限元與有限差分法應(yīng)用,試驗(yàn)報(bào)告

1、華中科技大學(xué)研究生課程考試答題本考生姓名顏小強(qiáng)考生學(xué)號(hào)D201677004系、年級(jí)數(shù)學(xué)與統(tǒng)計(jì)學(xué)院 2016級(jí)類(lèi) 別博二考試科目有限元法與有限差分法的應(yīng)用考試日期2016年10月30日題號(hào)得分題號(hào)得分總分：評(píng)卷人：注：1、無(wú)評(píng)卷人簽名試卷無(wú)效。2、必須用鋼筆或圓珠筆閱卷，使用紅色。用鉛筆閱卷無(wú)效題號(hào)回答內(nèi)容得 分Q1Please deduce the second-order forward difference, second-order backward difference and second-order central difference ofy= f(x), then deduce

2、 theprecision of y = f(x) second-order central difference based on Taylor seriesexpansion. (10 points)Answer:(1.1)Second-order forward difference formula:We know that the first order forward difference formula is:yf (x + Ax f (x )xAxSo, the second-order forward difference formula:2y 二4絲 iAx2Ax 3 Jf

3、f f (x + Ax )- f (x )x Ax,ff(x 十 4x) A/f(x)x Axx AxAx ；f (x +Ax + Ax)- f (x + Ax ) f (x +Ax )- f (x )=AxAxxAxf(x+2Ax)-2f(x+ix)+f(x)-Ax2A2yf (x + 2-x)-2f (x+-x)+f (x)x2Ax21回 答 內(nèi) 容(1.2)Second-order central difference formula:We know that the first order central difference formula is: TOC o 1-5 h z HYP

4、ERLINK l bookmark7 o Current Document .1.1 HYPERLINK l bookmark9 o Current Document f lx lx -fix - -Xy =.2.2xxSo, the second-order central difference formula:11,1f lx x -fix xf I”xx回 答 內(nèi) 容號(hào)f11 .1 .、x 十 Ax 十一Ax122 Jl-f11 .1 .、x 十一Ax Ax22 JxAx I)J,A，r &ecision of secoTaylor series expansionf (x +Ax) =

5、 f (x) +2f (x - Ax) = f (x) -4Adding equation (1) an f (x + Ax)cr f(x4 or,thus, second-order cenfx1 .1 五 ”x- -Ax + Ax22kf11A 1x Ax 22Ax xx +Ax 2f (x )+ f (x - Ax)x22y f (x )-2 f (x-Ax )+ f (x-24x )x2Ax2nd-order central difference formula:i for f(x+Ax) and f(x+Ax) are(Ax)2(Ax)3,叼 3 + 年 f + ； f (x) +O

6、Qx)4)(1)、xf (x) +-f ”(x)-箸 f (x) +O(9x)4)(2)d (2) we get+ f (x Ax) = 2 f (x) + (Ax)2 f (x) + O(Ax)4)產(chǎn)3=f.(x)+O(Ax)4)tral difference formula has two-order precision.3題號(hào)回答內(nèi)容得 分Q2Please combine the research direction of your subject, illustrate the typical application of the finite difference method in

7、 this research area3 5 points) Answer:My major is the mathematical calculation The exact direction is the numerical solution of partial differential equations and its application.First of all,FDM is widely used in the numerical solution of partial differential equations.Concrete exampleis as follow:

8、Finite difference method for solving boundary value problems一Au =cos3xsin 兀y,(x, y) w G = (0，n)父(0,1),h(x,0) =u(x,1) =0,0 MxM1Ux(0, y) =ux(n, y) =0,0 w y M1/ /c,_2、/c _、-.,1,1(exact solution is u=(9+n ) cos3xsinny .) .By step h1 = ,h2 = , we canget rectangular section and we can structure difference

