Ω,α 2 ( ) 0 u tan(α) x + 0 v y 1 ( ) 0 u 0 v (3) 0 tan(α) x + y ( ) 0 ) p 0 v + p 0 v, v V, tan(α) x + y q U = 1 (tan α x (u 0 ) y + y (u 0 ) y ), q L 2 (Ω α ), Ω α were (u 0, p 0 ) is a solution of Stokes system (S ) in Ω α and Ω 2 V = {v H 1 (Ω α ) 3 ; v = 0 on Γ Σ, v n = 0 on Σ p/k }. We will use tis formula later in Example Numerical experiments In tis section we will present numerical experiments tat provide better understanding of te problem. All experiments in a 2D case are done using FreeFem and visualization is made by Matematica 5.2. Triangulation and visualization of solutions in a 3D case are made using Gms, wile matrix assembly and solving of linear systems are done in FreeFem. In bot, a 2D and a 3D case, continuous piecewise quadratic and continuous piecewise linear elements for velocity and pressure, respectively, are used. Mes size is indicated by density of arrows in velocity field figures. Before stating results of numerical experiments we want to empasize one simple fact. Let p H (x, y) = ϱ g y be te ydrostatic pressure. Ten all solutions of system (NS ) ave form (u, q + p H ), were (u, q ) is a solution of omogenous system (NS ) (g = 0). In te sequel we will present results for (u, q ) since ydrostatic pressure can be added easily Stokes case Wen we consider Stokes case, we can, witout loss of generality, assume µ = 1 because if (u, p) is a solution of problem (S ) for µ = 1 ten (u/µ, p) is a solution of problem (S ) for general µ D case In te next few examples we will analyse dependence of function F α () on α. Geometrical parameters are L = 10, P p/k = ±5, d 1 = 1.6, d = 1.6 and we vary α and.

6 578 B. Mua and Z. Tutek Example 1. Tis example illustrates asymptotic beaviour of function F α. We take α = π 3 fixed and vary eigt from 0 to 16 wit step For every we solve problem (S π 3 ) in Ω π 3 and compute F π 3 (). We keep te maximal diameter of a triangle in triangulation approximately te same. In Figure 2 we can see tat function F π 3 approaces its asymptotic value very fast. F Π Figure 2. Grap of function F π 3 Example 2. Te second example sows a grap of function F α. For every, Stokes system (S ) in Ω α is solved and F α () computed. Notice tat in tis example we solve a series of Stokes problems (S ) in fixed domains and we do not solve full original problem (2). Figure 3. Graps of functions F 1.05, F 0.79, F 0.31 and F 0.11 In Figure 3 graps of F 1.05 (red), F 0.79 (green), F 0.31 (blue) and F 0.11 (violet) are given. One can notice tat for all α, F α ave te same qualitative beaviour, i.e. first, it is strictly monotone on some interval and ten it as a critical point after wic it asymptotically approaces some constant at infinity. Asymptotic decay can be proved using Leray s flow (see [4]). Tis example is a good illustration of Teorem 2 because we can see tat F attains te same value for different eigt. Since for every we can find related (u, p ), we ave two stationary states for some P 0. Example 3. Wit tis example we will illustrate non-uniqueness of a solution of problem (2) more precisely. We will sow two different flows wic act upon te

7 Numerical analysis of a free piston problem 579 piston wit te same force. We take α = π 3 and solve problem (S ) for 1 = and 2 = Ten we compute F π 3 (0.465) = and F π 3 (0.98) = Furtermore, we compute total fluid force on te piston in point 3 = 0.6, F π 3 (0.6) = Figure 4. Velocity field in domain Ω π Figure 5. Velocity field in domain Ω π Example 4. We take fixed outer pressure P V = and solve full problem (2) using Newton s metod. More precisely, we use Newton s metod for solving equations F α () = P 0 (α, d), were P 0 (α, d) = P V cos α d. We will vary parameters of geometry (α and d) and consider dependence of equilibrium eigt of te piston on geometry d Figure 6. Dependence of equilibrium eigt on diameter of te piston d

8 580 B. Mua and Z. Tutek Figures 6 and 7 sow dependence of equilibrium eigt of te piston on diameter of te piston d and angle α, respectively. One can see tat eigt is greater for greater d and greater α. Te solution is in general not unique, so a numerical solution depends on initial guess in Newton s metod. In tis example we took initial Α rad Figure 7. Dependence of equilibrium eigt on angle α guess on interval were F is decreasing. If we took initial guess on eigt were F stabilizes and g 0, equilibrium eigt would depend only on te asymptotic value of pressure as. Namely, we know tat F α asymptotically approaces to some constant cos αc α (see [4] and Figure 2) and terefore a balance of forces is approximately reduced to cos αc α ϱg = P 0 (α) for great enoug. In te next example we will analyse constant C α more closely. Example 5. As we ave proven in [4], for every given geometry and data tere exists constant C α suc tat lim F α () = cos αc α. Te goal of tis example is to analyse dependence of constants C α on parameter α. We take all parameters te same as in te previous example, = 3 (wic is large enoug) and vary parameter α between 0 and 1 3π wit step π 120 and compute F α (3). Since it is easy to verify symmetry property F α () = F α (), we do not need to compute negative α-s. Again, we empasize tat ere we do not solve full problem (2), but problem (S ) for some fixed parameters in order to provide better understanding of qualitative beaviour of total force F α. 0.5 CΑ Α rad Figure 8. Dependence of constant C α on angle α Figure 8 strongly suggests linear dependence of asymptotic force on angle α.

