一维对流扩散问题求解..doc
文本预览下载声明
Subjects：A property is transported by means of convection and diffusion through the one-dimensional domain. The governing equation is; boundary conditions are at x=0 and at x=L. Using five equally spaced cells and the central differencing scheme for convection and diffusion calculate the distribution of as a function of x for
(i)Case 1: case 1 u=0.1 m/s,
(ii) Case2: Case 1: u=2.5 m/s, and compare the results with the analytical solution.
(iii)Case 3: recalculate the solution for case 3 u=2.5m/s with 20 grid nodes and compare the results with the analytical solution, the following date apply: length L=1.0 m,.
Solution:
Programming language:
File name:二维稳态导热.cpp
Compile software: Microsoft Visual C++ 6.0
Code:
#include iostream
#includemath.h
#includestdlib.h
using namespace std;
#define Gn1 5 //网格节点数.
#define Gn2 20 //网格节点数.
#define P 1 //流体密度，单位：kg/m/m/m.
#define ProA 1 //A端传递特性.
#define ProB 0 //B端传递特性.
//////FDD///////追赶法过程
void TDMA1(double a[],double b[], double c[] ,double f[],double *p)
{ //矩阵形式
//b0 c0 0 0 //=f0
//a0 b1 c1 0 //=f1
//... ... ... ... ...
//0 0 ... b(N-1) c(N-2) //=f(N-1)
double d[Gn1-1],u[Gn1],ll[Gn1-1], y[Gn1],X[Gn1];
int i;
for(i=0; i=Gn1-2;i++)
{
d[i] = c[i];
}
u[0] = b[0];
for(i=1; i=Gn1-1;i++)
{
ll[i-1]=a[i-1]/u[i-1];
u[i]=b[i]-ll[i-1]*c[i-1];
}
y[0]=f[0];
for(i=1; i=Gn1-1;i++)
{
y[i]=f[i]-ll[i-1]*y[i-1];
}
X[Gn1-1]=y[Gn1-1]/u[Gn1-1];
for (i=Gn1-2; i=0;i--)
{
X[i]=(y[i]-c[i]*X[i+1])/u[i];
}
for (i=0; i=Gn1-1;i++)
*(p+i)=X[i];
}
void TDMA2(double a[],double b[], double c[] ,double f[],double *p)
{
double d[Gn2-1],u[Gn2],ll[Gn2-1], y[Gn2],X[Gn2];
int i;
for(i=0;i=Gn2-2;i++)
{
d[i]=c[i];
}
u[0] = b[0];
for(i=1;i=Gn2-1;i++)
{
ll[i-1]=a[i-1]/u[i-1];
u[i]=b[i]-ll[i-1]*c[i-1];
}
y[0] = f[0];
for(i=1;i=Gn2-1;i++)
{
y[i]=f[i]-ll[i-1]*y[i-1];
}
X[Gn2-1]=y[Gn2-1]/u[Gn2-1];
for (i=Gn2-2;i=0;i--)
{
X[
显示全部