One dimensional study of the depressurization process and solution of the Riemann Problem
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(2) One dimensional study of the depressurization process. Bertrand Mercier Nov 13, 2020 Introduction In pressurized water reactors, depressurization occurs when there is a breach on the primary circuit, this is also called flash evaporation although it takes some time, as we shall see. We shall study this phenomenon numerically. Then we shall need an equation of state for equilibrium two-phase mixtures (§1). We shall use 1D Lagrangian hydrodynamics and compare the Wilkins scheme [5] with the acoustic solver scheme [1] (§2 and §3). We have used non conservative schemes, but we prove in §4 that, for this isentropic process, it gives almost the same results as the conservative version of the acoustic solver scheme. We then consider the steam explosion in §5, for which we just need a simple stiffened gas equation of state, and that we solve with the conservative version of the acoustic solver scheme. In §6, we come back to the study of the depressurization process for a pressurized but sub-saturated liquid for which we need to combine both equations of state. In §7 we find the exact solution of the Riemann problem in the two-phase case (with tabulated equation of state) and finally, in §8 we give the Lagrangian solution with the acoustic solver of the Riemann Problem considered in §7. §1 Equation of state. In what follows the subscript (resp. ) stands for liquid (resp. steam) We want to define an equation of state for an equilibrium diphasic mixture (steam + liquid). Such a mixture is at the saturation temperature . Let denote the liquid mass fraction. We know that the saturation pressure and the Gibbs potential depend on only, not on , provided that 0 < < 1. We have the following result Faccanoni et al [6] or Corot [2 ,p.158] : Result : Let and be given : 1. Assume there exists
(3) ∗ , ∗ , ∗ , ∗ , ∗ with 0 < ∗ < 1 and solving the following system
(4) , = ,
(5) , = ,
(6) , = , . = + 1 − = + 1 − Then the equilibrium state is a diphasic mixture and the associated entropy is = ∗
(7) ∗ , ∗ + 1 − ∗ ∗ , ∗ 2. Otherwise , defines a monophasic state. ∎. If we have an EOS for each phase, this result is used to evaluate the specific volume, the internal energy and the entropy of each phase at saturation. However if we know them, this result can also be used to find , and , when and are given.. We use numerical values given by Faccanoni [4] and reproduced in Annex 1. There are 99 lines in the table. For 1 ≤ ≤ 99, the table gives a value for the saturation temperature and the 7 values , , , , , , . When = 1 − + we interpolate these values linearly so that e.g.. 1.
(8) = 1 − + . Method A : to compute , and , when and are given : Let ! =
(9) − $" − # % =
(10) − $" − #. To compute we just have to solve the equation ! = % . This is a non linear equation with one unknown which can be easily solved by • finding such that ! > % and ! < % ≤ ≤ • solving a second degree equation to find such that = 1 − + and then . (Indeed, we have to solve
(11) − " − # =
(12) − " − #. •. where the functions , , and are all linear in ) Let ∗ denote the common value of ! and % we let = ∗ + 1 − ∗ ∎. Method B : to compute , and , when and are given : In the same way, we solve ! = ' = ∗ ! =
(13) − $" − #. ' =
(14) − $" − #. The details are left to the reader. ∎. Method C : to compute , when and are given : This is still easier : being given, first we evaluate and then compute ∗ =
(15) − $" − #. So that = ∗ + 1 − ∗ . ∎. Test of our equation of state. We let = , : a well-known result in thermodynamics (see e.g. [4]) is that we should have ( (!. = −. To check that this is the case, we have selected = 4.4957 and 5.81494≤ ≤ 10.46689. We compute the derivative of w.r.t both by forward and backward difference. The results given in Fig.1 show a rather good agreement that make us confident with the validity of our equation of state. (1). 2.
(16) 678 16 15 14 13 12 11 10 9 5. 6. 7. 8. depsdtau1. 9 P. 10. 11. depsdtau2. Fig.1 pressure p vs − / ⁄/. 3/5. Sound speed When we select , as the primitive thermodynamic variables, we have (2) 1 = 2− /⁄/. provided we use international units for each variable. When we select , as the primitive thermodynamic variables, we shall see in §2 that (3) 1 = 2. /⁄/ − /⁄/. also in international units. In the following test we replace partial derivatives by finite differences, and we get the results given in Fig 2. We notice that the sound speed in a diphasic mixture is much lower than in the liquid phase, where it is of the order of 1800 m/s. This result is well known.. Sound speed (m/s) by 2 methods w.r.t mass rate 450 400 350 300 250 200 150 100 50 0 0. 0.2. 0.4 sonbis. 0.6. 0.8. 1. son. Fig 2 Sound speed evaluated either with (2) or (3) as a function of the steam mass fraction x (there are 2 superposed curves). 3.
(17) §2 The Wilkins scheme In Lagrangian coordinates, in 1D, the gas dynamics system can be written ( see e.g. B. Després [3] formula (2.21) page 20) : (4) (5). (! (; − : =0 (9 (< (; (= + : =0 (9 (< (> ( + : (< ? (9. =0. Where = 1⁄@ and A denotes the mass variable such that = @. : BC , @ = @C is the volumic mass, : denotes the tube section, and C is the space variable. Moreover D = + ? E ⁄2 Is the total energy per unit mass. In §2 and §3 we shall assume that the flow is isentropic, so we don’t need the energy equation (6) which, for smooth solutions, is equivalent to (6). (7). (' (9. =0. Proof When the solution is differentiable (6) implies (% (9. + ? (9 + : (< + : ? (< = 0. (% (9. + (9 = 0. (% (9. (;. + : (< = 0. (;. (=. (;. so that, using (5), we get (6b). (!. Using (4), we finally get : Which proves (7) since B = B + B from the second law of thermodynamics ∎. Remark 1 : 1. With (6b) we see that our system of equations is an hyperbolic system : with = , we have 0 −: 0. ( ( G?H + I: /⁄/ 0 : /⁄/ J (< G?H = 0 (9. 0 : 0 Provided that . /⁄/ − /⁄/ > 0 we can define K = 2. /⁄/ − /⁄/ , and see that this matrix has 3 real eigenvalues ±:K and 0. In the MA, NO the equation of characteristics is BA = ±:K BN. Since BA = @: BC, we see that @ BC = ±K BN or BC = ±1 BN with 1 = K = 2. /⁄/ − /⁄/. Which proves that 1 is the sound speed. Note that K is the impedance. 2. We could have selected : = 1 as Despres does in [3]. However if we do that we get BA = @ BC which is non homogeneous in terms of units. ∎ The Wilkins scheme [5]: We introduce a « mesh » for variable : … A QE < A Q < A < A < A E …. Similarly we discretize time with a timestep ∆N, so we shall compute the approximate solution at the following discrete times : 0 < ∆N < 2∆N … < S∆N < S + 1∆N < ⋯. 4.