9 Numerical analysis of a free piston problem 581 Example 6. We compute bifurcation point B for various α by solving equation (F α ) () = 0. Te bisection metod is used because computation of (F α ) is too expensive and would result in an additional numerical error. Tis time we take α < 0. We can directly compute results for α > 0 using symmetry property F α () = F α (). B Α rad Figure 9. Dependence of bifurcation point B on angle α F Α B B Figure 10. Value of function F α in bifurcation point Figure 9 sows tat te bifurcation point will be reaced earlier for α closer to 0. Tis is in accordance wit Figure 3 and related discussion afterwards. Figure 10 as a pysical interpretation of maximal total outer force in direction s on te piston tat te flow can support D case In a 3D, case numerical experiments suggest similar beaviour of solutions as in a 2D case, but of course computations are more complex and terefore we cannot use triangulations fine enoug to ensure a numerical error to be as low as in a 2D case. Here we present an example illustrating it. Example 7. Geometrical parameters are L = 10, P p/k = ±5, d 1 = 3, d = 2.

10 582 B. Mua and Z. Tutek F Figure 11. Grap of function F π 3 CΑ Α rad Figure 12. Dependence of C α on angle α Figure 11 sows analogous beaviour of function F α as in a 2D case. As we ave explained before, small variations after te bifurcation point are due to a numerical error. Figure 12 suggests tat linear dependence of constant C α on α is not a property of a 2D case, but it also olds in a 3D case Navier-Stokes case For every fixed, we will solve te Navier-Stokes problem (NS ) using Newton s metod described in [2]. In every step of Newton s metod we will solve te linearized Navier-Stokes system. We will use a solution of te Stokes system for initial guess in Newton s iterations. Terefore, in tis section we will only consider laminar flow for wic we also ave teoretical results (see Section 2). In te Navier-Stokes case we will lose symmetry properties tat we ad in te Stokes case and we will illustrate tose differences wit te following few numerical examples. Example 8. We will consider blood flow troug larger arterioles. Since in tis case vessel diameter is muc larger tan cell diameters, we can assume tat blood as constant viscosity µ = Pa s and it is Newtonian (see [5]). Oter data are: density ϱ = 1060 kg m, diameters of vessels d 3 1 = 0.05mm, d = 0.03mm, lengt L = 0.4mm (part of tis lengt is enoug to sow effects near te junction) and pressure drop δp = 125 P a m. Reynolds number of tis flow is Furtermore, we will consider te same flow in a 2D case and make analysis of dependence of total force on te angle and eigt.

11 Numerical analysis of a free piston problem 583 Figure 13. Solution (u,p) in Ω 0 Figure 14. Solution (u,p) in Ω 1 Figures 13 and 14 (ere x = (x, y), y = z) sow solution (u, p) of problem (NS ) in Ω 0 (case wen pipes are perpendicular) and Ω 1 (just orizontal pipe), respectively. One can see tat we ave taken a segment of arteriola tat is long enoug to allow a fully developed Poiseuille flow in te orizontal pipe. Te only significant difference between flows in Figures 13 and 14 is near te junction of te pipes and terefore only tis part is plotted. F Π 4 mn Figure 15. Grap of function F π 4

12 584 B. Mua and Z. Tutek CΑ mn Α rad Figure 16. Dependence of total force C α on angle α Figure 15 sows tat te function of total force F α as te different qualitative, but same asymptotic beaviour as in Stokes case. From Lemma 1 we see tat x 1 component of force F α is negligible due to te smallness of parameter µ (µ = kg/mms). However, due to te fact tat te flow is still laminar, we ave only a negligible non-linear effect from te fluid and beaviour of te total fluid pressure on te piston stays te same as in te Stokes case. Finally, Figure 16 sows tat linear dependence of total force on α was a linear effect and in Navier-Stokes case we do not ave tis effect anymore even in te case of a laminar flow. One can notice tat blood flow troug arterioles as a very low Reydnols number (around 0.07) and terefore non-linear effects are not so pronounced. If we take a more turbulent flow (Reynolds number around 800), yet still laminar (Figure 17), we can notice tat dependence of force on angle α as different qualitative beaviour. However, qualitative beaviour of F α in dependence of stays te same as in te case of te Stokes flow and te blood te flow (Figure 15) because flow is still laminar. CΑ Α rad Figure 17. Dependence constant C α on angle α References [1] C. Conca, F. Murat, O. Pironneau, Te Stokes and Navier-Stokes equations wit boundary conditions involving te pressure, Japan J. Mat 20(1994), [2] V. Girault, P. A. Raviart, Finite Elements Metods for Navier-Stokes Equations, Teory and Algoritams, Springer-Verlag, Berlin, 1986.

13 Numerical analysis of a free piston problem 585 [3] E. Marušić-Paloka, Rigorous justification of te Kircoff law for junction of tin pipes filled wit viscous fluid, Asymptot. Anal. 33(2003), [4] B. Mua, Z. Tutek, On a free piston problem for Stokes and Navier-Stokes equations, submitted, available at ttp://web.mat.r/$\ticksim$borism/preprints/free\_piston.ps. [5] J. T. Ottesen, M. S. Olufsen, J. K. Larsen, Applied Matematical Models in Human Pysiology, SIAM Monograps on Matematical Modeling and Computation, Piladelpia, [6] R. Temam, Navier-Stokes equations: teory and numerical analysis, AMS Celsea Publising, Providence, 2000.