有限差分作业-华中科技大学版

1、华中科技大学研究生课程考试答题本考生姓名 考生学号 系、年级考试日期题号得分题号得分总分:评卷人:注：1、无评卷人签名试卷无效。2、必须用钢笔或圆珠笔阅卷，使用红色。用铅笔阅卷无效。1Q1. PI ease deduce the sec on d-order forward differe nee, sec on d-order backward differe nee and sec on d-order cen tral differe nee of y=f(x), the n deduce the p recisi on of y=f(x) sec on d-order cen tral

2、 differe nce,based on Taylor series expansion.An swer:On e-order forward differe nee:迥=f（X +ixf（X）On e-order backward differe nee:y = f(X )- f (x-Ax )On e-order cen tral differe nee:Differe ntiate aga in, and we can get the sec on d-order differe nee as follows: Secon d-order forward differe nee: 2y

3、 = ( Ay )=A f (X + 也Xf(X )=if(X +Ax )-Af(X )f (x + Ax )- f(X 卩=f (X +2Axf (x+也X =f(X + 2也X )-2 f (x+x )+ f(X )Secon d-order backward differe nee:2y = A (Ay )=A f(X f ( X -纵)=if(X )也f(X 也X )=f(X )- f (x-x )卜f (X-Ax )- f (x-2ix)=f(X )2f(X-也X ) + f (x-2Ax )Secon d-order cen tral differe nee:2A y =A(3 )

4、f1)r1I f I x+ ix ( f I x ix K2(1 X +Ax/-if1 2丿XI 2丿X丄也X .2丿=f (X + Axf (X f (X f (X - Ax=f(X + Ax )2 f(X )+ f ( X - Ax )Accord ing to the Taylor series expansion:+0( Axi )f (x+Ax)= f (x)+AxLf O(x)+ f(4*x )234f(X 也X )= f(X )AxLf (X(X)(xfx )+0(2xj )Add the two formulas above, and we can get:f (x+4)-2f

5、 (x)+f(X-Ax)什CHA r=f (x)+0(ix)23x)So the sec on d-order cen tral differe nee quotie nt has sec on d-order p recisi on.Q2. PI ease write the FTCS format of the conv ecti on equati onT 丄Ec+a=0 ctex：(x,0)=I(x)And point out the trun catio n error p recisi on. An swer:At (Xi ,tn ), repl ace time derivati

6、ve with first-order forward differe nee quotie nt.HiAtRep lace sp ace derivative with first-order cen tral differe nee quotie nt,15n_24十I3尸23x)+111So the convection equation at (Xj ) can be approximate ash +a=0加2 AxGiven the initial conditions, we can get the FTCS format of the convection equatio n:

7、匚rx)Accord ing to the Taylor series expansion,匚(Xi,tn + 心t )匚(Xi,tn)+f 匚(Xi +Ax,tn )匚(Xi -心 X,tn )it仃总2匚+ I2 -2卫2丿 页V2Ri+Ogx)So the trun cati on error of the FTCS format is2 2F =O(At ) + O(Ax) )=O(At,(Ax)Therefore, the FTCS format has one-order precision of it and two-order precision of Ax .Q3. PI e

8、ase simulate the temp erature field of an H-sha ped cast ing using FDM. Th geometric con diti ons and in itial con diti ons are as follows:1) The material of the H-sha ped cast ing is ZG25, the en vir onment temp erature is 25 C and the mold material is res in sand;2) Pouring temperature is 15600 ;3

9、) Casti ng size as show n in Figure 2 and Figure 3, the mold thick ness is 40mm Requireme nts:1) Write out the 2D or 3D mathematical model that describes the temp erature field of the cast ing cooli ng p rocess;2) Deduce the FDM format of the mathematical model;3) Draw up the FDM grid map, describes

10、 it use data structure;4) P rovide the thermal prop erties of ZG25, resi n sand and the air;5) P rogram to simulate this p hysical p rocess,assu ming the cavity was filled very fast and the in itial temp erature eve niy distributed. PI ease pro vide the main40170Figure3. A 2D slice of the H-sha ped

11、castingf或:I4 00 r LZ-.：nil4QAn swer:1) The Fourier heat con duct ion differe ntial equati on on the three dime nsional occasi ons is:22-2 iflcxcy住丿ctIn which:den sity; Cpheat.If thereT temp erature; t time; x,y,z sp ace coord in ates; P sp ecific heat;扎 coefficie nt of heat con ducti on; L late ntis