9、 scheme.Grid node is x =ih1, yj =jh2,i,j =0,1,., N .Difference equation is as follow:Ui+j2與 +U-jUi,j 書(shū)2uj +Ui,jcccQvcinr、，一(2+2) =cos3x sin ny：,Ah2i,j=1,2,.,N,Boundary value condition is as follow:叫=即=0,i =0,., N,uj =u1j, j =1,., N -1,uNj =UN，j =1,.,N -1.Successively, We order N =4,8,16,32,and solve

10、 it with the elimination methods.小 j(x, yj) = ( 一, ), i, j =1,2,3 .We can make a list of exact solutions and numerical 4 4solutions.4題號(hào)回答內(nèi)容得 分Q3Please simulate the temperature field of anH-shaped casting using FDM.The geometric conditions and initial conditions are as follows:The material of the H-s

11、haped casting is ZG25, the environment temperature is 25c and the mold material is resin sand;Pouring temperature is 1560c ;Casting sizes are shown in Figure 2 and Figure 3, the mold thickness is 40mm.Requirements:Write out the 2D or 3D mathematical model that describes the temperature field of the

12、casting cooling process; (10 points)Deduce the FDM format of the mathematical model; (10 points)Draw the FDM grid map, and describe it using data structure; (10 points)Provide the thermal properties parameters of ZG25, resin sand and the air; (10 points)Program to simulate this physical process, ass

13、uming the cavity was filled very fast and the initial temperature evenly distributed. Please provide the main code of the program. (25 points)How long does it take while the highest temperature of the casting drops to 1450c ? (10 points)5Figure 1: 3D model of the H-shaped casting+03:A 2D slice of th

14、e回 答 內(nèi) 容Answer: TOC o 1-5 h z e 3D mathematical model is 一222 HYPERLINK l bookmark19 o Current Document -。1c Tc |c TcLPC=九 + +J 7c 2c 2c 2Ict I ex 0VGZ J dpWhere T is temperature (K), is density (kg/m3), C is heat capacity (J/(kJK),is thermal conductivity (W/(m ?K),L is the latent heat (J/kg)If we d

15、o not take the latent heat into consideration ,then the last term of the above equation should be omitted.discretize the Fourier differential equation of heat conduction based on FDM.And the heat exchange process is shown as follows.The deducing process :The quantity of heat which the element i has

16、absorbed is:Q=：0()3(丁The quantity of heat flow into I from the 6 element adjacent to it is:Qi-Tit)-tAccording to the law of conservation of energyRCpi(Ax)3(Tit-Tit) = Z JX，？ (T；工)&j%i jAfter rearrangement we get the explict schem of temperature evolution.t 6Tit =T/：:iC pi x j -re we introduce a func

17、tion of flag-1,air layerflag = 0, sand layerJZG 25Where ZG 25 would be cut into four small pieces to be handled easily. for(i=8;i=88;i+) for(j=8;j=28;j+) for(k=8;k=24;k+) flagijk=1;/ZG for(i=38;i=58;i+) for(j=8;j=28;j+) for(k=24;k=100;k+) flagijk=1;/ZG for(i=8;i=88;i+) for(j=8;j=28;j+) for(k=100;k=1

18、16;k+)flagijk=1;/ZG for(i=8;i=88;i+) for(j=8;j=28;j+)for(k=116;k=132;k+)if(i-48.0)*(i-48.0)+(j-18.0)*(j-18.0)3二營(yíng) jwMgl_ii_i照一j!l=:=Eiiw 副biiei廟一囤Bl一胃胃一胃一二 :r=:=；w TrnriiI BHiHB - - Bi_iB - BBliH BBiaBB - Ball _ 三胃一a一一酷凰-虱一UH山”-1三iHH J -Mi-i.BSI 疝=:Eii甥國(guó)10(3.4)The thermal properties parameters of ZG2

19、5,resin sand and the air are :heat conduct coefficient(W/(m K)density(Kg/m3)specific heat(J/(Kg K)ZG2527.27750470sand0.7316101054.9air0.02591.2050.001005(3.5)The results of this simulation is listing following:11Our program main include 5 sections below:4 invoking functions and the main functionInit