(18) Our target is to simulate the depressurization process of a saturated liquid, which is represented by a rarefaction wave and is then isentropic. U In the Wilkins scheme, pressures and volumic masses are cell centered, so we denote /E and U. /E their values at time S∆N on cell A , A . However, velocities are defined on a staggered grid : ? time S + 1/2∆N. The discrete analogs of (4) and (5) are :. U /E. (7) (8). Z Z[X !VWX/Y Q!VWX/Y. ∆9. − :. ZWX/Y Z[X/Y ;V Q;V. ∆9. Z[X/Y. ;VWX. + :. Z[X/Y. Q;V. <VWX Q<V. denotes the velocity at node A and. =0. Z Z =VWX/Y Q=V[X/Y. <VWX/Y Q<V[X/Y. =0. For practical purpose, it is useful to come back to the physical space : the mesh in the physical space is updated as follows : (9). \VZWX Q\VZ ∆9. = ? . U /E. Such a mesh is following the fluid. Given the new grid C U at time S + 1∆N, rather than using (7) (which means mass conservation) U we compute the new volume :.
(19) C − C U of cell number + 1/2. Its mass ] /E being constant we deduce (10). U . /E =. (11). ] . ZWX ^.
(20) \VWX Q\VZWX . _VWX/Y. .. U Knowing /E and the entropy in each cell, we can use the equation of state to get the pressure U /E On the other hand, we notice that (8) is equivalent to ZWX/Y. ;V. Z[X/Y. Q;V. ∆9. where ] = E
(21) ] . /E. U + : .
(22) . + ] Q. /E . /E. U − Q. /E . =0. = E A − A + A − A Q = E
(23) A + A − A + A Q . which is nice since (11) means that for node mass multiplied by acceleration is equal to the resulting force acting on it.. Boundary conditions We shall assume that the depressurization process begins at time t=0, with a breach on the left side of the tube C`` = 0 8N N = 0. The pressure ∗ is given on the left boundary C = C`U .. On the right side we assume that there is a wall so that we set ?a. U /E. = 0.. The numerical results obtained with the Wilkins scheme are given in §3.. §3 The acoustic solver scheme. [1, 3] We have seen on Fig.1 that, if we write = , , for fixed , , is a decreasing function of , so that the sound speed 1 = 2− /⁄/. exists. In this section, we shall assume the flow to be isentropic which means that is constant. Let @ = , , with = 1⁄@, we have / B. 1 / / = − E = − E = 1E b @ = / B@ @ /. /. 5.
(24) 1 = 2b @ The acoustic solver uses the following remarks on the linearized problem. We consider a small perturbation ? , @ of a permanent flow @` , ?` . Let ` = 1⁄@` , ` = @` et 1` = 2b @` . We shall have @` + @ ≅ @` + b @` @ = ` + 1`E @ Hence ≅ ` + 1`E @ and = 1`E @ . In the same way, we have B = − B@⁄@E so that = − @ ⁄@`E and = −1`E @`E. So that. Replacing ? by ?` + ? , @ by @` + @ in (4) and (5) We get ( ( + − : ?` + ? (9 ` (< ( ( ?` + ? + : ` + (9 (< (. (9 ( ? (9. − :. ( ? =0 (< ( − 1`E @`E : (<. and then. =0. =0 or. =0. or. ( (9 ( ? (9. + 1`E @`E : +. ( : (< . ( ? (<. =0. =0. which is nothing but the linear wave equation, which can be solved by the method of characteristics : we have ( (9 ( (9. + @` 1` ? + @` 1` : (< + @` 1` ? = 0 (. − @` 1` ? − @` 1` : (< − @` 1` ? = 0 (. There are two Riemann invariants : let K` = @` 1` : d = − K` ? which satisfies ( d (9. − K` : (< d = 0 and (. e = + K` ? which satisfies ( e (9. + K` : (< e = 0 (. Let C = A/@` : we also have ( d (9. − 1` (\ d = 0 (. dC, N = fC + 1` N In the same way eC, N = gC − 1` N The solution is the sum of two parts : the first one propagates at speed -1` ; the other one at speed + 1` .. so that. Remark 2: Assume that we have an interface in A = 0 such that @ A < 0 @` A = h i k @j A > 0 A < 0 We shall have ` A = h i k with j A > 0 i = i @i et j = j @j Since there is no shock, we must have continuity of the pressure, and the velocity, so that in particular i @i = j @j . 6.
