数值传热学第四章编程题

1、精品好资料学习推荐4-5迭代法求解节点温度。说明：此处给出的是C+程序代码，使用牛顿迭代法，迭代收敛精度1.0e-6；程序运行结果附后。/*NHT 4-5 newton*created on 2012-10-19 by Sanye*/#include #include #include using namespace std;int main() double funT=1.0,dfunT=1.0,temp1=1.0,temp2=1.0; double T=20.0;/primary value int i=0; /for TEST! coutprimary T= T ;=1.0e-6) i+

2、; if(i=1&(T=20.0)T=100.0;/in case unreasonable T; temp1=pow(T-20.0,0.25);temp2=pow(T-20.0,-0.75); funT=0.5*T-80+2*T*temp1-40*temp1; dfunT=0.5+2*temp1+0.5*T*temp2-10*temp2; T=T-funT/dfunT; cout step i;tT3-T3= funT/dfunT ;endl; couttotal steps: iendl; couttT1= 100tT2= 0.5*(T+130)tT3= T endl; return 0;

3、运行结果：4-12 编写TDMA算法程序验证其正确性。说明：此处给出的是C+程序代码，程序运行结果附后。/*TDMA.cpp *Created on 2012-10-19 by Sanye*/#include #include #include #include #include TDMA.husing namespace std;int main() int num=10;/num: number of T points; vector A,B,C,D,T_before,T_after; int rand_temp1,rand_temp2; /*Generate A,B,C,D,T_befo

4、re Randomly*/ for(int i=0;inum;+i) rand_temp1=rand()%10; rand_temp2=rand()%20; A.push_back(1.0+rand_temp1/2.0); B.push_back(rand_temp2/(rand_temp1+3.0)/5.0+1.0); C.push_back(rand_temp1*rand_temp2/10.0+2.0); T_before.push_back(sqrt(i*i+1.0)*11.0-20.0); B0=0.0;Cnum-1=0.0; for(int i=0;inum;+i) double t

5、emp; if(i=0) temp=A0*T_before0-B0*T_before1; else if(i=num-1) temp=Anum-1*T_beforenum-1-Cnum-1*T_beforenum-1-1;else temp=Ai*T_beforei-Bi*T_beforei+1-Ci*T_beforei-1; D.push_back(temp); /*Get T_after*/ T_after.resize(num); T_after=TDMA(A,B,C,D); /*Calculate T_err*/ vector T_err; for(int i=0;inum;+i)T_

6、err.push_back(T_afteri-T_beforei); /*Cout T*/ cout.setf(ios:fixed); cout.precision(4); for(int i=0;inum;+i) coutT_beforei+1= T_beforeitT_afteri+1= T_afteri tT_erri+1= T_erri endl; return 0;/* TDMA.h*/#ifndef TDMA_H_INCLUDED#define TDMA_H_INCLUDED#include using namespace std;vector TDMA(vector &A, vector &B, vector &C, vector &D)vector P,Q,T; int n=A.size(); P.resize(n); Q.resize(n); T.resize(n); /*Get P,Q*/