hardy cross method code

3
 #include <iostream> #include <conio.h> #include <math.h> #include <iomanip> using namespace std; const float PI=3.14159265358979323 8463; const float n=1.85; float area(float dia) { return PI*dia*dia/4; } int main() { float pipel[2][4],piped[2][4],pipeqass[2][4],vel[2][4],hf[2][4],hfbyq[2] [4],delq[2][4],delq1,d elq2,qdash[2][4],sighf[2 ]={0,0},sighfbyq[2]={0,0} ; int common[2][4]; float qpn,dp4,cpn; for(int i=0;i<2;++i) { cout<<"\nENTER THE VALUES OF PIPE IN LOOP"<<i+1; for(int j=0;j<4;++j) { cout<<"\nPIPE NO"<<j+1; cout<<"\nPIPE LENGTH(in m):"; cin>>pipel[i][j]; cout<<"\nPIPE DIAMETER(in mm):"; cin>>piped[i][j]; cout<<"\nIS IT A COMMON PIPE?(1-YES,0-NO)"; cin>>common[i][j]; cout<<"\nFIRST ASSUMED VALUE OF Q(in m^3/min)"; cin>>pipeqass[i][j]; vel[i][j]=(pipeqass[i][j]/60)/area(piped[i][j]/1000); qpn=pow(pipeqass[i][j]/60,n); if(pipeqass[i][j]>=0) { dp4=pow(piped[i][j]/1000,4.87); cpn=pow(100,n); hf[i][j]=10.622*(pipel[i][j])*(qpn)/(cpn*dp4); hfbyq[i][j]=hf[i][j]/pipeqass[i][j]; if(i==0) { sighf[0]+=hf[i][j]; sighfbyq[0]+=hfbyq[i][j]; } if(i==1) { sighf[1]+=hf[i][j]; sighfbyq[1]+=hfbyq[i][j]; } } else { dp4=pow(-piped[i][j]/1000,4.87); cpn=pow(100,n); hf[i][j]=10.622*(pipel[i][j])*(qpn)/(cpn*dp4); hfbyq[i][j]=hf[i][j]/pipeqass[i][j]; if(i==0) {

Upload: arun-rajagopal

Post on 08-Oct-2015

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DESCRIPTION

This is a c++ program coded by me to perform hardy cross method for pipe networks. Environmental Engineering. Civil Department.

TRANSCRIPT

#include #include #include #include using namespace std;const float PI=3.141592653589793238463;const float n=1.85;float area(float dia){return PI*dia*dia/4;}int main(){float pipel[2][4],piped[2][4],pipeqass[2][4],vel[2][4],hf[2][4],hfbyq[2][4],delq[2][4],delq1,delq2,qdash[2][4],sighf[2]={0,0},sighfbyq[2]={0,0};int common[2][4];float qpn,dp4,cpn;for(int i=0;i