(25) However, it is possible that 1i ≠ 1j and Ki = @i 1i ≠ Kj = @j 1j Now, let fj , gj (resp. fi , gi ) denote the Riemann invariants on the right side A > 0 (resp. on the left side A < 0 ) we have i A, N + Ki ?i A, N = gi A − Ki N i A, N − Ki ?i A, N = fi A + Ki N So that i A, N = gi A − Ki N + fi A + Ki N ?i A, N = gi A − Ki N − fi A + Ki N/Ki Similarly, on the leftside A < 0 we find j A, N + Kj ?j A, N = gj A − Kj N j A, N − Kj ?j A, N = fj A + Kj N j A, N = gj A − Kj N + fj A + Kj N ?j A, N = gj A − Kj N − fj A + Kj N/Kj To match both sides we apply velocity and pressure continuity in A = 0) gi −Ki N + fi Ki N = gj −Kj N + fj Kj N
(26) gi −Ki N − fi Ki N⁄Ki = gj −Kj N − fj Kj N⁄Kj Which gives a relation between incident and reflected waves at the interface. Note that in the particular case Ki = Kj we have gj = gi and fj = fi .∎. Definition of the « acoustic » scheme : Its purpose is to solve the system (4) (5) namely (! (; − : (< = 0 (9 (; (= + : (< = 0 (9. with a finite difference method. Like for the Wilkins scheme, we introduce a mesh for A and a time discretization with a time step ∆N.. Integrating (4) and (5) pour A ≤ A ≤ A and for S∆N ≤ N ≤ S + 1∆N we get (12). (13). ZWX Z !VWX/Y Q!VWX/Y. ∆9. ZWX Z ;VWX/Y Q;VWX/Y. ∆9. − :. ZWX/Y. ;VWX. + :. ZWX/Y. Q;V. <VWX Q<V. ZWX/Y. =VWX. ZWX/Y. Q=V. <VWX Q<V. =0. =0. U where /E denotes the (assumed to be constant) value of at time S∆N on the cell U A ≤ A ≤ A and the same for ? /E . There is no staggered grid here.. However, we have to evaluate some values ? or at the interface between two cells. We shall use Riemann invariants. We have seen that − K? propagates from right to left and that + K? propagates from left to right. To simplify notations we let U /E. U ?j = ? . /E. U , ?i = ? Q. /E. U /E. and ?∗ = ? . U /E. U U ∗ j = /E , i = Q /E and = We shall ask that ∗ + Ki ?∗ = i + Ki ?i ∗ − Kj ?∗ = j − Kj ?j By linear combination we get that Kj + Ki ∗ = Kj i + Ki j + Kj Ki ?i − ?j (14) Kj + Ki ?∗ = i − j + Ki ?i + Kj ?j (15) U /E. 7.
(27) In annex 2, the stability of this scheme is studied in the linear case it shows that we must have CFL ≤ 1 where CFL is the Courant-Friedrichs-Lewy number.. Boundary conditions In our depressurization problem we know that pressure ∗ on the left side is given. We then ask ∗ − Kj ?∗ = j − Kj ?j So this gives a way to evaluate ?∗ . On the right side we have seen that there is a wall so that we can fix ?∗ = 0, and determine ∗ through ∗ + Ki ?∗ = i + Ki ?i. Note that we have determined ? and inside and at the boundaries, we can make a forward step in time both for and ? by using (12) and (13). U Finally we use the EOS to determine /E in all cells. U /E. U /E. Results On Fig 3 we show the result of our depressurization process at time t=10ms both by the Wilkins scheme and by the acoustic solver scheme. The initial pressure is p=8.1 MPa. The external pressure is p=0.15 MPa. Note that these data match more with a RBMK reactor than a PWR. But we are also interested in RBMK. At time t=0 the fluid is purely liquid at saturation temperature T = 569 K. The initial length of the tube is 1.4 m and we chose a mesh of 140 cells so that ∆C = 1 1A. We can see that in 10 ms, only half of the tube has been depressurized. We also see that there is a significant expansion of the Lagrangian mesh on the left side where depressurization occurs. Note that the time step was 0.1 ms so that we had a CFL = 0.4. Depressurization t=10 ms. Pressure (MPa). 9 8 7 6 5 4 3 2 1 0. -12. -10. -8. -6 pSA. -4. -2. 0. 2. p WI. Fig 3 . Wilkins scheme vs Acoustic solver scheme for the depressurization process §4 Conservative scheme We shall know explain how to define a conservative scheme based on the acoustic scheme. To be able to handle shocks we need to apply conservation of energy (see (6)) Abscissa (m) 8.
(28) (> (9. + : (< ? = 0 (. For its discretization, we proceed as follows : (16). ZWX Z >VWX/Y Q>VWX/Y. ∆9. + :. ZWX/Y =;ZWX/Y VWX Q=;V. <VWX Q<V. =0. U /E U /E U To get ? and ? we use (14) and (15). Then we get D /E . We evaluate the internal energy by using : U U U /E = D /E −
(29) ? /E ⁄2 E. U U U Finally, from /E and /E we can evaluate the new pressure /E by using the equation of state as explained in §1.. Results for the depressurization problem On Fig 4 we show the results obtained with the conservative scheme compared with the Wilkins scheme. We observe that the results are very similar, which is good. Some post processing calculations show that the entropy is almost constant. For example if we look at cell nb 12 initially the entropy s=3.216 kJ/kg/° and after 10ms s = 3.22267 kJ/kg/° (note that the entropy increase is due to the numerical dissipation).. depressurization t=10 ms. Pressure (MPa) 9 8 7 6 5 4 3 2 1 0. -12. -10. -8. -6 pSA. -4 p WI. -2 pj. 0. 2. Abscissa (m). Fig 4 . Wilkins scheme vs conservative Acoustic solver scheme for the depressurization process. §5 Simulation of the “steam” explosion. 9.
(30) What we call a “steam” explosion is actually the propagation of a shock generated by a large energy deposition somewhere in the fluid. The energy is supposed to come from an increase of fission energy in the fuel. The shock is stronger when the fluid is purely liquid and this is what we shall consider. From general results in thermodynamics the fluid will remain liquid behind the shock so that we shall not use the diphasic equation of state that we have described in §1. Rather, we shall use a stiffened gas equation of state.[4] (17). = −mn + m − 1 − o/. For the liquid water, we select (see Corot [2] table 6.1) : m = 2.35, n = 1. D9 78, o = −1167D3 g/5. From what we saw in §2, we have. 1 = 2. /⁄/ − /⁄/. From (17) we get. /⁄/ = m − 1/. /⁄/ = − m − 1 − o/ E. 1 E = E . /⁄/ − /⁄/ = E −mm − 1 n ⁄ + mm − 1 − o/ E . = m −mn + n + m − 1 − o/ = m + n . With the above values and p=8 MPa and @ = 705kg/m3, we find 1 = 1833 A/. Our purpose is to estimate the shock wave velocity caused by the energy deposition and to estimate its duration.. We shall assume that our tube is 6m long and that the energy deposition takes place in the interval 5.1m ≤ C ≤ 6m. The energy deposition is such that the pressure which was equal to 8 MPa instantaneously increases to 200 MPa without modification of the volumic mass which is @ = 705kg/m3. As for boundary conditions we choose a wall both at x=0 and x=6m. This is a Riemann problem, the exact solution of which can be computed : there is a shock wave propagating to the left and a rarefaction wave propagating to the right. We find that the intermediate constant state corresponds to a 102 MPa pressure and a 71 m/s velocity. After 0.6 ms, the rarefaction wave reflects on the wall located at C = 6 A and the reflected wave propagates to the left.. We choose a mesh of 600 cells, with ∆C = 1 1A initially. On Fig. 5 we show the solution at time t= 1 ms and at a later time 2 ms, that is after the reflection of the rarefaction wave on the wall. We can see that the shock speed is 1890 m/s and its width 1.8 m. It propagates to the left. The reflected rarefaction wave propagates at the same speed.. 10.