12、 no inner heat source, and set =a , a is thermal PCpdiffusivity(m mathematical model:2/s), then we can get the three dimensional temperature field2) Unit i is a regular hexahedron element with a length of one side 也x , and it is tran sferri ng heat with the six un its adjace nt to it. The thermal eq

13、uilibriumis:relati on shi p diagram of three dime nsional differe nee unit i is show n below.In a very short time At, heat absorption Q of unit i is:Q =PiC piexTit-Tit)Heat summati on QSUM flow ing from the adjace nt unit 1,2, 3, 4, 5, 6 to unit iQsum 国豊込-T）加人几jAccord ing to the law of con servati o

14、n of en ergy:-6阿XF 说-丄N W仃jtT,)The n, 丁=Ti + Z PCpQx 口 处+竺 2人 2片Transform the formula mathematical model:t4above, and we canZ 1ECpQx j# 丝 +鱼2)气2召Tt +getthe FDM format of theTjt6zPjCpQx匸空X+鱼 2扎i2几jAtIn additi on, from the p hysical meaning, we can get the stabilizati on con diti on:1人XAx+2AjF=H=I 一 f

15、flfflffl 一 EH-tIH-?Fa 山一 一 mi噩|圜_曇一置 五-i jEi-Ei二三一= = n-3i5-H二ss-s-m=m-Bi-=-B-s-3-=-a-s-n-D-s-3-D-=-= B_E 蓋 E 器亜I囉 EL-fflfflffl m- - -Ffflffla fflfflffl fflfflffl-mfflffl Is US InP(kg /m3 )Cp(J/(kgLJK )A（w/（mK）T(K)ZG25775047027.21833.15Resin sand16101054.90.41298.15Air1.2050.0010050.0259298.15AtThe

16、thermal prop erties of ZG25, resi n sand and the air are show n in the table below: The mai n code of the p rogram is as follows: Mesh ing:for (int z = 0;z = z_bo un dary;z+)for (int y = -y_bo un dary;y = y_bo un dary;y+) for (int x = - x_bo un dary;x = x_bo un dary;x+,no de+) Nodezy + y_bo un daryx

17、 + x_bo un dary = no de;if (z = 0 II z = z_bo un dary )for (i nt k = 0;k 0 , in which, ai =ZRCpijTransform: At 0 .3) The FDM grid map is show n below:rfniiiiw*3nTnnmp*Mnimnti*Mninnmw4tmTimfPPmHmnTnH3tF mmiHiHiEnTnnffigs5TTnmnffHFSirHrnnHbri4*tininimFS?amnffTH；Fl 3 田 iiuljj rtmti 田 g Q 3 9 wtim田氐 田 i

18、mta=A ffl IUIIHtb Hi nil ffl Ri ilHI HI lUU ffl Illi Hl Hi Hmnrl R11 HIM 番 F 3mre5=F3iTnEE3imrKEEFTH&5S3imEmF&=FFliill：=i=i；iili：lillill=Fhiliiil：=f=il；iliA=liilli：l= m “I I irm m ii ni “m iiiiiri m n- ni iiii I-1 imrnnrninTnTinigqnTininTniraninnniFrnnninirngqinirnnnBgr iniiiiiHi占nnTnfnBmiinifUi” 0

19、& z = H_sha pe_cycleHigh / sp aceLe ngth )if (x*x+y*y) = Radius*Radius)Temp 0zy + y_bou ndaryx + x_bou ndary = 1833.15;/cast ingMaterialzy + y_bo un daryx + x_bo un dary = 0;Den sityzy + y_bo un daryx + x_bo un dary = 7750;Con dzy + y_bo un daryx + x_bo un dary = 470;Thermalzy + y_bo un daryx + x_bo

20、 un dary = 27.2;else if (x = -x_bo un dary | x = x_bo un dary | y = - y_bo un dary | y =y_bo un dary )for (i nt k = 0;k H shape cycleHigh / spaceLength & z H_sha pe_cycleHigh+H_sha pe_lo ng+H_sha pe_h_lo ng)/s paceLe ngth& z =(H_sha pe_cycleHigh+2*H_sha pe_lo ng+H_sha pe_h_lo ng)/s paceLe ngth)if(X

21、= -x_bo un dary | x = x_bo un dary | y = - y_bo un dary | y = y_bo un dary)- - for (i nt k = 0;k = -(H_sha pe_le ngth / sp aceLe ngth)/ 2 & x = -(H_sha pe_width / 2) / sp aceLe ngth ) & y = (H_sha pe_width / 2) /sp aceLe ngth)Temp 0zy + y_bou ndaryx + x_bou ndary = 1833.15;/cast ingMaterialzy + y_bo