20、() is used to initialize variables such as temperature and thermal properties parameters. void init() for(i=0;i=NX+1;i+)for(j=0;j=NY+1;j+)for(k=0;k=NZ+1;k+)if(flagijk=-1)/air layerCPijk=0.001005;rhoijk=1.205;thermijk=0.0259;temijk=25;tem_0ijk=25;if(flagijk=1)/ZGCPijk=470;rhoijk=7750;thermijk=27.2;te

21、mijk=1560;tem_0ijk=1560;if(flagijk=0)/sandCPijk=1054.9;rhoijk=1610;thermijk=0.73;temijk=25;tem_0ijk=25;calculate_temperature() is used to calculate the evolution of the temperature field.void caculate_temperature()double TEMPLE,t1,t2,t3,t4,t5,t6;TEMPLE=0;Tmax=0;for(i=1;i=NX;i+)for(j=1;j=NY;j+)for(k=

22、1;k1560) printf(x=%d,y=%dn,i,j);if(temijkTmax)Tmax=temijk; for(i=1;i=NX;i+)for(j=1;j=NY;j+)for(k=1;k=NZ+1;k+)tem_0ijk = temijk;data() is used to output the results file.void data( int n )sprintf( fName, temperature%d.dat, n);if( (fp = fopen(fName,w) = NULL ) return;fprintf( fp, VARIABLES = X, Y , Z,

23、 Tn);fprintf( fp, Zone, I=%d,J=%d,K=%d,F=BLOCKn, NX+2, NY+2,NZ+2 );for(k=0;k=NZ+1;k+)for( j=0; j=NY+1; j+)for(i=0; i=NX+1; i+) fprintf( fp, %dt, i);fprintf(fp,n);for(k=0;k=NZ+1;k+)for( j=0; j=NY+1; j+)for(i=0; i=NX+1; i+) fprintf( fp, %dt, j);13fprintf(fp,n);)for(k=0;k=NZ+1;k+)for( j=0; j=NY+1; j+)f

24、or(i=0; i=NX+1; i+)fprintf( fp, %dt, k);)fprintf(fp,n);)for(k=0;k=NZ+1;k+)for( j=0; j=NY+1; j+)for(i=0; i=NX+1; i+)fprintf( fp, %.5et, temijk);)fprintf(fp,n);)fclose(fp);)output() is used to draw the grid map.void output()FILE *file;file=fopen(flag.dat,w);if(file=0) printf(can not open flag file);fp

25、rintf(file,Title=flagn);fprintf(file,variables=x,y,z,flagn);fprintf(file,zone T=BOX,I=%d,J=%d,K=%dn,NX+2,NY+2,NZ+2);for(k=0;k=NZ+1;k+)for(j=0;j=NY+1;j+)for(i=0;i=NX+1;i+)fprintf(file,%d,%d,%d,%dn,i,j,k,flagijk);)fclose(file);)14(3.6)Through the results of the procedure,we can easily get the time of

26、decresing from 1560 to 1450 is 133.62 ,where it cost 6681 steps coupled with our Time interval of dt is 0.02The 6G7M The Tlie The 6673The 6674The 6675The 6676The 6677 The 6678The 8679 The ccea The GW 占二133-Pr-ess any1450-2241B7 14CB.204221 145S.184256 1450.164293 450-144322 X4S0.X24372 145B.104414 1

27、4B6.08445S 1450.06450 1450.0445S1 1456.624G06 145BQ4What do you think of this courseFirst of all,this course is an international course offered by the Institute of materials science, materials science, materials science, and other professional graduate students or doctoral students in Huazhong University of Science and Technology. This course emphasizes the connotation and extension of teaching guide, based on student-centered, teacher led the starting point to the curriculum as the carrier and platform, the positioning in the course with the help of a variety of