1、相變傳熱與流體流動(dòng)數(shù)值分析作業(yè)1學(xué) 院（系）： 能源與動(dòng)力學(xué)院 專 業(yè)： 能源與環(huán)境工程 學(xué) 生 姓 名： 學(xué) 號(hào)： 指 導(dǎo) 教 師： 完 成 日 期： 大連理工大學(xué)Dalian University of TechnologyThe Finite Volume Method for Diffusion ProblemsSubjects：I.Consider the problem of source-free heat conduction in an insulated rod.The equation governing one-dimensional steady state cond

2、uctive heat transfer is ddxkdTdx=0, where k equals 1000W/m/K. The ends are maintained at constant temperatures of 100 and 500,the cross-section area A is 0.01.II.A large plate of thickness L=2cm,the thermal conductivity k=0.5 W/m/K, uniform heat generation q=1000kW/m3.The face A and B are at tempera

3、tures of 100and200 respectively. The governing equation is ddxkdTdx+q=0.III.There is a cylindrical fin with uniform cross-section area A.The base is at a temperature of 100 and the end is insulated. The fin is exposed to an ambient temperature of 20. The governing equation is ddxkAdTdx-hp(T-T)=0.And

4、 m=hp/(kA)=25m-2,L=1m.Solution:/ 王佳琪-作業(yè)1.cpp : 定義控制臺(tái)應(yīng)用程序的入口點(diǎn)。/#include #include #include#includeusing namespace std;#define N 5 void main()double aN-1,bN,cN-1,SpN,lN-1,uN,fN,yN,TN; /追趕法使用的數(shù)組int i,j; /j 題號(hào)double k,L,A,TA,TB,q,m,x; /邊界變量coutj;if(j=1|j=2) coutk= ,L= ,A= ,TA= ,TB= ,q= kLATATBq; x=L/N; f

5、or(i=0;iN-1;i+) /定義數(shù)a=aw,c=ae ai=-k*A/x, ci=-k*A/x; for(i=0;iN;i+) /定義數(shù)組Sp if(i=0|i=N-1) Spi=-2*k*A/x; else Spi=0; for(i=0;iN;i+) /定義數(shù)組b=ap if(i=0) bi=-Spi-ci; else if(i=N-1) bi=-Spi-ai-1; else bi=-Spi-ci-ai-1; for(i=0;iN;i+) /定義數(shù)組f=Su if(i=0) fi=2*k*A*TA/x+q*A*x; else if(i=N-1) fi=2*k*A*TB/x+q*A*x;

6、 else fi=q*A*x; else coutm= ,L= ,TA=,TB= mLTATB;x=L/N; for(i=0;iN-1;i+) /定義數(shù)組a=aw, c=ae ai=-1/x, ci=-1/x; for(i=0;iN;i+) /定義數(shù)組Sp if(i=0) Spi=-2/x-m*x; else Spi=-m*x; for(i=0;iN;i+) /定義數(shù)組b=ap if(i=0) bi=-Spi-ci; else if(i=N-1) bi=-Spi-ai-1; else bi=-Spi-ci-ai-1; for(i=0;iN;i+) /定義數(shù)組f=Su if(i=0) fi=2*

7、TB/x+m*x*TA; else fi=m*x*TA; u0=b0;for(i=1;iN;i+) /追趕法求解三對(duì)角矩陣 li-1=ai-1/ui-1; ui=bi-li-1*ci-1; y0=f0;for(i=1;i=0;i-) Ti=(yi-ci*Ti+1)/ui;cout計(jì)算結(jié)果x=;for(i=0;iN;i+) /輸出數(shù)組T coutTi ;The result: Subject I.Subject II.Subject III.Discussion:The subjects 1 2 and 3 all can be dealing with 1D problem. But the subject 1 does not have a source term, subject 2 have a real source term and subject 3 have an equivalent source term. The subjects 1and 2 are coincided with the analytical solution. But the subject 3 is