22、 un daryx + x_bo un dary = 0;Den sityzy + y_bo un daryx + x_bo un dary = 7750;Con dzy + y_bo un daryx + x_bo un dary = 470;Thermalzy + y_bo un daryx + x_bo un dary = 27.2;elseTemp 0zy + y_bou ndaryx + x_bou ndary = 298.15; /moldMaterialzy + y_bo un daryx + x_bo un dary = 1;Den sityzy + y_bou ndaryx

23、+ x_bou ndary = 1610;Con dzy + y_bou ndaryx + x_bou ndary = 1054.9;Thermalzy + y_bo un daryx + x_bo un dary =0.41; elseif (x = -x_bo un dary | x = x_bo un dary | y = - y_bo un dary | y = y_bo un dary)for (i nt k = 0;k d;k+)Temp kzy + y_bou ndaryx + x_bou ndary = 298.15; /airMaterialzy + y_bo un dary

24、x + x_bo un dary = 2;Den sityzy + y_bo un daryx + x_bo un dary = 1.205;Con dzy + y_bou ndaryx + x_bou ndary = 0.001005;Thermalzy + y_bo un daryx + x_bo un dary = 0.0259; elseTemp 0zy + y_bou ndaryx + x_bou ndary = 298.15; /moldMaterialzy + y_bo un daryx + x_bo un dary = 1;Den sityzy + y_bou ndaryx +

25、 x_bou ndary = 1610;Cond zy + y_bou ndaryx + x_bou ndary = 1054.9;Thermalzy + y_bo un daryx + x_bo un dary = 0.41; Calculate:for (i nt i =1;i vd;i+)for (int z = 1;z z_bo un dary;z+)for (int y = -y_bo un dary + 1;y y_bo un dary;y+) for (int x = - x_bo un dary +1;x x_bo un dary;x+)Temp izy + y_bo un d

26、aryx + x_bo un dary = Temp i-1zy + y_bo un daryx+ x_bo un dary +1.0e6*(time_s pace / sp aceLe ngth) / (De nsityzy + y_bo un daryx +x_bo un dary * Con dzy + y_bo un daryx + x_bo un dary) *(Te mp i-1zy + y_bo un daryx + x_bo un dary+1 - Temp i-1zy +y_bo un daryx + x_bo un dary) /(s paceLe ngth / 2)/Th

27、ermalzy + y_bo un daryx + x_bo un dary+1 +(sp aceLe ngth / 2)/Thermalzy + y_bo un daryx + x_bo un dary)+ (Temp i-1zy + y_bo un daryx + x_bo un dary-1 - Temp i-1zy +y_bo un daryx + x_bo un dary) /(s paceLe ngth / 2)/Thermalzy + y_bo un daryx + x_bo un dary-1 +(sp aceLe ngth / 2)/Thermalzy + y_bo un d

28、aryx + x_bo un dary)+(Temp i-1zy + y_bo un dary+1x + x_bo un dary -Te mp i-1zy +y bo un daryx + x bo un dary) /(s paceLe ngth / 2)/Thermalzy + y_bo un dary+1x + x_bo un dary +(sp aceLe ngth / 2)/Thermalzy + y_bo un daryx + x_bo un dary)+(Temp i-1zy + y_bo un dary-1x + x_bo un dary - Temp i-1zy +y_bo

29、 un daryx + x_bo un dary) /(s paceLe ngth / 2)/Thermalzy + y_bo un dary-1x + x_bo un dary +(sp aceLe ngth / 2)/Thermalzy + y_bo un daryx + x_bo un dary)+(Te mp i-1z+1y + y_bo un daryx + x_bo un dary -Te mp i-1zy +y_bo un daryx + x_bo un dary) /(sp aceLe ngth / 2)/Thermalz-1y + y_bo un daryx + x_bo u

30、n dary+1 +(sp aceLe ngth / 2)/Thermalzy + y_bo un daryx + x_bo un dary);Save the result:fstream fs;char szFileName20;double count = 0.0;for(i nt time = 0;time d;time+=10,co un t+ )sprin tf(szFileName,G:%d. pl t,time);fs.open( szFileName,ios_base:out|ios:_Nore place); fsTitle=fi nite-eleme nt data for bricke ndl; fsvvVariables=x,y,z,Te mp,we ndl