(31) t=2ms vs t=1 ms. Pressure (MPa). 120 100 80 60 40 20 0 0. 1. 2. 3 p 1ms. 4 p 2ms. 5. 6. 7. Abscissa (m). Fig 5. Schock propagation in the “steam” explosion. §6 Depressurization for a sub-saturated liquid In [4] Faccanoni shows that, when and are selected as the primitive variables, the SG equation of state can be written = − n + m − 1 Qq exp − A⁄u Since we know that our depressurization process is isentropic, we can also write for two states ,. and ` , ` on the same isentrope we have q. +n = ` + n " !v # !. In what follows, we shall select ` = 5.664 678 and ` = 1.3083 3/5 which correspond to saturated liquid water at = 545 w, and = 2.9935 kJ/kg/K . To define our equation of state we just have to select n and m. In view of what we see in the NIST tables [7] we select n = 186 678 and m = 2.79, but other choices are possible. Our sub-saturated fluid will be initially at specific volume = 1.30098 3/5. Since this is close to ` the corresponding pressure will be close to ` so that, with a good approximation since both are ≪ n , we can linearize (18) so that !Q! (19) = ` − m v ` + n (18). !v. In a diagram , our isentrope is then a straight line. We complement our isentrope in the two-phase mixture domain by using the second method described in §1. We get the result shown on Fig. 6. 11.
(32) Isentrope =2.9935 kJ/kg/K. 678. 14 12 10 8 6 4 2 0. 0. 2. 4. 6. 8. 10. p diphasic. 12. 14. 16. 18. 3/5. p SG. Fig 6 Isentrope in a , diagram. The SG part is shown in red.. Obviously the isentrope is continuous but there is a strong slope discontinuity between both parts. This corresponds to a strong discontinuity of the sound speed 1 (see Fig 7). Note that (18) and (2) give that for the SG part : 1 = ym. (19). =v =z !v. ≅ 830 A⁄. c (m/s) 900 800 700 600 500 400 300 200 100 0 0. 2. 4. 6. 8. 10. 12. 3/5. Fig 7 Sound speed 1 along the isentrope = 2.9935 kJ/kg/K.. Applying (18), we shall start our depressurization process from = 8.656 678. As in §2 and 3, we consider a 1.4m long tube initially filled with subsaturated water at = and. = . The boundary conditions are = 0.15 678 on the left side of the tube and a wall (? = 0 on the right side.. The boundary conditions are = 0.15 678 on the left side of the tube and a wall (? = 0 on the right side. Using the acoustic solver described in §3 we find the pressure profiles given on Fig 8.. 12.
(33) 678. -0.4. -0.2. Depressurization of a sub-saturated liquid 10 9 8 7 6 5 4 3 2 1 0 0. 0.2. 0.4. 0.6 p1ms. 0.8. 1. p0.5ms. 1.2. 1.4. 1.6. abscissa (m) (m). Fig 7 Pressure profiles at t=0.5ms and t=1ms. We can see that a rarefaction wave rapidly propagates in the purely liquid phase. It is followed by a constant state at = and = .. depressurization starting from p1,τ1 or from p0,τ0 6 5 4 3 2 1 0 -1. -0.5. 0. 0.5 p2mssat. 1. 1.5. p2ms. Fig 8. Pressure profiles at t=2ms starting either from , (red) or from ` , ` .(blue) We conjecture that after this transient phase everything happens as if we started from a saturated liquid at = ` and = ` . To prove this conjecture numerically we show the pressure profiles at t=2ms, i-e after the transient phase. We get almost exactly the same results (see Fig.8). However, a careful examination of the results shows that the constant state , is not at u=0 m/s but at u=5 m/s. Then at a later time t=8ms we see a slight discrepancy on the pressure profiles with a weak rarefaction wave starting from the right wall.. 13.
(34) Pressure profiles at t=8ms 6. 5. 4. 3. 2. 1. 0 -4. -3. -2. -1 p8ms. 0. 1. 2. p8msvar. Fig 9. Pressure profiles at t=8ms starting either from , (red) or from ` , ` .(blue) If we compare Fig.9 (t=8ms) and Fig.4 (t=10 ms) we can see that depressurization occurs at about the same speed 50m/s to the right in both cases. Starting from subsaturated water at about the same pressure does not change significantly. §7 : Solution of the Riemann problem with real equation of state We shall first consider the case where we have the same diphasic fluid with two different states separated by a diaphragm which is to be removed at time t=0.. uL, pL, τL. uR, pR, τR. We then have ?j = ?i = 0 and we shall assume that j > i . We anticipate that we shall have a 1-shock (propagating to the left) and a 3-rarefaction wave propagating to the right. For t > 0 we shall have an intermediate constant state ?∗ , ∗ , itself subdivided in 2 parts separated by a contact discontinuity. On the left (resp. on the right) of the contact discontinuity, we shall have =. (resp. = E ). We have 4 unknowns ?∗ , ∗ , E , , and we need 4 scalar equations. First we shall use the fact that the following Riemann invariant is constant along a 3-rarefaction wave. We remind the reader that in Eulerian coordinates ? −. 0 /. /. G?H + I ! ? % J G?H = 0 /N 0 . ? /C Let us call λ , λE and λ{ the 3 eigenvalues of the matrix of this hyperbolic sytem they satisfy ? − λ{ + E . ! . ? − λ − E . . % . ? − λ = 0 So that 14.
