Meshless Methods for Scientific Computing Final Project

Transcription

1 Meshless Methods for Scientific Comuting Final Project D 洪啟耀 Introduction Floating island becomes an imortant study in recent years, because the lands we can use are limit, so eole start thinking how to use the sace on the river or ocean. In Korea, there is the biggest floating island in Seoul (Figure 1). Figure 1 Floating Island in Han River, Seoul I m interested in this toic, and I have done some exeriments about the floating island (Figure ). Figure Exeriments in NTU In this study, we try to use adjusted MFS to simulate the motion of floating island face to the incoming wave. J.J. Stoker in 195 has simulated the motion of the board in 1-D case, in this study we try to simulate the -D case and we will comare the result with J.J. Stoker s results. We try to understand this roblem ste by ste, we searate in few cases. In Case I, we consider the board is a illar; we only need to consider the motion outside of the illar. In Case II, we simulate the illar to the

2 board which means we need to consider the motion under the board, but the board is fixed on the surface. In Case III, we started to move the board with forced which means the board swings on the surface. In Final case, we let the board start moving with the incoming wave. Mathematical Derivation Because we assume the roblem is base on the wave, we can start with the wave equation (1). gh t 1 We assume the amlitude is much smaller than the deth of water ( h ). We searate the otential of wave into time and sace (). i t ( x, y) e () Where is the arameter with meaning of frequency. Take () into (1), we can get (3). (1) ( x, y) ( x, y) 0 gh (3) is the Helmholtz equation, and this is the ernal governing equation. (3) Case I In Case I, we consider the floating board as a illar, to simulate the motion of wave, we accumulative original wave and the reflect wave. For the original wave we can write the form in (4). 0 i t ik ik x x y y Am e e e (4) Take (4) into (3), we can get the relationshi between different arameters. kx ky (5) gh And the reflective wave art, the form should fit the governing equation, we use the same term of time, so the time s term will be eliminate during the rocess, we add the back time term at lot. We chose the fundamental solution from C.S. Chen (011), which can be written in (6) i () ex t( x, y) H0 ( r) (6) 4 Where () H0 is the second kind Hankle function. The solution in this case should

3 be the accumulation of these two solutions (7). 0 i ( xy, ) (7) The boundary condition can be consider as the flux cannot ass through the illar, so the condition can be write in (8). n 0 (8) Using MFS to solve this roblem, the source oints and the boundary oints are laced in Figure 3. We consider in the simle case, the boundary is consider as the cylinder and the we lace the source oints inside of boundary, which means there is not the boundary to constraint the cylinder. Figure 3 Case I oints distribution. Base on (7) and (8), we can solve the equation (9) 0 1 n n n x y x y ( nx ny ) 1( nx ny ) 0 We can write the equation in matrix form (10). nx n y 1 nx n y x y x y The numerical result can be shown in Figure 4. (9) (10)

4 Figure 4 Numerical Result (t=1~4) We can comare the result with (C.C. Mei, 006) result on the internet, which is shown in Figure 5. Figure 5 Numerical Result form C.C.Mei Comare the result, three different zones can be both obviously observed in our result and in C.C. Mei s result. Zone I: In front of the cylinder, standing waves resent because the scattered waves erform like the reflected waves in this zone. Zone III: Behind the cylinder, a shadow zone resents. Zone II: Besides these two zones, the incident waves dominate and the effect of the scattered waves is weak but can be observed as the oint source wave generated from the center of the cylinder. Another examination we make, we take gradient of our result, we can calculate the velocity, and the result can be illustrated in Figure 6.

5 Figure 6 Velocity distribution form numerical result We can find out the velocity distribution basically fit the hysical theorem, so we can say the method is ok. We can test n case with more comlex idea. Case II In this case, we consider the floating board is floating on the water surface, but the osition is fixed. The board will not move during the incoming wave. But the water can still ass through under the board, beside the governing equation we write in Case I, we need to add the new governing equation under the board. We consider the mass conservation under the board, which can be written in (11). u v 0 x y Where uvmeans, the velocity in x-direction and y-direction, base on the theory of otential flow, we can rewrite (11) into (1). x y int int 0 It is easy to find out the governing equation become the Lalace equation. We can find out the fundamental solution in C.S Chen s t book, which can be written in (13). (11) (1) 1 int int ; int ( x, y) log( r) (13) The boundary condition in this case also fit the theory of mass conservation and becomes (14). n n int Base on (13), (14), and the information in Case I. The boundary condition can be (14)

6 written in following equation (15). ( nx ny ) ( nx ny ) (15) x y x y As we can find out in (15), the relationshis between internal and ernal are deendent, so we cannot solve internal without ernal. We need to solve these two unknown in same time. Write (15) into matrix form (16). int int 1 nx ny ( nx ny ) nx n y x y x y x y (16) We can use MFS to solve this system. Different from MFS we use before, the matrix combine two kinds of arameter. Because there are two kinds of MFS, we need to lace two kinds of the source oints, but the same boundary oints, which cen be illustrated in Figure 7. Figure 7 Points distribution in Case II As we can figure out in this case we lace the boundary oints as a thin late, in urose to simulate the situation similar as 1-D case, so that we can comare our result with J.J. Stoker s analytical solution. We comare the velocity distribution in one dimension. The numerical result is shown in Figure 8. We consider the analytical solution form J.J. Stoker (17). ( x, y) x (17) int (17) which is the function of otential, so the velocity is gradient of the function. We calculate the velocity and comare the result with our result.

