相变传热与流体流动数值分析作业1

1、.相变传热与流体流动数值分析作业1学 院（系）： 能源与动力学院 专 业： 能源与环境工程 学 生 姓 名： 学 号： 指 导 教 师： 完 成 日 期： 大连理工大学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 con

2、ductive 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 temper

3、atures 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.An

4、d m=hp/(kA)=25m-2,L=1m.Solution:/ 王佳琪-作业1.cpp : 定义控制台应用程序的入口点。/#include <stdafx.h>#include <iostream>#include<math.h>#include<stdlib.h>using namespace std;#define N 5 void main()double aN-1,bN,cN-1,SpN,lN-1,uN,fN,yN,TN; /追赶法使用的数组int i,j; /j 题号double k,L,A,TA,TB,q,m,x; /边界变量co

5、ut<<"j= "cin>>j;if(j=1|j=2) cout<<"k= ,L= ,A= ,TA= ,TB= ,q= "<<endl; /边界初始化 cin>>k>>L>>A>>TA>>TB>>q; x=L/N; for(i=0;i<N-1;i+) /定义数a=aw,c=ae ai=-k*A/x, ci=-k*A/x; for(i=0;i<N;i+) /定义数组Sp if(i=0|i=N-1) Spi=-2*k*A/x;

6、else Spi=0; for(i=0;i<N;i+) /定义数组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;i<N;i+) /定义数组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; else fi=q*A*x; else cout<<"m= ,L= ,TA=,TB= "<<endl; /边界初始化cin>>m&g

7、t;>L>>TA>>TB;x=L/N; for(i=0;i<N-1;i+) /定义数组a=aw, c=ae ai=-1/x, ci=-1/x; for(i=0;i<N;i+) /定义数组Sp if(i=0) Spi=-2/x-m*x; else Spi=-m*x; for(i=0;i<N;i+) /定义数组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;i<N;i+) /定义数组f=Su if(i=0) fi=2*TB/x

8、+m*x*TA; else fi=m*x*TA; u0=b0;for(i=1;i<N;i+) /追赶法求解三对角矩阵 li-1=ai-1/ui-1; ui=bi-li-1*ci-1; y0=f0;for(i=1;i<N;i+) /求解数组y,T yi=fi-li-1*yi-1; TN-1=yN-1/uN-1; for(i=N-2;i>=0;i-) Ti=(yi-ci*Ti+1)/ui;cout<<"计算结果x="for(i=0;i<N;i+) /输出数组T cout<<Ti<<" "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 solutio