(35) ? − λ|? − λE − E . % − ! } = ? − λ|? − λE − 1 E } And we get the well-known result that λ = ? − 1, λE = ?, λ{ = ? + 1.. −. We check that ~{ = I 1 J is the eigenvector associated to λ{ indeed . −. −1 −. 0 0 0 I ! −1 % J I 1 J = I− E ! − 1 E + % E J = I0J . 0 −1 0 0 A function = , ?, is a 3-Riemann invariant iff ∇. ~{ = 0 i-e − ! + 1 ; + % = 0 Then = ? − is a 3-Riemann invariant iff 1 = b or b = 1⁄. As a second Riemann invariant we can choose the entropy which is constant in a rarefaction wave. Let j denote the entropy of the right state, we let 1j = 1 , j We can choose ! = ! 1j ⁄ B Riemann invariants :. v. Remark : By using Method B introduced in §1, we can tabulate the isentrope associated to j . More precisely, we compute a 5-column table such that we find , , , 1 and in the 5 columns. So that we have tabulated values for , but also for = By assuming linear interpolation, we can also evaluate b and b ∎. We now get our first two equations : (20) ?∗ − E −
(36) ?j − j = 0 (21) ∗ − E = 0 Now what happens along the 1-shock ? We have the Rankine-Hugoniot relations. Let denote the speed of the shock, we should have (22) @ ? − @i ?i = @ − @i @ ?E + ∗ − @i ?iE + i = @ ? − @i ?i (23) @ D + ∗ ?∗ − @i Di + i ?i = @ D − @i Di (24) where (noting ? = ?∗ D = + E ?E. Di = i + E ?iE. @ − @i "@ ?E + ∗ − @i ?iE + i # = @ ? − @i ?i E. First from (22) and (23) we get :. @E ?E + @iE ?iE − @ @i ?E + ?iE + @ − @i ∗ − i = @E ?E + @iE ?iE − 2@ ? @i ?i @ − @i ∗ − i = @ @i ?E − 2? ?i + ?iE = @ @i ? − ?i E ?∗ − ?i E = i − ∗ − i (25) Second from (22) and (24) we get : @ − @i
(37) @ D + ∗ ? − @i Di + i ?i = @ ? − @i ?i @ D − @i Di @E D ? − @ @i Di ?i + D ? + @iE Di ?i + @ − @i ∗ ? − i ?i = = @E D ? − @ @i ? Di + ?i D + @iE Di ?i @ − @i ∗ ? − i ?i = @ @i Di ?i + D ? − ? Di − ?i D = @ @i ? − ?i D − Di ? − ?i D − Di = i − ∗ ? − i ?i Finally, we get (remind that ? = ?∗) 15.
(38) ?∗ − ?i " + ?∗E − i − ?iE # = i − ∗ ?∗ − i ?i E E. Then (20)(21)(25)(26) are 4 scalar equations for our 4 unknowns ?∗ , ∗ , E , . We remind that there exists a function = , (see §1, method C) such that = ∗ , . In the following we shall denote by = and ! the partial derivatives of , which we can evaluate by using finite differences. (26). To solve this non linear system we could use Newton’s method, and solve f = 0 where ?∗ − E −
(39) ?j − j ?∗ ∗ − E f= = ∗ ?∗ − ?i E + − i ∗ − i E. E E ?∗ − ?i " + E ?∗ − i − E ?i # + − i ∗ ?∗ − i ?i . However, it is more intuitive to use the Hugoniot Curve We follow DESPRÉS B. Numerical Methods for Eulerian and Lagrangian Conservation Laws Springer International Publishing, 2017, p.155, to set up the Hugoniot relation : (26b) − i + E + i − i = 0. Since = , equation (26b) defines a (so called Hugoniot) curve in the plane ( , . We denote by = the relation so obtained between et .. Hugoniot vs isentrope. 6 5 4 3 2 1 0 0. 50. 100. 150 P isentrope. 200. 250. 300. 350. P hug. Fig.10 Hugoniot curve and Isentrope starting from the same point. On Fig. 10 we compare the isentrope starting from. i = 313.7083 3/5 ; 7i = 0.15 68 ; ?i = 0 and the Hugoniot curve (7 in MPa and in L/kg). We notice that both curves are very close to eachother around the point M i , 7i O, but this is a well known result To solve the Riemann problem defined by. j = 1.3083 3/5 ; 7j = 5.664 678 ; ?j = 0. i = 313.7083 3/5 ; 7i = 0.15 678 ; ?i = 0 we proceed as follows : 1. We build the isentrope starting from j ; 7j ; ?j 2. We build the Hugoniot curve starting from i ; 7i ; ?i 3. We define a function E → E such that E = ? − ? where 16.
(40) a. M E , 7∗ O is on the same isentrope as M j , 7j O b. ? = E +
(41) ?j − j = 0 c. M , 7∗ O is on the same Hugoniot curve as M i , 7i O d. ? = ?i − 2 i − 7∗ − 7i 4. We use the dichotomy method to solve E = 0. Example : With the above data, we get | E | < 10Q ` in 40 steps. We get :. E = 49.5734 L/kg. = 54.9847 L/kg 7∗ = 0.80977 MPa ?∗ = ? = ? = -413.137m/s We give below a plot of the solution of the Riemann problem at t = 2.5ms. (Note that we need first to use (22) to find = −500.938 m/s). tau ex 2.5ms. 350 300 250 200 150 100 50 0 -2. -1.5. -1. -0.5. 0. 0.5. tau ex 2.5ms. P ex 2.5 ms 6 5 4 3 2 1 0 -2. -1.5. -1. -0.5. 0. 0.5. P ex 2.5 ms. 17.