7 Figure 8. Comarisons of velocity We can find out the result is retty good, two results are very close. And we calculate the RMSE; the RMSE in this case is We can say in these comarisons we can rove the result we make is ok. Then we can take this method into more comlex art. Case III In this case, we consider the board is moving with forced. The movement of board can be illustrated in Figure 9. Figure 9 Movement of board In Figure 9, the height of board move is, and angular velocity. The movement should include another movement of gravity center in z-axis. The movement can be written in (18). i t ( z ( x x) ( y y) ) e (18) x In this case, we start need to consider the gravity center of the board. So we use green theory to calculate the osition of the gravity center, which we can write in (19). y

8 xda yda Gx Gy da da Similarly, the term resent the time can be eliminated in the rocess. Because of the new movement, the governing equation under the board should be udate (19). h int 0 t Because the equation become non-homogenous, we take (18) into (0), in order to find the articular solution. The solution of internal art can accumulative with articular solution and the fundamental solution. The articular solution can be written in (1) i 3 i 3 i x x x y y z (1) 6h 6h h And the fundamental solution is same as the Case II, which is written in (13). Because we add articular solution into equation, so the boundary condition should be rewritten in (). n n int x y x y x y x y int int ( nx ny ) 1( nx ny ) ( nx ny ) ( nx ny ) In this case, because we force the board to move, which means the angular velocity and the movement in z-direction are given. The articular solution will be another source term. The result can be written in matrix form (). int int 1 nx ny ( nx ny ) ( nx ny ) ( nx ny) x y x y x y x y The boundary oints and source oints we laced can be illustrated in Figure 10. (19) (0) () (3)

9 Figure 10 Points distribution for Case III In this case we laced the oints in irregular shae, in urose to figure out the reflect wave is irregular or not. To make sure our calculation is fit to the hysical theory. The result can be shown in Figure 11. Figure 11 Numerical result of Case III In Figure 11, we can figure out the reflect wave is not symmetrical and that is make sense. The wave still have the shadow zone, the zone after board still have a shadow zone when the board left u on the other side. However, we want to try the case which the movement all base on the wave case.

10 Case IV In this case, we let board movement only influence by the wave, so the angular velocity and z-direction movement will become unknown arameters. The governing equations are same as revious case (4). ( x, y) ( x, y) 0 gh (4) h int 0 t We need to add new condition equations to solve the system. We consider the Bernoulli equation of unsteady form (5). u t int g 0 In (5) we cancel the term with u, because it is higher order in our case that can be simlified. We can rewrite the equation in (6). t int g After we know how to resent ressure, we can consider the force balance in this case, we can write the force balance in this case (7). i t F m( z ) e ( ) da I x y x I y ( x) da ( y) da Where I, I means the moments in x,y direction, and m means the mass of board. x y In (6) we integrate the ressure in numerical way. Add these three equations into the system, and the boundary condition will be change because angular velocity and the movement of z-direction become unknown, so the boundary condition can be rewrite in (8) n n int x y x y x y x y int int ( nx ny ) 1( nx ny ) ( nx ny ) ( nx ny ) x y x y x y x y int int ( nx ny ) ( nx ny ) 1( nx ny ) ( nx ny ) We combine the equation together, and write in the matrix form, and we can solve the system at same time. The result can be illustrated in Figure 1. (5) (6) (7) (8)

11 Figure 1 Movement of board in different time (t = 1~4)

12 In Figure 1, we can observe the motion basically fit hysical theory. But we did not examine the result in more comact way. So we only can base on the case I~III, and say our result is good. Conclusion In this roject, we use the benefit of MFS which can change the condition easily and imrove our result ste by ste. We successfully change original MFS, which we add some more condition equation. These equations make our results become more reliable. Although we did not use radial basis function to be the articular solution, but the result can resent exactly henomena. In future, we want to examine our results in more comactly way. We can comare the results with 1-D result, just like we did in Case II, besides the velocity we can examine the results with ressure, in that case the result will be more reliable. We want to aly this numerical result in our exeriments, which can comare the difference between exeriment and numerical calculation. We will try to romote our result from the board to the box, in fact the floating island is made from box design, because the real situation is the gravity center and floating center is not the same oints. If we want to comare the numerical result with real cases, we need to calculate the floating center and gravity center; in that case we can comare the result with real cases.