(42) u ex 2.5 ms 0 -2. -1.5. -1. -0.5. -50 0. 0.5. -100 -150 -200 -250 -300 -350 -400 -450. We note that the rarefaction wave propagates relatively slowly (50 m/s) to the right. We also note that the amplitude of the contact is small : this is due to the fact that the left state has been chosen on the same isentrope as the right state. §8 Lagrangian solution of the Riemann Problem (saturated case) with the acoustic solver. Coarse mesh (∆C = 1 1A, 200 1). Coarse mesh SA solution vs exact solution 0. -2. -1.5. -1. -0.5. -50 0. 0.5. 1. -100 -150 -200 -250 -300 -350 -400 -450 u ex 2.5 ms. uj. Fine mesh (∆C = 0.4 1A, 600 1). 18.
(43) SA solver 600 cells vs exact solution 0 -2. -1.5. -1. -0.5. -50 0. 0.5. 1. -100 -150 -200 -250 -300 -350 -400 -450 Série1. Série2. The rarefaction wave is better reproduced with the fine mesh. The shock speed is also closer to the exact solution. Here is a plot of the specific volume (unit = L/kg).. tau SA vs tau ex 2.5 ms 350 300 250 200 150 100 50 0. -2. -1.5. -1. -0.5 Série1. 0. 0.5. 1. Série2. We note that on this specific case has a weak jump at the contact discontinuity. This is due to the fact that we have selected the right state and the left state with the same entropy. Case where we have air on the left and saturated water on the right. For air we choose a perfect gas equation of state. =! In such a case we use , = to evaluate qQ = ∗ , . in (26). Here are the results when we start from. j = 1.3083 3/5 ; 7j = 5.664 68 ; ?j = 0. i = 773.395 3/5 ; 7i = 0.1 68 ; ?i = 0 We get. 19.
(44) tau 900 800 700 600 500 400 300 200 100 0 -1. -0.8. -0.6. -0.4. -0.2. 0. 0.2. 0.4. Solution of the Riemann problem at t=1ms For the pressure we compare with the previous case. perfect gas vs diphasic water on the left 6 5 4 3 2 1 0 -1. -0.8. -0.6. -0.4 Pdiphasic. -0.2. 0. 0.2. 0.4. PGPatm. We also compare with the Lagrangian solution (same as in §6 but at time t=10 ms) we see that we get similar results at least in the right side. On the left side we see that, in the Lagrangian case, even though we impose given pressure (0.1MPa) on the left boundary, the rarefaction wave on the right is quite similar to the exact solution of the Riemann problem. Of course, the shock wave in the air is not reproduced with such boundary conditions.. 20.
(45) Lagrange SA vs Exact Solution t=10 ms 6 5 4 3 2 1 0 -10. -8. -6. -4 pSA. -2. 0. 2. 4. PGPatm. References [1] COROT T., MERCIER B., A new nodal solver for the two dimensional Lagrangian hydrodynamics, J. Computational Physics 353 (2018)1-25. [2] COROT T. Numerical simulation of shock waves in a bi-fluid flow: application to steam explosion. Conservatoire national des arts et métiers - CNAM, 2017. [3] DESPRÉS B. Numerical Methods for Eulerian and Lagrangian Conservation Laws Springer International Publishing, 2017 [4] G. FACCANONI. Étude d'un modèle fin de changement de phase liquide-vapeur. PhD thesis, École Polytechnique, 2009 [5] M.L. WILKINS, Finite difference scheme for calculating problems in two space dimensions and time, JCP 5 3 (1970) 406-414 [6] Gloria Faccanoni, Samuel Kokh, Grégoire Allaire. “Numerical Simulation with Finite Volume of Dynamic Liquid-Vapor Phase Transition.” FVCA5, Jun 2008, Aussois, France. pp.391-398. HAL-00976927 [7] https://www.nist.gov/system/files/documents/srd/NISTIR5078-Tab3.pdf. 21.
(46) Annexe 1 : Tabulated values for saturated water (from [4]) T (K) 335 338 341 344 347 350 353 356 359 362 365 368 371 374 377 380 383 386 389 392 395 398 401 404 407 410 413 416 419 422 425 428 431 434 437 440 443 446 449 452 455 458 461. P (Mpa) tauf (L/kg) tauv (L/kg) epsf(MJ/kg) epsv(MJ/kg) sf(kJ/kg/K) sv(kJ/kg/K) 0.021718 1.0181 7078.8000 0.25890 2.45830 0.8544 7.8787 0.024874 1.0198 6232.7000 0.27146 2.46220 0.8917 7.8319 0.028411 1.0215 5502.0000 0.28403 2.46610 0.9288 7.7862 0.032366 1.0233 4869.3000 0.29659 2.46990 0.9655 7.7414 0.036776 1.0251 4320.0000 0.30917 2.47380 1.0019 7.6977 0.041682 1.0270 3841.9000 0.32175 2.47760 1.0380 7.6549 0.047127 1.0290 3424.6000 0.33433 2.48140 1.0738 7.6131 0.053158 1.0310 3059.5000 0.34693 2.48510 1.1093 7.5722 0.059822 1.0330 2739.3000 0.35953 2.48890 1.1446 7.5321 0.067172 1.0351 2457.8000 0.37213 2.49260 1.1795 7.4929 0.075260 1.0373 2209.8000 0.38475 2.49620 1.2142 7.4545 0.084142 1.0395 1990.8000 0.39737 2.49990 1.2487 7.4169 0.093880 1.0418 1797.0000 0.41000 2.50350 1.2829 7.3801 0.104530 1.0441 1625.1000 0.42264 2.50700 1.3168 7.3440 0.116170 1.0465 1472.4000 0.43529 2.51060 1.3505 7.3086 0.128850 1.0490 1336.4000 0.44796 2.51410 1.3839 7.2738 0.142660 1.0515 1215.0000 0.46063 2.51750 1.4172 7.2398 0.157660 1.0540 1106.5000 0.47332 2.52090 1.4502 7.2063 0.173930 1.0566 1009.4000 0.48601 2.52430 1.4829 7.1735 0.191540 1.0593 922.2300 0.49872 2.52760 1.5155 7.1413 0.210600 1.0620 843.8900 0.51145 2.53090 1.5478 7.1097 0.231170 1.0648 773.3600 0.52419 2.53410 1.5800 7.0786 0.253500 1.0676 709.7600 0.53694 2.53730 1.6119 7.0480 0.277220 1.0705 652.3100 0.54971 2.54040 1.6436 7.0180 0.302890 1.0735 600.3300 0.56250 2.54350 1.6752 6.9884 0.330450 1.0765 553.2300 0.57531 2.54650 1.7065 6.9593 0.360010 1.0796 510.4900 0.58813 2.54950 1.7377 6.9307 0.391660 1.0827 471.6400 0.60097 2.55240 1.7687 6.9025 0.425510 1.0860 436.2800 0.61383 2.55520 1.7995 6.8748 0.461670 1.0892 404.0500 0.62672 2.55580 1.8301 6.8474 0.500250 1.0926 374.6300 0.63962 2.55580 1.8606 6.8205 0.541380 1.0960 347.7400 0.65255 2.56340 1.8909 6.7939 0.585160 1.0995 323.1400 0.66550 2.56600 1.9210 6.7677 0.631720 1.1030 300.5900 0.67847 2.56850 1.9510 6.7418 0.681180 1.1066 279.9000 0.69147 2.57090 1.9809 6.7163 0.733670 1.1103 260.9000 0.70450 2.57330 2.0106 6.6911 0.789320 1.1141 243.4300 0.71755 2.57560 2.0402 6.6662 0.848260 1.1179 227.3300 0.73063 2.57780 2.0696 6.6416 0.910630 1.1218 212.5000 0.74374 2.58000 2.0989 6.6173 0.976560 1.1258 198.8100 0.75688 2.58210 2.1281 6.5932 1.046200 1.1299 186.1600 0.77005 2.58400 2.1571 6.5694 1.119700 1.1341 174.4600 0.78325 2.58590 2.1861 6.5458 1.197200 1.1383 163.6300 0.79649 2.58780 2.2149 6.5225 22.
(47) 464 467 470 473 476 479 482 485 488 491 494 497 500 503 506 509 512 515 518 521 524 527 530 533 536 539 542 545 548 551 554 557 560 563 566 569 572 575 578 581 584 587 590 593 596 599. 1.278800 1.364700 1.455100 1.550100 1.649800 1.754500 1.864200 1.979200 2.099700 2.225700 2.357500 2.495300 2.639200 2.789400 2.946100 3.109500 3.279800 3.457100 3.641700 3.833800 4.033600 4.241200 4.456900 4.680800 4.913300 5.154500 5.404700 5.664000 5.932700 6.211000 6.499300 6.797600 7.106200 7.425500 7.755700 8.097000 8.449600 8.814000 9.190300 9.578900 9.980000 10.394000 10.821000 11.262000 11.716000 12.185000. 1.1427 1.1471 1.1517 1.1563 1.1610 1.1659 1.1708 1.1758 1.1810 1.1863 1.1917 1.1972 1.2029 1.2087 1.2147 1.2208 1.2270 1.2334 1.2400 1.2468 1.2537 1.2609 1.2682 1.2757 1.2835 1.2915 1.2998 1.3083 1.3171 1.3261 1.3355 1.3453 1.3553 1.3658 1.3766 1.3879 1.3996 1.4118 1.4246 1.4379 1.4519 1.4665 1.4820 1.4982 1.5154 1.5335. 153.5900 144.2800 135.6400 127.6000 120.1200 113.1500 106.6600 100.6100 94.9520 89.6680 84.7250 80.0980 75.7640 71.7000 67.8870 64.3060 60.9410 57.7760 54.7980 51.9930 49.3500 46.8570 44.5030 42.2810 40.1810 38.1950 36.3150 34.5350 32.8480 31.2480 29.7310 28.2890 26.9200 25.6180 24.3800 23.2000 22.0770 21.0050 19.9830 19.0070 18.0750 17.1830 16.3290 15.5110 14.7260 13.9730. 0.80976 0.82307 0.83642 0.84980 0.86323 0.87669 0.89020 0.90376 0.91736 0.93101 0.94470 0.95845 0.97226 0.98611 1.00000 1.01400 1.02800 1.04210 1.05630 1.07050 1.08490 1.09920 1.11370 1.12820 1.14290 1.15760 1.17240 1.18730 1.20230 1.21740 1.23270 1.24800 1.26350 1.27910 1.29480 1.31070 1.32680 1.34300 1.35940 1.37590 1.39720 1.40970 1.42690 1.44440 1.46220 1.48030. 2.58950 2.59110 2.59270 2.59410 2.59550 2.59670 2.59790 2.59890 2.59990 2.60070 2.60140 2.60200 2.60250 2.60290 2.60310 2.60320 2.60320 2.60300 2.60270 2.60220 2.60160 2.60080 2.59990 2.59880 2.59750 2.59600 2.59440 2.59250 2.59040 2.58820 2.58570 2.58290 2.57990 2.57670 2.57320 2.56940 2.56530 2.56090 2.55610 2.55100 2.54550 2.53960 2.53320 2.52640 2.51900 2.51110. 2.2436 2.2722 2.3007 2.3291 2.3574 2.3856 2.4138 2.4418 2.4698 2.4977 2.5256 2.5533 2.5810 2.6087 2.6363 2.6639 2.6914 2.7189 2.7464 2.7739 2.8013 2.8287 2.8561 2.8836 2.9110 2.9385 2.9660 2.9935 3.0210 3.0486 3.0763 3.1041 3.1319 3.1598 3.1878 3.2160 3.2443 3.2727 3.3014 3.3302 3.3592 3.3885 3.4181 3.4479 3.4781 3.5087. 6.4994 6.4765 6.4538 6.4313 6.4089 6.3868 6.3647 6.3428 6.3211 6.2994 6.2779 6.2565 6.2351 6.2139 6.1927 6.1715 6.1504 6.1293 6.1082 6.0872 6.0661 6.0450 6.0239 6.0027 5.9814 5.9601 5.9387 5.9171 5.8955 5.8736 5.8517 5.8295 5.8071 5.7845 5.7616 5.7385 5.7150 5.6912 5.6669 5.6423 5.6172 5.5916 5.5654 5.5386 5.5110 5.4827 23.
(48) 602 605 608 611 614 617 620 623 626 629. 12.669000 13.167000 13.681000 14.211000 14.757000 15.320000 15.901000 16.499000 17.116000 17.752000. We shall take : = 1. We start from. 1.5529 1.5735 1.5956 1.6193 1.6451 1.6731 1.7039 1.7382 1.7768 1.8211. 13.2490 12.5520 11.8800 11.2320 10.6050 9.9973 9.4067 8.8308 8.2670 7.7118. 1.49870 1.51740 1.53660 1.55620 1.57640 1.59720 1.61860 1.64100 1.66640 1.68910. 2.50260 2.49340 2.48350 2.47270 2.46100 2.44830 2.43430 2.41890 2.40190 2.38270. 3.5398 3.5713 3.6034 3.6362 3.6697 3.7041 3.7396 3.7765 3.8150 3.8556. 5.4536 5.4234 5.3922 5.3597 5.3258 5.2903 5.2528 5.2131 5.1705 5.1246. Annex 2 : linear stability analysis for the acoustic solver. ( ( + K E (< ? = (9 ( ( ? + (< = 0 (9 E. 0. With = −K . We have. ( (9 ( (9. − K? − K (< − K? = 0 (. + K? + K (< + K? = 0 (. Then the Riemann invariant ± K? is transported at speed ±K in the MA, NO space. Using the same notations as in §3, we shall ask that ∗ + K?∗ = i + K?i ∗ − K?∗ = j − K?j ∗ 2 = i + j + K?i − ?j 2K?∗ = i − j + K?i + ?j And then U . ? . X Y. X U Y. . and also. U = E
(49) Q. U = E
(50) Q. ZWX Z =VWX/Y Q=VWX/Y. ∆9. (3). ZWX Z ;VWX/Y Q;VWX/Y. ∆9. +. /E. /E. U − . U + X. U /E + E
(51) ? Q /E. U /E + E
(52) ? Q /E X. ZW ZW ; Y Q; Y K E VWX V . +. X X ZW ZW =VWXY Q=V Y. . =0. U + ? . U − ? . /E . /E . =0. We now replace i par j and assume (we let ℎ = ∆A and assume a uniform mesh) : (4). U ? . . U . E. U . /E. /E. = U exp " + E# ℎ = 7U exp " + E# ℎ. 1 U 1 1 1 1 1 7 exp − ℎ − exp + ℎ + U exp − ℎ + exp + ℎ 2K 2 2 2 2 2 1 U 1 1 K U 1 1 = 7 exp − ℎ + exp + ℎ + exp − ℎ − exp + ℎ 2 2 2 2 2 2. ?. U E. so that. U . ?. E. =. =. 1 U 1 7 exp ℎ exp − ℎ − exp ℎ + U exp ℎ exp − ℎ + exp ℎ 2K 2 2 2 2 2 24.
(53) U . . E. 1 K = 7U exp ℎ exp − ℎ + exp ℎ + U exp ℎ exp − ℎ − exp ℎ 2 2 2 2 2 2. − U ℎ 1 ℎ 7 expℎ 2S + U exp ℎ21 2K 2 2 2 1 ℎ K ℎ U E = 7U expℎ 21 − U exp ℎ2S 2 2 2 2 − ℎ ℎ U ? E = expℎ 7U S + U 1 K 2 2 ℎ ℎ U E = expℎ 7U 1 − K U S 2 2 − ℎ ℎ U ? E = exp + 1ℎ 7U S + U 1 K 2 2 ℎ ℎ U E = exp + 1ℎ 7U 1 − K U S 2 2 − ℎ ℎ U U ? E − ? E = exp + 1ℎ − expℎ 7U S + U 1 K 2 2 ℎ ℎ U U E − E = exp + 1ℎ − expℎ 7U 1 − K U S 2 2 i-e − ℎ ℎ U U ? E − ? E = exp + 1⁄2ℎ exp ℎ⁄2 − exp− ℎ⁄2 7U S + U 1 K 2 2 ℎ ℎ U U E − E = exp + 1⁄2ℎ exp ℎ⁄2 − exp− ℎ⁄2 7U 1 − K U S 2 2 ℎ − ℎ ℎ U U ? E − ? E = exp + 1⁄2ℎ 2 S . 7U S + U 1 2 K 2 2 ℎ ℎ ℎ U U E − E = exp + 1⁄2ℎ 2 S . 7U 1 − K U S 2 2 2 Let λ = ∆N⁄ℎ, we have Let. U . ?. E. =. U . U . U U /E = . /E. − λ K E G? E − ?. U U ? /E = ? . /E. − λ G E − . E. H. = exp " + E# ℎ 7U − λ K E 2 S "E # . " 7U S E + U 1 E #. hence. U . U . H. Q . . . = exp " + E# ℎ U − λ 2 S "E # . "7U 1 E − K U S E # . . . 7U = 7U − λ K E 2 S "E # . " 7U S E + U 1 E # . Q . . U = U − λ 2 S "E # . "7U 1 E − K U S E # . i-e. E. . 1 − 2λ KSE E U 7 " U # = −2 λ S 1 E. . E. . . . −2 λ K E S E 1 E . 1 − 2λ KSE E. . . 7U " U# . We now compute the eigenvalues ] of this 2x2 matrix which satisfy ℎ ℎ ℎ 1 − 2λ KSE − ] −2 λ K E S 1 2 2 2 = 0 ℎ ℎ ℎ −2 λ S 1 1 − 2λ KSE − ] 2 2 2. 25.
(54) or. E. ] = 1 − 2λ KSE ± 2 λ K S 1. We then have and. E. "1 − 2λ KSE − ]# = 4 E K E λE SE 1 E E. E. E. E. E. E E E E E E E 4λ KSE + 4K E λE "SE # + 4K E λE "S E 1 E # E E E E E E 4λ KS + 4K λ S = 1 − 4λ K1 − λ KSE E E E. |]|E = "1 − 2λ KSE # + "2 λ K S 1 # = =1−. =1−. Then, to get stability, it is necessary (and sufficient) that λ K < 1 i-e K ∆N⁄ℎ < 1. Since K = @` 1` and ℎ = ∆A = @` ∆C we get 1` ∆N⁄∆C < 1 which is known as the CFL condition.. 26.
(55)
Figure
+2
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